Chemical and Ionic Equilibrium - Result Question 30
####30. The initial rate of hydrolysis of methyl acetate $(1 M)$ by a weak acid $(H A, 1 M)$ is $1 / 100$ th of that of a strong acid $(H X, 1 M)$, at $25^{\circ} C$. The $K _a(H A)$ is
(2013 Adv.)
(a) $1 \times 10^{-4}$
(b) $1 \times 10^{-5}$
(c) $1 \times 10^{-6}$
(d) $1 \times 10^{-3}$
Show Answer
Solution:
- PLAN $R COO R^{\prime}+H _2 O \xrightarrow{H^{+}} R COOH+R^{\prime} OH$
Acid hydrolysis of ester is follows first order kinetics.
For same concentration of ester in each case, rate is dependent on $\left[H^{+}\right]$from acid.
$$ \text { Rate }=k\left[R COOR^{\prime}\right] $$
Also for weak acid, $H A \rightleftharpoons H^{+}+A^{-}$
$$ \begin{aligned} K _a & =\frac{\left[H^{+}\right]\left[A^{-}\right]}{[H A]} \ (\text { Rate }) _{H A} & =k\left[H^{+}\right] _{H A} \ (\text { Rate }) _{H X} & =k\left[H^{+}\right] _{H X} \ (\text { Rate }) _{H X} & =100(\text { Rate }) _{H A} \end{aligned} $$
$\therefore$ Also in strong acid, $\left[H^{+}\right]=[H X]=1 M$
$$ \begin{aligned} & \frac{(\text { Rate }) _{H X}}{(\text { Rate }) _{H A}}=100=\frac{\left[H^{+}\right] _{H X}}{\left[H^{+}\right] _{H A}}=\frac{1}{\left[H^{+}\right] _{H A}} \ & \therefore {\left[H^{+}\right] _{H A}=\frac{1}{100} } \ & H A \rightleftharpoons H^{+}+A^{-} \ & 1 \quad 0 \quad 0 \ &(1-x) \quad x \quad x \ & \therefore \quad \quad K _a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{[H A]}=\frac{0.01 \times 0.01}{0.99}=1.01 \times 10^{-4} \end{aligned} $$
