What is the percentage hydrolysis of ` CH_3COONa ` in 0.01 M Solution ? ` (K_a= 1.78 xx 10 ^(-5) and K_w = 1.0 xx 10 ^(-14))`

Answer

To find the percentage hydrolysis of sodium acetate (CH₃COONa) in a 0.01 M solution, we can use the formula for hydrolysis of salts:

\[ H = \sqrt{\frac{K_w}{K_a} \cdot C} \]

Where:
- \( H \) is the degree of hydrolysis (fraction of salt that hydrolyzes),
- \( K_w \) is the ionic product of water,
- \( K_a \) is the dissociation constant of the weak acid (acetic acid in this case),
- \( C \) is the concentration of the salt solution.

### Step 1: Identify the values
From the question, we have:
- \( K_a = 1.78 \times 10^{-5} \)
- \( K_w = 1.0 \times 10^{-14} \)
- \( C = 0.01 \, \text{M} \)

### Step 2: Substitute the values into the formula
Now, substituting the values into the formula:

\[ H = \sqrt{\frac{1.0 \times 10^{-14}}{1.78 \times 10^{-5}} \cdot 0.01} \]

### Step 3: Calculate \( \frac{K_w}{K_a} \)
First, calculate \( \frac{K_w}{K_a} \):

\[ \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.78 \times 10^{-5}} \approx 5.617 \times 10^{-10} \]

### Step 4: Multiply by concentration \( C \)
Now multiply this result by the concentration \( C \):

\[ \frac{K_w}{K_a} \cdot C = 5.617 \times 10^{-10} \cdot 0.01 = 5.617 \times 10^{-12} \]

### Step 5: Take the square root
Now take the square root:

\[ H = \sqrt{5.617 \times 10^{-12}} \approx 2.37 \times 10^{-6} \]

### Step 6: Convert to percentage
To find the percentage hydrolysis, we multiply \( H \) by 100:

\[ \text{Percentage Hydrolysis} = H \times 100 = 2.37 \times 10^{-6} \times 100 = 2.37 \times 10^{-4} \% \]

### Final Result
The percentage hydrolysis of sodium acetate in a 0.01 M solution is approximately:

\[ \text{Percentage Hydrolysis} \approx 0.000237 \% \]