Consider the titration of 50.0 mL of 0.315 M weak base B (Kb = 7.5 x 10⁻⁶) with 0.340 M HNO₃. What is the pH of the solution before any strong acid has been added?

Answer 1

B + H₂O ==> BH⁺ + OH^-

Kb = [BH⁺][OH^-] / [B]

7.5x10^-6 = (x)(x) / 0.315 - x (ignoring x in denominator for now)....

x² = 2.36x10^-6

x = 1.54x10^-3 M = [OH-] (note: above assumption was valid as this is ~ 0.5% of 0.315 M)

pOH = -log 1.54x10^-3 = 2.82

pH = 14 - pOH = 14 - 2.82

pH = 11.2