Types of Chemical Reactions and Solution Stoichiometry

Chapter 4- Types of Chemical Reactions and Solution

Stoichiometry

Pg. 133-187 in text

The Bottom Line:Chapter 4

• Aqueous reactions account for virtually all chemistry that takes place in living systems.

Water, the Common Solvent

• Shape- bent ~1050

• Electrons not distributed evenly

• Polar molecule• Likes dissolve likes

– Polar and ionic– Polar and polar– Nonpolar and

nonpolar

Likes Dissolve Likes

State whether each pair of substances will mix. State Why or why not.

a. NaNO3 and H2O

b. C6H14 and H2O

c. I2 and C6H14

d. I2 and H2O

Polar Water Molecules Interact with the Positive and Negative Ions of a Salt

Water Also Dissolves Many Nonionic Substances

Why is Ethanol (C2H5OH) very soluble in water?

• Ethanol contains a polar -OH bond like those in water

Polar Bond

Water Does Not Dissolve all Nonionic Substances

Water = Polar

Fats= Nonpolar

Likes dissolve likes

Hydrophilic / Hydrophobic

Solute / Solvent

• Solute:1. If it and the solvent are present in the same

phase, it is the one in lesser amount.2. If it and the solvent are present in different

phases, it is the one that changes phase.3. The one that dissolves into the solvent.

Aqueous solution means that water is the solvent.

Strong, Weak, Non-Electrolytes

Electrolyte Conductivity Degree of Dissociation

Examples

Strong Strong acids, many salts, strong bases

Weak Weak organic acids, weak bases

Non Highly nonpolar organic compounds like sugar, insoluble salts like AgCl

Acids

Strong acids -

hydrochloric and sulfuric acid

Weak acids –

acetic and formic acid

Bases

Strong bases –

sodium hydroxide

Weak bases –

ammonia

Strong Acids-Virtually every molecule

dissociates into given ions

Strong Bases- soluble compounds containing the hydroxide (-OH-) ion

Group IA metal hydroxides = LiOH, NaOH, KOH, RbOH, CsOH

Heavy group 2 metal hydroxides = Ca(OH)2, Sr(OH)2, Ba(OH)2

List whether each of the following is a strong, weak, or nonelectrolyte

a. HClO4

b. C6H12

c. LiOH

d. NH3

e. CaCl2f. HC2H3O2

Common Terms of Solution Concentration

Stock - routinely used solutions prepared in concentrated form.

Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl)

Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl)

The Composition of Solutions

• Molarity = moles of solute / liters of solution.

Example 4.1 pg. 96:Calculate the molarity of a solution prepared by bubbling

1.56 g of gaseous HCl into enough water to make 26.8 mL of solution.

Standard Solutions- a solution whose concentration is accurately known.

• Example 4.4 pg. 97:To analyze the alcohol content of a certain

wine, a chemist needs 1.00 L of an aqueous 0.200 M K2Cr2O7 (potassium dichromate) solution. Describe how to prepare this solution.

Ion Concentration in Solutions

Determine the molarity of Fe3+ ions and SO4

2- ions in a solution prepared by dissolving 48.05 g of Fe2(SO4)3 in enough water to make 800. mL of solution.

Dilution- water is added to a stock concentrated stock solution to achieve

desired molarity.Concept:

Moles of solute after dilution = moles of solute before dilution

M1V1 = M2V2

Example:

What volume of 12 M hydrochloric acid must be used to prepare 600. mL of a 0.30 M HCl solution?

Types of Chemical Reactions

o Precipitation reactionsAgNO3(aq) + NaCl(aq) AgCl(s) +NaNO3(aq)

o Acid-base reactionsNaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

o Oxidation-reduction reactionsFe2O3(s) + Al(s) Fe(l) + Al2O3(s)

Precipitation Reactions

• Remember: In virtually every case, when a solid containing ions dissolves in water, the ions separate and move around independently

MgSO4 (aq) + Na2CO3 (aq)

Predicting whether or not a precipitation reaction will occur depends on the solubility of the potential end products.

Memorize:1. Most nitrate (NO3

) salts are soluble.

2. Most alkali (group 1A) salts and NH4+ are soluble.

3. Most Cl, Br, and I salts are soluble (NOT Ag+, Pb2+, Hg2

2+)

4. Most sulfate salts are soluble (NOT BaSO4, PbSO4, HgSO4, CaSO4)

5. Most OH salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble)

6. Most S2, CO32, CrO4

2, PO43 salts are only slightly

soluble.

The Reaction of K2CrO4(aq) and Ba(NO3)2(aq)

Describing Reactions in Solution

The Reaction of KCI(aq) and AgNO3(aq)

Complete and balance the following reactions, determining, in each case, if a precipitate is formed.

• KCl (aq) + Pb(NO3)2 (aq) -->

• AgNO3 (aq) + MgBr2 (aq) -->

• Ca(OH)2 (aq) + FeCl3 (aq) -->

• NaOH (aq) + HCl (aq) -->

Describing Reactions in Solution

1. Molecular equation (reactants and products as compounds)

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

2. Complete ionic equation (all strong electrolytes shown as ions)

Describing Reactions in Solution (continued)

3. Net ionic equation (show only components that actually react)

Write the molecular, complete ionic, and net ionic forms

Aqueous nickel (II) chloride reacts with aqueous sodium hydroxide

Selective Precipitation

Precipitation reactions allow us to target specific substances, and separate and recover them from a solution.

Example:A solution contains Ca2+, Cu2+, and Pb2+.

What anions can we add, and in what order , to separate and recover each cation?

Stoichiometry of Precipitation Reactions

• Solving problems involving precipitates from solutions makes use of molarity, solubility rules, balancing equations, and limiting reactant calculations.

• Take a systematic approach!

Example:

What mass of precipitate is produced when 35.mL of a 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M KOH solution?

Performing Calculations for Acid-Base Reactions

1. List initial species and predict reaction.2. Write balanced net ionic reaction.3. Calculate moles of reactants.4. Determine limiting reactant.5. Calculate moles of required

reactant/product.6. Convert to grams or volume, as

required.

Key Titration Terms

Titrant - solution of known concentration used in titration

Analyte - substance being analyzed

Equivalence point –

Endpoint -

Acid-Base Reactions

• Acid = • Base =

Neutralization of a Strong Acid- products will always be water and a salt.

Example 1: How many mL of a 0.800M NaOH solution is needed to just neutralize 40.00 mL of a 0.600M HCl solution?

Example 2:

You wish to determine the molarity of a solution of sodium hydroxide. To do this, you titrate a 25.00 mL aliquot of your sample, which has had 3 drops of phenolphthalein indicator added so that it is pink, with 0.1067 M HCl. The sample turns clear (indicating that the NaOH (aq) has been precisely neutralized by the HCl solution) after the addition of 42.95 mL of the HCl. Calculate the molarity of your NaOH solution.

Oxidation-Reduction Reactions

Redox reactions involve a transfer of electrons. In order to determine if an electron has been transferred, one must be able to assign oxidation states.

Assigning oxidation states to an element in a molecule requires knowledge of a set of rules. These rules are outlined on page 120 of your text.

Rules for Assigning Oxidation States1. OS of an atom in an element is 0.

2. OS of a monatomic ion is the same as its charge.

3. In its covalent compounds with nonmetals,

hydrogen is assigned an OS of +1.

4. Oxygen is assigned an OS of -2 in its covalent compounds.

The exception to this rules occurs in peroxides

(compounds contains the O22- group), where each

oxygen is assigned an OS of -1.

5. In binary compounds the element with the greater attraction for the electrons in the bond is assigned a negative OS equal to its charge in its ionic compounds.

6. The sum of the oxidation states must

be zero for an electrically neutral compound and must be equal to the overall charge for an ionic species.

Determine the Oxidation States

Fe2O3 + 2Al --> Al2O3 + 2Fe

LEO the Lion Says GEROIL RIG

A Summary of an Oxidation-Reduction Process

LEO the lion says GER or OIL RIG

Fe2O3 + 2Al --> Al2O3 + 2Fe

Which Atoms Undergo Redox?

1. 2H2 (g) + O2 (g) --> 2H2O (g)

2. Zn (s) + Cu2+(aq) --> Zn2+ (aq) + Cu(s)

3. 2AgCl (s) + H2 (g) --> 2H+ (aq) + 2Ag(s) + 2Cl- (aq)

4. 2MnO4- (aq) + 16H+ (aq) + 5C2O4

2- (aq) --> 2Mn2+(aq) + 10 CO2 (g) + 8 H2O (l)

Balancing by Half-Reaction Method

1. Write separate reduction, oxidation reactions.

2. For each half-reaction:

Balance elements (except H, O)

Balance O using H2O

Balance H using H+

Balance charge using electrons

Balancing by Half-Reaction Method (continued)

3. If necessary, multiply by integer to equalize electron count.

4. Add half-reactions.

5. Check that elements and charges are balanced.

The Half-Reaction Method for Balancing Redox ReactionsHalf Reactions = the two parts of an

oxidation reduction reaction, one representing oxidation, the other reduction.

Balance the following equation in acid solution using the half reaction method.

Cu(s) + HNO3 (aq) --> Cu2+ (aq) + NO(g)

Cu(s) + HNO3 (aq) --> Cu2+ (aq) + NO(g)

1. Identify and write equations for the half reactions.

Copper is being oxidized: Cu --> Cu2+

Nitrogen is being reduced: HNO3 --> NO

2. Balance each half reaction.

(oxidation) Cu --> Cu2+ + 2e-

(reduction) HNO3 --> NO

a. Balance all atoms that are neither oxygen nor hydrogen.

b. Balance oxygens by adding water to the side that needs oxygen.

c. Balance hydrogens by adding H+ to the side that needs hydrogen.

d. Balance charges by adding electrons to the side that is more positive

3. Equalize the electron transfer. The same number of electrons must be gained as are lost in the reaction

4. Add the half reactions and cancel appropriately to get a complete redox reaction.

Cancel electrons on both sides and double check. Do we have the correct number of electrons on both sides?

Cr2O72- (aq) + NO (g) --> Cr3+ (aq) + NO3

- (aq)

Balancing Redox Equations in Basic Solutions

Balance the following equation (it is already balanced in acid) assuming it is now in basic solution.

Cr2O72-(aq) + 2NO (g) + 6H+ (aq) --> 2Cr3+ (aq) + 2NO3

- (aq) + 3H2O (l)

We need to get rid of excess H+, because OH- is the dominate acid base related species.

Solution: add 6 OH- to both sides of the equation.Cr2O7

2-(aq) + 2NO (g) + 6H+ (aq) + 6OH- --> 2Cr3+ (aq) + 2NO3- (aq) + 3H2O (l) + 6OH-

Cr2O72-(aq) + 2NO (g) + 6 H2O --> 2Cr3+ (aq) + 2NO3

- (aq) + 3H2O (l) + 6OH-

Cancel waters on both sides

Cr2O72-(aq) + 2NO (g) + 3 H2O --> 2Cr3+ (aq) + 2NO3

- (aq) + 6OH- (aq)

Simple Oxidation-Reduction Titrations

1. Balance the redox equation.

2. Determine the moles of titrant.

3. Use the balanced redox equation to determine the number of moles of unknown.

4. Convert from moles of unknown to grams, percent, molarity, or whatever.

The use of potassium permanganate (KMnO4) as an oxidizing agent is described in your text. A 0.0483M KMnO4 solution was used to titrate a solution containing 0.8329 g of impure calcium oxalate, CaC2O4. If 30.25 mL of the KMnO4 solution was required to reach the titration endpoint, calculate the percent purity of the CaC2O4.

MnO4- (aq) + C2O4

2- (aq) --> Mn2+ (aq) +CO2 (g)