`pK_(a)` value for acetic acid at an experimental temperature is 5. The percentage hydrolysis of `0.1` M sodium acetate solution will be

Options

A. `1.0 xx 10^(-4)`
B. `1.0 xx 10^(-2)`
C. `1.0xx 10^(-3)`
D. `1.0 xx 10^(-5)`

Answer

To find the percentage hydrolysis of a 0.1 M sodium acetate solution given that the pK_a value for acetic acid is 5, we can follow these steps:

### Step 1: Determine the value of K_a from pK_a
The relationship between pK_a and K_a is given by the formula:
\[ K_a = 10^{-pK_a} \]
Substituting the given pK_a value:
\[ K_a = 10^{-5} = 1 \times 10^{-5} \]

### Step 2: Calculate K_w
At 25°C, the ion product of water (K_w) is:
\[ K_w = 1 \times 10^{-14} \]

### Step 3: Calculate K_h (hydrolysis constant)
For the hydrolysis of a salt, the hydrolysis constant (K_h) can be calculated using the formula:
\[ K_h = \frac{K_w}{K_a} \]
Substituting the values we have:
\[ K_h = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} = 1 \times 10^{-9} \]

### Step 4: Set up the hydrolysis equilibrium expression
For the hydrolysis of sodium acetate (CH₃COONa), we can write the reaction:
\[ CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^- \]
Let the degree of hydrolysis be represented by \( h \). The equilibrium concentrations will be:
- [CH₃COO⁻] = 0.1 - h
- [CH₃COOH] = h
- [OH⁻] = h

### Step 5: Write the expression for K_h
Using the equilibrium concentrations, we can express K_h as:
\[ K_h = \frac{[CH_3COOH][OH^-]}{[CH_3COO^-]} = \frac{h \cdot h}{0.1 - h} = \frac{h^2}{0.1 - h} \]

### Step 6: Solve for h
Since K_h is very small compared to the initial concentration (0.1 M), we can approximate \( 0.1 - h \approx 0.1 \):
\[ K_h \approx \frac{h^2}{0.1} \]
Substituting the value of K_h:
\[ 1 \times 10^{-9} = \frac{h^2}{0.1} \]
Rearranging gives:
\[ h^2 = 1 \times 10^{-9} \times 0.1 = 1 \times 10^{-10} \]
Taking the square root:
\[ h = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \]

### Step 7: Calculate the percentage hydrolysis
The percentage hydrolysis is given by:
\[ \text{Percentage hydrolysis} = \left( \frac{h}{C} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage hydrolysis} = \left( \frac{1 \times 10^{-5}}{0.1} \right) \times 100 = 0.01\% \]

### Final Answer
The percentage hydrolysis of the 0.1 M sodium acetate solution is **0.01%**.

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