## Chemistry: The Molecular Science (5th Edition)

The percent ionization for acetic acid in this case is $0.036\%$.
- Write the $K_a$ expression: $K_a = \frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}$ $1.8 \times 10^{-5} = \frac{x * (0.050 + x)}{0.100 - x}$ - Now, calculate $x$: Considering "x" is very small: $1.8 \times 10^{-5} = \frac{x * 0.050}{0.100}$ $1.8 \times 10^{-6} = x * 0.050$ $x = \frac{1.8 \times 10^{-6}}{0.050} = 3.6 \times 10^{-5}M$ - Find the percent ionization: $\% ionization = \frac{x}{[CH_3COOH]_{initial}} \times 100\% = \frac{3.6 \times 10^{-5}}{0.100} \times 100\%$ $\% ionization = 3.6 \times 10^{-4} \times 100\% = 0.036\%$