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

Options

A. `3.89xx10^(-6)`
B. `3.89xx10^(-5)`
C. `3.89xx10^(-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: Understand the Hydrolysis Reaction
The salt KF dissociates in water to form K⁺ and F⁻ ions. The F⁻ ion can hydrolyze in water as follows:
\[ \text{F}^- + \text{H}_2\text{O} \rightleftharpoons \text{HF} + \text{OH}^- \]

### Step 2: Set Up the Initial Concentrations
Let the initial concentration of KF be \( C = 0.01 \, \text{M} \). At the start of the reaction:
- \([F^-] = C = 0.01 \, \text{M}\)
- \([HF] = 0\)
- \([OH^-] = 0\)

### Step 3: Define the Change in Concentrations
Let \( H \) be the degree of hydrolysis. At equilibrium, the concentrations will be:
- \([F^-] = C(1 - H) = 0.01(1 - H)\)
- \([HF] = CH = 0.01H\)
- \([OH^-] = CH = 0.01H\)

### Step 4: Write the Expression for Hydrolysis Constant (K_h)
The hydrolysis constant \( K_h \) can be expressed as:
\[ K_h = \frac{[HF][OH^-]}{[F^-]} \]
Substituting the equilibrium concentrations into this expression gives:
\[ K_h = \frac{(0.01H)(0.01H)}{0.01(1 - H)} = \frac{0.0001H^2}{0.01(1 - H)} \]
This simplifies to:
\[ K_h = \frac{0.01H^2}{1 - H} \]

### Step 5: Relate K_h to K_w and K_a
The hydrolysis constant \( K_h \) is related to the ion product of water \( K_w \) and the acid dissociation constant \( K_a \) of HF:
\[ K_h = \frac{K_w}{K_a} \]
Given \( K_w = 1 \times 10^{-14} \) and \( K_a = 6.6 \times 10^{-4} \), we can calculate \( K_h \):
\[ K_h = \frac{1 \times 10^{-14}}{6.6 \times 10^{-4}} \]

### Step 6: Calculate K_h
Calculating \( K_h \):
\[ K_h = \frac{1 \times 10^{-14}}{6.6 \times 10^{-4}} \approx 1.515 \times 10^{-11} \]

### Step 7: Substitute K_h into the Hydrolysis Equation
Now we can set the expression for \( K_h \) equal to the calculated value:
\[ \frac{0.01H^2}{1 - H} = 1.515 \times 10^{-11} \]

### Step 8: Assume H is Small
Assuming \( H \) is small compared to 1, we can approximate \( 1 - H \approx 1 \):
\[ 0.01H^2 \approx 1.515 \times 10^{-11} \]

### Step 9: Solve for H
Rearranging gives:
\[ H^2 \approx \frac{1.515 \times 10^{-11}}{0.01} = 1.515 \times 10^{-9} \]
Taking the square root:
\[ H \approx \sqrt{1.515 \times 10^{-9}} \approx 3.89 \times 10^{-5} \]

### Final Answer
Thus, the degree of hydrolysis of the 0.01 M solution of KF is approximately:
\[ H \approx 3.89 \times 10^{-5} \]

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