The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th of that of a strong acid (HX, 1M), at 25°C. The Ka of HA is

Engineering

Chemistry

Pseudo Order Reaction

pH Calculation for Acids and Bases

Question

The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100^th of that of a strong acid (HX, 1M), at 25°C. The K_a of HA is

1 × 10^–5

1 × 10^–3

1 × 10^–6

1 × 10^–4

Rate = K [Ester] [H⁺]

For strong acid [H⁺] = 1M

For weak acid [H+] = $\sqrt{K_{a} . C}$

$\frac{{Rate}_{1}}{{Rate}_{2}} = \frac{{[H^{+}]}_{1}}{{[H^{+}]}_{2}}$

$\frac{1}{1 / 100} = \frac{1}{\sqrt{K_{a} \times 1}}$

K_a = 10^–4 M

Concept Explanation: Acid-Catalyzed Hydrolysis and Acid Strength

The hydrolysis of methyl acetate is catalyzed by acids. The rate of this reaction depends on the concentration of H⁺ ions, as H⁺ acts as a catalyst. A strong acid (like HX) dissociates completely, providing a high [H⁺]. A weak acid (HA) dissociates only partially, providing a lower [H⁺]. The rate of hydrolysis is directly proportional to the [H⁺] from the acid.

Step-by-Step Solution:

Step 1: Understand the Rate Relationship

Given: The initial rate with weak acid HA is 1/100th of the rate with strong acid HX. Since both acids are at the same concentration (1M), and the rate is proportional to [H⁺], we can write:

\frac{r_{HA}}{r_{HX}} = \frac{[_{H⁺}^{]}}{HA} = \frac{1}{100}

Step 2: Find [H⁺] for the Strong Acid (HX)

A strong acid dissociates completely. For a 1M solution of HX:

[_{H⁺}^{]} HX = 1 M

Step 3: Find [H⁺] for the Weak Acid (HA)

From Step 1:

\frac{[_{H⁺}^{]}}{HA} = \frac{1}{100} \Rightarrow [_{H⁺}^{]} HA = 0.01 M = 1 \times 10^{- 2} M

Step 4: Apply the Formula for a Weak Acid

For a weak acid HA, the acid dissociation constant K_a is given by:

K_{a} = \frac{[_{H⁺}^{]} [_{A⁻}^{]}}{[_{HA}^{]}}

For a weak acid where dissociation is small, we can make the approximation that [H⁺] ≈ [A⁻]. The initial concentration [HA]₀ = 1 M. The concentration of undissociated acid [HA] is approximately equal to its initial concentration because dissociation is small ([H⁺] = 0.01 M is much less than 1 M). Therefore:

K_{a} \approx \frac{[_{H⁺}^{]} \cdot [_{H⁺}^{]}}{[_{HA]}} = \frac{{(1 \times 10^{- 2})}^{2}}{1} = 1 \times 10^{- 4}

Final Answer: The K_a of HA is $1 \times 10^{- 4}$

Key Formulae and Theory:

1. Rate of Acid-Catalyzed Reaction: For a reaction catalyzed by H⁺, the rate is often proportional to the catalyst concentration. $Rate \propto [_{H⁺}^{]}$

2. Strong Acid Dissociation: A strong acid dissociates completely in water. $HX \to H^{+} + X^{-} \Rightarrow [_{H⁺}^{]} = [_{HX]_{initial}}$

3. Weak Acid Dissociation (Approximation): For a weak acid HA with initial concentration C and small dissociation ( $K_{a} << 1$ ), the concentration of H⁺ ions is given by: $[_{H⁺}^{]} \approx \sqrt{K_{a} \cdot C} and K_{a} \approx \frac{{[_{H⁺}^{]}}^{2}}{C}$

Detailed Solution

Concept Explanation: Acid-Catalyzed Hydrolysis and Acid Strength

Step-by-Step Solution:

Related Topics:

Key Formulae and Theory:

Detailed Solution

Concept Explanation: Acid-Catalyzed Hydrolysis and Acid Strength

Step-by-Step Solution:

Related Topics:

Key Formulae and Theory:

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