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 solve the problem of finding the degree of hydrolysis for the reaction \( A^- + H_2O \rightleftharpoons HA + OH^- \) at a salt concentration of \( 0.001 \, M \) with \( K_a = 1 \times 10^{-5} \), we can follow these steps:

### Step 1: Understand the Reaction and Initial Conditions
The hydrolysis reaction involves the anion \( A^- \) reacting with water to form the weak acid \( HA \) and hydroxide ions \( OH^- \). The initial concentration of the salt (which provides \( A^- \)) is given as \( 0.001 \, M \) or \( 1 \times 10^{-3} \, M \).

### Step 2: Set Up the Equilibrium Expression
At equilibrium, let \( h \) be the degree of hydrolysis. The concentrations at equilibrium will be:
- \( [A^-] = C - h \) (where \( C = 1 \times 10^{-3} \, M \))
- \( [HA] = h \)
- \( [OH^-] = h \)

### Step 3: Write the Hydrolysis Constant Expression
The hydrolysis constant \( K_h \) can be expressed as:
\[ K_h = \frac{[HA][OH^-]}{[A^-]} = \frac{h \cdot h}{C - h} = \frac{h^2}{C - h} \]

### Step 4: Relate Hydrolysis Constant to Acid Dissociation Constant
The hydrolysis constant \( K_h \) can also be expressed in terms of the dissociation constant of water \( K_w \) and the acid dissociation constant \( K_a \):
\[ K_h = \frac{K_w}{K_a} \]
Given \( K_w = 1 \times 10^{-14} \) and \( K_a = 1 \times 10^{-5} \):
\[ K_h = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} = 1 \times 10^{-9} \]

### Step 5: Substitute \( K_h \) into the Hydrolysis Expression
Now, substituting \( K_h \) into the equilibrium expression:
\[ 1 \times 10^{-9} = \frac{h^2}{1 \times 10^{-3} - h} \]

### Step 6: Assume \( h \) is Small
Since \( h \) is expected to be small compared to \( C \), we can approximate \( 1 \times 10^{-3} - h \approx 1 \times 10^{-3} \):
\[ 1 \times 10^{-9} \approx \frac{h^2}{1 \times 10^{-3}} \]

### Step 7: Solve for \( h \)
Rearranging gives:
\[ h^2 = 1 \times 10^{-9} \times 1 \times 10^{-3} = 1 \times 10^{-12} \]
Taking the square root:
\[ h = \sqrt{1 \times 10^{-12}} = 1 \times 10^{-6} \]

### Step 8: Conclusion
The degree of hydrolysis \( h \) at a salt concentration of \( 0.001 \, M \) is \( 1 \times 10^{-6} \).