Consider the titration of 30.0 mL of 0.251 M weak base B (Kb = 1.3 x 10⁻¹⁰) with 0.150 M HI. What is the pH of the solution before any acid has been added?

Answer 1

Before any acid has been added, all you have is a solution of 0.251 M B (Kb = 1.31x10^-10)

B + H₂O ==> HB⁺ + OH^-

Kb = 1.31x10^-10 = [HB⁺][OH^-] / [B]

1.31x10^-10= (x)(x) / 0.251-x (assume x is small relative to 0.251 and ignore it in the denominator)

1.31x10^-10 = x² / 0.251

x² = 3.29x10^-11

x = 5.73x10^-6 M = [OH-] (note: this is insignificant compared to 0.251 M so above assumption was valid)

pOH = -log [OH-] = 5.24

pH = 14 - pOH

pH = 8.76