Answer
The percent ionization for acetic acid in this case is $0.036\%$.
Work Step by Step
- 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:
$\% ionization = \frac{x}{[CH_3COOH]_{initial}} \times 100\% = \frac{3.6 \times 10^{-5}}{0.100} \times 100\%$
$\% ionization = 3.6 \times 10^{-4} \times 100\% = 0.036\%$
