Answer
Percent dissociation = 1.1%
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ C_6H_5COOH ]& [ C_6H_5COO^- ]& [ H_3O^+ ]\\
Initial& 0.55 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.55 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ C_6H_5COO^- ][ H_3O^+ ]}{[ C_6H_5COOH ]}$$
$$K_a = \frac{(x)(x)}{[ C_6H_5COOH ]_{initial} - x}$$
3. Assuming $ 0.55 \gt\gt x:$
$$K_a = \frac{x^2}{[ C_6H_5COOH ]_{initial}}$$
$$x = \sqrt{K_a \times [ C_6H_5COOH ]_{initial}} = \sqrt{ 6.3 \times 10^{-5} \times 0.55 }$$
$x = 5.9 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 5.9 \times 10^{-3} }{ 0.55 } \times 100\% = 1.1 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 5.9 \times 10^{-3} $
