Question

The dibasic compound B (pK_b1 = 4.00, pK_b2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH after 3.1 mL of acid are added.

Find the pH after 20 mL of acid are added.

Answer 1

B(OH)₂ ---> B(OH)⁺ + OH^- pKb1 = 4

B(OH)⁺ ---> B²⁺ + OH^- pKb2 = 8

SInce Kb1 >> Kb2. We can neglect the [OH^-] from second dissociation

Kb1 = [B(OH)⁺] [OH^-] / [B(OH)₂]

x² / (0.1 - x) = 10^-4 M

Solving we get x = 0.0031 M

[B(OH)⁺] = 0.0031 M

Moles of B(OH)⁺ present = 0.0031 M * 100 mL = 0.31 mmol

Moles of B(OH)₂ present = 10 - 0.31 = 9.69 mmol

(i) Before first equivalence point:

Moles of acid added = 1 M * 3.1 mL = 3.1 mmol

Moles of B(OH)⁺ present = 0.31 + 3.1 mmol = 3.41 mmol

Moles of B(OH)₂ = 9.69 - 3.1 = 6.59 mmol

pOH = pKb1 + log([B(OH)⁺]/[B(OH)₂]) = 4 + log(3.41/6.59) = 3.7

pH = 14 - 3.7 = 10.3

(ii) For second equivalence point 20 mL of acid is required

Total volume of solution = 100 + 20 = 120 mL

B²⁺ + H₂O ---> B(OH)⁺ + H⁺

10/120 - y y y

Ka2 = Kw/Kb2 = 10^-14 / 10^-8 = 10^-6

Ka2 = y² / {(10/120) - y)}

Solving we get y = 0.000288

[H⁺] = 0.000288

pH = -log(0.000288) = 3.54