Question

Calculate $\left[O{H}^{-}\right]$ of 0.01 M solution of ammonium hydroxide solution. The ionization constant ($K_b$) for $N{H}_{4}OH=1.8\times {10}^{-5}$.

• A. $4.24\times {10}^{-4}mol{L}^{-1}$
• B. $4.24\times {10}^{-2}mol{L}^{-1}$
• C. $1.80\times {10}^{-4}mol{L}^{-1}$
• D. $1.80\times {10}^{-3}mol{L}^{-1}$

Answer 1

The correct option is A $4.24\times {10}^{-4}mol{L}^{-1}$

Given, $\left[N{H}_{4}OH\right]=0.01M,K_b=1.8\times {10}^{-5}$
The dissociation is given by
$N{H}_{4}OH\left(aq\right)\rightleftharpoons N{H}_4^{+}\left(aq\right)+O{H}^{-}\left(aq\right)C-C\alpha C\alpha C\alpha K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{N{H}_{4}OH}{K}_b=\frac{C^2{\alpha }^2}{C(1-\alpha )}{K}_b=\frac{C{\alpha }^2}{(1-\alpha )}1.8\times {10}^{-5}=\frac{0.01\times {\alpha }^2}{1-\alpha }\alpha =\sqrt{1.8\times {10}^{-3}}=4.24\times {10}^{-2}(\because 1-\alpha \approx 1as\alpha <<1)$
$\left[O{H}^{-}\right]=C\alpha =0.01\times 4.24\times {10}^{-2}=4.24\times {10}^{-4}$
Hence, concentration of $O{H}^{-}=4.24\times {10}^{-4}mol{L}^{-1}$