Question

Calculate $\left[O{H}^{-}\right]$ of a 0.01 M solution of ammonium hydroxide. The ionization constant for $N{H}_{4}OH\left(k_b\right)=1.6\times {10}^{-5}$ (Take $log2=0.30$)

• A. $4\times {10}^{-4}M$
• B. $1.6\times {10}^{-6}M$
• C. $1.6\times {10}^{-4}M$
• D. $4\times {10}^{-6}M$

Answer 1

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

We know,
$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$
$At,t=0C00$
$At,t=t_{eq}C-C\alpha C\alpha C\alpha$
Again, $K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}$
$\Rightarrow K_b=\frac{C{\alpha }^2}{1-\alpha }$
$\Rightarrow 1.6\times {10}^{-5}=0.01\times {\alpha }^2$
$\Rightarrow {\alpha }^2=1.6\times {10}^{-3}$
$\Rightarrow \alpha =4\times {10}^{-2}$
So,
$\left[O{H}^{-}\right]=C\alpha =0.01\times 4\times {10}^{-2}=4\times {10}^{-4}M$