Question

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

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

Answer 1

Answer: (B)

$0.018$%

$HCl\rightleftharpoons H^{+}+{Cl}^{-}$

initial $0.1$ $0$ $0$

$0$ $0.1$ $0.1$ after it's complete dissociation

$H_{2}O$ is also added in case of weak acid

${CH}_{3}COOH+H_2{O}_{\left(l\right)}\rightleftharpoons {CH}_3{COO}^{-}+H^{+}$

initial $0.01$ $0$ $0$ $0.1$ (from $HCl$ dissociation)

$0.01-x$ $0$ $x$ $x+0.1$ (after dissociation)

$K_a=\frac{\left[{CH}_3{COO}^{-}\right]\left[H^{+}\right]}{\left[{CH}_{3}COOH\right]}$ given, $K_a$ of acetic acid $=1.8\times {10}^{-5}$

$1.8\times {10}^{-5}=\frac{x(0.1+x)}{0.01}$

$1.8\times {10}^{-7}=x\left(0.1\right)+x^2$

$\searrow$ negligible

$x=\frac{1.8\times {10}^{-7}}{0.1}$

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

So, % ionisation of acid $=\frac{x}{0.01}\times 100=1.8\times {10}^{-2}$%

$=0.018$%

Percentage ionisation is $0.018$%