Question

A solution contains $0.09MHCl,0.09MCC{l}_{2}HCOOH$, and $0.1MC{H}_{3}COOH$. If total $\left[H^{+}\right]=0.1$ and $K_a$ for $C{H}_{3}COOH={10}^{-5}.K_a$ for $CC{l}_{2}HCOOH$ is:

• A. $1.35\times {10}^{-4}$
• B. $0.18\times {10}^{-2}$
• C. $0.18\times {10}^{-5}$
• D. $0.9\times {10}^{-4}$

Answer 1

The correct option is D $0.9\times {10}^{-4}$

For a weak acid, the expression for the hydrogen ion concentration is

$\left[H^{+}\right]=\sqrt{e\times K_a}$

Hence the total hydrogen ion concentration of the solution will be,

${\left[H^{+}\right]}_{total}={\left[H^{+}\right]}_{HCl}+{\left[H^{+}\right]}_{{CHCl}_{2}COOH}{\left[H^{+}\right]}_{{CH}_{3}COOH}$

$\Rightarrow {\left[H^{+}\right]}_{total}=0.1=0.09+\sqrt{C_{{CHCl}_{2}COOH}\times K_{aCH{Cl}_{2}COOH}}+\sqrt{C_{{CH}_{3}COOH}\times {Ka}_{{CH}_{3}COOH}}$

$\Rightarrow 0.01=\sqrt{0.09\times K_{a_{{CHCl}_{2}COOH}}}+\sqrt{0.1\times {10}^{-5}}$

Hence, $K_a$, ${CHCl}_{2}COOH=0.9\times {10}^{-4}$