Question

Calculate $\left[O{H}^{-}\right]$ and % age dissociation of $0.01M$ solution of ammonium hydroxide solution. The ionization constant for $N{H}_{3}OH$ is $K_b=1.8\times {10}^{-5}$.

Answer 1

$N{H}_{4}OH\leftrightharpoons N{H}_4^{+}+O{H}^{-}$

$k_b=\frac{\left[N{H}_4^{+}\right][+O{H}^{-}]}{N{H}_{4}OH]}$

From $ion$ method, $\left[N{H}_4^{+}\right]=\left[O{H}^{-}\right]$

$k_b=\frac{[O{H}^{-}{]}^2}{N{H}_{4}OH]}$ $\left[N{H}_{4}OH\right]=0.01;k_b=1.8\times {10}^{-5}$

$\left({10}^{-2}\right)(1.8\times {10}^{-5})=[OH{]}^2$

$N{H}_{4}OH\leftrightharpoons N{H}_4^{+}+O{H}^{-}{10}^{-2}00{10}^{-2}-xxxx=4.24\times {10}^{-4}$

$\%$ dissociation$=\frac{4.42\times {10}^{-4}}{{10}^{-2}}\times 100$

$=4.24\%$