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 \( NH_4Cl \), we can use the relationship between the degree of hydrolysis (h), the hydrolysis constant (\( K_h \)), and the concentration of the salt (c).

### Step-by-Step Solution:

1. **Understand the Concept of Degree of Hydrolysis**:
The degree of hydrolysis (h) is defined as the fraction of the salt that has undergone hydrolysis at equilibrium. It can be expressed mathematically as:
\[ h = \frac{\text{moles of salt hydrolyzed}}{\text{total moles of salt}} \]

2. **Use the Relationship**:
The relationship between the degree of hydrolysis, hydrolysis constant, and concentration of the salt is given by:
\[ h = \sqrt{\frac{K_h}{c}} \]
where:
- \( K_h \) is the hydrolysis constant.
- \( c \) is the concentration of the salt.

3. **Substitute the Given Values**:
We are given:
- \( K_h = 2.5 \times 10^{-9} \)
- \( c = 0.01 \, M \)

Now, substituting these values into the equation:
\[ h = \sqrt{\frac{2.5 \times 10^{-9}}{0.01}} \]

4. **Calculate the Value Inside the Square Root**:
First, calculate \( \frac{2.5 \times 10^{-9}}{0.01} \):
\[ \frac{2.5 \times 10^{-9}}{0.01} = 2.5 \times 10^{-7} \]

5. **Take the Square Root**:
Now, we take the square root of \( 2.5 \times 10^{-7} \):
\[ h = \sqrt{2.5 \times 10^{-7}} = \sqrt{2.5} \times 10^{-3.5} \]

Approximating \( \sqrt{2.5} \):
\[ \sqrt{2.5} \approx 1.58 \quad \text{(using a calculator)} \]
Therefore:
\[ h \approx 1.58 \times 10^{-3.5} = 1.58 \times 5 \times 10^{-4} = 5 \times 10^{-4} \]

6. **Final Result**:
Thus, the degree of hydrolysis of 0.01 M \( NH_4Cl \) is:
\[ h \approx 5 \times 10^{-4} \]