Question

Calculate the concentration of ${OH}^{-}$ ions of $0.01M$ ${NH}_{4}OH$ solution. The equilibrium constant of ${NH}_{4}OH$ is $1.8\times {10}^{-5}$.

Answer 1

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

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

$K=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[{NH}_{4}OH\right]}=\frac{C\alpha \times C\alpha }{C(1-\alpha )}\simeq C{\alpha }^2=0.01\times (\alpha {)}^2=1.8\times {10}^{-5}$

$\alpha =4.24\times {10}^{-2}$

Concentration of $O{H}^{-}$ ions=$C\alpha =4.24\times {10}^{-4}$