Question

Calculate the pH of $0.01MN{H}_{4}OH$ solution having 2% dissociation.

• A. 7.3
• B. 10.3
• C. 8.3
• D. 12.3

Answer 1

The correct option is B 10.3

$N{H}_{4}OH$ is a weak base and partially dissociated.
Let "C" be the initial concentration
$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$
Before:$C00$

After: $C(1-\alpha )C\alpha C\alpha$
$\text{Since}N{H}_{4}OH\text{is partially dissociated}\left[O{H}^{-}\right]=\text{Concentration}\left(C\right)\times \text{degree of dissociation}\left(\alpha \right)Here,\alpha =2\%=\frac{2}{100}C=0.01M$
$\therefore \left[O{H}^{-}\right]=C\alpha =0.01\times \frac{2}{100}=2\times {10}^{-4}MpOH=-\log \left[O{H}^{-}\right]=-\log (2\times {10}^{-4})=3.69pH=14-3.7=10.3$