Answer
The percent ionization is equal to 0.0063%
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HCN ]& [ CN^- ]& [ H_3O^+ ]\\
Initial& 0.125 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.125 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CN^- ][ H_3O^+ ]}{[ HCN ]}$$
$$K_a = \frac{(x)(x)}{[ HCN ]_{initial} - x}$$
3. Assuming $ 0.125 \gt\gt x:$
$$K_a = \frac{x^2}{[ HCN ]_{initial}}$$
$$x = \sqrt{K_a \times [ HCN ]_{initial}} = \sqrt{ 4.9 \times 10^{-10} \times 0.125 }$$
4. Calculate the percent ionization:
$$\frac{\sqrt{ 4.9 \times 10^{-10} \times 0.125 } }{ 0.125 } \times 100\% = 0.0063 \%$$
- Since the percent ionization is less than 5%, the assumption was valid.
