Question

The pH of a 0.35 M solution of a weak base is 10.58. What is the Kb of the base?

Answer 1

B(aq) + H₂O(l) $\to$ HB⁺ (aq) + OH^- (aq)

I _____ 0.35M __________0____________ 0

C_______- x __________+ x __________ + x

E _____0.35M - x _________x____________x

pH = 10.58

[B] = 0.35M

k_b = ?

pH = -log[H₃O⁺]

[H₃O⁺] = 10^10.58 = 2.63×10^-11M

k_w = [H₃O⁺][OH^-]

[OH^-] = $\frac{k_w}{[H_3O^+]}$ , k_W = 1.0×10^-14 (Constant)

[OH^-] = $\frac{1.0×10^-¹⁴}{2.63×10^-¹¹}$ = 3.80×10^-4 = x

[HB⁺] = 3.80×10^-4 = x

Now we find k_b

k_b = $\frac{[HB^+][OH^-]}{[B]}$

k_b = $\frac{(3.80×10^-⁴)(3.80×10^-⁴)}{0.35}$

k_b = 4.12×10^-7M