10. Ph and Poh Calculations with Weak Acids and Bases

10. pH and pOH Calculations with Weak Acids and Bases

a) Weak Acids

i) Example: What is the pH of 0.400 M H2S?

Not 100% dissociated so [H2S] ¹ [H3O+]; so can’t just pH = -log[0.400]

The weak acid will form an equilibrium:

H2S + H2O (l) HS- + H3O+

I 0.400 0 0

C -X +X +X .

E 0.400 – X X X

Ka = 9.1 x 10-8 = [HS-][H3O+] = X2 .

[H2S] 0.400 – X

Assume 0.400 –X = 0.400, because it is a weak acid and % dissociation < 5%

9.1 x 10-8 = X2 X = 1.91 x 10-4 = [H3O+]

0.400

pH = -log[1.91 x 10-4] = 3.719

ii) Example: What is pH of 0.200 M CH3COOH?

Not 100% dissociated so [CH3COOH] ¹ [H3O+]; so can’t just pH = -log[0.200]

The weak acid will form an equilibrium:

CH3COOH + H2O (l) CH3COO- + H3O+

I 0.200 0 0

C -X +X +X .

E 0.200 – X X X

Ka = 1.8 x 10-5 = [CH3COO-][H3O+] = X2 .

[CH3COOH] 0.200 – X

Assume 0.200 –X = 0.200, because it is a weak acid and % dissociation < 5%

1.8 x 10-5 = X2 X = 1.90 x 10-3 = [H3O+]

0.200

pH = -log[1.90 x 10-3] = 2.721

iii) Example: What is Ka for an unknown acid which gives a pH of

2.50 for a 0.300M solution?

Equilibrium [H3O+] = antilog(-2.50) = 3.2 x 10-3 M

HA + H2O (l) A- + H3O+

I 0.300 0 0

C -3.2 x 10-3 +3.2 x 10-3 +3.2 x 10-3 .

E 0.297 3.2 x 10-3 3.2 x 10-3

Ka = [H3O+][A-] = (3.2 x 10-3)2 = 3.4 x 10-5

[HA] 0.297

iv) Example: What is pH for a solution containing 0.300 M NH4NO3?

NH4NO3 ® NH4+ + NO3- ; NO3- is a spectator and NH4+ acts as a weak acid

NH4+ + H2O H3O+ + NH3

I 0.300 0 0

C -X +X +X .

E 0.300 - X X X

Ka = 5.6 x 10-10 = [NH3][H3O+] = X2 .

[NH4+] 0.300 – X

Assume 0.300 –X = 0.300, because it is a weak acid and % dissociation < 5%

5.6 x 10-5 = X2 X = 1.30 x 10-5 = [H3O+]

0.300

pH = -log[1.30 x 10-5] = 4.886

b) Weak Bases

i) Example: What is the pOH and pH of 0.150 M NH3?

Not 100% dissociated so [NH3] ¹ [OH-]

NH3 + H2O (l) OH- + NH4+

I 0.150 0 0

C -X +X +X .

E 0.150 – X X X

Must 1st find the Kb. Kb(NH3) = Kw = 1.00 x 10-14 = 1.78 x 10-5

Ka(NH4+) 5.6 x 10-10

Kb = [OH-][NH4+] = 1.78 x 10-5 = (X)2 .

[NH3] 0.150 – X

Assume 0.150 – X = 0.150

1.78 x 10-5 = X2 X = 1.63 x 10-3 M = [OH-]

0.150

pOH = -log(1.63 x 10-3) = 2.788; pH = 14 – 2.788 = 11.212

ii) Example: What is the pH of a solution containing 0.100 M NaNO2?

NaNO2 ® Na+ + NO2- ; Na+ is a spectator and NO2- acts as a weak base

NO2- + H2O OH- + HNO2

I 0.100 0 0

C -X +X +X .

E 0.100 - X X X

Kb(NO2-) = Kw = 1.00 x 10-14 = 2.17 x 10-11

Ka(HNO2) 4.6 x 10-4

Kb = [OH-][HNO2] = 2.17 x 10-11 = (X)2 .

[NO2-] 0.100 – X

Assume 0.100 – X = 0.100

2.17 x 10-11 = X2 X = 1.47 x 10-6 M = [OH-]

0.100

pOH = -log(1.47 x 10-6) = 5.833; pH = 14 – 5.833 = 8.167

iii) Example: A 0.400 M solution of an unknown base, A-, has a pH

of 11.50. What is the Kb for A-?

pOH = 14 = 11.50 = 2.50

Equilibrium [OH-] = antilog (-2.50) = 3.2 x 10-3 M

A- + H2O OH- + HA

I 0.400 0 0

C -3.2 x 10-3 +3.2 x 10-3 +3.2 x 10-3 .

E 0.397 3.2 x 10-3 3.2 x 10-3

Kb = [OH-][HA] = (3.2 x 10-3)2 = 2.6 x 10-5

[A-] 0.397

Do Questions: #74-83 page 152; #84-93 page 153-154