Chapter 16 Aqueous Ionic Equilibrium. pH / pOH Calculations Strong acids Strong bases Weak acids Weak bases Salts.

Chapter 16

Aqueous Ionic Equilibrium

pH / pOH Calculations

• Strong acids

• Strong bases

• Weak acids

• Weak bases

• Salts

NaAc(aq) → Na+(aq) + Ac−(aq)

Calculate pH for a mixture of HAc and Ac−

HAc(aq) ⇌ H+(aq) + Ac−(aq)acid

Ac−(aq) + H2O(l) ⇌ HAc(aq) + OH−(aq)base

NH4Cl(aq) → NH4+(aq) + Cl−(aq)

Calculate pH for a mixture of NH3 and NH4+

NH4+(aq) H⇌ +(aq) + NH3(aq)

acid

NH3(aq) + H2O(l) NH⇌ 4+(aq) + OH−(aq)

base

Calculate the pH of a solution that contains 0.100 M HAc

and 0.100 M NaAc. Ka for HAc is 1.8 x 10−5.

Example 16.1, page 716

pH = 4.74

weak acid + its conjugate base = buffer solution

weak base + its conjugate acid = buffer solution

Adding H+ or OH−, pH does not change too much

Calculate the change in pH that occurs when 0.010 mol solid

NaOH is added to 1.0 L of the buffered solution described in

Example 16.1. Compare this pH change with that which

occurs when 0.010 mol solid NaOH is added to 1.0 L of water.

Assume there is no volume change after solid NaOH is added.


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[A ]pH = pK + log

[HA]

Henderson-Hasselbalch equationfor buffer solutions

pH for buffer solutions: ICE → exact answer

pH for buffer solutions: approximation

Calculate the pH of a solution that contains 0.50 mol/L HAc

and 0.50 mol/L NaAc. Ka for HAc is 1.8 x 10−5.

pH = 4.74

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[A ]pH = pK + log

[HA]

Example 16.1, page 716. revisited

Calculate the pH of a buffer solution that is 0.050 M in

benzoic acid (HC7H5O2) and 0.150 M in sodium benzoate (NaC7H5O2).

For benzoic acid, Ka = 6.5 10–5.

Example 16.2, page 718.

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ai

[A ]pH = pK + log

[HA]

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[A ]pH = pK + log

[HA]

Use the Henderson–Hasselbalch equation to calculate

the pH of a buffer solution that is 0.50 M in NH3 and

0.20 M in NH4Cl. For ammonia, pKb = 4.75.

Example 16.4, page 725.

Calculate the change in pH that occurs when 0.010 mol gaseous

HCl is added to 1.0 L of each of the following solutions:

Solution A: 5.00 mol/L HAc and 5.00 mol/L NaAc

Solution B: 0.050 mol/L HAc and 0.050 mol/L NaAc

Ka for HAc is 1.8 x 10−5.

HAc Ac− HAc Ac−

H+

HAc Ac−

H+

Solution A

HAc Ac−

Solution B

-i

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[A ]pH = pK + log

[HA]

Buffer capacity

To maximize buffer capacity:

1) High [HA] and [A−]

2) [HA] = [A−] (example page 725-726)

pH = pKa

Titration

Titrate strong acid with strong base

Calculate the pH when the following quantities of

0.100 mol/L NaOH solution have been added to

50.0 mL of 0.100 mol/L HCl solution

a)0 mL; b) 49.0 mL;c) 50.0 mL; d) 51.0 mL;e) 60.0 mL.

a)1.000; b) 3.00;c) 7.00; d) 11.00;e) 11.96

0 10 20 30 40 50 60 700

1

2

3

4

5

6

7

8

9

10

11

12

13

14

pH

mL of NaOH

equivalence point

Titrate weak acid with strong base

Calculate the pH when the following quantities of

0.100 mol/L NaOH solution have been added to

50.0 mL of 0.100 mol/L HAc solution. Ka of HAc

is 1.8 x 10−5.

a)0 mL; b) 49.0 mL; c) 50.0 mL;d) 51.0 mL; e) 60.0 mL.

A very similar example: page 733 -- 738

a)2.87; b) 6.43;c) 8.72; d) 11.00;e) 12.00

Chapter 17

The pH Curves for the Titrations of 50.0-mL Samples of 0.10 M Acids with Various Ka Values

with 0.10 M NaOH

Solubility Equilibrium

CaF2 (s) Ca2+ (aq) + 2F− (aq)H2O

Solubility equilibrium is established

Aqueous solution of CaF2 is saturated

Ksp = [Ca2+][F−]2

solubility product constant

“insoluble” salts

Write down the dissociation reactions and the expression of Ksp

Solid silver chromate is added to pure water at 25 °C.

Equilibrium is achieved between undissolved Ag2CrO4(s)

and its aqueous solution. Silver ion concentration is

1.3 x 10−4 mol/L. Calculate Ksp for this compound.


Solubility equilibrium is established

Aqueous solution of CaF2 is saturated

Solubility: the concentration of a saturated solution.

Unit: mol/L or ∙ ∙ ∙

solubility ≠ solubility product constant Ksp

Solubility ↔ Ksp

value of Kc or Kp

equilibrium concentrations or pressures

Copper(I) bromide has a measured solubility of 2.0 x 10−4 mol/L

at 25 °C. Calculate its Ksp.

Ksp = 4.0 x 10−8

Bismuth sulfide (Bi2S3) has a measured solubility of

1.0 x 10−15 mol/L at 25 °C. Calculate its Ksp.

Ksp = 1.1 x 10−73

Try example 16.9 on page 746

The Ksp for copper(II) iodate, Cu(IO3)2, is 1.4 x 10−7 at 25 °C.

Calculate its solubility at 25 °C.

3.3 x 10−3 M


Calculate the solubility of CaF2 (Ksp = 1.46 x 10−10) in a

0.100 mol/L NaF solution.



Increase [Ca2+] or [F−] → equilibrium shifts to left → solubility ↓

common ion effect

For Mg(OH)2 Ksp = 1.8 x 10−11. What is the pH of a saturated

solution of Mg(OH)2? What is its solubility?

Suppose that solid Mg(OH)2 is equilibrated with a solution

buffered at a more acidic pH of 9.00. Ksp of Mg(OH)2 is 1.8 x 10−11

What are the [Mg2+] and solubility?


Q = [Ca2+][F−]2

If Q > Ksp, precipitate will form

If Q < Ksp, precipitate will not form

A solution is prepared by adding 750.0 mL of 4.00 x 10−3 mol/L

Ce(NO3)3 to 300.0 mL of 2.00 x 10−2 mol/L KIO3. Will Ce(IO3)3

(Ksp = 1.9 x 10−10) precipitate from this solution?


1 Ka for acetic acid (CH3COOH) is 1.8 x 10-5 while Ka for hypochlorous (HClO) ion is 3.0 x 10-8. A. Which acid is the stronger acid?B. Which is the stronger conjugate base? Acetate ion (CH3COO-) or chlorous (ClO-) ion?C. Calculate kb values for CH3COO- and ClO-. 2. a. Calculate the pH of a 1.50 L solution containing 0.750 mole of HCN and 0.62 mole of KCN. Ka = 4.0 x 10-10

b. If 0.015 mole of KOH was added, calculate the pH of the solution.

c. If 0.015 mole of HBr was added, calculate the pH of the solution.

Chapter 16 Aqueous Ionic Equilibrium. pH / pOH Calculations Strong acids Strong bases Weak acids Weak bases Salts.

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