Question

1. The Ka of a monoprotic weak acid is 5.90 × 10-3. What is the percent ionization of a 0.148 M solution of this acid?

2. If the Kb of a weak base is 4.5 × 10-6, what is the pH of a 0.26 M solution of this base?

3. Assuming complete dissociation, what is the pH of a 3.00 mg/L Ba(OH)2 solution?

Answer 1

Question No.1

Given Ka = 5.90 × 10-3.

0.148 M solution of acid

Solve for [H+] using the Ka equation
Let X = [H+]

HA ----à [A-] + [H+]

Ka = [A-] [H+]/[HA]

Ka = X² / (0.148 - X)
X² = 5.9 X 10^-3 (0.148 - X)

Solving for x = 0.0267

[H+] = 0.0267

% dissociation = 0.0267/0.148*100 = 18.04%

Question No. 2

Kb = [HB+][OH-]/ [B]

4.5 X 10^-6 = x^2/ 0.26
x^2 = 4.5 x 10^-6
x = 1.08 x 10^-3= [OH-]

Kw = [H+][OH-]
Kw/ [OH-] = [H+]
1.0x10^-14/ 1.08 x 10^-3 =
1.08 x 10^-11 = [H+]
pH = -log[H+] = 10.96

Question No. 3

Assuming complete dissociation, what is the pH of a 3.00 mg/L Ba(OH)2 solution

0.003/171.34 = 1.75 X 10^-5

Ba(OH)2 -> Ba^2+ + 2OH-
Multiply the molarity by two (since when it dissociate, you get 2 OH- ions per Ba(OH)2)

Number of moles of OH = 2 X 1.75 X 10^-5 = 3.5 X 10 ^-5

pOH = -log [OH]= -log 3.5 X 10 ^-5 = 4.46

pH= 14- pOH =14-4.46= 9.5