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?
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.
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