Question

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

• A. $7.84\times {10}^{-4}M$
• B. $0.88\times {10}^{-5}M$
• C. $1.56\times {10}^{-5}M$
• D. $4.24\times {10}^{-4}M$

Answer 1

The correct option is D $4.24\times {10}^{-4}M$

$N{H}_{4}OH(aq.)\rightleftharpoons N{H}_4^{+}(aq.)+O{H}^{-}(aq.)Initial:C00Equilibrium:C-C\alpha C\alpha C\alpha K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}=\frac{\left[C\alpha \right]\left[C\alpha \right]}{C(1-\alpha )}{K}_b=\frac{C{\alpha }^2}{1-\alpha }1.8\times {10}^{-5}=\frac{0.01\times {\alpha }^2}{1-\alpha }$

Since $N{H}_{4}OH$ is a weak base $\therefore \alpha <<11-\alpha \approx 1$

$0.01\times {\alpha }^2=1.8\times {10}^{-5}$
$\alpha =\sqrt{1.8\times {10}^{-3}}=4.24\times {10}^{-2}$
$\left[O{H}^{-}\right]=C\alpha =0.01\times 4.24\times {10}^{-2}\left[O{H}^{-}\right]=4.24\times {10}^{-4}M$