Question

Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCl. $K_a$ of acetic acid is $1.8\times {10}^{-5}$:

• A. 0.18%
• B. 0.018%
• C. 1.8%
• D. 18%

Answer 1

The correct option is B 0.018%

$\left[H^{+}\right]={10}^{-1}$

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$

$initially:0.01M$

$Afterdissociation:0.01-xxx+{10}^{-1}$

$Ka=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}=\frac{x(x+{10}^{-1})}{0.01-x}=1.8\times {10}^{-5}$

Approximation: The acid is weak, hence $0.01-x\approx 0.01$ and $x+0.1\approx 0.1$

$\therefore Ka=\frac{x\times 0.1}{0.01}=1.8\times {10}^{-5}$

$\therefore x=1.8\times {10}^{-6}$

$\therefore \%$ ionization $=\frac{x\times 100}{0.01}=\frac{1.8\times {10}^{-6}}{0.01}\times 100=0.018\%$