Question

Calculate the ratio $\frac{{\alpha }_1}{{\alpha }_2}$ of a solution obtained by mixing equal volume of 0.02 M $HOCl$ ( a weak acid) and 0.2 M $C{H}_{3}COOH$ solution.
where,
${\alpha }_1$ is the degree of dissociation of $HOCl$
${\alpha }_2$ is the degree of dissociation of $C{H}_{3}COOH$
Given that:
$K_{a_1}\left(HOCl\right)=2\times {10}^{-7}$
$K_{a_1}\left(C{H}_{3}COOH\right)=2\times {10}^{-5}$

• A. 0.2
• B. 0.02
• C. 0.01
• D. 0.5

Answer 1

The correct option is C 0.01

Denoting 'a' for $HOCl$ and 'b' for $C{H}_{3}COOH$.
Let the equal volumes of solution a and b be V , i.e. $V_a=V_b=V$
So, new concentration after mixing is given as:
for $HOCl$ ,
$C_1=\frac{C_a{V}_a}{V_a+V_b}=\frac{0.02\times V}{V+V}{C}_1=0.01M$
Similarly for $C{H}_{3}COOH$,
$C_2=\frac{C_b\times V_b}{V_a+V_b}=\frac{0.2\times V}{V+V}{C}_2=0.1M$
Now,

$\begin{array}{ccccc}HOCl\left(aq\right)& \rightleftharpoons & H^{+}\left(aq\right)& +& OC{l}^{-}\left(aq\right)\\ C_1(1-{\alpha }_1)& & C_1{\alpha }_1+C_2{\alpha }_2& & C_1{\alpha }_1\end{array}$

$\begin{array}{ccccc}C{H}_{3}COOH& \rightleftharpoons & H^{+}& +& C{H}_{3}CO{O}^{-}\\ C_2(1-{\alpha }_2)& & C_1{\alpha }_1+C_2{\alpha }_2& & C_2{\alpha }_2\end{array}$

Dissociation constant for $HOCl$ i.e. $K_{a_1}$ :
$K_{a_1}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{C_1(1-{\alpha }_1)}$

Dissociation constant for $C{H}_{3}COOH$ i.e. $K_{a_2}$ :
$K_{a_2}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)}{C_2(1-{\alpha }_2)}$
Since , both are weak acids, so ${\alpha }_{1}and{\alpha }_2$ are very small in comparison to unity for weak monoprotic acids. So, $1-{\alpha }_1\approx 1and1-{\alpha }_2\approx 1$.

so the above expressions become:

$K_{a_1}\times C_1=(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)$
$K_{a_2}\times C_2=(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)$
Dividing both the above equations, we get:
$\frac{K_{a_1}}{K_{a_2}}=\frac{{\alpha }_1}{{\alpha }_2}$

$\frac{{\alpha }_1}{{\alpha }_2}=\frac{2\times {10}^{-7}}{2\times {10}^{-5}}\frac{{\alpha }_1}{{\alpha }_2}=0.01$