Question

How much water must be added to $300\text{mL}$ of $0.2\text{M}$ solution of $C{H}_{3}COOH(K_a=1.8\times {10}^{-5})$ for the degree of ionisation $\left(\alpha \right)$ of the acid to double?

• A. $600\text{mL}$
• B. $900\text{mL}$
• C. $1200\text{mL}$
• D. $1500\text{mL}$

Answer 1

The correct option is B $900\text{mL}$

$\begin{array}{cccccc}& C{H}_{3}COOH& \leftrightharpoons & C{H}_{3}CO{O}^{-}& +& H^{+}\\ t=0& C& & 0& & 0\\ t=t& C(1-\alpha )& & C\alpha & & C\alpha \end{array}$
As $\alpha$ is negligble with respect to $1$, then we have
$K_a=C_1{\alpha }_1^2=C_2{\alpha }_2^2\Rightarrow C_2=C_1{\left(\frac{{\alpha }_1}{{\alpha }_2}\right)}^2\Rightarrow C_2=0.2\times \left(\frac{1}{4}\right)=0.05$
We know that $C_1{V}_1=C_2{V}_2\Rightarrow 300\times 0.2=0.05\times V_2\Rightarrow V_2=1200\text{mL}$
Volume of water added $=1200-300=900\text{mL}$