Question

Ionisation constant of $CH _{3} COOH$ is $1.7 \times 10^{-5}$ and concentration of $H ^{+}$ions is $3.4 \times 10^{-4}$. Then, find out initial concentration of $CH _{3} COOH$ molecules.

• A. $ 3.4 \times 10^{-4} $
• B. $ 3.4 \times 10^{-3} $
• C. $ 6.8 \times 10^{-4} $
• D. $ 6.8 \times 10^{-3} $

Answer 1

Answer: (D)

$CH _{3} COOH' \rightleftharpoons CH _{3} COO ^{-}+ H ^{+} $
$ K_{a} =\frac{\left[ CH _{3} COO ^{-}\right]\left[ H ^{+}\right]}{\left[ CH _{3} COOH \right]} $
Given that,
$\left[ CH _{3} COO ^{-}\right]=\left[ H ^{+}\right]=3.4 \times 10^{-4} M$
$K_{a}$ for $CH _{3} COOH =1.7 \times 10^{-5}$
$CH _{3} COOH$ is weak acid, so in it $\left[ CH _{3} COOH \right]$ is equal to initial concentration. Hence,
$1.7 \times 10^{-6} =\frac{\left(3.4 \times 10^{-4}\right)\left(3.4 \times 10^{-4}\right)}{\left[ CH _{3} COOH \right]} $
${\left[ CH _{3} COOH \right] } =\frac{3.4 \times 10^{-4} \times 3.4 \times 10^{-4}}{1.7 \times 10^{-5}}$
$=6.8 \times 10^{-3} M$