If at `25^(@)` the ionization constant of acetic acid is `2 xx 10^(5)` the hydrolysis constant of sodium acetate will be

Options

A. `5 xx 10^(-8)`
B. `5 xx 10^(-9)`
C. `5 xx 10^(-10)`
D. `4 xx 10^(-10)`

Answer

To find the hydrolysis constant (Kh) of sodium acetate given the ionization constant (Ka) of acetic acid, we can use the relationship between these constants. Here’s a step-by-step solution:

### Step 1: Understand the Ionization of Acetic Acid
Acetic acid (CH₃COOH) ionizes in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
The ionization constant (Ka) for this reaction is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]

### Step 2: Given Values
From the problem, we know:
- \( K_a = 2 \times 10^{-5} \) at 25°C
- The ion product of water (\( K_w \)) at 25°C is \( 1 \times 10^{-14} \).

### Step 3: Relationship Between Ka and Kh
The hydrolysis constant (Kh) for sodium acetate can be calculated using the formula:
\[ K_h = \frac{K_w}{K_a} \]

### Step 4: Substitute the Values
Now, substituting the known values into the equation:
\[ K_h = \frac{1 \times 10^{-14}}{2 \times 10^{-5}} \]

### Step 5: Calculate Kh
Now perform the division:
\[ K_h = \frac{1}{2} \times 10^{-14 + 5} = \frac{1}{2} \times 10^{-9} = 0.5 \times 10^{-9} = 5 \times 10^{-10} \]

### Final Answer
Thus, the hydrolysis constant (Kh) of sodium acetate is:
\[ K_h = 5 \times 10^{-10} \]

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