Question

Calculate $\left[O{H}^{-}\right]$ of $0.03M$ ammonium hydroxide solution. The ionization constant for $N{H}_{4}OH$ is $1.8\times {10}^{-5}M$.

• A. $7.35\times {10}^{-4}M$
• B. $0.88\times {10}^{-5}M$
• C. $4.24\times {10}^{-4}M$
• D. $2.4\times {10}^{-5}M$

Answer 1

The correct option is A $7.35\times {10}^{-4}M$

$N{H}_{4}OH(aq.)\rightleftharpoons N{H}_4^{+}(aq.)+O{H}^{-}(aq.)Initial:C00Equilibrium:C-C\alpha C\alpha C\alpha K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}=\frac{\left[C\alpha \right]\left[C\alpha \right]}{C(1-\alpha )}{K}_b=\frac{C{\alpha }^2}{1-\alpha }1.8\times {10}^{-5}=\frac{0.03\times {\alpha }^2}{1-\alpha }$

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

$0.03\times {\alpha }^2=1.8\times {10}^{-5}$
$\alpha =\sqrt{6\times {10}^{-4}}=2.45\times {10}^{-2}$
$\left[O{H}^{-}\right]=C\alpha =0.03\times 2.45\times {10}^{-2}\left[O{H}^{-}\right]=7.35\times {10}^{-4}M$