Degree of hydrolysis of `0.25 M CH_(3)COOHNa` is `10%` what will be the degree of hydrolysis if concentration of `CH_(3)COONa` is made `0.01M`

Options

A. `10%`
B. `1%`
C. `50%`
D. `75%`

Explanation

To solve the problem step by step, we will use the relationship between the degree of hydrolysis and concentration of the salt.

### Step 1: Understand the relationship
The degree of hydrolysis (H) is related to the concentration (C) of the salt by the formula:

\[ \frac{H_1}{H_2} = \sqrt{\frac{C_2}{C_1}} \]

Where:
- \(H_1\) = degree of hydrolysis at concentration \(C_1\)
- \(H_2\) = degree of hydrolysis at concentration \(C_2\)

### Step 2: Identify the known values
From the problem, we have:
- \(C_1 = 0.25 \, M\) (initial concentration)
- \(H_1 = 10\% = 0.1\) (degree of hydrolysis at \(C_1\))
- \(C_2 = 0.01 \, M\) (new concentration)

### Step 3: Substitute the known values into the formula
Now, we can substitute the known values into the formula:

\[ \frac{0.1}{H_2} = \sqrt{\frac{0.01}{0.25}} \]

### Step 4: Calculate the right side of the equation
First, calculate the fraction:

\[ \frac{0.01}{0.25} = 0.04 \]

Now take the square root:

\[ \sqrt{0.04} = 0.2 \]

### Step 5: Substitute back to find \(H_2\)
Now we can substitute this value back into the equation:

\[ \frac{0.1}{H_2} = 0.2 \]

### Step 6: Solve for \(H_2\)
Cross-multiply to solve for \(H_2\):

\[ 0.1 = 0.2 \cdot H_2 \]

Now, divide both sides by 0.2:

\[ H_2 = \frac{0.1}{0.2} = 0.5 \]

### Step 7: Convert \(H_2\) to percentage
To express \(H_2\) as a percentage:

\[ H_2 = 0.5 \times 100\% = 50\% \]

### Final Answer
The degree of hydrolysis when the concentration of \(CH_3COONa\) is made \(0.01M\) is **50%**.

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