Chemical and Ionic Equilibrium 2 Question 70
69. What is the $\mathrm{pH}$ of a $1.0 \mathrm{M}$ solution of acetic acid? To what volume must one litre of this solution be diluted so that the $\mathrm{pH}$ of the resulting solution will be twice the original value? Given, $K_{a}=1.8 \times 10^{-5}$
$(1990,4 \mathrm{M})$
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Solution:
- $\mathrm{CH}{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}{3} \mathrm{COO}^{-}+\mathrm{H}^{+}$
When concentration of $\mathrm{CH}_{3} \mathrm{COOH}$ is $1.0 \mathrm{M}$, ’ $\alpha$ ’ is negligible,
$$ \begin{aligned} {\left[\mathrm{H}^{+}\right] } & =\sqrt{K_{a} C}=4.24 \times 10^{-3} \mathrm{M} \ \mathrm{pH} & =-\log \left(4.24 \times 10^{-3}\right)=2.37 \end{aligned} $$
Now, let us assume that solution is diluted to a volume where concentration of $\mathrm{CH}_{3} \mathrm{COOH}$ (without considering ionisation) is $x$.
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Also, desired $\mathrm{pH}=2 \times 2.37=4.74$
$$ \left[\mathrm{H}^{+}\right]=1.8 \times 10^{-5}=x \alpha $$
$$ \begin{aligned} K_{a} & =1.8 \times 10^{-5}=\frac{1.8 \times 10^{-5} \alpha}{1-\alpha} \ \alpha & =0.5 \text { and } x=3.6 \times 10^{-5} \mathrm{M} \ \text { Volume (final) } & =1 / 3.6 \times 10^{-5}=27.78 \times 10^{3} \mathrm{~L} \end{aligned} $$
