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The initial rate of hydrolysis of methyl...

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 `K_(a)(HA)` is

A

`1xx10^(-4)`

B

`1xx10^(-5)`

C

`1xx10^(-6)`

D

`1xx10^(-3)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the acid dissociation constant \( K_a \) of the weak acid \( HA \) given the initial rates of hydrolysis of methyl acetate in the presence of a weak acid and a strong acid. ### Step-by-Step Solution: 1. **Understanding the Given Information**: - The initial rate of hydrolysis of methyl acetate with weak acid \( HA \) (1 M) is \( \frac{1}{100} \) of that with strong acid \( HX \) (1 M). - This means: \[ \text{Rate}_{HA} = \frac{1}{100} \times \text{Rate}_{HX} \] 2. **Expressing the Rates**: - The rate of hydrolysis in the presence of a strong acid \( HX \) can be expressed as: \[ \text{Rate}_{HX} = k \cdot [H^+]_{HX} \] - For the strong acid \( HX \), since it is a strong acid, it completely dissociates: \[ [H^+]_{HX} = 1 \text{ M} \] - Thus, \[ \text{Rate}_{HX} = k \cdot 1 = k \] 3. **Calculating the Rate for Weak Acid \( HA \)**: - For the weak acid \( HA \), the concentration of \( H^+ \) ions is less than the concentration of the acid due to partial dissociation. Let \( \alpha \) be the degree of dissociation. - The concentration of \( H^+ \) from the weak acid is: \[ [H^+]_{HA} = C \cdot \alpha = 1 \cdot \alpha = \alpha \] - Therefore, the rate of hydrolysis with weak acid is: \[ \text{Rate}_{HA} = k \cdot [H^+]_{HA} = k \cdot \alpha \] 4. **Relating the Rates**: - From the information given: \[ k \cdot \alpha = \frac{1}{100} k \] - Dividing both sides by \( k \) (assuming \( k \neq 0 \)): \[ \alpha = \frac{1}{100} \] 5. **Calculating \( K_a \)**: - The dissociation constant \( K_a \) for the weak acid \( HA \) is given by: \[ K_a = C \cdot \alpha^2 \] - Substituting \( C = 1 \) M and \( \alpha = \frac{1}{100} = 0.01 \): \[ K_a = 1 \cdot (0.01)^2 = 1 \cdot 10^{-4} = 10^{-4} \] 6. **Final Answer**: - Therefore, the \( K_a \) of the weak acid \( HA \) is: \[ K_a = 10^{-4} \]
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