Question

Find the percentage ionization of $0.2$M acetic acid solution, whose disassociation constant is $1.8\times {10}^{-5}$

Answer 1

Let x M is the concentration of acetic acid that is ionised.
$C{H}_3-COOH\rightleftharpoons C{H}_3-CO{O}^{-}+H^{+}$
Out of 0.2 M acetic acid, x M acetic acid ionises to give x M acetate ions and x M protons. $(0.2-x)M$ acetic acid will remain.
$K_a=\frac{[C{H}_3-CO{O}^{-}]\left[H^{+}\right]}{[C{H}_3-COOH]}$
$1.8\times {10}^{-5}=\frac{x\times x}{0.2-x}$

Since $K_a$ is very small, $0.2-x$ can be approximated to 0.2.

$1.8\times {10}^{-5}=\frac{x\times x}{0.2}$

$x=0.00189M$

The percent ionisation

$=\frac{x}{0.2}\times 100$

$=\frac{0.00189}{0.2}\times 100$

$=0.95$