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 solve the problem of finding the degree of hydrolysis of a 0.01 M solution of the salt \( A^- \) in the equilibrium reaction:

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

with the acid dissociation constant \( K_a = 1.0 \times 10^{-4} \), 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, which is \( 1.0 \times 10^{-14} \).

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

Substituting the values into the equation:

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

### Step 3: Set up the equilibrium expression

Let \( h \) be the degree of hydrolysis. The initial concentration of \( A^- \) is \( 0.01 \, M \) (or \( 1.0 \times 10^{-2} \, M \)). At equilibrium, the concentrations will be:

- \( [A^-] = 0.01 - h \)
- \( [HA] = h \)
- \( [OH^-] = h \)

Since \( h \) is small compared to the initial concentration, we can approximate:

\[ [A^-] \approx 0.01 \]

### Step 4: Write the expression for \( K_h \)

The expression for \( K_h \) is given by:

\[ K_h = \frac{[HA][OH^-]}{[A^-]} = \frac{h \cdot h}{0.01} = \frac{h^2}{0.01} \]

### Step 5: Substitute \( K_h \) into the equation

Now substituting \( K_h \) into the equation:

\[ 1.0 \times 10^{-10} = \frac{h^2}{0.01} \]

### Step 6: Solve for \( h^2 \)

Rearranging gives:

\[ h^2 = 1.0 \times 10^{-10} \times 0.01 = 1.0 \times 10^{-12} \]

### Step 7: Solve for \( h \)

Taking the square root:

\[ h = \sqrt{1.0 \times 10^{-12}} = 1.0 \times 10^{-6} \]

### Step 8: Calculate the degree of hydrolysis

The degree of hydrolysis \( h \) can also be expressed as a percentage:

\[ \text{Degree of hydrolysis} = \frac{h}{\text{Initial concentration}} \times 100 = \frac{1.0 \times 10^{-6}}{1.0 \times 10^{-2}} \times 100 = 0.01\% \]

### Final Answer

The degree of hydrolysis of the 0.01 M solution of the salt \( A^- \) is \( 1.0 \times 10^{-6} \) or \( 0.01\% \).

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