Additional Aspects of
Aqueous Equilibria
Chapter 17
The Common Ion Effect and Solubility
The presence of a common ion decreases
the solubility of the salt
AgBr (s)
Ag+ (aq) + Br- (aq)
If we add some solid NaBr to this solution:
NaBr (s)
Na+ (aq) + Br- (aq)
Buffered solution ≡ solution has the ability to resist changes in
pH upon the addition of small amounts of either acid or base.
Consists of:
1. A weak acid or a weak base and
2. The salt of the weak acid or weak base
Both must be present!
Fig 17.2 If a small amount of hydroxide is added to
an equimolar solution of HF in NaF, for example, the
HF reacts with the OH− to make F− and water
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Fig 17.2 Similarly, if acid is added, the F− reacts with
it to form HF and water
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Which of the following are buffer systems?
(a) KF/HF
(b) KBr/HBr,
(c) Na2CO3/NaHCO3
(a) KF 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
What is the pH of a solution containing 0.30 M HCOOH
and 0.52 M HCOOK? Ka for HCOOH = 1.8 X 10-4.
Mixture of weak acid and conjugate base!
HCOOH (aq)
Initial (M)
Change (M)
Equilibrium (M)
H+ (aq) + HCOO- (aq)
0.30
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
Can we approximate? Is [HCOOH]o > 100 Ka? YES!
0.30 – x  0.30
0.52 + x  0.52
x = [H+] = 1.04 X 10-4 M
pH = 3.98
Consider mixture of salt NaA and weak acid HA.
NaA (s)
HA (aq)
[H+]
Na+ (aq) + A- (aq)
H+ (aq) + A- (aq)
Ka [HA]
≈
[A-]
-log [H+] ≈ -log Ka - log
[H+][A-]
Ka =
[HA]
Henderson-Hasselbalch
[HA]
[A-]
-]
[A
-log [H+] ≈ -log Ka + log
[HA]
[A-]
pH ≈ pKa + log
[HA]
Equation
[conjugate base]
pH ≈ pKa + log
[acid]
pKa = -log Ka
What is the pH of a solution containing 0.30 M HCOOH
and 0.52 M HCOOK? Ka for HCOOH = 1.8 X 10-4.
Mixture of weak acid and conjugate base!
HCOOH (aq)
H+ (aq) + HCOO- (aq)
[HCOO-]
pH ≈ pKa + log
[HCOOH]
HCOOH pKa = 3.74
[0.52]
pH ≈ 3.74 + log
= 3.98
[0.30]
Limitations in Using HendersonHasselbalch Equation
• H-H equation is an approximation
• Fails when [HA] < 100 Ka or [BH] < 100 Kb
• pH of buffer made ≠ pH calculated
• Uncertainty in Ka and in Kb
• Approximations made (no use of quadratic eqn.)
• Ion-ion interactions when M > 0.01 M (no activities)
Buffer capacity ≡ the number of moles of a strong
acid or strong base that causes 1.00 L
of a buffer to change pH ± 1.00 unit.
[conjugate base]
pH ≈ pKa + log
[acid]
pKa for the acid
should lie within ± 1
unit of desired pH
Want:
log
[NaA]
[HA]
 0
i.e., [HA] ≈ [NaA]
Chemistry In Action: Maintaining the pH of Blood
Fig 17.4 Red blood cells
pH = 7.35 – 7.45
Carbonate- bicarbonate buffer
EXAM #4
• Covers Chapters 16 and 17.1 – 17.2
• 40 Multiple Choice (80%, all qualitative)
• Two calculations: (20%)
• Calculate pH of a weak acid solution
• Calculate pH of a buffer
• All equations provided
• Constants & Periodic Table provided
Practice problems:
1) Calculate the pH of a 0.060 M HF solution. Ka = 7.1 x 10-4.
2) What is the pH of a buffer of 0.15 M NH3 / 0.35 M NH4Cl?
Kb for NH3 = 1.8 x 10-5.