Answer
Percent Ionization = 0.85%
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HC_2H_3O_2 ]& [ C_2H_3{O_2}^- ]& [ H_3O^+ ]\\
Initial& 0.250 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.250 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_3{O_2}^- ][ H_3O^+ ]}{[ HC_2H_3O_2 ]}$$
$$K_a = \frac{(x)(x)}{[ HC_2H_3O_2 ]_{initial} - x}$$
3. Assuming $ 0.250 \gt\gt x:$
$$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.250 }$$
$x = 2.1 \times 10^{-3} $
4. Calculate the percent ionization:
$$\frac{ \sqrt{ 1.8 \times 10^{-5} \times 0.250 } }{ 0.250 } \times 100\% = 0.85 \%$$
