Answer
Percent dissociation = $1.9\%$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ CH_3COOH ]& [ CH_3COO^- ]& [ H_3O^+ ]\\
Initial& 0.050 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.050 -x& x& x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ CH_3COO^- ][ H_3O^+ ]}{[ CH_3COOH ]}$$
$$K_a = \frac{(x)(x)}{[ CH_3COOH ]_{initial} - x}$$
3. Assuming $ 0.050 \gt\gt x:$
$$K_a = \frac{x^2}{[ CH_3COOH ]_{initial}}$$
$$x = \sqrt{K_a \times [ CH_3COOH ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.050 }$$
$x = 9.5 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 9.5 \times 10^{-4} }{ 0.050 } \times 100\% = 1.9 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 9.5 \times 10^{-4} $