Question

$0.02M$ solution of ${NH}_{4}OH$ is $3$% dissociated. Calculate the dissociation constant of ${NH}_{4}OH$.

Answer 1

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

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

$K=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[{NH}_{4}OH\right]}=\frac{C\alpha \times C\alpha }{C(1-\alpha )}\simeq C{\alpha }^2=0.02\times (0.03{)}^2=1.8\times {10}^{-5}$