Answer
The pH of this aqueous solution is $5.17$.
Work Step by Step
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [CN^-] = x$
-$[HCN] = [HCN]_{initial} - x = 0.095 - x$
For approximation, we consider: $[HCN] = 0.095$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CN^-]}{ [HCN]}$
$Ka = 4.9 \times 10^{- 10}= \frac{x * x}{ 9.5 \times 10^{- 2}}$
$Ka = 4.9 \times 10^{- 10}= \frac{x^2}{ 9.5 \times 10^{- 2}}$
$ 4.65 \times 10^{- 11} = x^2$
$x = 6.82 \times 10^{- 6}$
5% test: $\frac{ 6.82 \times 10^{- 6}}{ 9.5 \times 10^{- 2}} \times 100\% = 0.00718\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = x = 6.82 \times 10^{- 6} $
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 6.82 \times 10^{- 6})$
$pH = 5.17$
