Question

A solution contains $HCl$, ${Cl}_{2}HCCOOH$ and ${CH}_{3}COOH$ at concentrations $0.09M$ in $HCl$, $0.09M$ in ${Cl}_{2}HCCOOH$ and $0.1M$ in ${CH}_{3}COOH$, $pH$ for the solution is $1$. Ionization constant of ${CH}_{3}COOH={10}^{-5}$. What is the magnitude of $K$ for dichloroacetic acid?

• A. $k{a}_2=1.25\times {10}^{-2}$
• B. $k{a}_2=1.25\times {10}^{-1}$
• C. $k{a}_2=1.25\times {10}^{-4}$
• D. $k{a}_2=1.25\times {10}^{-3}$

Answer 1

The correct option is A $k{a}_2=1.25\times {10}^{-2}$

Given,

$HCl⟶0.09M$
${CH}_{3}COOH⟶C_1=0.1M$, ${\alpha }_1,K_{a_1}={10}^{-5}$
${Cl}_{2}CHCOOH⟶C_2=0.09M$, ${\alpha }_2$, $K_{a_2}=?$
$pH=1,\left[H^{+}\right]=0.1$
$0.1=0.09+C_1{\alpha }_1+C_2{\alpha }_2$

${CH}_{3}COOH\rightleftharpoons {CH}_{3}CO{O}^{-}+H^{+}$
$C_1$
$C_1-C_1{\alpha }_1$ $C_1{\alpha }_1$ $0.1$

${Cl}_{2}HCOOH\rightleftharpoons {Cl}_{2}HCCO{O}^{-}+H^{+}$
$C_2$
$C_2-C_2{\alpha }_2$ $C_2{\alpha }_2$ $0.1$

$K_{a_1}=\frac{\left(C_1{\alpha }_1\right)\left(0.1\right)}{C_1(1-{\alpha }_1)}\simeq {\alpha }_1\times 0.1={10}^{-5}$

${\alpha }_1={10}^{-4}$
Putting this in equation $\left(i\right)$, we get
${10}^{-4}\times 10.1+0.09{\alpha }_2=0.01$
${\alpha }_2=0.111$
$K_{a_2}=\frac{\left(C_2{\alpha }_2\right)\left(0.1\right)}{C_2(1-{\alpha }_2)}$

$=\frac{\left(0.111\right)\left(0.1\right)}{1-0.111}$

$=1.248\times {10}^{-2}$