The degree of hydrolysis of 0.01 M `NH_(4)CI` is `(K_(h) = 2.5 xx 10^(-9))`

Options

A. `5 xx 10^(-5)`
B. `5 xx 10^(-4)`
C. `5 xx 10^(-3)`
D. `5 xx 10^(-7)`

Answer

To find the degree of hydrolysis of 0.01 M NH4Cl, we can follow these steps:

### Step 1: Understand the Hydrolysis Reaction
NH4Cl is a salt formed from a weak base (NH4OH) and a strong acid (HCl). In water, NH4+ ions will undergo hydrolysis to form NH4OH and H+ ions:
\[ \text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4\text{OH} + \text{H}^+ \]

### Step 2: Write the Expression for Hydrolysis Constant (Kh)
The hydrolysis constant (Kh) is given by:
\[ K_h = \frac{[\text{NH}_4\text{OH}][\text{H}^+]}{[\text{NH}_4^+]} \]

### Step 3: Set Up Initial Concentrations
Let the initial concentration of NH4Cl be \( C = 0.01 \, M \). At equilibrium, if \( x \) is the degree of hydrolysis, the concentrations will be:
- \([\text{NH}_4^+] = C - x\)
- \([\text{NH}_4\text{OH}] = x\)
- \([\text{H}^+] = x\)

### Step 4: Substitute into the Kh Expression
Substituting these values into the Kh expression gives:
\[ K_h = \frac{x \cdot x}{C - x} = \frac{x^2}{C - x} \]

### Step 5: Simplify the Expression
Since \( K_h \) is very small compared to \( C \) (which is 0.01 M), we can assume \( C - x \approx C \):
\[ K_h \approx \frac{x^2}{C} \]

### Step 6: Solve for x (Degree of Hydrolysis)
Now we can rearrange the equation to solve for \( x \):
\[ x^2 = K_h \cdot C \]
\[ x = \sqrt{K_h \cdot C} \]

### Step 7: Substitute the Values
Given \( K_h = 2.5 \times 10^{-9} \) and \( C = 0.01 \):
\[ x = \sqrt{(2.5 \times 10^{-9}) \cdot (0.01)} \]
\[ x = \sqrt{2.5 \times 10^{-11}} \]
\[ x = 5 \times 10^{-6} \]

### Step 8: Conclusion
The degree of hydrolysis of 0.01 M NH4Cl is \( 5 \times 10^{-6} \).

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