The degree of hydrolysis in hydrolytic equilibrium `A^(-) + H_(2)O to HA + OH^(-)` at salt concentration of 0.001 M is :`(K_(a)=1xx10^(-5))`

Options

A. `1xx10^(-3)`
B. `1xx10^(-4)`
C. `5xx10^(-4)`
D. `1xx10^(-6)`

Explanation

To find the degree of hydrolysis (H) in the given hydrolytic equilibrium \( A^- + H_2O \rightleftharpoons HA + OH^- \) at a salt concentration of 0.001 M, we will follow these steps:

### Step 1: Write the expression for the hydrolysis constant (K_H)
The hydrolysis constant \( K_H \) can be calculated using the relationship:
\[ K_H = \frac{K_w}{K_a} \]
where \( K_w \) is the ion product of water (at 25°C, \( K_w = 1 \times 10^{-14} \)) and \( K_a \) is the acid dissociation constant of the weak acid \( HA \).

### Step 2: Substitute the values of \( K_w \) and \( K_a \)
Given \( K_a = 1 \times 10^{-5} \):
\[ K_H = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} = 1 \times 10^{-9} \]

### Step 3: Use the formula for the degree of hydrolysis
The degree of hydrolysis \( H \) can be calculated using the formula:
\[ H = \sqrt{\frac{K_H}{C}} \]
where \( C \) is the concentration of the salt, which is given as 0.001 M (or \( 1 \times 10^{-3} \) M).

### Step 4: Substitute the values into the formula
Now substituting \( K_H \) and \( C \) into the formula:
\[ H = \sqrt{\frac{1 \times 10^{-9}}{1 \times 10^{-3}}} \]
\[ H = \sqrt{1 \times 10^{-6}} = 1 \times 10^{-3} \]

### Step 5: Conclusion
Thus, the degree of hydrolysis \( H \) is:
\[ H = 1 \times 10^{-3} \]

### Final Answer
The degree of hydrolysis in hydrolytic equilibrium at a salt concentration of 0.001 M is \( 1 \times 10^{-3} \).

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