ch. 17: aqueous ionic equilibriumfaculty.sdmiramar.edu/nsinkaset/powerpoints/chap17.pdf ·...

Ch. 17: Aqueous Ionic Equilibrium

Dr. Namphol Sinkaset Chem 201: General Chemistry II

I. Chapter Outline

I. Introduction II. Buffers III. Titrations and pH Curves IV. Solubility Equilibria and Ksp V. Complex Ion Equilibria

I. Last Aspects of Equilibria • In this chapter, we cover some final

topics concerning equilibria. • Buffers are designed to take advantage

of Le Châtelier’s Principle. • Solubility can be reexamined from an

equilibrium point of view. • Complex ions are introduced and their

formation explained using equilibrium ideas.

II. pH Resistive Solutions

• A solution that resists changes in pH by neutralizing added acid or base is called a buffer.

• Many biological processes can only occur within a narrow pH range.

• In humans, the pH of blood is tightly regulated between 7.36 and 7.42.

II. Creating a Buffer

• To resist changes in pH, any added acid or base needs to be neutralized.

• This can be achieved by using a conjugate acid/base pair.

II. How the Buffer Works

• For a buffer comprised of the conjugate acid/base pair of acetic acid/acetate: OH-

(aq) + CH3COOH(aq) H2O(l) + CH3COO-(aq)

H+(aq) + CH3COO-

(aq) CH3COOH(aq) • As long as we don’t add too much OH- or

H+, the buffer solution’s pH will not change drastically.

II. Calculating the pH of Buffers

• The pH of a buffer can be calculated by approaching the problem as an equilibrium in which there are two initial concentrations.

• Since an acid and its conjugate base are both in solution, problem can be solved from a Ka or Kb point of view.

II. Sample Problem 17.1

• A solution was prepared in which [CH3COONa] = 0.11 M and [CH3COOH] = 0.090 M. What is the pH of the solution? Note that the Ka for acetic acid is 1.8 x 10-5.

II. Sample Problem 17.2

• A student dissolves 0.12 mole NH3 and 0.095 mole NH4Cl in 250 mL of water. What’s the pH of this buffer solution? Note that the Kb for ammonia is 1.8 x 10-5.

II. A Special Buffer Equation

• Buffers are used so widely that an equation has been developed for it.

• This equation can be used to perform pH calculations of buffers.

• More importantly, it can be used to calculate how to make solutions buffered around a specific pH.

II. Henderson-Hasselbalch Eqn.

II. The H-H Equation

• The Henderson-Hasselbalch equation works for a buffer comprised of conjugate acid/base pairs.

• It works provided the “x is small” approximation is valid.

II. Sample Problem 17.3

• A student wants to make a solution buffered at a pH of 3.90 using formic acid and sodium formate. If the Ka for formic acid is 1.8 x 10-4, what ratio of HCOOH to HCOONa is needed for the buffer?

II. Sample Problem 17.4

• A researcher is preparing an acetate buffer. She begins by making 100.0 mL of a 0.010 M CH3COOH solution. How many grams of CH3COONa does she need to add to make the pH of the buffer 5.10 if the Ka for acetic acid is 1.8 x 10-5?

II. Upsetting the Buffer

• A buffer resists changes to pH, but it is not immune to change.

• Adding strong acid or strong base will result in small changes of pH.

• We calculate changes in pH of a buffer by first finding the stoichiometric change and then performing an equilibrium calculation.

II. Illustrative Problem

• A 1.00 L buffer containing 1.00 M CH3COOH and 1.00 M CH3COONa has a pH of 4.74. It is known that a reaction carried out in this buffer will generate 0.15 mole H+. If the pH must not change by more than 0.2 pH units, will this buffer be adequate?

II. Stoichiometric Calculation • When the H+ is formed, it will react

stoichiometrically with the base. • We set up a different type of table to find new

initial concentrations. • We use ≈0.00 mol because [H+] is negligible.

H+ + CH3COO- CH3COOH

Initial 0.15 mol 1.00 mol 1.00 mol Change -0.15 mol -0.15 mol +0.15 mol Final ≈0.00 mol 0.85 mol 1.15 mol

II. Equilibrium Calculation • We use the new buffer concentrations in an

equilibrium calculation to find the new pH. • Again, use ≈0.00 M because [H3O+] is negligible.

CH3COOH + H2O H3O+ + CH3COO-

Initial 1.15 M ---- ≈0.00 M 0.85 M

Change -x ---- +x +x

Equil. 1.15 – x ---- x 0.85 + x

Solve to get [H3O+] = 2.44 x 10-5 M, and pH = 4.613.

II. How the Buffer Works

II. A Simplification for Buffers

• In buffer problems, # of moles can be used in place of concentration.

• Why?*

II. Sample Problem 17.5

• Calculate the pH when 10.0 mL of 1.00 M NaOH is added to a 1.0-L buffer containing 0.100 mole CH3COOH and 0.100 mole CH3COONa. Note that the Ka for acetic acid is 1.8 x 10-5.

II. Buffers Using a Weak Base

• Up until now, we’ve been creating buffers with a weak acid and its conjugate base.

• Can also make a buffer from a weak base and its conjugate acid.

• Henderson-Hasselbalch still applies, but we need to get pKa of the conjugate acid.

II. Sample Problem 17.6

• Calculate the pH of a 1.0-L buffer that is 0.50 M in NH3 and 0.20 M in NH4Cl after 30.0 mL of 1.0 M HCl is added. Note that the Kb for NH3 is 1.8 x 10-5.

II. Making Effective Buffers • There are a few parameters to keep in

mind when making a buffer: The relative [ ]’s of acid and conjugate base

should not differ by more than factor of 10. The higher the actual [ ]’s of acid and

conjugate base, the more effective the buffer. The effective range for a buffer system it +/-

1 pH unit on either side of pKa.

II. Buffer Capacity • Buffer capacity is defined as the amount

of acid or base that can be added to a buffer without destroying its effectiveness.

• A buffer is destroyed when either the acid or conjugate base is used up.

• Buffer capacity increases w/ higher concentrations of buffer components.

III. Titrations • In an acid-base titration, a basic (or

acidic) solution of unknown [ ] is reacted with an acidic (or basic) solution of known [ ].

• An indicator is a substance used to visualize the endpoint of the titration.

• At the equivalence point, the number of moles of acid equals the number of moles of base.

III. Visual of a Titration

III. Titration/pH Curves

• A titration or pH curve is a plot of how the pH changes as the titrant is added.

• It is possible to calculate the pH at any point during a titration.

• Multiple pH’s can be calculated, and the results plotted to create the theoretical titration curve.

III. Find the Equivalence Point! • The keys to these types of problems

are writing the titration equation and finding the equivalence point of the titration.

• The calculation then depends on what region of the titration curve you are in:

1) Before titration begins 2) Pre-equivalence 3) Equivalence point 4) Post-equivalence

III. Illustrative Problem

• Sketch the pH curve for the titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH.

III. Illustrative Problem Solution

1) Write the titration equation. HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

2) Calculate the equivalence point. What volume of NaOH is needed to completely react with HCl?

III. Illustrative Problem Solution

3) Calculate initial pH before titration. Since HCl is strong, 0.100 M HCl has

[H3O+] = 0.100 M, and pH = 1.000. 4) Calculate the pH of some points in the

pre-equivalence region. As NaOH is added, the neutralization

reaction OH-(aq) + H3O+

(aq) H2O(l) takes place.

We calculate the pH after addition of 5.00 mL of NaOH.

III. Illustrative Problem Solution

• To calculate the pH after 5.00 mL, we need to calculate initial moles of acid and the number of moles of base.

• We put these moles into a reaction chart.

III. Illustrative Problem Solution H3O+ + OH- 2H2O

Initial 0.00250 mol 0.000500 mol 0 mol

Change -0.000500 mol -0.000500 mol +0.001000 mol

Final 0.00200 mol 0 mol 0.001000 mol

The H3O+ leftover is in a larger volume so we calculate its concentration.

III. Illustrative Problem Solution • We do the same thing for some other

points: 10.0 mL, 15.0 mL, and 20.0 mL. • Results summarized below.

Volume (ml) Mol H3O+ [H3O+] (M) pH 5.00 0.00200 0.06667 1.176 10.0 0.00150 0.04286 1.368 15.0 0.00100 0.02500 1.602 20.0 0.00050 0.01111 1.954

III. Illustrative Problem Solution

5) Calculate the pH at the equivalence point. For a strong-strong titration, pH always

equals 7.00 at the equivalence point! 6) Calculate the pH of some points in the

post-equivalence region. In this region, the pH depends on the

excess OH- added.

III. Illustrative Problem Solution

• To find the excess added, calculate how many mL past the equivalence point have been added, convert to moles, and divide by total volume. For 30.0 mL:

pH can then be found from pOH.

III. Illustrative Problem Solution • Again, calculate for additional points like

35.0, 40.0, and 50.0 mL. Results summarized below.

Volume (mL) [OH-] (M) pOH pH 30.0 0.009091 2.0414 11.959 35.0 0.01667 1.7781 12.222 40.0 0.02307 1.6370 12.363 50.0 0.03333 1.4771 12.523

III. Illustrative Problem Solution

• Now we plot the data points and sketch the pH titration curve!

III. Sample Problem 17.7

• A 0.0500 L sample of 0.0200 M KOH is being titrated with 0.0400 M HI. What is the pH after 10.0 mL, 25.0 mL, and 30.0 mL of the titrant have been added?

III. Weak Acid/Base Titrations • The situation becomes a little more

complicated when a weak acid/base is titrated with a strong base/acid.

• Again, the keys are to identify the titration reaction and the equivalence point.

• The method of calculating the pH will then depend on the region of the titration curve.

III. Four Different Regions • For a weak acid titrated with a strong

base, there are 4 regions as well: 1) Before titration: a weak acid problem! Why?* 2) Pre-equivalence: a buffer! Why?* 3) Equiv. point: a weak base problem! Why?* 4) Post-equivalence: a dilution problem! Why?*

• For a weak base titrated with a strong acid, everything is just rewritten w/ conjugates!

III. Illustrative Problem

• A 50.00 mL sample of 0.02000 M CH3COOH is being titrated with 0.1000 M NaOH. Calculate the pH before the titration begins, after 3.00 mL of the titrant have been added, at the equivalence point, and after 10.20 mL of the titrant have been added. Note that Ka = 1.75 x 10-5 for acetic acid.

III. Illustrative Problem Solution

1) First, we need to write the titration eqn. CH3COOH(aq) + OH-

(aq) CH3COO-(aq) + H2O(l)

2) Next, we calculate the equivalence point.

III. Illustrative Problem Solution 3) Before titration begins, it’s just a weak

acid problem.

CH3COOH + H2O H3O+ + CH3COO-

Initial 0.02000 M --- 0 0 Change -x --- +x +x Equil. 0.02000-x --- x x

Solving this (with simplification), we get [H3O+] = 5.916 x 10-4, so pH = 3.228.

III. Illustrative Problem Solution

4) After 3.00 mL of 0.1000 M NaOH have been added, we will have a mixture of CH3COOH and CH3COO- in solution. Since it’s a buffer, we can use the

Henderson-Hasselbalch eqn.

III. Illustrative Problem Solution • In the H-H equation, we need a ratio of

CH3COO- to CH3COOH. • After adding 3.00 mL, we are 3.00/10.00

to equivalence. We can use a relative concentration chart.

CH3COOH + OH- CH3COO- + H2O

Relative Initial 1 +3.00/10.00 0 --- Change -3.00/10.00 -3.00/10.00 +3.00/10.00 --- Relative Final 7.00/10.00 0 3.00/10.00 ---

III. Illustrative Problem Solution

• Now we just plug the relative final row into the H-H equation.

III. Illustrative Problem Solution 5) At the equivalence point, all CH3COOH has

been converted to CH3COO-. Initial moles of CH3COOH = moles of CH3COO-

at the equivalence point, but the volume has increased.

Must calculate [CH3COO-] at equiv. pt.

III. Illustrative Problem Solution • Now we solve a weak base problem.

CH3COO- + H2O OH- + CH3COOH

Initial 0.016667 M --- 0 0 Change -x --- +x +x Equil. 0.016667-x --- x x

Using Kb = 5.714 x 10-10 and the simplification, x = 3.089 x 10-6. Thus, pOH = 5.5102 and pH = 8.490. Note that pH does not equal 7.00!!

III. Illustrative Problem Solution

6) At 10.20 mL of added titrant, we are 0.20 mL past equivalence, and the pH depends only on excess OH-. Thus:

Of course, this means that pOH = 3.479 and pH = 10.52.

III. Sample Problem 17.8

• A 25.00 mL sample of 0.08364 M pyridine is being titrated with 0.1067 M HCl. What’s the pH after 4.63 mL of the HCl has been added? Note that pyridine has a Kb of 1.69 x 10-9.

III. Sample Weak/Strong Curves • Important aspects about pH curves for

weak/strong titrations: At equivalence, pH does not equal 7.00. At ½ equivalence, pH = pKa.

III. Polyprotic Acid Titration

• If the Ka’s are different enough, you will see multiple equivalence points.

• Since protons come off one at a time, 1st equiv. pt. refers to the 1st proton, 2nd to the 2nd, etc.

III. Detecting the Equiv. Pt.

• During a titration, the equivalence point can be detected with a pH meter or an indicator.

• The point where the indicator changes color is called the endpoint.

• An indicator is itself a weak acid that has a different color than its conjugate base.

III. Phenolphthalein

III. Indicators

• The indicator has its own equilibrium: HIn(aq) + H2O(l) In-

(aq) + H3O+(aq)

• The color of an indicator depends on the relative [ ]’s of its protonated and deprotonated forms. If pH > pKa of HIn, color will be In-. If pH = pKa of HIn, color will be in between. If pH < pKa of HIn, color will be HIn.

III. Selecting an Indicator

IV. Solubility

• In 1st semester G-chem, you memorized solubility rules and regarded compounds as either soluble or insoluble.

• Reality is not as clear cut – there are degrees of solubility.

• We examine solubility again from an equilibrium point of view.

IV. Solubility Equilibrium

• If we apply the equilibrium concept to the dissolution of CaF2(s), we get: CaF2(s) Ca2+

(aq) + 2F-(aq)

• The equilibrium expression is then: Ksp = [Ca2+][F-]2

• Ksp is the solubility product constant, and just like any other K, it tells you how far the reaction goes towards products.

IV. Some Ksp Values

IV. Calculating Solubility

• Recall that solubility is defined as the amount of a compound that dissolves in a certain amount of liquid (g/100 g water is common).

• The molar solubility is obviously the number of moles of a compound that dissolves in a liter of liquid.

• Molar solubilities can easily be calculated using Ksp values.

IV. Ksp Equilibrium Problems • Calculating molar solubility is essentially

just another type of equilibrium problem. • You still set up an equilibrium chart and

solve for an unknown. Pay attention to stoichiometry!

Fe(OH)3(s) Fe3+(aq) + 3OH-

(aq)

Initial --- 0 0 Change --- +S +3S

Equil. --- S 3S

IV. Sample Problem 17.9

• Which is more soluble: calcium carbonate (Ksp = 4.96 x 10-9) or magnesium fluoride (Ksp = 5.16 x 10-11)?

IV. The Common Ion Effect

• The solubility of Fe(OH)2 is lower when the pH is high. Why?* Fe(OH)2(s) Fe2+

(aq) + 2OH-(aq)

• common ion effect: the solubility of an ionic compound is lowered in a solution containing a common ion than in pure water.

IV. Sample Problem 17.10

• Calculate the molar solubility of lead(II) chloride (Ksp = 1.2 x 10-5) in pure water and in a solution of 0.060 M NaCl.

IV. pH and Solubility

• As seen with Fe(OH)2, pH can have an influence on solubility.

• In acidic solutions, need to consider if H3O+ will react with cation or anion.

• In basic solutions, need to consider if OH- will react with cation or anion.

IV. Sample Problems 17.11

a) Which compound, FeCO3 or PbBr2, is more soluble in acid than in base? Why?

b) Will copper(I) cyanide be more soluble in acid or water? Why?

c) In which type of solution is AgCl most soluble: acidic or neutral?

IV. Precipitation • Ksp values can be used to predict when

precipitation will occur. • Again, we use a Q calculation. If Q < Ksp, solution is unsaturated. Solution

dissolve additional solid. If Q = Ksp, solution is saturated. No more

solid will dissolve. If Q > Ksp, solution is supersaturated, and

precipitation is expected.

IV. Sample Problems 17.12

a) Will a precipitate form if 100.0 mL 0.0010 M Pb(NO3)2 is mixed with 100.0 mL 0.0020 M MgSO4?

b) The concentration of Ag+ in a certain solution is 0.025 M. What concentration of SO4

2- is needed to precipitate out the Ag+? Note that Ksp = 1.2 x 10-5

for silver(I) sulfate.

V. Complex Ions • In aqueous solution, transition metal

cations are usually hydrated. e.g. Ag+

(aq) is really Ag(H2O)2+

(aq). The Lewis acid Ag+ reacts with the Lewis

base H2O. • Ag(H2O)2

+(aq) is a complex ion.

A complex ion has a central metal bound to one or more ligands. A ligand is a neutral molecule or an ion that

acts as a Lewis base with the central metal.

V. Formation Constants

• Stronger Lewis bases will replace weaker ones in a complex ion. e.g. Ag(H2O)2

+(aq) + 2NH3(aq)

Ag(NH3)2+

(aq) + 2H2O(l) For simplicity, it’s common to write

Ag+(aq) + 2NH3(aq) Ag(NH3)2

+(aq)

Since this is an equilibrium, we can write an equilibrium expression for it.

V. Formation Constants

• Kf is called a formation constant. • What does the large Kf mean?

Ag+(aq) + 2NH3(aq) Ag(NH3)2

+(aq)

V. Sample Formation Constants

V. Calculations w/ Kf’s

• Since Kf’s are so large, calculations with them are slightly different.

• We assume the equilibrium lies essentially all the way to the right, and we calculate the “leak back.”

• This changes how we set up our equilibrium chart.

V. Illustrative Problem

• Calculate the concentration of Ag+ ion in solution when 0.085 g silver(I) nitrate is added to a 250.0 mL solution that is 0.20 M in KCN.

V. Illustrative Problem Solution

1) First, we must identify the complex ion. In solution, we will have Ag+, NO3

-, and K+, and CN-.

The complex ion must be made from Ag+ and CN-.

Looking at table of Kf’s, we find that Ag(CN)2

- has Kf = 1 x 1021.

V. Illustrative Problem Solution

2) Next, we need initial concentrations. Already know that [CN-] = 0.20 M. We calculate the [Ag+].

V. Illustrative Problem Solution

3) Now we set up our equilibrium chart. Since Kf is so big, we assume the

reaction essentially goes to completion.

Ag+(aq) + 2CN-

(aq) Ag(CN)2-(aq)

Initial 0.00200 M 0.20 M 0 Change ≈ -0.00200 ≈ -0.00400 ≈ +0.00200

Equil. x 0.196 0.00200

V. Illustrative Problem Solution 4) Finally, we solve for x.

Thus, [Ag+] = 5 x 10-23 M. It is very small, so our approximation is valid. Note that book would use 0.20 for [CN-] in the calculation.

V. Sample Problem 17.13

• A 125.0-mL sample of a solution that is 0.0117 M in NiCl2 is mixed with a 175.0-mL sample of a solution that is 0.250 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+

(aq) remains?

V. Complex Ions & Solubility • Formation of complex ions enhances the

solubility of some normally insoluble ionic compounds.

• Typically, Lewis bases will enhance solubility. • e.g. Adding NH3 to a solution containing

AgCl(s) will cause more AgCl(s) to dissolve. Why?*

AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 1.77 x 10-10

Ag+(aq) + 2NH3(aq) Ag(NH3)2

+(aq) Kf = 1.7 x 107

V. Complex Ion Formation

V. Metal Hydroxides

• All metal hydroxides can act as bases. e.g. Fe(OH)3(s) + 3H3O+

(aq) Fe3+(aq) +

6H2O(l) • Some metal hydroxides can act as

acids and bases; they are amphoteric. In addition to the above, Al(OH)3(s) can also

absorb hydroxide. Al(OH)3(s) + OH-

(aq) Al(OH)4-(aq)

V. Aluminum Hydroxide

• In acidic solutions: Al(H2O)6

3+(aq) + H2O(l) Al(H2O)5(OH)2+

(aq) + H3O+(aq)

• As OH- is added, solution becomes neutral. Al(H2O)5(OH)2+

(aq) + 2OH-(aq) Al(H2O)3(OH)3(s) +

2H2O(l) • In basic solutions: Al(H2O)3(OH)3(s) + OH-

(aq) Al(H2O)2(OH)4-(aq)

• Thus, solubility is very pH dependent.

V. Aluminum Hydroxide

V. Amphoteric Hydroxides

• There are not that many metal hydroxides that are amphoteric.

• Only Al3+, Cr3+, Zn2+, Pb2+, and Sn2+.