1. Acid-Base Equilibria
Chapter 16

2. Conjugate Acid-Base Pairs
Eqn 16.7 p 670

3. Conjugate Acid-Base Pairs
Eqn 16.8 p 671

4. Fig 16.4 Relative strengths of some conjugate acid-base pairs
The stronger
the acid…
the weaker its
conjugate base.

5. The Ion Product of Water
Kc =
[H+][OH-]
[H2O]
H2O (l) H+ (aq) + OH- (aq)
[H2O] = 55.6 M
= constant
Kc[H2O] = Kw = [H+][OH-]
Ion-product constant (Kw) - the product of the molar concentrations of H+ and OH- ions at a particular temperature.
Solution is:
[H+] = [OH-]
neutral
At 25°C:
Kw = [H+][OH-] = 1.0 x 10-14
[H+] > [OH-]
acidic
[H+] < [OH-]
basic

6. pH – A Measure of Acidity
pH = −log [H+]
pH = −log [H3O+]
Solution is:
At 25°C
neutral
[H+] = [OH-]
[H+] = 1 x 10-7
pH = 7
acidic
[H+] > [OH-]
[H+] > 1 x 10-7
pH < 7
basic
[H+] < [OH-]
[H+] < 1 x 10-7
pH > 7 pH
[H+]

7. Fig 16.5 H+ concentrations and pH of common substances
pOH = -log [OH-]
[H+][OH-] = Kw = 1.0 x 10-14
−log [H+] – log [OH-] = 14.00
pH + pOH = 14.00

8. The pH of rainwater collected in a certain region of the northeastern United States on a particular day was What is the H+ ion concentration of the rainwater?
pH = -log [H+]
[H+] = 10-pH =
= 1.5 x 10-5 M
The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?
pH + pOH = 14.00
pOH = -log [OH-]
= -log (2.5 x 10-7)
= 6.60
pH = – pOH = – 6.60 = 7.40

9. Other “p” Functions
The “p” in pH tells us to take the negative base - 10 logarithm of the quantity (in this case, hydronium ions)
Some similar examples:
pOH = -log [OH−]
pKw = -log Kw
pCl = -log [Cl−]

10. How Do We Measure pH? For less accurate measurements: Litmus paper
“Red” paper turns blue above ~pH = 8
“Blue” paper turns red below ~pH = 5
Indicator:

11. How Do We Measure pH? Fig 16.6 Digital pH meter
For more accurate measurements:
pH meter, which measures the voltage in the solution
Fig Digital pH meter

12. Strong Electrolyte – 100% dissociation
Strong Acids and Bases
Strong Electrolyte – 100% dissociation
Strong Acids and Strong Bases are strong electrolytes (p 130)
HCl (aq) + H2O (l) H3O+ (aq) + Cl− (aq)
NaOH (aq) Na+ (aq) + OH− (aq)
Table 4.2

13. Weak Acids
Weak electrolytes - only partially ionized in aqueous solution
Weak Acids are weak electrolytes
HF (aq) + H2O (l) H3O+ (aq) + F- (aq)
HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)
Most acidic substances are weak acids

14. Weak Acids (HA) and Acid Ionization Constants
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
HA (aq) H+ (aq) + A- (aq)
Ka =
[H+][A-]
[HA]
Ka ≡ acid ionization constant
weak acid
strength Ka

15. Table 16.2 Some Weak Acids in Water at 25 °C
16.5

16. Calculating Ka from the pH
Sample Exercise p 682
The pH of a 0.10 M solution of formic acid, HCOOH, at 25C
is Calculate Ka for formic acid at this temperature.
HCOOH (aq) H+ (aq) + HCOO- (aq)
Ka =
[H+][HCOO-]
[HCOOH]
pH = -log [H+]
2.38 = -log [H+]
-2.38 = log [H+]
= 10log [H+] = [H+]
4.2  10-3 = [H+] = [HCOO-]

17. Calculating Ka from the pH
4.2  10-3 = [H+] = [HCOO-]
HCOOH (aq) H+ (aq) + HCOO- (aq)
[HCOOH], M
[H3O+], M
[HCOO-], M
Initially
0.10
Change
- 4.2  10-3
+ 4.2  10-3
At Equilibrium
 10-3
= = 0.10
4.2  10-3
Ka =
[H+][HCOO-]
[HCOOH]
[4.2  10-3] [4.2  10-3]
[0.10]
= 1.8  10-4 =

18. For a monoprotic acid HA:
Concentration ionized
Original concentration
x 100%
Percent ionization =
For a monoprotic acid HA:
Percent ionization =
[H+]
[HA]0
x 100%
[HA]0 = initial concentration
Fig 16.9
The more dilute the acid,
the greater the percent
ionization:

19. What is the pH of a 0.50 M HF solution (at 25°C)?
Ka =
[H+][F-]
[HF]
= 6.8 x 10-4
HF (aq) H+ (aq) + F- (aq)
HF (aq) H+ (aq) + F- (aq)
Initial (M)
0.50
0.00
0.00
Change (M) -x +x +x
Equilibrium (M) x x x
Ka = x2 x
= 6.8 x 10-4
If [HF] > 100 Ka
0.50 – x  0.50
Ka  x2
0.50
= 6.8 x 10-4
x2 = 3.40 x 10-4
x = M
[H+] = [F-] = M
pH = -log [H+] = 1.72
[HF] = 0.50 – x = 0.48 M

20. When can I use the approximation?
If [HF] > 100 Ka
Then 0.50 – x  0.50
Let’s determine error introduced:
0.018 M
0.50 M
x 100% = 3.6%
Less than 5%
Approximation ok.
x = 0.018
What is the pH of a 0.05 M HF solution (at 25°C)?
Ka  x2
0.05
= 6.8 x 10-4
x = M
0.006 M
0.05 M
x 100% = 12%
More than 5%
Approximation not ok.
Must solve for x exactly using quadratic equation...

21. Solving weak acid ionization problems:
Identify the major species that can affect the pH.
In most cases, the autoionization of water can be ignored.
Ignore [OH¯] because it is determined by [H+].
Use ICE table to express the equilibrium concentrations in terms of single unknown x.
Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly.
Calculate concentrations of all species and/or pH of the solution.

22. What is the pH of a 0.122 M monoprotic acid whose
Ka is 5.7 x 10-3?
HA (aq) H+ (aq) + A- (aq)
Initial (M)
0.122
0.00
0.00
Change (M) -x +x +x
Equilibrium (M) x x x
Ka = x2 x
= 5.7 x 10-3
Is [HF] > 100 Ka ?
NO!!
Approximation not ok.

23. Ka = x2 x
= 5.7 x 10-3
x X 10-3 x – 6.95 X 10-4 = 0
-b ± b2 – 4ac  2a
x =
ax2 + bx + c =0
x =
x =
HA (aq) H+ (aq) + A- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.122
0.00 -x +x x x
[H+] = x = M
pH = -log[H+] = 1.625

24. Polyprotic Acids Have more than one ionizable proton
If difference between the Ka1 and subsequent Ka values > 103, the pH generally depends only on the first dissociation.