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AP*Chemistry The Chemistry of Acids and Bases

*AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. © 2008 by René McCormick. All rights reserved.

"ACID"--Latin word acidus, meaning sour. (lemon) "ALKALI"--Arabic word for the ashes that come from burning certain plants; water solutions feel slippery and taste bitter. (soap)

Acids and bases are extremely important in many everyday applications: our own bloodstream, our environment, cleaning materials, and industry. (Sulfuric acid is an economic indicator!) ACID-BASE THEORIES

ARRHENIUS DEFINITION acid--donates a hydrogen ion (H+) in water base--donates a hydroxide ion in water (OH−)

This theory was limited to substances with those "parts"; ammonia is a MAJOR exception!

BRONSTED-LOWRY DEFINITION acid--donates a proton in water base--accepts a proton in water

This theory is better; it explains ammonia as a base! This is the main theory that we will use for our acid/base discussion.

LEWIS DEFINITION acid--accepts an electron pair base--donates an electron pair

This theory explains all traditional acids and bases plus a host of coordination compounds and is used widely in organic chemistry. Uses coordinate covalent bonds. THE BRONSTED-LOWRY CONCEPT OF ACIDS AND BASES Using this theory, you should be able to write weak acid/base dissociation equations and identify acid, base, conjugate acid and conjugate base.

conjugate acid-base pair--A pair of compounds that differ by the presence of one H+ unit. This idea is critical when it comes to understanding buffer systems. Pay close attention now and it will pay off later!

HNO3 + H2 → H3O+ + NO3− neutral compound as an acid

acid base CA CB NH4

+ + H2O H3O+ + NH3 cation as an acid acid base CA CB H2PO4

− + H2O H3O+ + HPO42− anion as an acid

acid base CA CB

The Chemistry of Acids & Bases 2

In each of the acid examples---notice the formation of H3O+ -- this species is named the hydronium ion. It lets you know that the solution is acidic!

( hydronium, H3O+--H+ riding piggy-back on a water molecule; water is polar and the + charge of the “naked” proton is greatly attracted to Mickey's chin!)

NH3 + H2O NH4

+ + OH− neutral compound base acid CA CB CO3

2− + H2O HCO3− + OH− anion

base acid CA CB PO4

3− + H2O HPO42− + OH− anion

base acid CA CB Notice the formation of OH− in each of the alkaline examples. This species is named the hydroxide ion. It lets you know that the resulting solution is basic! You try!! Exercise 1 a) In the following reaction, identify the acid on the left and its CB on the right. Similarly identify the base on the left and its CA on the right. HBr + NH3 → NH4

+ + Br− b) What is the conjugate base of H2S? c) What is the conjugate acid of NO3

-?

ACIDS DONATE ONLY ONE PROTON AT A TIME!!!

monoprotic--acids donating one H+ (ex. HC2H3O2) diprotic--acids donating two H+'s (ex. H2C2O4) polyprotic--acids donating many H+'s (ex. H3PO4)

polyprotic bases--accept more than one H+; anions with −2 and −3 charges (ex. PO4

3− ; HPO42−)


Amphiprotic or amphoteric --molecules or ions that can behave as EITHER acids or bases;

water, anions of weak acids (look at the examples above—sometimes water was an acid, sometimes it acted as a base)

Exercise 2 Acid Dissociation (Ionization) Reactions Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids. a. Hydrochloric acid (HCl) b. Acetic acid (HC2H3O2) c. The ammonium ion (NH4

+) d. The anilinium ion (C6H5NH3

+) e. The hydrated aluminum(III) ion [Al(H2O)6]3+

A: HCl(aq) H+(aq) + Cl-(aq)

B: HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

C: NH4+(aq) H+(aq) + NH3(aq)

D: C6H5NH3+(aq) H+(aq) + C6H5NH2(aq)

E: Al(H2O)63+(aq) H+(aq) + [Al(H2O)5OH]2+(aq)


RELATIVE STRENGTHS OF ACIDS AND BASES Strength is determined by the position of the "dissociation" equilibrium.

Strong acids/strong bases 1. dissociates completely in water 2. have very large dissociation or K values

Weak acids/weak bases 1. dissociate only to a slight extent in water

2. dissociation constant is very small

Do Not confuse concentration with strength!

STRONG ACIDS: Memorize these SIX Hydrohalic acids: HCl, HBr, HI—note HF is missing! Nitric: HNO3 Sulfuric: H2SO4 Perchloric: HClO4

Strong Weak


The more oxygen present in the polyatomic ion of an oxyacid, the stronger its acid WITHIN that group. That’s a trend, but not an explanation. So, why? First, notice that the H of the acid is bound to an oxygen and NOT any other nonmetal present. Oxygen is very electronegative and attracts the electrons of the O−H bonds toward itself. If you add more oxygens, then this effect is magnified and there is increasing electron density in the region of the molecule that is opposite the H. The added electron density weakens the bond, thus less energy is required to break the bond and the acid dissociates more readily which we describe as “strong”.

STRONG BASES Hydroxides OR oxides of IA and IIA metals (except Mg and Be)

o Solubility plays a role (those that are very soluble are strong!)

THE STRONGER THE ACID THE WEAKER ITS CB, the converse is also true.


WEAK ACIDS AND BASES:

The vast majority of acid/bases are weak. Remember, this means they do not ionize much.

That means a equilibrium is established and it lies far to the left (reactant favored). The equilibrium expression for acids is known as the Ka (the acid dissociation constant). It is set up the same way as any other equilibrium expression. Many common weak acids are oxyacids, like phosphoric acid and nitrous acid. Other common weak acids are organic acids—those that contain a carboxyl group, the COOH group, like acetic acid and benzoic acid.

For weak acid reactions: HA + H2O H3O+ + A−

Ka = [H3O+][A−] < < 1 [HA]

Write the Ka expression for acetic acid. (Note: Water is a pure liquid and is thus, left out of the equilibrium expression.)

Weak bases (bases without OH−) react with water to produce a hydroxide ion. Common examples

of weak bases are ammonia (NH3), methylamine (CH3NH2), and ethylamine (C2H5NH2). The lone pair on N forms a bond with a H+. Most weak bases involve N.


The equilibrium expression for bases is known as the Kb. for weak base reactions: B + H2O HB+ + OH− Kb = [HB+][OH−] << 1 [B]

♦ Write the Kb expression for ammonia.

♦ Notice that Ka and Kb expressions look very similar. The difference is that a base produces the hydroxide ion in solution, while the acid produces the hydronium ion in solution.

♦ Another note on this point: H+ and H3O+ are both equivalent terms here. Often water is left completely out of the equation since it does not appear in the equilibrium. This has become an accepted practice. (* However, water is very important in causing the acid to dissociate.)

Exercise 3 Relative Base Strength Using table 14.2, arrange the following species according to their strength as bases: H2O, F−, Cl−, NO2

−, and CN−.

Cl- < H2O < F- < NO2

- < CN-


WATER, THE HYDRONIUM ION, AUTO-IONIZATION, AND THE pH SCALE

Fredrich Kohlrausch, around 1900, found that no matter how pure water is, it still conducts a minute amount of electric current. This proves that water self-ionizes.

• Since the water molecule is amphoteric, it may dissociate with itself to a slight extent. • Only about 2 in a billion water molecules are ionized at any instant!

H2O(l) + H2O(l) H3O+(aq) + OH−(aq)

• The equilibrium expression used here is referred to as the autoionization constant for water, Kw

• In pure water or dilute aqueous solutions, the concentration of water can be considered to be a constant (55.6 M), so we include that with the equilibrium constant and write the expression as:

Kw = [H3O+][OH−] = 1.008 × 10−14 @ 25°C = Ka × Kb

♦ Knowing this value allows us to calculate the OH− and H+ concentration for various

situations. ♦ [OH−] = [H+] solution is neutral (in pure water, each of these is 1.0 × 10−7) ♦ [OH−] > [H+] solution is basic ♦ [OH−] < [H+] solution is acidic

Exercise 5 Autoionization of Water At 60°C, the value of Kw is 1 × 10−13.

a. Using Le Chatelier’s principle, predict whether the reaction below is exothermic or endothermic.

2H2O(l) H3O+(aq) + OH−(aq)

b. Calculate [H+] and [OH−] in a neutral solution at 60°C.

A: endothermic

B: [H+] = [OH−] = 3 × 10-7 M


The pH Scale Used to designate the [H+] in most aqueous solutions where [H+] is small.

pH = −log [H+] pOH = − log [OH−] pH + pOH = 14

If pH is between zero and 6.999, the solution is acidic, if pH is 7.000, the solution is neutral and if the pH is above 7.000, the solution is basic.

Reporting the correct number of sig. figs on a pH is problematic since it is a logarithmic scale. The rule is to report as many decimal places on a pH as there are in the least accurate measurement you are given.

Example: The problem states a 1.15 M solution blah, blah, blah. That is your cue to report a pH with 3 decimal places. If the problem had stated a 1.2 M solution blah, blah, blah, then you would report your calculated pH to 2 decimal places. How did this ever get started? If you care…read the next bullet…otherwise go directly to Exercise 6!

In the old days, before calculators (Can you imagine?), students used log tables to work problems involving logarithms. If the logarithm was 7.45, then the “7” was the characteristic and the “.45” part was the mantissa. In fact, it is the mantissa that communicates the accuracy of the measurement. The characteristic is simply a place holder.

Exercise 6 Calculating [H+] and [OH−] Calculate either the [H+] or [OH−] from the information given for each of the following solutions at 25°C, and state whether the solution is neutral, acidic, or basic. a. 1.0 × 10−5 M OH−

b. 1.0 × 10−7 M OH−

c. 10.0 M H+

A: [H+] = 1.0 × 10-9 M, basic B: [H+] = 1.0 × 10-7 M, neutral

C: [OH−] = 1.0 × 10-15 M, acidic


Exercise 7 Calculating pH and pOH Calculate pH and pOH for each of the following solutions at 25°C. a. 1.0 × 10−3 M OH−

b. 1.0 M H+

A: pH = 11.00

pOH = 3.00 B: pH = 0.00

pOH = 14.00 Exercise 8 Calculating pH The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate pOH, [H+], and [OH−] for the sample.

pOH = 6.59 [H+] = 3.9 × 10-8 M

[OH−] = 2.6 × 10-7 M Exercise 9 pH of Strong Acids a. Calculate the pH of 0.10 M HNO3. b. Calculate the pH of 1.0 × 10−10 M HCl.

A: pH = 1.00 B: pH = 10.00


Exercise 10 The pH of Strong Bases Calculate the pH of a 5.0 × 10−2 M NaOH solution.

pH = 12.70

Calculating pH of Weak Acid Solutions Calculating pH of weak acids involves setting up an equilibrium. Always start by writing the balanced equation, setting up the acid equilibrium expression (Ka), defining initial concentrations, changes, and final concentrations in terms of x, substituting values and variables into the Ka expression and solving for x. Use the RICE TABLE method you learned in general equilibrium!

Example Calculate the pH of a 1.00 × 10−4 M solution of acetic acid. The Ka of acetic acid is 1.8 × 10−5

.

Reaction HC2H3O2 H+ + C2H3O2−

Initial 1.00 × 10−4 0 0 Change −x +x +x Equilibrium (1.00 × 10−4 ) − x x x

22 3 2 5

42 3 2

H C H O1.8 10

HC H O 1 10axK

x

+ −−

−

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= = × =× −

See that –x term in the denominator? That is your invitation to cross multiply and distribute the Ka value across the term so that you get x2 = 1.8 × 10−9 – 1.8 × 10−5x ; collect like terms and use either the solver on your graphing calculator or a quadratic formula solving program you’ve loaded on your calculator (all of this to avoid arithmetic mistakes!) to solve for x. You should determine that x = [H+] = 3.44 × 10−4 and that the pH = − log (3.44 × 10−4) = 3.46 (2 SF). Often, the –x term in a Ka expression can be neglected. That simplifies the math tremendously since you are now spared the tedium of having to use the quadratic formula. How do you know when to neglect x? Easy. Look at the original concentration and compare it to 100 Ka (or 100 Kb). IF the initial concentration is large by comparison, you can neglect subtracting the x term. We could not neglect x in the example we just worked since 100 Ka for acetic acid would equal 1.8 × 10−3 or 0.0018 which is too close to our initial acid concentration of 0.0001.


Need proof? Suppose our initial concentration had been 0.10 M for the acetic acid in the example problem we just worked. For acetic acid, 100Ka = 1.8 × 10−3 or 0.0018. That’s essentially subtracting zero from 0.10 M. Aside from that if you did subtract it, you’d still follow the “least decimal place” subtraction sig. fig. rule and report 0.10 M as your answer to the subtraction. OK, I’ll humor you and apply the quadratic formula to the example we just worked changing the initial concentration to 0.10 M. First I’ll do the quadratic formula and then I’ll work it by “neglecting x”. Example Calculate the pH of a 10.10 M solution of acetic acid. The Ka of acetic acid is 1.8 × 10−5

.

Reaction HC2H3O2 H+ + C2H3O2−

Initial 0.10 0 0 Change −x +x +x Equilibrium 0.10 − x x x

22 3 2 5

2 3 2

H C H O1.8 10

HC H O 0.10axK

x

+ −−

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= = × =−

By cross multiplying and NOT neglecting x, you get x2 = 1.8 × 10−6 – 1.8 × 10−5x ; collect like terms and use either the solver on your graphing calculator or a quadratic formula solving program you’ve loaded on your calculator (all of this to avoid arithmetic mistakes!) to solve for x. You should determine that x = [H+] = 0.0013327 (way too many sig. figs, I know!) and that the pH = − log (0.0013327) = 2.88 (2 SF) Had we “neglected x”, the math simplifies to

22 3 2 5

2 3 2

H C H O1.8 10

HC H O 0.10axK

x

+ −−

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= = × =−

and x2 = 1.8 × 10−6

So, take the square root of each side to get 2 6 1.8 10

0.00134 H

x

x

−

+

= ×

⎡ ⎤= = ⎣ ⎦

pH = − log (.001341) = 2.87 (2 SF) which is mighty, mighty close, so it is a really good approximation. So, what’s the good news? The AP exam does not have equilibrium problems that require the quadratic formula. Feel better? No promises about your homework, though! ☺


Exercise 11 The pH of Weak Acids The hypochlorite ion (OCl−) is a strong oxidizing agent often found in household bleaches and disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is a much stronger base than Cl-, for example) and forms the weakly acidic hypochlorous acid (HOCl, Ka = 3.5 × 10-8). Calculate the pH of a 0.100 M aqueous solution of hypochlorous acid.

pH = 4.23 Determination of the pH of a Mixture of Weak Acids


Only the acid with the largest Ka value will contribute an appreciable [H+]. Determine the pH based

on this acid and ignore any others. Exercise 12 The pH of Weak Acid Mixtures Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 × 10−10) and 5.00 M HNO2 (Ka = 4.0 × 10−4). Also calculate the concentration of cyanide ion (CN−) in this solution at equilibrium.

pH = 1.35 [CN−] = 1.4 × 10−8 M

Exercise 13 Calculating Percent Dissociation Calculate the percent dissociation of acetic acid (Ka = 1.8 × 10−5) in each of the following solutions. a. 1.00 M HC2H3O2 b. 0.100 M HC2H3O2

A: = 0.42 %

B: = 1.3 %


Exercise 14 Calculating Ka from Percent Dissociation Lactic acid (HC3H5O3) is a waste product that accumulates in muscle tissue during exertion, leading to pain and a feeling of fatigue. In a 0.100 M aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of Ka for this acid.

Ka= 1.4 × 10−4

Determination of the pH of a weak base is very similar to the determination of the pH of a weak acid.

Follow the same steps. Remember, however, that x is the [OH−] and taking the negative log of x will give you the pOH and not the pH!

Exercise 15 The pH of Weak Bases I Calculate the pH for a 15.0 M solution of NH3 (Kb = 1.8 × 10−5).

pH = 12.20

Exercise 16 The pH of Weak Bases II


Calculate the pH of a 1.0 M solution of methylamine (Kb = 4.38 × 10−4).

pH = 12.32

Calculating pH of polyprotic acids

Acids with more than one ionizable hydrogen will ionize in steps. Each dissociation has its own Ka value.

The first dissociation will be the greatest and subsequent dissociations will have much smaller equilibrium constants. As each H+ is removed, the remaining acid gets weaker and therefore has a smaller Ka. As the negative charge on the acid increases it becomes more difficult to remove the positively charged proton. Example: Consider the dissociation of phosphoric acid.

H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4

− (aq) Ka1 = 7.5 × 10-3

H2PO4-(aq) + H2O(l) H3O+

(aq) + HPO42−

(aq) Ka2 = 6.2 × 10-8

HPO42-

(aq) + H2O(l) H3O+(aq) + PO4

3−(aq) Ka3 = 4.8 × 10-13

Looking at the Ka values, it is obvious that only the first dissociation will be

important in determining the pH of the solution.


Except for H2SO4, polyprotic acids have Ka2 and Ka3 values so much weaker than their Ka1 value

that the 2nd and 3rd (if applicable) dissociation can be ignored. The [H+] obtained from this 2nd and 3rd dissociation is negligible compared to the [H+] from the 1st dissociation. Because H2SO4 is a strong acid in its first dissociation and a weak acid in its second, we need to consider both if the concentration is more dilute than 1.0 M. The quadratic equation is needed to work this type of problem.

Exercise 17 The pH of a Polyprotic Acid Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentrations of the species H3PO4, H2PO4

-, HPO42-, and PO4

3-.

pH = 0.72 [H3PO4] = 4.8 M

[H2PO4-] = 0.19 M

[HPO42-] = 6.2 × 10−8 M

[PO43-] = 1.6 × 10−19 M

Exercise 18 The pH of a Sulfuric Acid Calculate the pH of a 1.0 M H2SO4 solution.

pH = 0.00


Exercise 19 The pH of a Sulfuric Acid Calculate the pH of a 1.0 × 10−2 M H2SO4 solution.

pH = 1.84

ACID-BASE PROPERTIES OF SALTS: HYDROLYSIS (Splitting of water)

Salts are produced when an acid and base react. Salts are not always neutral. Some hydrolyze with water to produce aqueous solutions with pHs other than 7.00.

Neutral Salts--Salts that are formed from the cation of a strong base reacting with the anion of a strong acid are neutral. Beware of solubility issues! One salt such is NaNO3. Think about which acid reacted with which base to form the salt…if both the acid and base are strong, then the salt is neutral.

Basic Salts--Salts that are formed from the cation of a strong base reacting with the anion of a weak

acid are basic. Again, beware of solubility issues! The anion hydrolyzes the water molecule to produce hydroxide ions and thus a basic solution. K2C2H3O2 should be basic since C2H3O2

− is the CB of the weak acid HC2H3O2

, while K+ does not hydrolyze appreciably. C2H3O2

− + H2O OH− + HC2H3O2

strong base weak acid

Acidic Salts- Salts that are formed from the cation of a weak base reacting with the anion of a strong acid are acidic. The cation hydrolyzes the water molecule to produce hydronium ions and thus an acidic solution. NH4Cl should be weakly acidic, since NH4

+ hydrolyzes to give an acidic solution, while Cl− does not hydrolyze.

NH4+ + H2O H3O+ + NH3

strong acid weak base

If both the cation and the anion contribute to the pH situation, compare Ka to Kb.

If Kb is larger, basic! The converse is also true.


Here’s how to think this through: 1. Look at the salt and ask yourself which acid and which base reacted to form it?

2. Ask yourself “strong or weak?” for each.

3. Embrace the fact that “strong wins” and predict whether the salt is acidic or basic based

on that victory. a. If you predict basic, write ⎯⎯→←⎯⎯ OH−

b. If you predict acidic, write ⎯⎯→←⎯⎯ H+

4. Relish in the fact that “strong is a spectator”. Which means the remaining ion of the salt is the reactant along with water.

5. Write water as HOH to make it easier to see how the hydroxide or hydrogen ion was formed.

Example: Question: What is the qualitative pH of Fe(NO3)3?

1. Which acid reacted? Nitric 2. Strong or weak? Strong 3. Which base reacted? Iron(III) hydroxide 4. Strong or weak? Weak

Strong wins! ∴acidic so we write: + HOH ⎯⎯→←⎯⎯ H+ to get started Strong is also a spectator so, cross out the spectator and the remaining ion is the other reactant; Fe(NO3)3 Complete hydrolysis reaction: Fe3+ + 3 HOH ⎯⎯→←⎯⎯ 3 H+ + Fe(OH)3

Exercise 20 The Acid-Base Properties of Salts Predict whether an aqueous solution of each of the following salts will be acidic, basic, or neutral. Prove with appropriate equations. a. NaC2H3O2

b. NH4NO3 c. Al2(SO4)3


Exercise 21 Salts as Weak Bases Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2 × 10−4.

pH = 8.31 Exercise 22 Salts as Weak Acids I Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is 1.8 × 10−5.

pH = 5.13 Exercise 23 Salts as Weak Acids II Calculate the pH of a 0.010 M AlCl3 solution. The Ka value for Al(H2O)6

3+ is 1.4 × 10−5.

pH = 3.43


THE LEWIS CONCEPT OF ACIDS AND BASES

acid--can accept a pair of electrons to form a coordinate covalent bond base--can donate a pair of electrons to form a coordinate covalent bond

Yes, this is the dot guy and the structures guy--he was extremely busy making your life difficult! BF3--most famous of all!!

Exercise 24 Tell whether each of the following is a Lewis acid or base: Draw structures as proof. a) PH3 b) BCl3 c) H2S d) SF4


Exercise 25 Lewis Acids and Basis For each reaction, identify the Lewis acid and base. a. Ni2+(aq) + 6NH3(aq) → Ni(NH3)6

2+(aq) b. H+(aq) + H2O(aq) H3O+(aq)

A: Lewis acid = nickel(II) ion Lewis base = ammonia

B: Lewis acid = proton Lewis base = water molecule

AP* Chemistry Buffers Made Easy

(At least as easy as they are going to get!)


There are many ways to solve buffer problems. Some folks prefer using simply the equilibrium expression, some prefer Henderson-Hasselbalch and some I’ve converted to my way, which is probably closer to the latter than the former. Some of the methods are more time intensive than others. Never lose sight of the fact that your students need a method that works for THEM. This method should be reliable and quick since they will always be under a time crunch! After moderate success with every method I could find, this handout presents the approach I’ve found the most student-friendly. Is it the method WE learned? NO! If you’ve had trouble teaching buffers, or your students have had trouble working buffer problems, give this a try…. First, we need to know what a buffer is, what it does, and how it does it. A buffer is just a case of the common ion effect. IS: A buffer is a solution of a weak acid or base and its salt [which is its conjugate]. DOES: A buffer resists a change in pH. HOW IT WORKS: Since a buffer consists of both an acid or base and its conjugate which is its complement, an acid and a base are present in all buffer solutions. If a small amount of strong acid is added to the buffer, there is a base component ready and waiting to neutralize the “invader”. There are a few background concepts/skills at which students need to be proficient before any method of solving buffer problems becomes effective:

• Students must know the conjugate A/B concept to be successful at buffer problems. This means they understand that HA has A- as it’s conjugate and that NaA is not only the salt, but that any soluble salt releases the conjugate

• pH = − log [H+] • Ka × Kb = Kw • Realize that any titration involving a weak A/B is a buffer problem. If I titrate HA with NaOH,

then as soon as the first drop splashes into the container, I’ve made NaA− which promptly dissociates into A–

I use one and only one formula to solve buffer problems, this lessens student [and in my case, teacher] confusion! Here it is:

AcidH

BaseaK+⎡ ⎤⎣ ⎦⎡ ⎤

⎣ ⎦ ⎡ ⎤⎣ ⎦=

2 Buffers Made Easy

If acid is added to the buffer, simply add acid to the numerator AND subtract the same quantity from the base since it was self-sacrificing and neutralized the acid. If base is added, simply add the base to the denominator and subtract from the numerator. Add or subtract in moles NOT molarity! Moles = M ×V When equal concentrations (or moles) of Acid and Base are present [which occurs at the ½ equivalence point of a titration] the ratio of acid to base equals ONE and therefore, the pH = pKa. IF you are asked to construct a buffer of a specific pH and given a table of Ka’s, choose a Ka with an exponent close to the desired pH and use equal concentrations of the acid and base. Let’s try it! Here’s a common calculate the pH of a buffer before and after an “invader” is added. Convert to moles when doing the adding and subtracting! The Ka of formic acid is 1.8 × 10–4 Exercise 1 Calculate the pH of 0.500 L of a buffer solution composed of 0.50 M formic acid and 0.70 M sodium formate before and after adding 10.0 mL of 1.00 M HCl. PREPARING BUFFER SOLUTIONS: Use 0.10 M to 1.0 M solutions of reagents & choose an acid whose Ka is near the [H3O+] concentration we want. The pKa should be as close to the pH desired as possible. Adjust the ratio of weak A/B and its salt to fine tune the pH. Exercise 2 Using an acetic acid\sodium acetate buffer solution, what ratio of acid to conjugate base will you need to maintain the pH at 5.00? Explain how you would make up such a solution. It is the relative # of moles of acid/CB or base/CA that is important since they are in the same solution and share the same solution volume.

3 Buffers Made Easy

This allows companies to make concentrated versions of buffer and let the customer dilute--this will not affect the # of moles present--just the dilution of those moles. What if you’re given a Kb? Exercise 3 A buffered solution contains 0.25 M NH3 ( Kb = 1.8 × 10–5) and 0.40 M NH4Cl. a) Calculate the pH of this solution. b) Calculate the pH when 0.10 mol of gaseous HCl is added to 1.0 L of the buffered solution. Exercise 4 A chemist needs a solution buffered at pH 4.30 and can choose from the following list of acids and their soluble salt: a. chloroacetic acid Ka = 1.35 × 10–3 b. propanoic acid Ka = 1.3 × 10–5 c. benzoic acid Ka = 6.4 × 10–5 d. hypochlorus acid Ka = 3.5 × 10–8 Calculate the ratio of A/B required for each system to yield a pH of 4.30. Which system works best?

4 Buffers Made Easy

ACID-BASE TITRATION CURVES Only when the acid AND base are both strong is the pH at the equivalence point 7. Any other conditions and you get to do a nasty equilibrium problem. It’s really a stoichiometry problem with a limiting reactant. The “excess” is responsible for the pH Weak acid + strong base eq. pt. > pH 7 Strong acid + weak base eq. pt. < pH 7 There is a distinction between the equivalence point and the end point. The end point is when the indicator changes color. If you’ve made a careful choice, the equivalence point, when the number of moles of acid = number of moles of base, will be achieved at the same time. Exercise 5 For the titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH calculate the pH of the solution at the following selected points of the titration: a) NO NaOH has been added: b) 10.0 mL of NaOH has been added: c) 20.0 mL (total as opposed to additional) of NaOH has been added: d) 50.0 mL (total) of NaOH has been added: e) 100.0 mL (total) of NaOH has been added: f) 200.0 mL (total) of NaOH has been added:

FIVE POINTS OF INTEREST ALONG A TITRATION CURVE for weak acids/bases: 1. The pH before the titration begins. Treat as usual, the acid or base in the flask determines the

pH. If weak, a RICE table is in order. 2. The pH on the way to the equivalence point. You are in the “land of buffer” as soon as the

first drop from the buret makes a splash and reacts to form the salt. Whatever is in the

burette is the “added” part. Use AcidH

BaseaK+⎡ ⎤⎣ ⎦⎡ ⎤

⎣ ⎦ ⎡ ⎤⎣ ⎦= to solve for the hydrogen ion

concentration and subsequently the pH. Either the acid or the base [whichever is in the burette] starts at ZERO.

3. The pH at the midpoint of the titration (½ equivalence point): Acid

HBaseaK+⎡ ⎤⎣ ⎦ on the

way to the equivalence point as explained above; once the midpoint is reached, [H+] = Ka since ½ of the acid or base has been neutralized, AND the resulting solution in the beaker is composed of the half that remains AND the salt. That means the A/B term is a big tall ONE so, it follows that pH = pKa.

⎡ ⎤⎣ ⎦ ⎡ ⎤⎣ ⎦

=

4. The pH at the equivalence point.—you are simply calculating the pH of the salt, all the acid

or base is now neutralized [to salt + water!]. Write the hydrolysis reaction as the “R” of your RICE table. [Examples follow the fifth point of interest…]

5. The pH beyond the equivalence point—it’s stoichiometry again with a limiting reactant.

Calculate the molarity of the EXCESS and solve for either pH directly (excess H+) or pOH (excess OH−) and subtract it from 14 to arrive at pH. Be sure to track the total volume when calculating the molarity!

LEARNING TO WRITE SALT HYDROLYSIS REACTIONS Recall that when quantities of acid and base react, a neutralization reaction takes place and the products of such a reaction are salt and water. The trouble is that not all salts are neutral. However, reasoning through the pH of a salt is easy as is writing the subsequent hydrolysis reaction. Ponder the following points:

• “Hydrolysis” translates into “water splitting” but the salt is also split. The rest of this will be much easier if you think of H2O as HOH and as a reactant. Essentially, a hydrolysis reaction is the reverse of a neutralization reaction and is written in net ionic form for all soluble salts.

• IF a strong base is reacted with a weak acid, “Strong wins!” and the salt is basic, thus OH– forms as a product.

• IF a strong acid is reacted with a weak base, “Strong wins!” and the salt is acidic, thus H+ forms as a product.

• The “Strong” portion of the acid or base is source of the “Spectator” ion. • The remaining ion (or neutral molecule) is the reactant that reacts with the water.

5 Buffers Made Easy

Example 1 Write the hydrolysis reaction for the salt, NaCN.

Explanation: First, recognize NaCN as a salt if that information is not given—the polyatomic ion is a dead give away! Next, ask yourself which acid reacted with which base to generate the salt in the first place? Which base? NaOH. Which acid? HCN. Now, ask yourself: Is the base strong or weak? NaOH, strong. Is the acid strong or weak? HCN, weak. “Strong wins!”, so it is a basic salt, AND “Strong is also the Spectator.” So, sodium ion is not in the net ionic hydrolysis equation, leaving CN– as the reacting species. Ah, you are ready to write the hydrolysis reaction. Start the same way every time you try to write a hydrolysis reaction…draw the arrow in the center of your answer space! → It’s a hydrolysis reaction, water is a reactant and splits into H+ and OH–, so write it as HOH until you get the hang of this process. Your reaction looks like this so far: + HOH → Recap what you now know: NaCN is a basic salt. Since it is basic, OH– is a product. Your reaction improves… + HOH → OH– More knowledge: Na+ is a spectator (since “Strong, Spectator”—it’s an alliteration, get it? Strong and Spectator both start with an “S”.) leaving CN– as the other reactant. Your reaction improves yet again…

CN– + HOH → OH– Now, think…think…think. IF HOH released OH– as a product, then H+ is left to react with the negative cyanide ion. Your final improvement to the reaction finishes it off…

CN– + HOH → OH– + HCN Why is this such a big deal? A hydrolysis reaction is what you must plop into your RICE table to solve for the pH (or pOH) of an acid-base titration where either participant is weak. BECAUSE, at the equivalence point moles acid = moles base = moles salt formed. Just be sure and calculate the molarity of the salt by tracking the total volume before putting its concentration into the rice table.

6 Buffers Made Easy

7 Buffers Made Easy

Example 2 Write the hydrolysis reactions for the following salts:

a) KNO2 b) NH4NO3 c) CH3NH3Cl d) NaClO2

a) Basic salt NO2

– + HOH → OH– + HNO2

b) Acidic salt NH4+ + HOH → H+ + NH4OH

c) Acidic salt CH3NH3+ + HOH → H+ + CH3NH3OH

e) Basic salt ClO2–

+ HOH → OH– + HClO2

Exercise 6 Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka = 6.2 × 10–10) when dissolved in water. If a 50.0 mL sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution a) after 8.00 mL of 0.100 M NaOH has been added. b) at the halfway point of the titration. c) at the equivalence point of the titration.

CHOOSING INDICATORS FOR TITRATIONS Choose and indicator with a Ka near that of the acid you’re titrating AND whose color changes strongly at the equivalence point. Those blasted “e” words! moles acid equal moles base at the equivalence point…the color changes at the end point. Choose your indicator wisely, and the two coincide! Exercise 7 Use table 15.8 to decide which indicator would be best to use in the titration of ammonia with hydrochloric acid. Exercise 8 Bromthymol blue has a Ka value of 1.0 × 10–7, is yellow in its HIn form and blue in its In- form. Suppose we put a few drops of this indicator in a strongly acidic solution. If the solution is then titrated with NaOH, at what pH will the indicator color change first be visible? Shall we try this with some retired AP questions? [Go to the AP Buffers Free Response Questions handout.]

AP* Buffer Equilibrium Free Response Questions page 1

(1) AP® is a registered trademark of the College Board. The College Board was not involved in the production of and does not endorse this product. (2) Test Questions are Copyright © 1984-2008 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited.

Essay Questions 1972 Given a solution of ammonium chloride, what additional reagent or reagents are needed to prepare a buffer from the ammonium chloride solution? Explain how this buffer solution resists a change in pH when:

(a) Moderate amounts of strong acid are added. (b) Moderate amounts of strong base are added. (c) A portion of the buffer solution is diluted with an equal volume of water.

1983 (a) Specify the properties of a buffer solution. Describe the components and the composition of effective

buffer solutions. (b) An employer is interviewing four applicants for a job as a laboratory technician and asks each how to

prepare a buffer solution with a pH close to 9. Archie A. says he would mix acetic acid and sodium acetate solutions. Beula B. says she would mix NH4Cl and HCl solutions. Carla C. says she would mix NH4Cl and NH3 solutions. Dexter D. says he would mix NH3 and NaOH solutions.

Which of these applicants has given an appropriate procedure? Explain your answer, referring to your discussion in part (a). Explain what is wrong with the erroneous procedures. (No calculations are necessary, but the following acidity constants may be helpful: acetic acid, Ka = 1.8 × 10–5 NH4

+, Ka = 5.6 × 10–10)

1988

A 30.00 milliliter sample of a weak monoprotic acid was titrated with a standardized solution of NaOH. A pH meter was used to measure the pH after each increment of NaOH was added, and the curve above was constructed. (a) Explain how this curve could be used to determine the molarity of the acid. (b) Explain how this curve could be used to determine the dissociation constant Ka of the weak monoprotic

acid. (c) If you were to repeat the titration using a indicator in the acid to signal the endpoint, which of the following

indicators should you select? Give the reason for your choice. Methyl red Ka = 1 × 10–5 Cresol red Ka = 1 × 10–8 Alizarin yellow Ka = 1 × 10–11

(d) Sketch the titration curve that would result if the weak monoprotic acid were replaced by a strong

monoprotic acid, such as HCl of the same molarity. Identify differences between this titration curve and the curve shown above.

AP* Buffer Equilibrium Free Response Questions page 3 1992 The equations and constants for the dissociation of three different acids are given below.

HCO3– H+ + CO3

2– Ka = 4.2 × 10–7 H2PO4

– H+ + HPO42– Ka = 6.2 × 10–8

HSO4– H+ + SO4

2– Ka = 1.3 × 10–2 (a) From the systems above, identify the conjugate pair that is best for preparing a buffer with a

pH of 7.2. Explain your choice. (b) Explain briefly how you would prepare the buffer solution described in (a) with the conjugate pair you

have chosen. (c) If the concentrations of both the acid and the conjugate base you have chosen were doubled, how would

the pH be affected? Explain how the capacity of the buffer is affected by this change in concentrations of acid and base.

(d) Explain briefly how you could prepare the buffer solution in (a) if you had available the solid salt of the

only one member of the conjugate pair and solution of a strong acid and a strong base. 1998

An approximately 0.1-molar solution of NaOH is to be standardized by titration. Assume that the following materials are available.

Clean, dry 50 mL buret Analytical balance 250 mL Erlenmeyer flask Phenolphthalein indicator solution Wash bottle filled with distilled water

Potassium hydrogen phthalate, KHP, a pure solid monoprotic acid (to be used as the primary standard)

(a) Briefly describe the steps you would take, using materials listed above, to standardize the NaOH solution.

(b) Describe (i.e., set up) the calculations necessary to determine the concentration of the NaOH solution.


(c) After the NaOH solution has been standardized, it is used to titrate a weak monoprotic acid, HX. The equivalence point is reached when 25.0 mL of NaOH solution has been added. In the space provided at the right, sketch the titration curve, showing the pH changes that occur as the volume of NaOH solution added increases from 0 to 35.0 mL. Clearly label the equivalence point on the curve.

(d) Describe how the value of the acid-dissociation constant, Ka, for the weak acid HX could be determined from the titration curve in part (c).

(e) The graph below shows the results obtained by titrating a different weak acid, H2Y, with the standardized NaOH solution. Identify the negative ion that is present in the highest concentration at the point in the titration represented by the letter A on the curve.

AP* Buffer Equilibrium Free Response Questions page 5 2000 A volume of 30.0 mL of 0.10 M NH3(aq) is titrated with 0.20 M HCl(aq). The value of the base-dissociation constant, Kb, for NH3 in water is 1.8 × 10−5 at 25°C.

(a) Write the net-ionic equation for the reaction of NH3(aq) with HCl(aq). (b) Using the axes provided below, sketch the titration curve that results when a total of

40.0 mL of 0.20 M HCl(aq) is added dropwise to the 30.0 mL volume of 0.10 M NH3(aq).

(c) From the table below, select the most appropriate indicator for the titration. Justify your choice.

(d) If equal volumes of 0.10 M NH3(aq)and 0.10 M NH4Cl(aq) are mixed, is the resulting solution acidic, neutral, or basic? Explain.


Problems 1970 (a) What is the pH of a 2.0 molar solution of acetic acid. Ka acetic acid = 1.8 × 10–5? (b) A buffer solution is prepared by adding 0.10 liter of 2.0 molar acetic acid solution to 0.1 liter of a

1.0 molar sodium hydroxide solution. Compute the hydrogen ion concentration of the buffer solution.

(c) Suppose that 0.10 liter of 0.50 molar hydrochloric acid is added to 0.040 liter of the buffer prepared in (b). Compute the hydrogen ion concentration of the resulting solution.

1977 The value of the ionization constant, Ka, for hypochlorous acid, HOCl, is 3.1 × 10–8. (a) Calculate the hydronium ion concentration of a 0.050 molar solution of HOCl. (b) Calculate the concentration of hydronium ion in a solution prepared by mixing equal volumes of

0.050 molar HOCl and 0.020 molar sodium hypochlorite, NaOCl. (c) A solution is prepared by the disproportionation reaction below. Cl2 + H2O HCl + HOCl

Calculate the pH of the solution if enough chlorine is added to water to make the concentration of HOCl equal to 0.0040 molar.

1982 A buffer solution contains 0.40 mole of formic acid, HCOOH, and 0.60 mole of sodium formate, HCOONa, in 1.00 liter of solution. The ionization constant, Ka, of formic acid is 1.8 × 10–4. (a) Calculate the pH of this solution. (b) If 100. milliliters of this buffer solution is diluted to a volume of 1.00 liter with pure water, the pH

does not change. Discuss why the pH remains constant on dilution. (c) A 5.00 milliliter sample of 1.00 molar HCl is added to 100. milliliters of the original buffer solution.

Calculate the [H3O+] of the resulting solution. (d) A 800. milliliter sample of 2.00 molar formic acid is mixed with 200. milliliters of 4.80 molar NaOH.

Calculate the [H3O+] of the resulting solution. 1984 Sodium benzoate, C6H5COONa, is the salt of the weak acid, benzoic acid, C6H5COOH. A 0.10 molar solution of sodium benzoate has a pH of 8.60 at room temperature. (a) Calculate the [OH−] in the sodium benzoate solution described above. (b) Calculate the value for the equilibrium constant for the reaction:

C6H5COO– + H2O C6H5COOH + OH− (c) Calculate the value of Ka, the acid dissociation constant for benzoic acid. (d) A saturated solution of benzoic acid is prepared by adding excess solid benzoic acid to pure water at

room temperature. Since this saturated solution has a pH of 2.88, calculate the molar solubility of benzoic acid at room temperature.

AP* Buffer Equilibrium Free Response Questions page 7 1991 The acid ionization constant, Ka, for propanoic acid, C2H5COOH, is 1.3 × 10–5. (a) Calculate the hydrogen ion concentration, [H+], in a 0.20 molar solution of propanoic acid. (b) Calculate the percentage of propanoic acid molecules that are ionized in the solution in (a). (c) What is the ratio of the concentration of propanoate ion, C2H5COO−, to that of propanoic acid in

a buffer solution with a pH of 5.20? (d) In a 100. milliliter sample of a different buffer solution, the propanoic acid concentration is 0.35 molar

and the sodium propanoate concentration is 0.50 molar. To this buffer solution, 0.0040 mole of solid NaOH is added. Calculate the pH of the resulting solution.

1993

CH3NH2 + H2O CH3NH3+ + OH–

Methylamine, CH3NH2, is a weak base that reacts according to the equation above. The value of the ionization constant, Kb, is 5.25 × 10–4. Methylamine forms salts such as methylamonium nitrate, (CH3NH3

+)(NO3–).

(a) Calculate the hydroxide ion concentration, [OH–], of a 0.225 molar aqueous solution of methylamine. (b) Calculate the pH of a solution made by adding 0.0100 mole of solid methylamonium nitrate to

120.0 milliliters of a 0.225 molar solution of methylamine. Assume no volume change occurs. (c) How many moles of either NaOH or HCl (state clearly which you choose) should be added to the solution

in (b) to produce a solution that has a pH of 11.00? Assume that no volume change occurs. (d) A volume of 100. milliliters of distilled water is added to the solution in (c). How is the pH of the solution

affected? Explain. 1996

HOCl OCl− + H+

Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. The acid-dissociation constant, Ka, for the reaction represented above is 3.2 × 10−8.

(a) Calculate the [H+] of a 0.14-molar solution of HOCl.

(b) Write the correctly balanced net ionic equation for the reaction that occurs NaOCl is dissolved in water and calculate the numerical value of the equilibrium constant for the reaction.

(c) Calculate the pH of a solution made by combining 40.0 milliliters of 0.14-molar HOCl and 10.0 milliliters of 0.56-molar NaOH.

(d) How many millimoles of solid NaOH must be added to 50.0 milliliters of 0.20-molar HOCl to obtain a buffer solution that has a pH of 7.49? Assume that the addition of the solid NaOH results in a negligible change in volume.

(e) Household bleach is made by dissolving chlorine gas in water, as represented below.

Cl2(g) + H2O H+ + Cl− + HOCl(aq)

Calculate the pH of such a solution if the concentration of HOCl in the solution is 0.065 molar.

AP* Buffer Equilibrium Free Response Questions page 8 2001 Answer the following questions about acetylsalicylic acid, the active ingredient in aspirin.

(a) The amount of acetylsalicylic acid in a single aspirin tablet is 325 mg, yet the tablet has a mass of 2.00 g. Calculate the mass percent of acetylsalicylic acid in the tablet.

(b) The elements contained in acetylsalicylic acid are hydrogen, carbon, and oxygen. The combustion of 3.000 g of the pure compound yields 1.200 g of water and 3.72 L of dry carbon dioxide, measured at 750. mm Hg and 25°C. Calculate the mass, in g, of each element in the 3.000 g sample.

(c) A student dissolved 1.625 g of pure acetylsalicylic acid in distilled water and titrated the resulting solution to the equivalence point using 88.43 mL of 0.102 M NaOH(aq). Assuming that acetylsalicylic acid has only one ionizable hydrogen, calculate the molar mass of the acid.

(d) A 2.00 × 10−3 mole sample of pure acetylsalicylic acid was dissolved in 15.00 mL of water and then titrated with 0.100 M NaOH(aq). The equivalence point was reached after 20.00 mL of the NaOH solution had been added. Using the data from the titration, shown in the table below, determine (i) the value of the acid dissociation constant, Ka , for acetylsalicylic acid and (ii) the pH of the solution after a total volume of 25.00 mL of the NaOH solution had been

added (assume that volumes are additive).


2002B

HC3H5O3(aq) H+(aq) + C3H5O3−(aq)

Lactic acid, HC3H5O3, is a monoprotic acid that dissociates in aqueous solution, as represented by the equation above. Lactic acid is 1.66 percent dissociated in 0.50 M HC3H5O3 (aq) at 298 K. For parts (a) through (d) below, assume the temperature remains at 298 K.

(a) Write the expression for the acid-dissociation constant, Ka , for lactic acid and calculate its value.

(b) Calculate the pH of 0.50 M HC3H5O3.

(c) Calculate the pH of a solution formed by dissolving 0.045 mole of solid sodium lactate, NaC3H5O3, in 250. mL of 0.50 M HC3H5O3. Assume that volume change is negligible.

(d) A 100. mL sample of 0.10 M HCl is added to 100. mL of 0.50 M HC3H5O3. Calculate the molar concentration of lactate ion, C3H5O3

−, in the resulting solution.

2003

C6H5NH2(aq) + H2O(l) C6H5NH3+(aq) + OH–(aq)

Aniline, a weak base, reacts with water according to the reaction represented above. (a) Write the equilibrium constant expression, Kb , for the reaction represented above.

(b) A sample of aniline is dissolved in water to produce 25.0 mL of a 0.10 M solution.

The pH of the solution is 8.82. Calculate the equilibrium constant, Kb , for this reaction.

(c) The solution prepared in part (b) is titrated with 0.10 M HCl. Calculate the pH of the solution when 5.0 mL of the acid has been added.

(d) Calculate the pH at the equivalence point of the titration in part (c).

(e) The pKa values for several indicators are given below. Which of the indicators listed is most suitable for this titration? Justify your answer.

Indicator pKa

Erythrosine 3 Litmus 7

Thymolphthalein 10


2005

HC3H5O2(aq) C3H5O2–(aq) + H+(aq) Ka = 1.34 × 10–5

Propanoic acid, HC3H5O2 , ionizes in water according to the equation above.

(a) Write the equilibrium-constant expression for the reaction.

(b) Calculate the pH of a 0.265 M solution of propanoic acid.

(c) A 0.496 g sample of sodium propanoate, NaC3H5O2 , is added to a 50.0 mL sample of

a 0.265 M solution of propanoic acid. Assuming that no change in the volume of the solution occurs, calculate each of the following.

(i) The concentration of the propanoate ion, C3H5O2

–(aq), in the solution (ii) The concentration of the H+(aq) ion in the solution

The methanoate ion, HCO2

–(aq), reacts with water to form methanoic acid and hydroxide ion, as shown in the following equation.

HCO2–(aq) + H2O(l) HCO2H(aq) + OH–(aq)

(d) Given that [OH–] is 4.18 × 10–6 M in a 0.309 M solution of sodium methanoate, calculate each of

the following.

(i) The value of Kb for the methanoate ion, HCO2–(aq)

(ii) The value of Ka for methanoic acid, HCO2H

(e) Which acid is stronger, propanoic acid or methanoic acid? Justify your answer. 2006B

C6H5COOH(s) C6H5COO– (aq) + H+(aq) Ka = 6.46 × 10–5 Benzoic acid, C6H5COOH, dissociates in water as shown in the equation above. A 25.0 mL sample of an

aqueous solution of pure benzoic acid is titrated using standardized 0.150 M NaOH.

(a) After addition of 15.0 mL of the 0.150 M NaOH, the pH of the resulting solution is 4.37. Calculate each of the following.

(i) [H+] in the solution (ii) [OH–] in the solution (iii) The number of moles of NaOH added (iv) The number of moles of C6H5COO– (aq) in the solution (v) The number of moles of C6H5COOH in the solution


(b) State whether the solution at the equivalence point of the titration is acidic, basic, or neutral. Explain your reasoning.

In a different titration, a 0.7529 g sample of a mixture of solid C6H5COOH and solid NaCl is dissolved in water and titrated with 0.150 M NaOH. The equivalence point is reached when 24.78 mL of the base solution is added.

(c) Calculate each of the following.

(i) The mass, in grams, of benzoic acid in the solid sample (ii) The mass percentage of benzoic acid in the solid sample

AP* Chemistry Demystifying Titration Curves


This handout is meant to accompany the instructional video found at either http://www.vimeo.com/19007364 or http://www.edutube.org/en/video/ap-chemistry-interpreting-titration-curves. This won’t make a lick of sense unless you watch the video containing the explanations!

AP* Acid-Base Equilibrium Free Response Questions page 1

(1) AP® is a registered trademarK of the College Board. The College Board was not involved in the production of and does not endorse this product. (2) Test Questions are Copyright © 1984-2008 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited.

Essay Questions 1986

H2SO3 HSO3− HClO4 HClO3 H3BO3

Oxyacids, such as those above, contain an atom bonded to one or more oxygen atoms; one or more of these oxygen atoms may also be bonded to hydrogen. (a) Discuss the factors that are often used to predict correctly the strengths of the oxyacids listed above. (b) Arrange the examples above in the order of increasing acid strength. 1990 Give a brief explanation for each of the following. (a) For the diprotic acid H2S, the first dissociation constant is larger than the second dissociation constant

by about 105 (K1 = 105 K2). (b) In water, NaOH is a base but HOCl is an acid. (c) HCl and HI are equally strong acids in water but, in pure acetic acid, HI is a stronger acid than HCl. (d) When each is dissolved in water, HCl is a much stronger acid than HF. 1994 A chemical reaction occurs when 100. milliliters of 0.200-molar HCl is added dropwise to 100. milliliters of 0.100-molar Na3PO4 solution. (a) Write the two net ionic equations for the formation of the major products. (b) Identify the species that acts as both a Brönsted acid and as a Brönsted base in the equation in (a),

Draw the Lewis electron-dot diagram for this species. (c) Sketch a graph using the axes provided, showing the shape of the titration curve that results when

100. milliliters of the HCl solution is added slowly from a buret to the Na3PO4 solution. Account for the shape of the curve.

(d) Write the equation for the reaction that occurs if a few additional milliliters of the HCl solution are added

to the solution resulting from the titration in (c).

AP* Acid-Base Equilibrium Free Response Questions page 2 2000 A volume of 30.0 mL of 0.10 M NH3(aq) is titrated with 0.20 M HCl(aq). The value of the base-dissociation constant, Kb , for NH3 in water is 1.8 × 10−5 at 25°C. (a) Write the net-ionic equation for the reaction of NH3(aq) with HCl(aq). (b) Using the axes provided below, sketch the titration curve that results when a total of 40.0 mL of 0.20 M HCl(aq) is added dropwise to the 30.0 mL volume of 0.10 M NH3(aq) .

(c) From the table below, select the most appropriate indicator for the titration. Justify your choice.

(d) If equal volumes of 0.10 M NH3(aq) and 0.10 M NH4Cl(aq) are mixed, is the resulting solution acidic, neutral, or basic? Explain.

2010B

Alizarin yellow 10 – 12

(e) What is the difference between the equivalence point of a titration and the end point

of a titration?

(f) On the grid provided on the next page, sketch the titration curve that would result if the solutions in the beaker and buret were reversed (i.e., if 40.0 mL of the solution used in the buret in the previous titration were titrated with the solution that was in the beaker).

5. A solution of 0.100 M HCl and a solution of 0.100 M NaOH are prepared. A 40.0 mL sample of one of the solutions is added to a beaker and then titrated with the other solution. A pH electrode is used to obtain the data that are plotted in the titration curve shown above.

(a) Identify the solution that was initially added to the beaker. Explain your reasoning.

(b) On the titration curve above, circle the point that corresponds to the equivalence point.

(c) At the equivalence point, how many moles of titrant have been added?

(d) The same titration is to be performed again, this time using an indicator. Use the information in the table below to select the best indicator for the titration. Explain your choice.

Indicator pH Range of Color Change

Methyl violet 0 – 1.6

Methyl red 4 – 6

Alizarin yellow 10 – 12

(e) What is the difference between the equivalence point of a titration and the end point of a titration?

(f) On the grid provided below, sketch the titration curve that would result if the solutions in the beaker and buret were reversed (i.e., if 40.0 mL of the solution used in the buret in the previous titration were titrated with the solution that was in the beaker).

rmccormick

Rectangle

rmccormick

Rectangle

AP* Acid-Base Equilibrium Free Response Questions page 4

Problems 1987 The percentage by weight of nitric acid, HNO3, in a sample of concentrated nitric acid is to be determined. (a) Initially a NaOH solution was standardized by titration with a sample of potassium hydrogen

phthalate, KHC8H4O4, a monoprotic acid often used as a primary standard. A sample of pure KHC8H4O4 weighing 1.518 grams was dissolved in water and titrated with the NaOH solution. To reach the equivalence point, 26.90 milliliters of base was required. Calculate the molarity of the NaOH solution. (Molecular weight: KHC8H4O4 = 204.2)

(b) A 10.00 milliliter sample of the concentrated nitric acid was diluted with water to a total volume of 500.00 milliliters. Then 25.00 milliliters of the diluted acid solution was titrated with the standardized NaOH solution prepared in part (a). The equivalence point was reached after 28.35 milliliters of the base had been added. Calculate the molarity of the concentrated nitric acid.

(c) The density of the concentrated nitric acid used in this experiment was determined to be 1.42 grams per milliliter. Determine the percentage by weight of HNO3 in the original sample of concentrated nitric acid.

1996

Concentrated sulfuric acid (18.4-molar H2SO4) has a density of 1.84 grams per milliliter. After dilution with water to 5.20-molar, the solution has a density of 1.38 grams per milliliter and can be used as an electrolyte in lead storage batteries for automobiles.

(a) Calculate the volume of concentrated acid required to prepare 1.00 liter of 5.20-molar H2SO4.

(b) Determine the mass percent of H2SO4 in the original concentrated solution.

(c) Calculate the volume of 5.20-molar H2SO4 that can be completely neutralized with 10.5 grams of sodium bicarbonate NaHCO3.

(d) What is the molality of the 5.20-molar H2SO4?

AP* Thermodynamics Free Response Questions page 1


Essay Questions 1991

BCl3(g) + NH3(g) Cl3BNH3(s) The reaction represented above is a reversible reaction. (a) Predict the sign of the entropy change, ΔS, as the reaction proceeds to the right. Explain your prediction. (b) If the reaction spontaneously proceeds to the right, predict the sign of the enthalpy change, ΔH. Explain your

prediction. (c) The direction in which the reaction spontaneously proceeds changes as the temperature is increased above a

specific temperature. Explain. (d) What is the value of the equilibrium constant at the temperature referred to in (c); that is, the specific

temperature at which the direction of the spontaneous reaction changes? Explain. 1993

2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l) The reaction represented above is spontaneous at 25°C. Assume that all reactants and products are in their standard state. (a) Predict the sign of ΔS° for the reaction and justify your prediction. (b) What is the sign of ΔG° for the reaction? How would the sign and magnitude of ΔG° be affected by an increase

in temperature to 50°C? Explain your answer. (c) What must be the sign of ΔH° for the reaction at 25°C? How does the total bond energy of the reactants

compare to that of the products? (d) When the reactants are place together in a container, no change is observed even though the reaction is known

to be spontaneous. Explain this observation. 1994

2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(g) At 298 K, the standard enthalpy change, ΔH° for the reaction represented above is -145 kilojoules. (a) Predict the sign of the standard entropy change, ΔS°, for the reaction. Explain the basis for your prediction. (b) At 298 K, the forward reaction (i.e., toward the right) is spontaneous. What change, if any, would occur in the

value of ΔG° for this reaction as the temperature is increased? Explain your reasoning using thermodynamic principles.

(c) What change, if any, would occur in the value of the equilibrium constant, Keq, for the situation described in (b)? Explain your reasoning.

(d) The absolute temperature at which the forward reaction becomes nonspontaneous can be predicted. Write the equation that is used to make the prediction. Why does this equation predict only an approximate value for the temperature?

AP* Thermodynamics Free Response Questions page 2 1997 For the gaseous equilibrium represented below, it is observed that greater amounts of PCl3 and Cl2 are produced as the temperature is increased.

PCl5(g) PCl3(g) + Cl2(g)

(a) What is the sign of ΔS° for the reaction? Explain.

(b) What change, if any, will occur in ΔG° for the reaction as the temperature is increased. Explain your reasoning in terms of thermodynamic principles.

(c) If He gas is added to the original reaction mixture at constant volume and temperature, what will happen to the partial pressure of Cl2? Explain.

(d) If the volume of the original reaction is decreased at constant temperature to half the original volume, what will happen to the number of moles of Cl2 in the reaction vessel? Explain.

1999

Answer the following questions in terms of thermodynamic principles and concepts of kinetic molecular theory.

(a) Consider the reaction represented below, which is spontaneous at 298 K.

CO2(g) + 2 NH3(g) → CO(NH2)2(s) + H2O(l); ΔH°298 = −134 kJ

(i) For the reaction, indicate whether the standard entropy change, ΔS°298, is positive, or negative, or zero. Justify your answer.

(ii) Which factor, the change in enthalpy, Δ H°298, or the change in entropy, ΔS°298, provides the principal driving force for the reaction at 298 K? Explain.

(iii) For the reaction, how is the value of the standard free energy change, ΔG°, affected by an increase in temperature? Explain.

(b) Some reactions that are predicted by their sign of ΔG° to be spontaneous at room temperature do not proceed at a measurable rate at room temperature.

(i) Account for this apparent contradiction.

(ii) A suitable catalyst increases the rate of such a reaction. What effect does the catalyst have on ΔG° for the reaction? Explain.

AP* Thermodynamics Free Response Questions page 3 2000

O3(g) + NO(g) → O2(g) + NO2(g) Consider the reaction represented above.

(a) Referring to the data in the table below, calculate the standard enthalpy change, ΔH°, for the reaction at 25°C. Be sure to show your work.

(b) Make a qualitative prediction about the magnitude of the standard entropy change, ∆S°, for the reaction at25°C. Justify your answer.

(c) On the basis of your answers to parts (a) and (b), predict the sign of the standard free-energy change, ∆G°, for the reaction at 25°C. Explain your reasoning.

(d) Use the information in the table below to write the rate-law expression for the reaction, and explain how you obtained your answer.

(e) The following three-step mechanism is proposed for the reaction. Identify the step that must be the slowest in order for this mechanism to be consistent with the rate-law expression derived in part (d). Explain.

Step I: O3 + NO → O + NO3

Step II: O + O3 → 2 O2

Step III: NO3 + NO → 2 NO2


2003

Answer the following questions that relate to the chemistry of nitrogen. (a) Two nitrogen atoms combine to form a nitrogen molecule, as represented by the following equation.

2 N(g) → N2(g) Using the table of average bond energies below, determine the enthalpy change, ΔH, for the reaction. (b) The reaction between nitrogen and hydrogen to form ammonia is represented below.

N2(g) + 3 H2(g) → 2 NH3(g) ΔH° = −92.2 kJ

Predict the sign of the standard entropy change, ΔS°, for the reaction. Justify your answer. (c) The value of ΔG° for the reaction represented in part (b) is negative at low temperatures but positive at high

temperatures. Explain. (d) When N2(g) and H2(g) are placed in a sealed container at a low temperature, no measurable amount of

NH3(g) is produced. Explain.

2004B

N2(g) + 2 H2(g) N2H4(g) ΔH° 298 = +95.4 kJ mol–1; ΔS° 298 = −176 J K–1 mol–1 Answer the following questions about the reaction represented above using principles of thermodynamics. (a) On the basis of the thermodynamic data given above, compare the sum of the bond strengths of the

reactants to the sum of the bond strengths of the product. Justify your answer. (b) Does the entropy change of the reaction favor the reactants or the product? Justify your answer. (c) For the reaction under the conditions specified, which is favored, the reactants or the product? Justify

your answer. (d) Explain how to determine the value of the equilibrium constant, Keq, for the reaction. (Do not do any

calculations.) (e) Predict whether the value of Keq for the reaction is greater than 1,equal to 1, or less than 1. Justify your

prediction.

Bond Average Bond Energy (kJ mol–1)

N − N 160 N = N 420 N ≡ N 950


Problems 1990

Standard Free Energies of Formation at 298 K

Substance ΔG°f 298 K, kJ mol−1 C2H4Cl2(g) −80.3 C2H5Cl(g) −60.5 HCl(g) −95.3 Cl2(g) 0

Average Bond Dissociation Energies at 298 K

Bond Energy, kJ mol−1 C−H 414 C−C 347 C−Cl 377 Cl−Cl 243 H−Cl 431

The tables above contain information for determining thermodynamic properties of the reaction below.

C2H5Cl(g) + Cl2(g) → C2H4Cl2(g) + HCl(g)

(a) Calculate the ΔH° for the reaction above, using the table of average bond dissociation energies. (b) Calculate the ΔS° for the reaction at 298 K, using data from either table as needed. (c) Calculate the value of Keq for the reaction at 298 K. (d) What is the effect of an increase in temperature on the value of the equilibrium constant? Explain

your answer.

1992 Cl2(g) + 3 F2(g) → 2 ClF3(g)

ClF3 can be prepared by the reaction represented by the equation above. For ClF3 the standard enthalpy of formation, ΔHf°,is −163.2 kilojoules/mole and the standard free energy of formation, ΔGf°, is −123.0 kilojoules/mole. (a) Calculate the value of the equilibrium constant for the reaction at 298K. (b) Calculate the standard entropy change, ΔS°, for the reaction at 298K. (c) If ClF3 were produced as a liquid rather than as a gas, how would the sign and the magnitude of ΔS

for the reaction be affected? Explain. (d) At 298K the absolute entropies of Cl2(g) and ClF3(g) are 222.96 joules per mole ·Kelvin and 281.50

joules per mole · Kelvin, respectively. (i) Account for the larger entropy of ClF3(g) relative to that of Cl2(g). (ii) Calculate the value of the absolute entropy of F2(g) at 298K.

AP* Thermodynamics Free Response Questions page 6 1995 Propane, C3H8, is a hydrocarbon that is commonly used as fuel for cooking. (a) Write a balanced equation for the complete combustion of propane gas, which yields CO2(g) and H2O(l) . (b) Calculate the volume of air at 30°C and 1.00 atmosphere that is needed to burn completely 10.0 grams of

propane. Assume that air is 21.0 percent O2 by volume. (c) The heat of combustion of propane is -2,220.1 kJ/mol. Calculate the heat of formation, ΔHf° of propane

given that ΔHf° of H2O(l) = −285.3 kJ/mol and ΔHf° of CO2(g) = −393.5 kJ/mol. (d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00 kilograms of

water (specific heat = 4.18 J/g.K), calculate the increase in temperature of water. 1998

C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)

When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to the equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer the questions that follow.

Substance Standard Heat of Formation, ΔH°f, at 25°C (kJ/mol)

Absolute Entropy, S°, at 25°C (J/mol-K)

C(graphite) 0.00 5.69

CO2(g) −393.5 213.6 H2(g) 0.00 130.6 H2O(l) −285.85 69.91 O2(g) 0.00 205.0 C6H5OH(s) ? 144.0

(a) Calculate the molar heat of combustion of phenol in kilojoules per mole at 25°C.

(b) Calculate the standard heat of formation, ΔH°f, of phenol in kilojoules per mole at 25°C.

(c) Calculate the value of the standard free-energy change, ΔG° for the combustion of phenol at 25°C.

(d) If the volume of the combustion container is 10.0 liters, calculate the final pressure in the container when the temperature is changed to 110°C. (Assume no oxygen remains unreacted and that all products are gaseous.)

AP* Thermodynamics Free Response Questions page 7 2003B In an experiment, a sample of an unknown, pure gaseous hydrocarbon was analyzed. Results showed that the sample contained 6.000 g of carbon and 1.344 g of hydrogen. (a) Determine the empirical formula of the hydrocarbon. (b) The density of the hydrocarbon at 25°C and 1.09 atm is 1.96 g L–1.

(i) Calculate the molar mass of the hydrocarbon. (ii) Determine the molecular formula of the hydrocarbon.

In another experiment, liquid heptane, C7H16 (l), is completely combusted to produce CO2(g) and H2O(l), as represented by the following equation.

C7H16(l) + 11 O2(g) → 7 CO2(g) + 8 H2O(l) The heat of combustion, ΔH°comb, for one mole of C7H16(l) is −4.85 × 103 kJ. (c) Using the information in the table below, calculate the value of ΔH°f for C7H16 (l) in kJ mol–1.

(d) A 0.0108 mol sample of C7H16 (l) is combusted in a bomb calorimeter.

(i) Calculate the amount of heat released to the calorimeter.

(ii) Given that the total heat capacity of the calorimeter is 9.273 kJ °C –1, calculate the temperature change of the calorimeter.

2004

2 Fe(s) + O2(g) → Fe2O3(s) ΔH°f = −824 kJ mol–1 Iron reacts with oxygen to produce iron(III) oxide, as represented by the equation above. A 75.0 g sample of Fe(s) is mixed with 11.5 L of O2(g) at 2.66 atm and 298 K.

(a) Calculate the number of moles of each of the following before the reaction begins.

(i) Fe(s) (ii) O2(g)

(b) Identify the limiting reactant when the mixture is heated to produce Fe2O3(s). Support your answer

with calculations.


(c) Calculate the number of moles of Fe2O3(s) produced when the reaction proceeds to completion.

(d) The standard free energy of formation, ΔG°f , of Fe2O3(s) is −740. kJ mol at 298 K.

(i) Calculate the standard entropy of formation, ΔS°f , of Fe2O3(s) at 298 K. Include units with

your answer. (ii) Which is more responsible for the spontaneity of the formation reaction at 298 K, the standard

enthalpy of formation, ΔH°f , or the standard entropy of formation, ΔS°f ? Justify your answer.

The reaction represented below also produces iron(III) oxide. The value of ΔH° for the reaction is −280. kJ per mole of Fe2O3(s) formed.

2 FeO(s) + O2(g) → Fe2O3(s)

Calculate the standard enthalpy of formation, , ΔH°f, of FeO(s).

2006B Answer the following questions about the thermodynamics of the reactions represented below. Reaction X: I2(s) + Cl2(g) ICl(g) ΔH°f = 18 kJ mol–1, ΔS°298 = 78 J K–1 mol–1 Reaction Y: I2(s) + Br2(l) IBr(g) ΔH°f = 41 kJ mol–1, ΔS°298 = 124 J K–1 mol–1 (a) Is reaction X, represented above, spontaneous under standard conditions? Justify your answer with a

calculation.

(b) Calculate the value of the equilibrium constant, Keq, for reaction X at 25°C.

(c) What effect will an increase in temperature have on the equilibrium constant for reaction X ? Explain your answer.

(d) Explain why the standard entropy change is greater for reaction Y than for reaction X.

(e) Above what temperature will the value of the equilibrium constant for reaction Y be greater than 1.0 ? Justify your answer with calculations.

(f) For the vaporization of solid iodine, I2(s) → I2(g), the value of ΔH°298 is 62 kJ mol–1. Using this information, calculate the value of ΔH°298 for the reaction represented below.

I2(g) + Cl2(g) 2 ICl(g)

AP* Chemistry Entropy and Free Energy

Domain of thermodynamics(the final and initial states)

WHAT DRIVES A REACTION TO BE THERMODYNAMICALLY FAVORABLE?

ENTHALPY (∆H) – heat exchange (exothermic reactions are generally favored) ENTROPY (∆S) – dispersal (disorder) of the matter and energy of a system (more dispersal/disorder is favored) Nature tends toward chaos! Think about your room at the end of the week. Your mom knows this concept all too well.

Thermodynamically favored processes or reactions are those that involve both a decrease in the internal energy of the components (ΔH° < 0) and an increase in entropy of the components (ΔS° > 0). These processes are necessarily “thermodynamically favored” (ΔG° < 0) or negative. If you are using an older textbook, you may see these reactions referred to as “spontaneous”. Avoid that language as you answer free-response questions!

Thermodynamically favored processes occur without outside intervention once the energy of activation has been reached. Thermodynamics does not predict nor take into account the rate of the reaction. That is the realm of kinetics. Some reactions are very fast (like combustion of hydrogen) other reactions are very slow (like graphite turning to diamond) but both are thermodynamically favored.

The first law of thermodynamics: Energy can never be created nor destroyed. Therefore, the energy of the universe is constant. This is simply a statement of the law of conservation of energy you’ve know about for quite some time.

The second law of thermodynamics: the universe is constantly increasing the dispersal of matter and energy. Rudolph Clausius “discovered” it and gave it its symbol.

The third law of thermodynamics: the entropy of a perfect crystal at 0 K is zero. [Not a lot of perfect crystals out there so, entropy values are RARELY ever zero—even elements] So what? This means the absolute entropy of a substance can then be determined at any temp. higher than zero K. (Handy to know if you ever need to defend why G & H for elements = 0. . . . BUT S does not!)

*AP is a registered trademark of the College Board, which was not involved in the production of this product. Special thanks to the contributionsof Lisa McGaw and David Wentz. © 2013 by René McCormick. All rights reserved.

What is entropy? It’s difficult to narrow this concept down to a single definition, but let’s try anyway! Entropy is a thermodynamic function that describes the number of arrangements (positions and/or energy levels) that are available to a system existing in a given state. Entropy is closely related to probability. The key concept is that the more ways a particular state can be achieved; the greater is the likelihood (probability) of finding that state. In English…nature spontaneously proceeds toward the states that have the highest probabilities of existing. Let’s use a simple example, an ideal gas expanding into an evacuated bulb. WHY is this process thermodynamically favorable? Simple, the driving force is probability. Because there are more ways of having the gas evenly spread throughout the container than there are ways for it to be in any other possible state, the gas disperses attaining a uniform distribution. The consequences are dramatic for large numbers of molecules, as you can see at right. Let’s simplify! How many possible microstates exist for a sample of four ideal gas molecules in two bulbs of equal volume? (Why ideal gas molecules? There are no pesky attractive forces to influence their motion.)

Predicting the entropy of a system is based on physical evidence:

• The greater the dispersal of matter and/or energy in a system, the larger the entropy. • The entropy of a substance always increases as it changes from solid to liquid to gas. • When a pure solid or liquid dissolves in a solvent, the entropy of the substance increases (Carbonates are an

exception! Carbonates interact with water and actually bring MORE order to the system.) • When a gas molecule escapes from a solvent, the entropy increases • Entropy generally increases with increasing molecular complexity (crystal structure: KCl vs. CaCl2) since there are

more MOVING electrons! • Reactions increasing the number of moles of particles often increase entropy.

In general, the greater the number of arrangements, the higher the entropy of the system!

2 Entropy & Free Energy

Exercise 1 Predicting Entropy Changes Predict the sign of the entropy change for each of the following processes. Justify your answers. a) Solid sugar is added to water to form a solution. b) Iodine vapor condenses on a cold surface to form crystals.

a) positive b) negative

Calculating Entropy from tables of standard values: (You already know how to solve problems like this, just pay close attention to the UNITS!)

ΔS is + when dispersal/disorder increases (favored) ΔS is – when dispersal/disorder decreases NOTE: Units are usually J/(molrxn • K) (not kJ!)

Exercise 2 Calculate the entropy change at 25°C, in J/(molrxn • K) for: 2 SO2(g) + O2(g) → 2 SO3(g) Given the following data: SO2(g) 248.1 J/(mol• K) O2(g) 205.3 J/(mol• K) SO3(g) 256.6 J/(mol • K)

‒188.3 J/(molrxn • K)

BIG MAMMA, verse 2: ∆S°rxn = Σ ∆S° (products) - Σ ∆S° (reactants)


ENTROPY CHANGES AS THEY RELATE TO REVERSIBLE PHASE CHANGES

Phase changes occur at constant temperature and represent a system which is also in equilibrium ∴ΔG = 0.

o

rxn

heat transferred J = expressed in temperature at which change occurs mol Ksurroundings

q HST T

−∆∆ = =

⋅

**Where the heat supplied (endothermic) (q > 0) or evolved (exothermic) (q < 0) is divided by the temperature in Kelvins ** It is important here to note if the reaction is endothermic or exothermic. The actual significance of this is really dependent on the temperature at which the process occurs. (i.e., If you gave a millionaire $100 it would not make much difference in his happiness; if you gave a poor college student $100 it would create a totally different expression of happiness!)

EX: water ( @ 100 °C) water (g @ 100°C)

The entropy will increase for the forward reaction (vaporizing) since the reaction produces water in a less condensed state, thus the molecules are more dispersed.

Exercise 3 Determining ∆Ssurr In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores:

Sb2S3(s) + 3 Fe(s) → 2 Sb(s) + 3 FeS(s) ∆H = ‒125 kJ/molrxn Carbon is used as the reducing agent for oxide ores:

Sb4O6(s) + 6 C(s) → 4 Sb(s) + 6 CO(g) ∆H = 778 kJ/molrxn Calculate ∆Ssurr for each of these reactions at 25°C and 1 atm.

419 J/K •molrxn & ‒2.61 × 103 J/K•molrxn

4

Entropy & Free Energy

ENTROPY SUMMARY: ∆S = + MORE DISPERSAL (FAVORED CONDITION) ∆S = ‒ LESS DISPSERSAL Whether a reaction will occur spontaneously may be determined by looking at the ∆S of the universe.

ΔS system + ΔS surroundings = ΔS universe IF ΔS universe is +, then reaction is thermodynamically favorable IF ΔS universe is ‒, then reaction is NOT thermodynamically favorable

Consider 2 H2 (g) + O2 (g) → H2O (g) ignite & rxn is fast! ΔSsystem = ‒88.9 J/( molrxn • K)

Therefore, the entropy declines mainly due to 3 moles of gas → 2 moles of gas which is a more condensed or less dispersed state!

To confirm we need to know entropy of surroundings!

ΔSsurroundings = q surroundings T

ΔHsystem = ‒483.6 kJ/mol

The First Law of Thermodynamics demands that this energy is transferred from the system to the surroundings so... …Note that the change (increase or decrease) in the entropy of the surroundings is the result of energy flow into or from the surroundings as a function of temperature, such that,

;surrounings systemsurroundings

H HS

T T∆ −∆

∆ = =

∴ ‒ΔHsystem = ΔHsurroundings OR ‒ (‒483.6 kJ) = +483.6 kJ

Now, ΔSsurroundings = ΔHsurroundings = + 483.6 kJ/molrxn = 1,620 J/( molrxn • K)

T 298 K Now we can find ΔSuniverse

ΔSsystem + ΔSsurroundings = ΔSuniverse

J J J88.9 1620 1530(mol K) (mol K) (mol K)rxn rxn rxn

− + =• • •

Even though the entropy of the system declines, the entropy change for the surroundings is SO VERY large that the overall change for the universe is positive. Bottom line: A process is thermodynamically favorable in spite of a negative entropy change as long as it is EXTREMELY exothermic. In other words, sufficient exothermicity offsets system ordering.


What is FREE ENERGY?

The calculation of Gibbs free energy, ΔG is what ultimately decides whether a reaction is thermodynamically favored or not. A NEGATIVE sign on ∆G indicates the reaction is thermodynamically favored.

∆G can be calculated several ways and links thermochemistry, entropy, equilibrium and electrochem together!

1. “Big Mamma” Equation, verse 3: ∆G°rxn = Σ ∆G° (products) − Σ ∆G° (reactants)

You already know how to calculate enthalpy and entropy, just substitute free energy values usingtables of standard values! So, calculating the standard molar free energy of formation is simply thesame song, 3rd verse. BOTH ΔHf° and ΔGf° = 0 for elements in their standard state and both bearunits of kJ/molrxn. You have no idea how handy this is going to be with regard to solving homeworkproblems and acing quizzes & exams! The Big Mamma equations are simply different versions ofHess’s Law.

But, aw shucks! You’ve got to stop and look up S° values rather them being zero as well. (Note thelack of a “delta”. That’s not a typo! ) Only a perfect diamond at absolute zero has a S° value = 0.

2. “Granddaddy Equation” for calculating Gibbs Free Energy for a system at constanttemperature : ∆G = ∆H - T∆S

By far, one of the most beneficial equations to learn for the AP exam! Case in point, taking favoredequilibrium conditions where ΔG = 0, into consideration, the equation rearranges rather quickly toallow you to determine the absolute temperature at which a process becomes thermodynamicallyfavorable. Shall we? (At least indulge me!)

0 at equilibrium (negative signs cancel)

G H T SH T S

H T ST S H

HTS

∆ ° = ∆ ° − ∆ °∴ = ∆ ° − ∆ °−∆ ° = − ∆ °∴ ∆ ° = ∆ °

∆ °∴ =

∆ °

3. Hess’s Law of Summation for a “new” reaction when given a series of chemical reactions and theΔG° for each reaction. Hess’s Law of Summation works exactly the same as in the enthalpycalculations; arrange a series of chemical equations for which you know the ΔG°rxn to obtain the“goal equation”. If you need to reverse an equation, then you change the sign of ΔG°rxn and cross offcommon moles of substances as you sum the equations to deliver the goal equation. If you double anequation to obtain the goal, double the value of ΔG°rxn, if you halve a reaction halve the value ofΔG°rxn for that reaction, etc.

4. “Rat Link” equation for calculating ∆G ° at standard conditions using the given temperature andequilibrium constant, K: ∆G° = ‒RTlnK

Be sure to use the “energy R” 8.3145 J/mol•K(I predict you’ll use this one the most!) In this case, the system is at equilibrium, so ∆G° = 0 and K


represents the equilibrium constant under standard conditions.

reactants

productsp

PKP

= still raised to power of coefficients

Knowing that ( ) nP cK K RT ∆= , where Δn is equal to the change in the number of moles of gas for the

reaction.

5. Solving for ∆G ° using the “minus nunfe” equation given the standard cell potential, Faraday’sconstant and the number of moles of electrons involved. Sounds far scarier than it is!∆G° = ‒ nFE° where n = number of moles of electrons transferred in a balanced redox reaction, F isFaraday’s constant 96,485 Coulombs/mole e− and E° is the standard cell potential for theelectrochemical process. It’s also handy to know that 1 volt = 1 joule/coulomb so you’re units workout as they should.

Exercise 4 2 H2O() + O2(g) → 2 H2O2()

Calculate the free energy of formation for the oxidation of water to produce hydrogen peroxide given the following information ∆Go

f values: H2O() ‒56.7 kcal/molrxn O2(g) 0 kcal/molrxn H2O2() ‒27.2 kcal/molrxn

59.0 kcal/molrxn

Exercise 5

2 SO2(g) + O2(g) → 2 SO3(g)

The reaction above was carried out at 25°C and 1 atm. Calculate ∆H°, ∆S°, and ∆G° using the following data:

∆H° = ‒198 kJ/molrxn; ∆S° = ‒187 J/K•molrxn; ∆G° = ‒142 kJ/molrxn


Exercise 6 Cdiamond(s) + O2(g) → CO2(g) ∆G°= ‒397 kJ Cgraphite(s) + O2(g) → CO2(g) ∆G°= ‒394 kJ

Calculate ∆G° for the reaction Cdiamond(s)→Cgraphite(s)

+3 kJ/molrxn

Exercise 7

The overall reaction for the corrosion (rusting) of iron by oxygen is 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)

Using the following data, calculate the equilibrium constant for this reaction at 25°C.

K = 10261

The Gibbs equation can also be used to calculate the phase change temperature of a substance. During the phase change, the system is in equilibrium, thus the value of ∆G° is zero.


Exercise 8 Calculate the thermodynamic boiling point of H2O() → H2O(g) given the following information:

ΔHvap = +44 kJ/molrxn ΔSvap = 118.8 J/(K•molrxn)

370 K

Exercise 9 For the reaction 2 CO(g) + O2(g) → 2 CO2(g), the ΔG° for the reaction is −257.2 kJ/molrxn. Calculate the equilibrium constant at 25°C.

1.27 × 1045

SUMMARY:

• If ∆G is NEGATIVE, the reaction is thermodynamically favorable and Ecell would be POSITIVE


• If ∆G is ZERO, the reaction is at equilibrium and Ecell would also be ZERO

If ∆G is POSITIVE, the reaction is NOT thermodynamically favorable and Ecell would be NEGATIVE•

Conditions of ∆G: ∆H ∆S Result negative positive thermodynamically favorable at all temperatures positive positive thermodynamically favorable at high temperatures negative negative thermodynamically favorable at low temperatures positive negative NOT thermodynamically favorable, EVER

Relationship of ∆G to K and E: ∆G K E 0 at equilibrium; K = 1 0 negative >1, products favored positive positive <1, reactants favored negative

COMMON MISCONCEPTIONS ABOUT THERMODYNAMIC FAVORIBILITY

“IF ΔG > 0 (positive thus NOT thermodynamically), the process cannot occur” Not true! External sources of energy can be used to drive change in these cases.

Consider the following:

1. Electric current is applied to charge a battery. You probably describe the equilibrium condition for abattery as “dead”. It’s not dead, it just reached equilibrium (and you can recharge it, but never to fullcapacity again).

2. Light and photon absorption in the photosystems of a chloroplast causing photoionization duringphotosynthesis.

3. Coupling a thermodynamically unfavorable reaction with one that is favorable as in the conversion ofATP to ADP in biological systems.

“IF ΔG is large and negative, the process must proceed at a measurable rate. Not true!

Consider the following:

There are kinetic controls related to high activation energies such as

1. reactant molecules being held together by strong covalent bonds or strong IMFs2. a required collision orientation3. a required energy associated with collisions correctly oriented


AP* Chemistry ELECTROCHEMISTRY Terms to Know: Electrochemistry – the study of the interchange of chemical and electrical energy Voltaic or Galvanic Cell – IS a battery but not a dry cell; generates useful electrical energy Electrolytic Cell – requires useful electrical energy to drive a thermodynamically unfavorable reaction

OIL RIG – oxidation is loss, reduction is gain (of electrons) Oxidation – the loss of electrons, increase in charge Reduction – the gain of electrons, reduction of charge Oxidation number – the assigned charge on an atom Electrochemistry Involves TWO MAIN TYPES Of Electrochemical Cells:

1. Galvanic (voltaic) cells – which are thermodynamically favorable chemical reactions (battery)

2. Electrolytic cells – which are thermodynamically unfavorable and require external e− source (a direct current or DC power source)

BOTH of these fit into the category entitled Electrochemical cells

*AP is a registered trademark of the College Board, which was not involved in the production of this product. © 2013 by René McCormick. All rights reserved.

GALVANIC or VOLTAIC CELL “ANATOMY”

• Anode – the electrode where oxidation occurs. After a period of time, the anode may appear to become

smaller as it falls into solution. (Zn in our illustration below)

• Cathode – the electrode where reduction occurs. After a period of time it may appear larger, due to ions from solution plating onto it. (Cu in our illustration below)

• Inert electrodes – used when a gas is involved OR ion to ion involved such as Fe3+ being reduced to Fe2+ rather than Fe0; made of Pt (expensive) or graphite (cheap)

• Salt bridge – used to maintain electrical neutrality in a galvanic cell; may be filled with agar which contains a neutral salt

• Electron flow – ALWAYS through the wire from anode to cathode (alphabetical order)

• Voltmeter – measures the cell potential (emf) in volts.

Examine the diagram above. Take note of the following mnemonic devices (easy ways to remember “stuff”): All of the following refer to the construction of a thermodynamically favorable cell – one that can act as a battery: AN OX – oxidation occurs at the anode (may show mass decrease) RED CAT – reduction occurs at the cathode (may show mass increase) FAT CAT – The electrons in a voltaic or galvanic cell ALWAYS flow

From the Anode To the CAThode

Ca+hode – the cathode is + in galvanic (voltaic) cells, so it stands to reason the anode is negative Salt Bridge – bridge between cells whose purpose is to provide ions to balance the charge.

Usually made of a salt filled agar (KNO3) or a porous disk may be present instead. EPA – in an electrolytic cell, there is a positive anode.

. Electrochemistry

2

Galvanic cells involve oxidation-reduction or redox reactions. Balance this redox reaction:

MnO4− + Fe2+ → Mn2+ + Fe3+ [acidic]

RED:

OX:

Overall rxn:

• If we place MnO4− and Fe2+ in the same container, the electrons are transferred directly when the reactants

collide. No useful work is obtained from the chemical energy involved which is instead, released as heat!• We can harness this energy if we separate the oxidizing agent from the reducing agent, thus requiring the e−

transfer to occur through a wire! We can harness the energy that way to run a motor, light a bulb, etc.• Sustained electron flow cannot occur in the picture above.

Why not? As soon as electrons flow, a separation of charge occurs which in turn stops the flow of electrons.How do we fix it? Add a salt bridge or allow flow through a porous disk.

• Salt Bridge – its job is to balance thecharge using an electrolyte [usually in a U-shaped tube filled with agar that has the saltdissolved into it before it gels]. It connectsthe two compartments, ions flow from it,AND it keeps each “cell” neutral.

Use KNO3 as the salt when constructingyour own diagram so that no precipitationoccurs!

• porous disk or cup – also allows both cellsto remain neutral by allowing ions to flow

• cell potential – Ecell, Emf, or εcell—it is a measure of the electromotive force or the “pull” of the electrons asthey travel from the anode to the cathode [more on that later!]� volt (V) – the unit of electrical potential; equal to 1 joule of work per coulomb of charge transferred� voltmeter – measures electrical potential; some energy is lost as heat [resistance] which keeps the

voltmeter reading a tad lower than the actual or calculated voltage. Digital voltmeters have lessresistance. If you want to get picky and eliminate the error introduced by resistance, you attach avariable-external power source called a potentiometer. Adjust it so that zero current flows—theaccurate voltage is then equal in magnitude but opposite in sign to the reading on the potentiometer.

Here’s a nice animation from YouTube: http://www.youtube.com/watch?v=raOj8QGDkPA

. Electrochemistry

3

ORIGIN OF STANDARD REDUCTION POTENTIALS

• Each half-reaction has a cell potential• Each potential is measured against a standard, which is the

standard hydrogen electrode [consists of a piece of inertPlatinum that is bathed by hydrogen gas at1 atm].

The hydrogen electrode is assigned a value of 0.00 Vmuch like the isotope C-12 is assigned an atomic mass ofexactly 12.000 amu and all other atomic masses are measured relative to it.

• standard conditions –1 atm for gases, 1.0M for solutions and 25°C for all (298 K)• naught, ° – we use the naught to indicate standard conditions [Experiencing a thermo flashback?]

That means Ecell, Emf, or εcell become Ecello , Emfo , or εcell

o when measurements are taken at standard conditions. You’ll soon learn how these change when the conditions are non-standard!

• The diagram to the right illustrates what really happens when aGalvanic cell is constructed from zinc sulfate and copper(II)sulfate using the respective metals as electrodes.

� Notice that 1.0 M solutions of each salt are used� Notice an overall voltage of 1.10 V for the process

So, how do we construct a fully functional Galvanic or Voltaic cell?

First, we must make wise choices depending on materials available and cost. Knowing how to choose wisely is our next lesson! We need to interpret the data given on the table of standard reduction potentials as we engineer our Galvanic or Voltaic cell.

Interpreting a Table of Standard Electrode Potentials

� Elements that have the most positive reduction potentials are easily reduced (in general, non-metals)

� Elements that have the least positive reduction potentials are easily oxidized (in general, metals)

� The reduction potential table can also be used as an activity series. Metals having less positive reduction potentials are more active and will replace metals with more positive potentials.

. Electrochemistry

4

Let the engineering begin! The MORE POSITIVE reduction potential gets to indeed be reduced IF you are trying to set up a cell that can act as a galvanic or voltaic cell (a battery in other words).

There once was a table of reduction potentials in the reference tables of the AP Chemistry exam. Currently, we expect the data will be given in either a small table or simply embedded within the text of the question. For your homework, you’ll need to consult a table similar to this one.

Calculating Standard Cell Potential Symbolized by E°cell OR Emf° OR εcell° [I’ll mix and match!]

1. Decide which element is oxidized or reduced using the table of reduction potentials. Once again,THE Metal with the MORE POSITIVE REDUCTION POTENITA L gets to be REDUCED.So, it stands to reason that the other metal is oxidized!

2. Write both equations AS IS from the chart with their associated voltages.

3. Reverse the equation that will be oxidized and change the sign of its voltage [this is now E°oxidation]

4. Balance the two half reactions **do not multiply voltage values** Why not?

A volt is equivalent to a J/coulomb or J

c which is a ratio.

5. Add the two half reactions and the voltages together.

6. E°cell = E°oxidation + E°reduction ° means standard conditions: 1atm, 1M, 25°C

. Electrochemistry

5

ANIONS from the salt move to the anode while CATIONS from the salt move to the cathode!

Exercise 1 a. Consider a galvanic cell based on the reaction

Al3+(aq) + Mg(s) → Al(s) + Mg2+(aq) Give the balanced cell reaction and calculate E° for the cell. b. A galvanic cell is based on the reaction [you’ll need a more complete table of reduction potentials!]

MnO4−(aq) + H+(aq) + ClO3

−(aq) → ClO4−(aq) + Mn2+(aq) + H2O(l)

Give the balanced cell reaction and calculate E° for the cell.

A: 0.71 V B: 0.32 V

. Electrochemistry

6

(J)( )

(C)

workemf V

chargeε= =

(J)

(C)

work ww q

charge qε ε

−= − = ∴− =

Exercise 2 Calculate the cell voltage for the galvanic cell that would utilize silver metal and involve iron(II) ion and iron(III) ion. Draw a diagram of the galvanic cell for the reaction and label completely.

E°cell= 0.03 V

CELL POTENTIAL, ELECTRICAL WORK & FREE ENERGY

It is time to combine the thermodynamics and the electrochemistry, not to mention a wee bit of physics.

• The work that can be accomplished when electrons are transferred through a wire depends on the “push” oremf which is defined in terms of a potential difference [in volts] between two points in the circuit.

• Thus one joule of work is produced [or required] when one coulomb of charge is transferred between twopoints in the circuit that differ by a potential of one volt

� IF work flows OUT of the system, it is assigned a MINUS sign (makes sense since Joules were LOST)

� When a cell produces a current, the cell potential is positive and the current can be used to do workTHEREFORE ε and work have opposite signs!

� faraday(F)—the charge on one MOLE of electrons = 96,485 coulombs (Think 96,500 whenanswering multiple choice questions )

� q = # moles of electrons × F

� For a process carried out at constant temperature and pressure, wmax [neglecting the very small amountof energy that is lost as friction or heat] is equal to ΔG, therefore….

∆Go = −nFEo

. Electrochemistry

7

G = Gibb’s free energy n = number of moles of electrons

F = Faraday constant 96,485 J

V•mol

So, it follows that: −Eo implies thermodynamically unfavorable. +Eo implies thermodynamically favorable (would be a good battery!) Exercise 3 Using the table of standard reduction potentials, calculate ∆G° for the reaction

Cu2+(aq) + Fe(s) → Cu(s) + Fe

2+(aq)

Explain whether or not this reaction is thermodynamically favorable. Exercise 4 Using the table of standard reduction potentials, predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au3+ solution.

= no

. Electrochemistry

8

DEPENDENCE OF CELL POTENTIAL ON CONCENTRATION

Voltaic cells at NONstandard conditions: LeChatlier’s principle can be applied. An increase in the concentration of a reactant will favor the forward reaction and the cell potential will increase. The converse is also true!

Exercise 5 For the cell reaction

2Al(s) + 3Mn2+(aq) → 2Al3+(aq) + 3Mn(s) E°cell =

predict whether Ecell is larger or smaller than E°cell for the following cases and justify your answer.

a. [Al 3+ ] = 2.0 M, [Mn2+ ] = 1.0 Mb. [Al 3+ ] = 1.0 M, [Mn2+] = 3.0 M

a. smallerb. larger

NMSI Disclaimer: The curriculum framework for the course excludes the quantitative treatment of the Nernst Equation, but we feel it is in the best interest of the student to learn this concept for two reasons:

1) It makes students more college ready, especially the students that will earn a qualifying score and obtain full credit for the universitycourse.

2) At least half of the students in any given AP Chemistry course are mathematically inclined and may actually understand a concept betterthrough the application of the mathematics of Q and how it relates back to LeChâtelier’s Principle.

For a more quantitative approach at nonstandard conditions use the Nernst Equation:

lnRT

E E QnF

= °−

R = Gas constant 8.315 J/K• mol F = Faraday constant

Q = reaction quotient = [ ]

[ ]

products

reactants

coefficient

coefficient

E = Energy produced by reaction T = Temperature in Kelvins n = # of electrons exchanged in BALANCED redox equation

Rearranged, another useful form

NERNST EQUATION : 0.0592

logE E Qn

= °− @ 25°C (298K)

. Electrochemistry

9

As E declines with reactants being converted into products, E eventually reaches zero. Zero potential means reaction is at equilibrium [dead battery]. Also note, Q = K AND ∆G = 0 as well.

CONCENTRATION CELLS We can construct a cell where both compartments contain the same components BUT at different concentrations Notice the difference in the concentrations pictured at left. Because the right compartment contains 1.0 M Ag+ and the left compartment contains 0.10 M Ag+, there will be a driving force to transfer electrons from left to right. Silver will be deposited on the right electrode, thus lowering the concentration of Ag+ in the right compartment. In the left compartment the silver electrode dissolves [producing Ag+ ions] to raise the concentration of Ag+ in solution.

Exercise 6 Determine the direction of electron flow and designate the anode and cathode for the cell represented here.

left � right

Exercise 7 Determine Eo

cell and Ecell based on the following half-reactions:

VO2+ + 2H+ + e− → VO

2+ + H2O E°= 1.00 V Zn2+ + 2e− → Zn E° = −0.76V

Where, T = 25°C [VO2

+] = 2.0 M [H+] = 0.50 M [VO2+] = 1.0 × 10−2 M [Zn2+] = 1.0 × 10−1 M

Eocell = 1.76 V

Ecell = 1.89 V

. Electrochemistry

10

AP* Electrochemistry Free Response Questions page 1

(1) AP® is a registered trademark of the College Board. The College Board was not involved in the production of and does not endorse this product. (2) Test Questions

are Copyright © 1984-2008 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are

permitted to reproduce the questions. Web or Mass distribution prohibited.

1. Which metal is the cathode?

necessary to make the cell operate?

(c) How does the potential of this cell change if the concentration of Cr(NO3)3 is changed to 3.00-molar

Galvanic (Voltaic) Cells 1988 An electrochemical cell consists of a tin electrode in an acidic solution of 1.00 molar Sn

2+ connected by a salt

bridge to a second compartment with a silver electrode in an acidic solution of 1.00 molar Ag+.

(a) Write the equation for the half-cell reaction occurring at each electrode. Indicate which half-reaction

occurs at the anode.

(b) Write the balanced chemical equation for the overall spontaneous cell reaction that occurs when the

circuit is complete. Calculate the standard voltage, E°, for this cell reaction. (c) Calculate the equilibrium constant for this cell reaction at 298 K.

(d) A cell similar to the one described above is constructed with solutions that have initial concentrations of

1.00 molar Sn2+

and 0.0200 molar Ag+. Calculate the initial voltage, E, of this cell.

1993 A galvanic cell is constructed using a chromium electrode in a 1.00-molar solution of Cr(NO3)3 and a copper

electrode in a 1.00-molar solution of Cu(NO3)2. Both solutions are at 25°C.

(a) Write a balanced net ionic equation for the spontaneous reaction that occurs as the cell operates. Identify

the oxidizing agent and the reducing agent.

(b) A partial diagram of the cell is shown below.

2. What additional component is

3. What function does the component in (ii) serve?

at 25°C? Explain.





1996

Sr(s) + Mg2+ → Sr2+ + Mg(s)

Consider the reaction represented above that occurs at 25°C. All reactants and products are in their standard

states. The value of the equilibrium constant, Keq, for the reaction is 4.2 × 1017

at 25°C.

(a) Predict the sign of the standard cell potential, E°, for a cell based on the reaction. Explain your prediction.

(b) Identify the oxidizing agent for the spontaneous reaction.

(c) If the reaction were carried out at 60°C instead of 25°C, how would the cell potential change? Justify

your answer.

(d) How would the cell potential change if the reaction were carried out at 25°C with a 1.0-molar solution of

Mg(NO3)2 and a 0.10-molar solution of Sr(NO3)2? Explain.

(e) When the cell reaction in (d) reaches equilibrium, what is the cell potential?

1998

Answer the following questions regarding the electrochemical cell shown above.

(a) Write the balanced net-ionic equation for the spontaneous reaction that occurs as the cell operates,

and determine the cell voltage.

(b) In which direction do anions flow in the salt bridge as the cell operates? Justify your answer.

(c) If 10.0 mL of 3.0-molar AgNO3 solution is added to the half-cell on the right, what will happen to

the cell voltage? Explain.

(d) If 1.0 grams of solid NaCl is added to each half-cell, what will happen to the cell voltage? Explain.

(e) If 20.0 mL of distilled water is added to both half-cells, the cell voltage decreases. Explain.





2001

Answer the following questions that refer to the galvanic cell shown in the diagram above.

(a) Identify the anode of the cell and write the half-reaction that occurs there.

(b) Write the net ionic equation for the overall reaction that occurs as the cell operates and calculate the

value of the standard cell potential, E°cell.

(c) Indicate how the value of Ecell would be affected if the concentration of Ni(NO3)2(aq) was changed

from 1.0 M to 0.10 M and the concentration of Zn(NO3)2(aq) remained at 1.0 M. Justify your

answer.

(d) Specify whether the value of Keq for the cell reaction is less than 1, greater than 1, or equal to 1.

Justify your answer.





2002B

The diagram below shows the experimental setup for a typical electrochemical cell that contains two

standard half-cells. The cell operates according to the reaction represented by the following equation.

Zn(s) + Ni2+

(aq) → Ni(s) + Zn2+

(aq)

(a) Identify M and M2+ in the diagram and specify the initial concentration for M2+ in solution.

(b) Indicate which of the metal electrodes is the cathode. Write the balanced equation for the reaction

that occurs in the half-cell containing the cathode.

(c) What would be the effect on the cell voltage if the concentration of Zn2+

was reduced to 0.100 M

in the half-cell containing the Zn electrode?

(d) Describe what would happen to the cell voltage if the salt bridge was removed. Explain.





2003B

Answer the following questions about electrochemistry.

(a) Several different electrochemical cells can be constructed using the materials shown below. Write

the balanced net-ionic equation for the reaction that occurs in the cell that would have the greatest

positive value of E°cell.

(b) Calculate the standard cell potential, E°cell, for the reaction written in part (a).

(c) A cell is constructed based on the reaction in part (a) above. Label the metal used for the anode on

the cell shown in the figure below.





2004B

The following questions refer to the electrochemical cell shown in the diagram above.

(a) Write a balanced net ionic equation for the spontaneous reaction that takes place in the cell.

(b) Calculate the standard cell potential, E°, for the reaction in part (a).

(c) In the diagram above,

(i) label the anode and the cathode on the dotted lines provided, and

(ii) indicate, in the boxes below the half-cells, the concentration of AgNO3 and the concentration of

Zn(NO3)2 that are needed to generate E°.

(d) How will the cell potential be affected if KI is added to the silver half-cell? Justify your answer.





2005

AgNO3(s) → Ag+(aq) + NO3

−(aq)

The dissolving of AgNO3(s) in pure water is represented by the equation above.

(a) Is ΔG for the dissolving of AgNO3(s) positive, negative, or zero? Justify your answer.

(b) Is ΔS for the dissolving of AgNO3(s) positive, negative, or zero? Justify your answer.

(c) The solubility of AgNO3(s) increases with increasing temperature.

(i) What is the sign of ΔH for the dissolving process? Justify your answer.

(ii) Is the answer you gave in part (a) consistent with your answers to parts (b) and (c) (i) ? Explain.

The compound NaI dissolves in pure water according to the equation NaI(s) → Na+(aq) + I

−(aq) . Some of the

information in the table of standard reduction potentials given below may be useful in answering the questions that follow.

(d) An electric current is applied to a 1.0 M NaI solution.

(i) Write the balanced oxidation half-reaction for the reaction that takes place.

(ii) Write the balanced reduction half-reaction for the reaction that takes place.

(iii) Which reaction takes place at the anode, the oxidation reaction or the reduction reaction?

(iv) All electrolysis reactions have the same sign for ∆G°. Is the sign positive or negative? Justify your

answer.





Electrolytic Cells 1991

Explain each of the following.

(a) When an aqueous solution of NaCl is electrolyzed, Cl2(g) is produced at the anode, but no Na(s) is

produced at the cathode.

(b) The mass of Fe(s) produced when 1 faraday is used to reduce a solution of FeSO4 is 1.5 times the mass

of Fe(s) produced when 1 faraday is used to reduce a solution of FeCl3.

Zn + Pb2+ (1.0 M) → Zn2+ (1.0 M) + Pb

(c) The cell that utilized the reaction above has a higher potential when [Zn2+] is decreased and [Pb2+] held

constant, but a lower potential when [Pb2+] is decreased and [Zn2+] is held constant.

(d) The cell that utilizes the reaction given in (c) has the same cell potential as another cell in which [Zn2+]

and [Pb2+] are each 0.10 M.

1992 An unknown metal M forms a soluble compound, M(NO3)2.

(a) A solution of M(NO3)2 is electrolyzed. When a constant current of 2.50 amperes is applied for

35.0 minutes, 3.06 grams of the metal M is deposited. Calculate the molar mass of M and identify the

metal.

(b) The metal identified in (a) is used with zinc to construct a galvanic cell, as shown below. Write the net

ionic equation for the cell reaction and calculate the cell potential, E°.

(c) Calculate the value of the standard free energy change, ∆G°, at 25°C for the reaction in (b).

(d) Calculate the potential, E, for the cell shown in (b) if the initial concentration of ZnSO4 is 0.10-molar,

but the concentration of the M(NO3)2 solution remains unchanged.





1997 In an electrolytic cell, a current of 0.250 ampere is passed through a solution of a chloride of iron, producing

Fe(s) and Cl2(g).

(a) Write the equation for the reaction that occurs at the anode.

(b) When the cell operates for 2.00 hours, 0.521 gram of iron is deposited at one electrode.

Determine the formula of the chloride of iron in the original solution.

(c) Write the balanced equation for the overall reaction that occurs in the cell. (d) How many liters of Cl2(g), measured at 25 °C and 750 mmHg, are produced when the cell

operates as described in part (b)?

(e) Calculate the current that would produce chlorine gas at a rate of 3.00 grams per hour.

2000

Answer the following questions that relate to electrochemical reactions.

(a) Under standard conditions at 25°C, Zn(s) reacts with Co2+(aq) to produce Co(s).

(i) Write the balanced equation for the oxidation half reaction.

(ii) Write the balanced net-ionic equation for the overall reaction.

(iii)Calculate the standard potential, E°, for the overall reaction at 25°C.

(b) At 25°C, H2O2 decomposes according to the following equation.

2 H2O2(aq) → 2 H2O(l) + O2(g) E° = 0.55 V

(i) Determine the value of the standard free energy change, ∆G°, for the reaction at 25°C.

(ii) Determine the value of the equilibrium constant, Keq , for the reaction at 25°C.

(iii) The standard reduction potential, E°, for the half reaction O2 (g) + 4 H+(aq) + 4 e− → 2 H2O (l) has a

value of 1.23 V. Using this information in addition to the information given above, determine the value

of the standard reduction potential, E°, for the half reaction below.

O2 (g) + 2 H+(aq) + 2 e− → H2O2(aq)

(c) In an electrolytic cell, Cu(s) is produced by the electrolysis of CuSO4(aq). Calculate the maximum mass

of Cu(s) that can be deposited by a direct current of 100. amperes passed through 5.00 L of

2.00 M CuSO4(aq) for a period of 1.00 hour.





2007B

2 H2(g) + O2(g) → 2 H2O(l)

In a hydrogen-oxygen fuel cell, energy is produced by the overall reaction represented above.

(a) When the fuel cell operates at 25C and 1.00 atm for 78.0 minutes, 0.0746 mol of O2(g) is consumed.

Calculate the volume of H2(g) consumed during the same time period. Express your answer in liters measured

at 25°C and 1.00 atm.

(b) Given that the fuel cell reaction takes place in an acidic medium,

(i) write the two half reactions that occur as the cell operates,

(ii) identify the half reaction that takes place at the cathode, and

(iii) determine the value of the standard potential, E, of the cell.

(c) Calculate the charge, in coulombs, that passes through the cell during the 78.0 minutes of operation as

described in part (a).

STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25 C∞

Half-reaction (V)E∞

2F ( ) 2g e-+ Æ 2F- 2.87 3+Co e-+ Æ 2Co + 1.82 3+Au 3e-+ Æ Au( )s 1.50

2Cl ( ) 2g e-+ Æ 2Cl- 1.36 +

2O ( ) 4H 4g e-+ + Æ 22H O( )l 1.23

2Br ( ) 2l e-+ Æ 2Br- 1.07 2+2Hg 2e-+ Æ 2+

2Hg 0.92 2+Hg 2e-+ Æ Hg( )l 0.85 +Ag e-+ Æ Ag( )s 0.80 2+

2Hg 2e-+ Æ 2Hg( )l 0.79 3+Fe e-+ Æ 2+Fe 0.77

2I ( ) 2s e-+ Æ 2I- 0.53 +Cu e-+ Æ Cu( )s 0.52 2+Cu 2e-+ Æ Cu( )s 0.34 2+Cu e-+ Æ +Cu 0.15 4+Sn 2e-+ Æ 2+Sn 0.15 +S( ) 2H 2s e-+ + Æ 2H S( )g 0.14 +2H 2e-+ Æ 2H ( )g 0.00 2+Pb 2e-+ Æ Pb( )s – 0.13 2+Sn 2e-+ Æ Sn( )s – 0.14 2+Ni 2e-+ Æ Ni( )s – 0.25 2+Co 2e-+ Æ Co( )s – 0.28 2+Cd 2e-+ Æ Cd( )s – 0.40 3+Cr e-+ Æ 2+Cr – 0.41 2+Fe 2e-+ Æ Fe( )s – 0.44 3+Cr 3e-+ Æ Cr( )s – 0.74 2+Zn 2e-+ Æ Zn( )s – 0.76

22H O( ) 2l e-+ Æ 2H ( ) + 2OHg - – 0.83 2+Mn 2e-+ Æ Mn( )s – 1.18 3+Al 3e-+ Æ Al( )s – 1.66 2+Be 2e-+ Æ Be( )s – 1.70 2+Mg 2e-+ Æ Mg( )s – 2.37 +Na e-+ Æ Na( )s – 2.71 2+Ca 2e-+ Æ Ca( )s – 2.87 2+Sr 2e-+ Æ Sr( )s – 2.89 2+Ba 2e-+ Æ Ba( )s – 2.90 +Rb e-+ Æ Rb( )s – 2.92 +K e-+ Æ K( )s – 2.92 +Cs e-+ Æ Cs( )s – 2.92 +Li e-+ Æ Li( )s – 3.05



1. Which metal is the cathode?

necessary to make the cell operate?

(c) How does the potential of this cell change if the concentration of Cr(NO3)3 is changed to 3.00-molar

Galvanic (Voltaic) Cells 1988 An electrochemical cell consists of a tin electrode in an acidic solution of 1.00 molar Sn2+ connected by a salt bridge to a second compartment with a silver electrode in an acidic solution of 1.00 molar Ag+.

(a) Write the equation for the half-cell reaction occurring at each electrode. Indicate which half-reaction occurs at the anode.

(b) Write the balanced chemical equation for the overall spontaneous cell reaction that occurs when the circuit is complete. Calculate the standard voltage, E°, for this cell reaction.

(c) Calculate the equilibrium constant for this cell reaction at 298 K. (d) A cell similar to the one described above is constructed with solutions that have initial concentrations of

1.00 molar Sn2+ and 0.0200 molar Ag+. Calculate the initial voltage, E, of this cell. 1993 A galvanic cell is constructed using a chromium electrode in a 1.00-molar solution of Cr(NO3)3 and a copper electrode in a 1.00-molar solution of Cu(NO3)2. Both solutions are at 25°C.

(a) Write a balanced net ionic equation for the spontaneous reaction that occurs as the cell operates. Identify the oxidizing agent and the reducing agent.

(b) A partial diagram of the cell is shown below.

2. What additional component is3. What function does the component in (ii) serve?

at 25°C? Explain.



1996

Sr(s) + Mg2+ → Sr2+ + Mg(s) Consider the reaction represented above that occurs at 25°C. All reactants and products are in their standard states. The value of the equilibrium constant, Keq, for the reaction is 4.2 × 1017 at 25°C. (a) Predict the sign of the standard cell potential, E°, for a cell based on the reaction. Explain your prediction. (b) Identify the oxidizing agent for the spontaneous reaction. (c) If the reaction were carried out at 60°C instead of 25°C, how would the cell potential change? Justify

your answer. (d) How would the cell potential change if the reaction were carried out at 25°C with a 1.0-molar solution of

Mg(NO3)2 and a 0.10-molar solution of Sr(NO3)2? Explain. (e) When the cell reaction in (d) reaches equilibrium, what is the cell potential? 1998

Answer the following questions regarding the electrochemical cell shown above.

(a) Write the balanced net-ionic equation for the spontaneous reaction that occurs as the cell operates, and determine the cell voltage.

(b) In which direction do anions flow in the salt bridge as the cell operates? Justify your answer. (c) If 10.0 mL of 3.0-molar AgNO3 solution is added to the half-cell on the right, what will happen to

the cell voltage? Explain. (d) If 1.0 grams of solid NaCl is added to each half-cell, what will happen to the cell voltage? Explain. (e) If 20.0 mL of distilled water is added to both half-cells, the cell voltage decreases. Explain.



2001

Answer the following questions that refer to the galvanic cell shown in the diagram above. (a) Identify the anode of the cell and write the half-reaction that occurs there. (b) Write the net ionic equation for the overall reaction that occurs as the cell operates and calculate the

value of the standard cell potential, E°cell. (c) Indicate how the value of Ecell would be affected if the concentration of Ni(NO3)2(aq) was changed

from 1.0 M to 0.10 M and the concentration of Zn(NO3)2(aq) remained at 1.0 M. Justify your answer.

(d) Specify whether the value of Keq for the cell reaction is less than 1, greater than 1, or equal to 1. Justify your answer.



2002B

The diagram below shows the experimental setup for a typical electrochemical cell that contains two standard half-cells. The cell operates according to the reaction represented by the following equation.

Zn(s) + Ni2+(aq) → Ni(s) + Zn2+(aq) (a) Identify M and M2+ in the diagram and specify the initial concentration for M2+ in solution.

(b) Indicate which of the metal electrodes is the cathode. Write the balanced equation for the reaction that occurs in the half-cell containing the cathode.

(c) What would be the effect on the cell voltage if the concentration of Zn2+ was reduced to 0.100 M in the half-cell containing the Zn electrode?

(d) Describe what would happen to the cell voltage if the salt bridge was removed. Explain.



2003B

Answer the following questions about electrochemistry. (a) Several different electrochemical cells can be constructed using the materials shown below. Write

the balanced net-ionic equation for the reaction that occurs in the cell that would have the greatest positive value of E°cell.

(b) Calculate the standard cell potential, E°cell, for the reaction written in part (a). (c) A cell is constructed based on the reaction in part (a) above. Label the metal used for the anode on

the cell shown in the figure below.



2004B

The following questions refer to the electrochemical cell shown in the diagram above. (a) Write a balanced net ionic equation for the spontaneous reaction that takes place in the cell. (b) Calculate the standard cell potential, E°, for the reaction in part (a). (c) In the diagram above,

(i) label the anode and the cathode on the dotted lines provided, and (ii) indicate, in the boxes below the half-cells, the concentration of AgNO3 and the concentration of

Zn(NO3)2 that are needed to generate E°. (d) How will the cell potential be affected if KI is added to the silver half-cell? Justify your answer.



2005

AgNO3(s) → Ag+(aq) + NO3−(aq)

The dissolving of AgNO3(s) in pure water is represented by the equation above.

(a) Is ΔG for the dissolving of AgNO3(s) positive, negative, or zero? Justify your answer. (b) Is ΔS for the dissolving of AgNO3(s) positive, negative, or zero? Justify your answer. (c) The solubility of AgNO3(s) increases with increasing temperature.

(i) What is the sign of ΔH for the dissolving process? Justify your answer. (ii) Is the answer you gave in part (a) consistent with your answers to parts (b) and (c) (i) ? Explain.

The compound NaI dissolves in pure water according to the equation NaI(s) → Na+(aq) + I−(aq) . Some of the information in the table of standard reduction potentials given below may be useful in answering the questions that follow.

(d) An electric current is applied to a 1.0 M NaI solution.

(i) Write the balanced oxidation half-reaction for the reaction that takes place. (ii) Write the balanced reduction half-reaction for the reaction that takes place. (iii) Which reaction takes place at the anode, the oxidation reaction or the reduction reaction? (iv) All electrolysis reactions have the same sign for ∆G°. Is the sign positive or negative? Justify your

answer.



Electrolytic Cells 1991

Explain each of the following.

(a) When an aqueous solution of NaCl is electrolyzed, Cl2(g) is produced at the anode, but no Na(s) is produced at the cathode.

(b) The mass of Fe(s) produced when 1 faraday is used to reduce a solution of FeSO4 is 1.5 times the mass of Fe(s) produced when 1 faraday is used to reduce a solution of FeCl3.

Zn + Pb2+ (1.0 M) → Zn2+ (1.0 M) + Pb

(c) The cell that utilized the reaction above has a higher potential when [Zn2+] is decreased and [Pb2+] held constant, but a lower potential when [Pb2+] is decreased and [Zn2+] is held constant.

(d) The cell that utilizes the reaction given in (c) has the same cell potential as another cell in which [Zn2+] and [Pb2+] are each 0.10 M.

1992 An unknown metal M forms a soluble compound, M(NO3)2.

(a) A solution of M(NO3)2 is electrolyzed. When a constant current of 2.50 amperes is applied for 35.0 minutes, 3.06 grams of the metal M is deposited. Calculate the molar mass of M and identify the metal.

(b) The metal identified in (a) is used with zinc to construct a galvanic cell, as shown below. Write the net ionic equation for the cell reaction and calculate the cell potential, E°.

(c) Calculate the value of the standard free energy change, ΔG°, at 25°C for the reaction in (b). (d) Calculate the potential, E, for the cell shown in (b) if the initial concentration of ZnSO4 is 0.10-molar,

but the concentration of the M(NO3)2 solution remains unchanged.



1997 In an electrolytic cell, a current of 0.250 ampere is passed through a solution of a chloride of iron, producing Fe(s) and Cl2(g).

(a) Write the equation for the reaction that occurs at the anode. (b) When the cell operates for 2.00 hours, 0.521 gram of iron is deposited at one electrode.

Determine the formula of the chloride of iron in the original solution. (c) Write the balanced equation for the overall reaction that occurs in the cell. (d) How many liters of Cl2(g), measured at 25 °C and 750 mmHg, are produced when the cell

operates as described in part (b)? (e) Calculate the current that would produce chlorine gas at a rate of 3.00 grams per hour.

2000 Answer the following questions that relate to electrochemical reactions. (a) Under standard conditions at 25°C, Zn(s) reacts with Co2+(aq) to produce Co(s).

(i) Write the balanced equation for the oxidation half reaction. (ii) Write the balanced net-ionic equation for the overall reaction. (iii)Calculate the standard potential, E°, for the overall reaction at 25°C.

(b) At 25°C, H2O2 decomposes according to the following equation.

2 H2O2(aq) → 2 H2O(l) + O2(g) E° = 0.55 V

(i) Determine the value of the standard free energy change, ∆G°, for the reaction at 25°C. (ii) Determine the value of the equilibrium constant, Keq , for the reaction at 25°C. (iii) The standard reduction potential, E°, for the half reaction O2 (g) + 4 H+(aq) + 4 e− → 2 H2O (l) has a

value of 1.23 V. Using this information in addition to the information given above, determine the value of the standard reduction potential, E°, for the half reaction below.

O2 (g) + 2 H+(aq) + 2 e− → H2O2(aq)

(c) In an electrolytic cell, Cu(s) is produced by the electrolysis of CuSO4(aq). Calculate the maximum mass

of Cu(s) that can be deposited by a direct current of 100. amperes passed through 5.00 L of 2.00 M CuSO4(aq) for a period of 1.00 hour.



2007B

2 H2(g) + O2(g) → 2 H2O(l) In a hydrogen-oxygen fuel cell, energy is produced by the overall reaction represented above.

(a) When the fuel cell operates at 25C and 1.00 atm for 78.0 minutes, 0.0746 mol of O2(g) is consumed. Calculate the volume of H2(g) consumed during the same time period. Express your answer in liters measured at 25°C and 1.00 atm.

(b) Given that the fuel cell reaction takes place in an acidic medium, (i) write the two half reactions that occur as the cell operates, (ii) identify the half reaction that takes place at the cathode, and (iii) determine the value of the standard potential, E, of the cell.

(c) Calculate the charge, in coulombs, that passes through the cell during the 78.0 minutes of operation as described in part (a).

AP* Stoichiometry Free Response Questions page 1

(1) AP® is a registered trademark of the College Board. The College Board was not involved in the production of and does not endorse this product. (2) Test

Questions are Copyright © 1984-2008 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes,

classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited.

1982

Water is added to 4.267 grams of UF6. The only products are 3.730 grams of a solid containing only

uranium, oxygen and fluorine and 0.970 gram of a gas. The gas is 95.0% fluorine, and the remainder is

hydrogen.

(a) From these data, determine the empirical formula of the gas.

(b) What fraction of the fluorine of the original compound is in the solid and what fraction in the gas after

the reaction?

(c) What is the formula of the solid product?

(d) Write a balanced equation for the reaction between UF6 and H2O. Assume that the empirical formula

of the gas is the true formula.

1986

Three volatile compounds X, Y, and Z each contain element Q. The percent by weight of element Q in each

compound was determined. Some of the data obtained are given below.

Percent by weight Molecular

Compound of Element Q Weight

X 64.8% ?

Y 73.0% 104.

Z 59.3% 64.0

(a) The vapor density of compound X at 27°C and 750. mm Hg was determined to be 3.53 grams per liter.

Calculate the molecular weight of compound X.

(b) Determine the mass of element Q contained in 1.00 mole of each of the three compounds.

(c) Calculate the most probable value of the atomic weight of element Q.

(d) Compound Z contains carbon, hydrogen, and element Q. When 1.00 gram of compound Z is

oxidized and all of the carbon and hydrogen are converted to oxides, 1.37 grams of CO2 and 0.281

gram of water are produced. Determine the most probable molecular formula of compound Z.

1991

The molecular formula of a hydrocarbon is to be determined by analyzing its combustion products and

investigating its colligative properties.

(a) The hydrocarbon burns completely, producing 7.2 grams of water and 7.2 liters of CO2 at standard

conditions. What is the empirical formula of the hydrocarbon?

(b) Calculate the mass in grams of O2 required for the complete combustion of the sample of the

hydrocarbon described in (a).

(c) The hydrocarbon dissolves readily in CHCl3. The freezing point of a solution prepared by mixing

100. grams of CHCl3 and 0.600 gram of the hydrocarbon is −64.0°C. The molal freezing-point

depression constant of CHCl3 is 4.68°C/molal and its normal freezing point is −63.5°C. Calculate

the molecular weight of the hydrocarbon.

(d) What is the molecular formula of the hydrocarbon?


1993

I. 2 Mn2+

+ 4 OH− + O2(g) → 2 MnO2(s) + 2 H2O

II. MnO2(s) + 2 I− + 4 H

+ → Mn

2+ + I2(aq) + 2 H2O

III. 2 S2O32−

+ I2(aq) → S4O62−

+ 2 I−

The amount of oxygen, O2, dissolved in water can be determined by titration. First, MnSO4 and NaOH

are added to a sample of water to convert all of the dissolved O2 to MnO2, as shown in equation I

above. Then H2SO4 and KI are added and the reaction represented by equation II proceeds. Finally, the

I2 that is formed is titrated with standard sodium thiosulfate, Na2S2O3, according to equation III.

(a) According to the equation above, how many moles of S2O32− are required for analyzing

1.00 mole of O2 dissolved in water?

(b) A student found that a 50.0-milliliter sample of water required 4.86 milliliters of 0.0112-molar

Na2S2O3 to reach the equivalence point. Calculate the number of moles

of O2 dissolved in this sample.

(c) How would the results in (b) be affected if some I2 were lost before the S2O32− was added? Explain.

(d) What volume of dry O2 measured at 25°C and 1.00 atmosphere of pressure would have to be

dissolved in 1.00 liter of pure water in order to prepare a solution of the same concentration as that

obtained in (b)? (cont.)

(e) Name an appropriate indicator for the reaction shown in equation III and describe the change you

would observe at the end point of the titration.

1998

An unknown compound contains only the three elements C,H, and O. A pure sample of the compound

is analyzed and found to be 65.60 percent C and 9.44 percent H by mass.

(a) Determine the empirical formula of the compound.

(b) A solution of 1.570 grams of the compound in 16.08 grams of camphor is observed to freeze

at a temperature 15.2 Celsius degrees below the normal freezing point of pure camphor.

Determine the molar mass and apparent molecular formula of the compound. (The molal

freezing-point depression constant, Kf, for camphor

is 40.0 kg⋅K⋅mol−1.)

(c) When 1.570 grams of the compound is vaporized at 300 °C and 1.00 atmosphere, the gas

occupies a volume of 577 milliliters. What is the molar mass of the compound based on this

result?

(d) Briefly describe what occurs in solution that accounts for the difference between the results

obtained in parts (b) and (c).


2000

Answer the following questions about BeC2O

4(s) and its hydrate.

(a) Calculate the mass percent of carbon in the hydrated form of the solid that has the formula BeC2O

4 • 3H2O

(b) When heated to 220.°C, BeC2O

4 • 3 H

2O(s) dehydrates completely as represented below.

BeC2O

4 • 3 H

2O(s) → BeC

2O

4(s) + 3 H

2O(g)

If 3.21 g of BeC2O

4 • 3 H

2O(s) is heated to 220.°C, calculate

(i) the mass of BeC2O

4(s) formed, and,

(ii) the volume of the H2O(g) released, measured at 220.°C and 735 mm Hg.

(c) A 0.345 g sample of anhydrous BeC2O

4 , which contains an inert impurity, was dissolved in sufficient

water to produce 100. mL of solution. A 20.0 mL portion of the solution was titrated with KMnO4(aq).

The balanced equation for the reaction that occurred is as follows.

16 H+

(aq) + 2 MnO4

−

(aq) + 5 C2O

4

2−

(aq) → 2 Mn2+

(aq) + 10 CO2(g) + 8 H

2O(l).

The volume of 0.0150 M KMnO4(aq) required to reach the equivalence point was 17.80 mL.

(i) Identify the reducing agent in the titration reaction. (ii) For the titration at the equivalence point, calculate the number of moles of each of the

following that reacted.

• MnO4

−

(aq)

• C2O

4

2−

(aq)

(iii) Calculate the total number of moles of C2O42−

(aq) that were present in the 100. mL of

prepared solution.

(iv) Calculate the mass percent of BeC2O

4(s) in the impure 0.345 g sample.

2003B

Answer the following questions that relate to chemical reactions.

(a) Iron(III) oxide can be reduced with carbon monoxide according to the following equation.

Fe2O

3(s) + 3 CO(g) → 2 Fe(s) + 3 CO

2(g)

A 16.2 L sample of CO(g) at 1.50 atm and 200.°C is combined with 15.39 g of Fe2O

3(s).

(i) How many moles of CO(g) are available for the reaction?

(ii) What is the limiting reactant for the reaction? Justify your answer with calculations.

(iii) How many moles of Fe(s) are formed in the reaction?


(b) In a reaction vessel, 0.600 mol of Ba(NO3)

2(s) and 0.300 mol of H

3PO

4(aq) are combined with

deionized water to a final volume of 2.00 L. The reaction represented below occurs.

3 Ba(NO

3)

2(aq) + 2 H

3PO

4(aq) → Ba

3(PO

4)2(s) + 6 HNO

3(aq)

(i) Calculate the mass of Ba

3(PO

4)2(s) formed.

(ii) Calculate the pH of the resulting solution.

(iii) What is the concentration, in mol L–1

, of the nitrate ion, NO3

–

(aq), after the reaction

reaches completion?

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