Hi can anyone show me the solution of this please

Question

The [OH-] of a 0.290 M weak base is 1.25 x 10^ -2 M. Calculate Kb.

Answer 1

Let the weak base be represented by B

In water we have

B + H2O ==> BH⁺ + OH^-

Kb for B = [BH⁺][OH^-] / [B]

If [OH-] = 1.25x10^-2 M, then [BH⁺] also = 1.25x10^-2 M

Kb = (1.25x10^-2)(1.25x10^-2) / 0.290 - 1.25x10^-2

Since 1.25x10^-2 is less than 5% of 0.290, we can ignore it in the denominator and avoid using a quadratic

Hence, we can simplify to

Kb = (1.25x10^-2)(1.25x10^-2) / 0.290

Kb = 5.39x10^-4