Answer
The percent ionization for this acid is $1.34\%$ for a 0.100M solution.
Work Step by Step
- Considering a pure acetic acid solution:
$\% ionization = \frac {[H_3O^+]}{[CH_3COOH]_{initial}} \times 100\% $
$\% ionization = \frac {0.00134M}{0.100M} \times 100\% $
$\% ionization = 0.0134 \times 100\% = 1.34\%$
