Rate of Hydrolysis of Methyl Acetate Catalyzed by Weak Acid

The acid-catalyzed hydrolysis of methyl acetate is a pseudo-first-order reaction, where the rate of reaction is directly proportional to the concentration of hydrogen ions ($H^+$) present in the solution:

$\text{Rate} = k \cdot [\text{CH}_3\text{COOCH}_3] \cdot [H^+]$

Since the concentration of methyl acetate ($1\text{ M}$) is kept constant in both cases, the ratio of the rates is equal to the ratio of the concentration of $H^+$ ions produced by each acid:

$\frac{\text{Rate}_{HA}}{\text{Rate}_{HX}} = \frac{[H^+]_{HA}}{[H^+]_{HX}}$

Given that the strong acid $HX$ ($1\text{ M}$) undergoes complete dissociation: $[H^+]_{HX} = C = 1\text{ M}$

For the weak acid $HA$ ($1\text{ M}$), the dissociation equilibrium is: $HA \rightleftharpoons H^+ + A^-$

The concentration of $H^+$ ions produced by the weak acid is given by: $[H^+]_{HA} = \sqrt{K_a \cdot C} = \sqrt{K_a \cdot 1} = \sqrt{K_a}$

According to the given condition, the rate of hydrolysis with $HA$ is $\frac{1}{100}^{\text{th}}$ of that with $HX$:

$\frac{\text{Rate}_{HA}}{\text{Rate}_{HX}} = \frac{1}{100}$

Substituting the expressions for $H^+$ ion concentrations:

$\frac{\sqrt{K_a}}{1} = \frac{1}{100}$

$\sqrt{K_a} = 10^{-2}$

Squaring both sides gives the dissociation constant $K_a$:

$K_a = (10^{-2})^2 = 1 \times 10^{-4}$

Thus, the correct option is A.