Question

The concentration of $HCN$ and $NaCN$ in a solution is $0.01M$ each. Calculate the concentration of hydroxyl ions if the dissociation constant of $HCN$ is $7.2\times {10}^{-10}$.

• A. $14\times {10}^{-7}M$
• B. $14\times {10}^{-5}M$
• C. $1.4\times {10}^{-5}M$
• D. $1.4\times {10}^{-7}M$

Answer 1

The correct option is C $1.4\times {10}^{-5}M$

HCN ia weak acid and it dissociate as $H^{+}$ and $C{N}^{-}$ ions
(i) $\begin{array}{ccccc}HCN& \rightleftharpoons & C{N}^{-}& +& H^{+}\\ (0.01-X)M& & XM& & XM\end{array}$

(ii) $\begin{array}{cccccc}& NaCN& \rightleftharpoons & C{N}^{-}& +& N{a}^{+}\\ \text{Initial}& 0.01M& & 0M& & 0M\\ \text{Final}& 0.0M& & 0.01M& & 0.01M\end{array}$
So at equilibrium, $\left[H^{+}\right]=XM$
$[C{N}^{-}{]}_{\text{total}}=[C{N}^{-}{]}_{\text{acid}}+[C{N}^{-}{]}_{\text{salt}}$
$=XM+0.01M=0.01M$
$(\text{X+0.01 is close to 0.01 because HCN is a weak acid )}$
$\left[HCN\right]=0.01M=0.01M$
$pH=p{K}_a+{\text{log}}_{10}\frac{\left[C{N}^{-}\right]}{\left[HCN\right]}$
$=-\text{log}7.2\times {10}^{-10}+\text{log}\frac{0.01M}{0.01M}$
$-\text{log}\left[H^{+}\right]=-\text{log}7.2\times {10}^{-10}$
$\left[H^{+}\right]=7.2\times {10}^{-10}M$
$\left[O{H}^{-}\right]=\frac{K_w}{\left[H^{+}\right]}$
$=\frac{1\times {10}^{-14}}{7.2\times {10}^{-10}}=1.4\times {10}^{-5}M$