The degree of hydrolysis of 0.1 M `RNH_(3)Cl` solution is `1.0%.` If the concentration of `RNH_(3)Cl` is made 0.4 M, what is the new degree of hydrolysis (in percentage)?

Options

A. `0.01`
B. `0.001`
C. `0.2`
D. `0.5`

Answer

To solve the problem, we need to find the new degree of hydrolysis when the concentration of `RNH₃Cl` is increased from 0.1 M to 0.4 M. We will use the relationship between the degree of hydrolysis and concentration.

### Step-by-Step Solution:

1. **Understand the Degree of Hydrolysis Formula**:
The degree of hydrolysis (H) is given by the formula:
\[ H = \sqrt{\frac{K_H}{C}} \]
where \( K_H \) is the hydrolysis constant and \( C \) is the concentration of the solution.

2. **Identify Initial Conditions**:
For the initial concentration \( C_1 = 0.1 \, \text{M} \) and the degree of hydrolysis \( H_1 = 1\% \):
\[ H_1 = \sqrt{\frac{K_H}{C_1}} = 1\% \]

3. **Set Up the New Conditions**:
When the concentration is increased to \( C_2 = 0.4 \, \text{M} \), we need to find the new degree of hydrolysis \( H_2 \):
\[ H_2 = \sqrt{\frac{K_H}{C_2}} \]

4. **Relate the Degrees of Hydrolysis**:
We can express the relationship between the two degrees of hydrolysis:
\[ \frac{H_2}{H_1} = \sqrt{\frac{C_1}{C_2}} \]

5. **Substitute Known Values**:
Substitute \( C_1 = 0.1 \, \text{M} \) and \( C_2 = 0.4 \, \text{M} \):
\[ \frac{H_2}{1\%} = \sqrt{\frac{0.1}{0.4}} \]

6. **Calculate the Square Root**:
Simplifying the fraction:
\[ \frac{0.1}{0.4} = \frac{1}{4} \]
Therefore:
\[ \sqrt{\frac{1}{4}} = \frac{1}{2} \]

7. **Find the New Degree of Hydrolysis**:
Now substituting back:
\[ H_2 = 1\% \times \frac{1}{2} = 0.5\% \]

### Final Answer:
The new degree of hydrolysis when the concentration of `RNH₃Cl` is increased to 0.4 M is **0.5%**.