Question

If the Kb of a weak base is 4.9 × 10-6, what is the pH of a 0.49 M solution of this base?

Answer 1

Consider a weak base MOH.

MOH---->M⁺+OH^-

Kb=[M⁺][OH^-]​​​^​​​​/[MOH]

We know that [M⁺]=[OH^-]

Kb= [OH]²/[MOH]

[MOH]=0.49(given)

Kb=4.9×10^-6

[OH]²=Kb×[MOH] =4.9×10^-6×0.49=2.4×10^-6.

[OH]=√2.4×10^-6M =1.5×10^-3

pOH= -log([OH^-])=-log(1.5×10^-3) =2.80

Therefore pH = 14-2.80=11. 2

pH=11.2

As pH + pOH =14.