1. Applications of Acid-Base Equilibria
Acids and Bases
Part II
Applications of Acid-Base Equilibria

2. Common Ion Effect
Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2x10-4) and 1.0 M NaF.
1.0 M
1.0 M co
change
ceq

3. Imagine 1 M HF solution at equilibrium, then NaF is added.
Previous Example:
Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2x10-4).
1 M HF
1 M HF + 1 M NaF x
4.2x10-3
7.2x10-4 pH
1.57
3.14 a%
0.42%
0.072%
Le Chatelier
Imagine 1 M HF solution at equilibrium, then NaF is added.
What happens? Common Ion Effect

4. Buffered Solution
is a solution that persists the change in its pH when H+ or OH- is added.
Solution of weak acid and the salt of its conjugate base: HAc/NaAc
Solution of weak base abd the salt of its conjugate acid: NH3/NH4+
Blood is a buffered solution.

5. A buffered solution contains 0. 50 M acetic acid HC2H3O2, Ka=1
A buffered solution contains 0.50 M acetic acid HC2H3O2, Ka=1.8x10-5) and 0.50 M sodium acetate (NaC2H3O2). Calculate the pH of this solution.
0.5 M
0.5 M co
change
ceq

6. Calculate the change in pH that occurs when 0
Calculate the change in pH that occurs when mol solid NaOH is added to 1.0 L of the above buffered solution.
0.01 mol
0.01 mol
n = M x V
(before reaction)
0.5x1.0
=0.5 mol
change
-0.01
+0.01
0.49
mol
0.51 co
0.49
0.51
change -x x
ceq
0.49-x
0.49+x

7. Compare this pH change with that which occurs when 0
Compare this pH change with that which occurs when mol solid NaOH is added to 1.0 L water.
0.01 mol
0.01 mol

9. Henderson-Hasselbalch Equation

10. co
[HA]o
[A-]o
change -x x
ceq
[HA]o-x
[A-]o+x
Calculate the pH of a solution containing 0.75 M lactic acid (Ka=1.4x10-4) and 0.25 M sodium lactate.

11. A buffered solution contains 0. 25 M NH3 (Kb=1. 8x10-5) and 0
A buffered solution contains 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.
NH3
NH4+

12. Calculate the pH of the solution that results when 0
Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution containing 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl.
0.1 mol
0.1 mol
n = M x V
(before reaction)
0.25x1.0
=0.25 mol
0.40x1.0
=0.40 mol
change
-0.1
+0.1
0.15
mol
0.50

13. Acid addition pH ↓
Base addition pH ↑

14. Buffering Capacity
It represents the amount of H+ or OH the buffer can absorb without a significant change in pH.
High buffering capacity: solution absorbs large amounts of acid or base without significant change in pH.
Low buffering capacity: small amounts of acid or base can produce a significant change in pH.
Factors affecting the buffering capacity:
1) The concentrations of buffer components
The more concentrated the components of a buffer , the greater the buffer capacity.
2) The ratio [A-]/[HA]
The closer this ratio is to 1, the higher the buffering capacity.

15. 0.01 mol solid NaOH added to 1 L buffer.
The relation between buffer capacity and pH change.
0.01 mol solid NaOH added to 1 L buffer.

16. maximum buffer capacity
Factors affecting the buffering capacity:
The concentrations of buffer components
2) The ratio [A-]/[HA]
The closer this ratio is to 1, the higher the buffering capacity.
maximum buffer capacity

17. Problem
A chemist needs a solution buffered at pH= 4.30 and can choose from the following acids and their sodium salts:
chloroacetic acid, Ka= 1.35 x 10-3
propanoic acid, Ka= 1.3 x 10-5
3. benzoic acid, Ka= 6.4 x 10-5
4. hypochlorous acid, Ka= 3.5 x 10-8
Calculate the ratio [A-]/[HA] required for the best system to yield a pH of 4.30.
pKa
2.87
4.89
4.19
7.46

18. Strong acid – Strong base titration
HCl(aq) + NaOH(aq)  H2O + NaCl(aq)
net ionic equation:
H+(aq) + OH-(aq)  H2O

19. HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
Titration of
50 mL M HNO3 with M NaOH
A. No NaOH has been added.
B. 10 mL NaOH has been added.
HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)

21. HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
C. 20 mL NaOH (total) has been added.
HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
D. 50 mL NaOH (total) has been added.

22. HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
So on until:
E. 100 mL NaOH (total) has been added.
HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
Stoichiometric point or point of equivalence

23. HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
F. 150 mL NaOH (total) has been added.
HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)

24. HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
F. 200 mL NaOH (total) has been added.
HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)

26. Weak acid – Strong base titration
Titration of
50 mL M HAc with M NaOH
A. No NaOH has been added.
pH of 0.10 M HAc?

27. HAc(aq) + OH-(aq)  H2O + Ac-(aq)
B. 10 mL NaOH has been added.
HAc(aq) + OH-(aq)  H2O + Ac-(aq)

28. HAc(aq) + OH-(aq)  H2O + Ac-(aq)
C. 25 mL NaOH has been added.
HAc(aq) + OH-(aq)  H2O + Ac-(aq)

29. HAc(aq) + OH-(aq)  H2O + Ac-(aq)
D. 40 mL NaOH has been added.
HAc(aq) + OH-(aq)  H2O + Ac-(aq)

30. HAc(aq) + OH-(aq)  H2O + Ac-(aq)
E. 50 mL NaOH has been added.
HAc(aq) + OH-(aq)  H2O + Ac-(aq) co
change
ceq

32. HAc(aq) + NaOH(aq)  H2O + NaAc(aq)
F. 60 mL NaOH (total) has been added.
HAc(aq) + NaOH(aq)  H2O + NaAc(aq)
G. 75 mL NaOH (total) has been added.

35. HCN(aq) + OH-(aq)  H2O + CN-(aq)
Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka =6.2x10-10) when dissolved in water. If a 50.0-mL sample of M HCN is titrated with M NaOH, calculate the pH of the solution
a. after 8.00 mL of M NaOH has been added.
HCN(aq) + OH-(aq)  H2O + CN-(aq)

36. HCN(aq) + OH-(aq)  H2O + CN-(aq)
b. at the halfway point of the titration.
HCN(aq) + OH-(aq)  H2O + CN-(aq)
halfway point of the titration:

37. HCN(aq) + OH-(aq)  H2O + CN-(aq)
c. at the equivalence point of the titration.
HCN(aq) + OH-(aq)  H2O + CN-(aq) co
change
ceq
pH=10.96

39. Titration Curves of Polyprotic Acids
pH = pKa2
pOH from Kb equil’m of A2–
pH = pKa1
V/2 V
3V/2 2V

40. A chemist dissolves 2. 00 mmol of a solid acid in 100
A chemist dissolves 2.00 mmol of a solid acid in mL water and titrates the resulting solution with M NaOH. After 20.0 mL NaOH has been added, the pH is What is the Ka value for the acid?

41. Weak base – Strong acid titration
The pH curve for the titration of mL of M NH3 with 0.10 M HCl. Note the pH at the equivalence point is less than 7, since the solution contains the weak acid NH4+. .

42. HIn (aq)  H+ (aq) + In– (aq)
Acid –Base Indicators
Organic Dyes
weak acid-weak base conjugate pair
HIn (aq)  H+ (aq) + In– (aq)
Acid form base form
(one color) (another color)
If you put a small amount of HIn in solution
Will be one color if acidic (phenolphthalein = colorless)
Another color if basic (phenolphthalein = bright magenta)
Color will change when pH rises or falls

43. HIn (aq)  H+ (aq) + In– (aq)
Color Change
Bromothymol Blue
HIn (aq)  H+ (aq) + In– (aq)
Acid form base form
(one color) (another color)
If , the color of In- will be observed.
If , the color of HIn will be observed.

44. If , the color of In- will be observed.
If , the color of HIn will be observed.
Color Change occurs in the range pH=pKIn1

45. Bromothymol Blue KIn=1.0x10-7
Color Change occurs in the range pH=pKa1
pH=71=6-8
pH<6
pH=6-8
pH>8

46. Colors and approximate pH range of some common acid-base indicators.

47. Best indicator when pH at the point of equiv. is nearest to pKIn
Is this indicator suitable for your titration?
titration of 50 mL of 0.1 M HC2H3O2 with 0.1 M NaOH.
100.0 mL of 0.10 M HCl with 0.10 M NaOH.
To find if a given indicator is suitable, calculate pH at the point of equiv.
Best indicator when pH at the point of equiv. is nearest to pKIn