1. Acid-Base Equilibria and Solubility Equilibria
Chapter 16
Semester 1/2017
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2. Chapter 16
Semester 1 / 2016
16.1 Homogeneous versus Heterogeneous Solution Equilibria
16.3 Buffer Solution
16.6 Solubility Equilibria
16.8 The Common Ion Effect and Solubility

3. 16.1 Homogeneous Versus Heterogeneous Solution Equilibria
At equilibrium a weak acid solution contains nonionized acid as well as H+ ions and the conjugate base. All of these species are dissolved so the system is said to be Homogeneous.
If we consider equilibrium of the dissolution and precipitation of slightly soluble substances, the system is said to be Heterogeneous.

4. 16.3 Buffer Solutions A buffer solution is a solution of:
A weak acid or a weak base and
The salt of the weak acid or weak base (Both must be present!)
Definition: A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO- (aq) CH3COOH (aq)
CH3COO- combines with H+ ions from strong acid to produce
weak acid (weak dissociation)
Add strong base
OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)
CH3COOH combines with OH- ions from strong base to produce
H2O(weak dissociation)
16.3

5. HCl + CH3COO- CH3COOH + Cl-
HCl H+ + Cl-
HCl + CH3COO CH3COOH + Cl-
16.3

6. Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3
(a) HF is a weak acid and F- is its conjugate base
buffer solution
(b) HBr is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is it conjugate acid
buffer solution
16.3

7. NH4+ (aq) H+ (aq) + NH3 (aq)
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution?
moles of NH3 =0.024= (0.30*80)/1000
NH4+ (aq) H+ (aq) + NH3 (aq)
moles of NH4+ = 0.029=(0.36*80)/1000
pH = pKa + log
[NH3]
[NH4+]
pH = log
[0.30]
[0.36]
pKa = 9.25
= 9.17
moles of OH- =0.001=(0.05*20)/1000
start (moles)
0.029
0.001
0.024
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
end (moles)
0.028
0.0
0.025
final volume = 80.0 mL mL = 100 mL
[NH4+] =
0.028
0.10
[NH3] =
0.025
0.10
pH = log
[0.25]
[0.28]
= 9.20
16.3

8. Chemistry In Action: Maintaining the pH of Blood
16.3

9. 16.6 Solubility Equilibria
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
Ksp is the solubility product constant
MgF2 (s) Mg2+ (aq) + 2F- (aq)
Ksp = [Mg2+][F-]2
Ag2CO3 (s) Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO33-]2
Dissolution of an ionic solid in aqueous solution:
Q < Ksp
Unsaturated solution
No precipitate
Q = Ksp
Saturated solution
No precipitate
Q > Ksp
Supersaturated solution
Precipitate will form
16.6

10. 16.6

11. Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution.
Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.
16.6

12.  What is the solubility of silver chloride in g/L ?
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = 1.6 x 10-10
Initial (M)
Change (M)
Equilibrium (M)
0.00
0.00
Ksp = [Ag+][Cl-] +s +s
Ksp = s2
s = Ksp  s s
s = 1.3 x 10-5
[Ag+] = 1.3 x 10-5 M
[Cl-] = 1.3 x 10-5 M
1.3 x 10-5 mol AgCl
1 L soln
g AgCl
1 mol AgCl x
Solubility of AgCl =
= 1.9 x 10-3 g/L
16.6

13. 16.6

14. The ions present in solution are Na+, OH-, Ca2+, Cl-.
If 2.00 mL of M NaOH are added to 1.00 L of M CaCl2, will a precipitate form?
The ions present in solution are Na+, OH-, Ca2+, Cl-.
Only possible precipitate is Ca(OH)2 (solubility rules).
Is Q > Ksp for Ca(OH)2?
[Ca2+]0 = M
[OH-]0 = 4.0 x 10-4 M
Q = [Ca2+]0[OH-]0 2
= 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6
Q < Ksp
No precipitate will form
16.6

15. 16.8 The Common Ion Effect and Solubility
Consider a solution containing two dissolved substances that share a common ion (AgCl and AgNO3)
AgCl (s) Ag+ (aq) + Cl- (aq)
AgNO3 (s) Ag+ (aq) + NO3- (aq)
Ag+ ions come from two sources: i.e AgCl & AgNO3
The total increase in [Ag+] ion concentration will make the ion product
greater than the solubility product:
Q = [Ag+]total[Cl-] > Ksp, To reestablish equilibrium, some
AgCl will precipitate out of the solution, until the ion product
is equal to Ksp  Adding a common ion decreases the solubility of the salt (AgCl) in solution.