Question

What is the percentage degree of dissociation of $0.05MN{H}_{4}OH$ at $25{}^{o}C$ solution having $pH=10$ ?

• A. $0.002$
• B. $0.2$
• C. $0.02$
• D. $2.0$

Answer 1

The correct option is B $0.2$

The dissociation reaction of weak base, $N{H}_{4}OH$ is given below :
$N{H}_{4}OH\left(aq\right)\rightleftharpoons N{H}_4^{+}\left(aq\right)+O{H}^{-}\left(aq\right)Initial:C00Change:-C\alpha +C\alpha +C\alpha Final:C-C\alpha C\alpha C\alpha$
Given,
$pH=10$
$\left[N{H}_{4}OH\right]=0.05$
$\left[H^{+}\right]={10}^{-10}$
We know at $25{}^{o}C$
$\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}\therefore \left[O{H}^{-}\right]=\frac{1\times {10}^{-14}}{{10}^{-10}}={10}^{-4}=C\alpha \therefore \alpha =\frac{\left[O{H}^{-}\right]}{C}=\frac{{10}^{-4}}{0.05}=2\times {10}^{-3}$
%$\alpha =2\times {10}^{-3}\times 100=0.2$%