General Chemistry

2. Chapter 15
Acid-Base Equilibria

3. Common Ion Effect

4. Section 15.1
Solutions of Acids or Bases Containing a
Common Ion
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Consider mixture of CH3COONa (strong electrolyte) and
CH3COOH (weak acid).
CH3COONa (s) Na+ (aq) + CH3COO- (aq)
CH3COOH (aq) H+ (aq) + CH3COO- (aq)
common
ion
What will happen?
Based on the Le Chatelier's principle the reaction shits to the left.

5. Section 15.1
Solutions of Acids or Bases Containing a
Common Ion
Example
NaCN(aq) + H2O(l) Na+(aq) + CN-(aq)
HCN(aq) + H2O(l) H3O+(aq) + CN-(aq)
 A solution of HCN and NaCN is less acidic than a
solution of HCN alone. Why?
Ka =
[H3O+][CN-]
[HCN]

6. a) What is the pH of a solution containing 0.30 M HCOOH?
b) What is the pH of a solution containing 0.30 M HCOOH and 0.52 M
HCOOK? Ka = 1.8 × 10-4
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30 0.00
-x +x
0.30 - x
0.00
+x
x x
a)
Ka = x= 0.0073
pH= 2.13
x2
0.3-x

7. a) What is the pH of a solution containing 0.30 M HCOOH?
b) What is the pH of a solution containing 0.30 M HCOOH and 0.52 M
HCOOK? Ka = 1.8 × 10-4
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30 0.00
-x +x
0.30 - x
0.52
+x
x 0.52 + x
0.30 – x  0.30
0.52 + x  0.52
b) Mixture of weak acid and conjugate base!
Ka = x= 1.03 × 10-4
pH = 3.98
0.52x
0.3

8. Section 15.1
Solutions of Acids or Bases Containing a
Common Ion
Buffer Solution

9. Section 15.2
Buffered Solutions
Key Points about Buffered Solutions
 A buffer solution is a solution of:
1. A weak acid or a weak base and
2. The salt of the weak acid or weak base
Example: mixture of CH3COOH and CH3COONa
 Buffered Solution – resists a change in pH.
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10. 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) CO3
2- is a weak base and HCO3
- is its conjugate acid
buffer solution

11. Section 15.2
Buffered Solutions
Buffered Solution – resists a change in pH.
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12. Addition of HCl H+ + Cl-
CH3COOH (aq) CH3COOH + H+
resists a change in pH

13. Section 15.2
Buffered Solutions
Henderson–Hasselbalch Equation

14. Section 15.2
Buffered Solutions
Henderson–Hasselbalch Equation
 For a particular buffering system (conjugate acid–base
pair), all solutions that have the same ratio [A–] / [HA]
will have the same pH.
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 a
A
pH = p + log
HA

  K

15. Section 15.2
Buffered Solutions
What is the pH of a buffer solution that is 0.45 M acetic
acid (HC2H3O2) and 0.85 M sodium acetate (NaC2H3O2)?
The Ka for acetic acid is 1.8 × 10–5.
pH = –logKa + log([C2H3O2
–] / [HC2H3O2])
= –log(1.8 × 10–5) + log(0.85 M / 0.45 M)
= 5.02
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EXERCISE!

16. Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer
system.
NH4
+ (aq) H+ (aq) + NH3 (aq)
Ka= 5.6 x 10-10
pH = pKa + log
[NH3]
[NH4
+]
pKa = 9.25
pH = 9.25 + log
[0.30]
[0.36]
= 9.17

17. Adding Strong Acid to a Buffered Solution

18. Section 15.2
Buffered Solutions
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19. The pH of a buffer solution containing 1 M CH3COOH and
1 M CH3COONa is 4.742. Ka = 1.82×10-5
a) What is the pH of a solution after 0.01 mole of HCl (g)
has been added to one liter of buffer?
b) What is the pH of a solution after 0.01 mole of NaOH
(s) has been added to one liter of buffer?
CH3COOH (aq) H+ (aq) + CH3COO- (aq)
Initial (M)
after H+ addition:
1 1.82×10-5
1+ 0.01 ?
1
1-0.01
pH = pKa + log
[CH3COO-]
[CH3COOH]
pH = 4.74 + log
[0.99]
[1.01]
= 4.731
a)
4.742
Compare to

20. Buffering Capacity

21. Section 15.3
Buffering Capacity
 The amount of protons or hydroxide ions the buffer can
absorb without a significant change in pH.
 Determined by the magnitudes of [HA] and [A–].
 A buffer with large capacity contains large
concentrations of the buffering components.
Buffer capacity: HA(1M)/A-(1M)> HA(0.5M)/A-(0.5M)
The most effective buffering will occur when [HA] is equal
to [A-] or pH= pKa
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22. Preparing a Buffer

23.  pKa of the weak acid to be used in the buffer should be
as close as possible to the desired pH.
 Example: Prepare a solution buffered at pH 4.30 using
benzoic acid-sodium benzoate.
pH = pKa + log
[A-]
[BA]
= 1.27
[A-]
[BA]
4.3 = 4.19 + log
[A-]
[BA]
[Benzoic acid]=1 M and [Sodium benzoate]=1.27 M