Chemical and Ionic Equilibrium - Result Question 126
69. What is the $pH$ of a $1.0 M$ solution of acetic acid? To what volume must one litre of this solution be diluted so that the $pH$ of the resulting solution will be twice the original value? Given, $K _a=1.8 \times 10^{-5}$
(1990,4 M)
Show Answer
Answer:
Correct Answer: 69. $(27.78 \times 10^{3})$
Solution:
- $CH _3 COOH \rightleftharpoons CH _3 COO^{-}+H^{+}$
When concentration of $CH _3 COOH$ is $1.0 M$, ’ $\alpha$ ’ is negligible,
$ \begin{aligned} {\left[H^{+}\right] } & =\sqrt{K _a C}=4.24 \times 10^{-3} M \\ 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 $CH _3 COOH$ (without considering ionisation) is $x$.
$\underset{x(1-\alpha)}{\mathrm{CH}_3 \mathrm{COOH}} \rightleftharpoons \underset{x \alpha}{\mathrm{CH}_3 \mathrm{COO}^{-}}+\underset{x \alpha}{\mathrm{H}^{+}}$
$K_a=\frac{x \alpha^2}{1-\alpha}$
Also, desired $\mathrm{pH}=2 \times 2.37=4.74$
$\left[\mathrm{H}^{+}\right]=1.8 \times 10^{-5}=x \alpha$
$\begin{aligned} K_a & =1.8 \times 10^{-5}=\frac{1.8 \times 10^{-5} \alpha}{1-\alpha} \\ \alpha & =0.5 \text { and } x=3.6 \times 10^{-5} \mathrm{M} \\ \text { Volume (final) } & =1 / 3.6 \times 10^{-5}=27.78 \times 10^3 \mathrm{~L} . \end{aligned}$
BETA
