Question

A solution of $0.01M$ concentration of ${NH}_{4}OH$ is $2.6$% dissociated. Calculate $\left[H^{+}\right],\left[{OH}^{-}\right],\left[{NH}_4^{+}\right],\left[{NH}_{4}OH\right]$ and pH of solution.

Answer 1

${NH}_{4}OH⟶{N{H}_4}^{+}+{OH}^{-}$

Before dissociation $1$ $0$ $0$

After dissociation $1-\alpha$ $\alpha$ $\alpha$

$\left[{OH}^{-}\right]=C.\alpha =C\sqrt{K_b/C}=\sqrt{K_{b}C}\Rightarrow K_b=C{\alpha }^2=0.01\times {\left(0.026\right)}^2=6.76\times {10}^{-6}$
$\left[O{H}^{-}\right]=\left[N{H}_4^{+}\right]=\sqrt{K_{b}C}=2.6\times {10}^{-4}$

$\therefore \left[H^{+}\right]={10}^{-14}/2.6\times {10}^{-4}=3.846\times {10}^{-11}M$

$\therefore pH=-\log \left[H^{+}\right]=-\log (3.846\times {10}^{-11})=10.415$