In the equilibrium `A^(-)+ H_(2)O hArr HA + OH^(-) (K_(a) = 1.0 xx 10^(-4))`. The degree of hydrolysis of `0.01 M solution of the salt is

Options

A. `10^(-3)`
B. `10^(-4)`
C. `10^(-5)`
D. `10^(-6)`

Answer

To find the degree of hydrolysis of a 0.01 M solution of the salt in the equilibrium reaction:

\[ A^- + H_2O \rightleftharpoons HA + OH^- \]

where \( K_a = 1.0 \times 10^{-4} \), we can follow these steps:

### Step 1: Determine the Hydrolysis Constant \( K_h \)

The hydrolysis constant \( K_h \) can be calculated using the relation:

\[ K_h = \frac{K_w}{K_a} \]

where \( K_w \) is the ionization constant of water, which is \( 1.0 \times 10^{-14} \) at 25°C.

### Step 2: Calculate \( K_h \)

Substituting the values:

\[ K_h = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} = 1.0 \times 10^{-10} \]

### Step 3: Use the Hydrolysis Constant to Find 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 solution, which is \( 0.01 \, M \).

### Step 4: Substitute the Values

Substituting \( K_h \) and \( C \):

\[ h = \sqrt{\frac{1.0 \times 10^{-10}}{0.01}} = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4} \]

### Conclusion

The degree of hydrolysis of the 0.01 M solution of the salt is:

\[ h = 1.0 \times 10^{-4} \]