Calculate the degree of hydrolysis of the 0.01 M solution of salt` (KF)(Ka(HF)=6.6xx10^(-4))` :-

Options

A. `3.87xx10^(-6)`
B. `3.87xx10^(-5)`
C. `3.87xx10^(-2)`
D. None of these

Explanation

To calculate the degree of hydrolysis of a 0.01 M solution of the salt KF, we can follow these steps:

### Step 1: Identify the nature of the salt
KF is a salt formed from a weak acid (HF) and a strong base (KOH). This means that KF will undergo hydrolysis in water, leading to the formation of HF and KOH.

### Step 2: Write the hydrolysis reaction
The hydrolysis of KF can be represented as:
\[ \text{KF} \, + \, \text{H}_2\text{O} \, \rightleftharpoons \, \text{K}^+ \, + \, \text{HF} \, + \, \text{OH}^- \]

### Step 3: Set up the equilibrium expression
Let the initial concentration of KF be \( C = 0.01 \, \text{M} \). If \( H \) is the degree of hydrolysis, then at equilibrium:
- Concentration of HF = \( H \)
- Concentration of KOH = \( H \)
- Concentration of KF = \( C - H \)

The equilibrium expression for the hydrolysis constant \( K_H \) can be written as:
\[ K_H = \frac{[\text{HF}][\text{OH}^-]}{[\text{KF}]} = \frac{H \cdot H}{C - H} = \frac{H^2}{C - H} \]

### Step 4: Approximate \( C - H \)
Since \( H \) is expected to be small compared to \( C \), we can approximate \( C - H \approx C \). Therefore, the equation simplifies to:
\[ K_H \approx \frac{H^2}{C} \]

### Step 5: Relate \( K_H \) to \( K_a \)
For a weak acid strong base salt, the relationship between the hydrolysis constant \( K_H \) and the acid dissociation constant \( K_a \) is given by:
\[ K_H = \frac{K_w}{K_a} \]
where \( K_w \) is the ion product of water, approximately \( 1.0 \times 10^{-14} \) at 25°C.

Given \( K_a \) for HF is \( 6.6 \times 10^{-4} \), we can calculate \( K_H \):
\[ K_H = \frac{1.0 \times 10^{-14}}{6.6 \times 10^{-4}} \]

### Step 6: Calculate \( K_H \)
Calculating \( K_H \):
\[ K_H \approx 1.515 \times 10^{-11} \]

### Step 7: Substitute \( K_H \) back into the equilibrium expression
Now we can substitute \( K_H \) back into our earlier expression:
\[ 1.515 \times 10^{-11} = \frac{H^2}{0.01} \]

### Step 8: Solve for \( H \)
Rearranging gives:
\[ H^2 = 1.515 \times 10^{-11} \times 0.01 \]
\[ H^2 = 1.515 \times 10^{-13} \]
Taking the square root:
\[ H = \sqrt{1.515 \times 10^{-13}} \]
\[ H \approx 1.23 \times 10^{-7} \]

### Step 9: Calculate the degree of hydrolysis
The degree of hydrolysis \( \alpha \) can be calculated as:
\[ \alpha = \frac{H}{C} = \frac{1.23 \times 10^{-7}}{0.01} \]
\[ \alpha \approx 1.23 \times 10^{-5} \]

### Final Result
The degree of hydrolysis of the 0.01 M solution of KF is approximately \( 1.23 \times 10^{-5} \).

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