1. Acid-Base Equilibria and Solubility Equilibria
Chapter 16
Semester 2/2016
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2. Chapter 16
Semester 2 / 2016
16.1 Homogeneous versus Heterogeneous Solution Equilibria The Common Ion Effect
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. 16.2 common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.
The presence of a common ion suppresses the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).
CH3COONa (s) Na+ (aq) + CH3COO- (aq)
Common ion
CH3COOH (aq) H+ (aq) + CH3COO- (aq)

5. Henderson-Hasselbalch equation
Consider mixture of salt NaA(CH3COONa) and weak acid HA.(CH3COOH)
CH3COONa(s) Na+ (aq) + CH3COO- (aq)
Ka =
[H+][CH3COO-]
[CH3COOH]
CH3COOH(aq) H+ (aq) + CH3COO-(aq)
[H+] =
Ka [CH3COOH]
[CH3COO-]
Henderson-Hasselbalch equation
-log [H+] = -log Ka - log
[CH3COOH]
[CH3COO-]
-log [H+] = -log Ka + log
[CH3COO-]
[CH3COOH]
pH = pKa + log
[CH3COO-]
[CH3COOH]
pH = pKa + log
[conjugate base]
[acid]
pKa = -log Ka

6. 16.3 Buffer Solutions A buffer solution is a solution of:
weak acid AND weak acid salt (weak acid & strong base salt example (CH3COOH+CH3COONa) 2.Weak base and weak base salt (weak base and strong acid) Example (NH4OH+NH4Cl)
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) cannot change pH odf
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

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

8. 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

9. Calculate the pH of the (1M CH3COOH+1M CH3COONa) buffer system
Calculate the pH of the (1M CH3COOH+1M CH3COONa) buffer system. What is the pH after the addition of 0.1mol of HCl to 1L of the buffer solution?
pH = pKa + log
[Salt]
[Acid]
16.3

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

11. Example: Describe how you would prepare a “phosphate buffer” with a pH of about 7.40.
For a buffer to function effectively, the concentrations of the acid component must be roughly equal to the conjugate base component. According to Equation (16.4), when the desired pH is close to the pKa of the acid, that is, when
pH ≈ pKa,

12. Because phosphoric acid is a triprotic acid, we write the three stages of ionization as follows. The Ka values are obtained from Table 15.5 and the pKa values are found by applying Equation (16.3).

13. The most suitable of the three buffer systems is pKa = 7
The most suitable of the three buffer systems is pKa = 7.21 because the pKa of the acid is closest to the desired pH. From the Henderson-Hasselbalch equation we write

14. Thus, one way to prepare a phosphate buffer with a pH of 7
Thus, one way to prepare a phosphate buffer with a pH of 7.40 is to dissolve disodium hydrogen phosphate (Na2HPO4) and sodium dihydrogen phosphate (NaH2PO4) in a mole ratio of 1.5:1.0 in water. For example, we could dissolve 1.5 moles of Na2HPO4 and 1.0 mole of NaH2PO4 in enough water to make up a 1-L solution.

15. 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

16. 16.6

17. 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

18.  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

19. 16.6

20. 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

21. 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.