Let's break down this problem step-by-step:

Understanding the Reaction:

The hydrolysis of methyl acetate (MeCOOCH₃) is catalyzed by both strong acids (like HX) and weak acids (like HA). The reaction proceeds as follows:

$$ MeCOOCH_3 + H_2O \xrightarrow[or \, HA]{H^+ \, or \, HX} MeCOOH + CH_3OH $$

Rate of Reaction and Acid Strength:

The rate of acid-catalyzed hydrolysis depends on the concentration of H⁺ ions.

• Strong acids ionize completely. So, a 1M solution of HX provides 1M H⁺ ions.


• Weak acids ionize partially. So, a 1M solution of HA will provide significantly less than 1M H⁺ ions.

Information from the Problem:

We are told that the initial rate of hydrolysis with HA is 1/100^th of that with HX. Since the rate is directly proportional to [H⁺], this means the [H⁺] from HA is 1/100^th of that from HX.

Calculating K_a:

1. [H⁺] from HX: Since HX is a strong acid and fully ionizes, [H⁺] = 1M.


2. [H⁺] from HA: This is 1/100^th the [H⁺] from HX, so [H⁺] = (1/100) \* 1M = 10^-2 M.


3. Setting up the K_a expression:

$$ K_a = \frac{[H^+][A^-]}{[HA]} $$

Since the initial concentration of HA is 1M and it ionizes to a small extent, we can approximate:

• [HA] ≈ 1 M


• [H⁺] = [A^-] = 10^-2 M

1. Calculating K_a:

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

Answer:

Therefore, the K_a of the weak acid HA is 1 × 10^-4 (Option A).