Answer
Percent ionization $= 0.006261\%$
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.
-$[H_3O^+] = [CN^-] = x$
-$[HCN] = [HCN]_{initial} - x = 0.125 - x$
For approximation, we consider: $[HCN] = 0.125M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CN^-]}{ [HCN]}$
$Ka = 4.9 \times 10^{- 10}= \frac{x * x}{ 0.125}$
$Ka = 4.9 \times 10^{- 10}= \frac{x^2}{ 0.125}$
$ 6.125 \times 10^{- 11} = x^2$
$x = 7.826 \times 10^{- 6}$
Percent ionization: $\frac{ 7.826 \times 10^{- 6}}{ 0.125} \times 100\% = 0.006261\%$
%ionization < 5% : Right approximation.
