The hydrolysis constant of `CH_(3)COONa` given by

Options

A. `K_(h) = (K_(w))/(K_(a))`
B. `K_(h) = (K_(w))/(K_(b))`
C. `K_(h) = (K_(w))/(K_(a).K_(b))`
D. `K_(h) = K_(a) + K_(b)`

Explanation

To find the hydrolysis constant (K_H) of sodium acetate (CH₃COONa), we need to understand the relationship between the hydrolysis constant and the dissociation constant of the weak acid from which the salt is derived. Sodium acetate is the salt of acetic acid (CH₃COOH), which is a weak acid, and sodium hydroxide (NaOH), which is a strong base.

### Step-by-Step Solution:

1. **Identify the Components**:
Sodium acetate (CH₃COONa) is formed from acetic acid (CH₃COOH) and sodium hydroxide (NaOH).

2. **Understand Hydrolysis**:
When sodium acetate is dissolved in water, it dissociates completely into sodium ions (Na⁺) and acetate ions (CH₃COO⁻). The sodium ion does not hydrolyze because it comes from a strong base (NaOH), but the acetate ion can hydrolyze.

3. **Write the Hydrolysis Reaction**:
The hydrolysis of the acetate ion can be represented as:
\[ CH₃COO⁻ + H₂O \rightleftharpoons CH₃COOH + OH⁻ \]
This reaction shows that acetate ions react with water to form acetic acid and hydroxide ions.

4. **Define the Hydrolysis Constant (K_H)**:
The hydrolysis constant (K_H) for this reaction can be expressed as:
\[ K_H = \frac{[CH₃COOH][OH⁻]}{[CH₃COO⁻]} \]

5. **Relate K_H to K_a**:
The dissociation constant (K_a) of acetic acid is given by:
\[ K_a = \frac{[H⁺][CH₃COO⁻]}{[CH₃COOH]} \]
We can relate K_H to K_a using the ion product of water (K_w):
\[ K_w = [H⁺][OH⁻] \]
By rearranging the equations, we find that:
\[ K_H = \frac{K_w}{K_a} \]

6. **Conclusion**:
Therefore, the hydrolysis constant of sodium acetate (K_H) is given by the relationship:
\[ K_H = \frac{K_w}{K_a} \]

### Final Answer:
The hydrolysis constant of CH₃COONa is given by:
\[ K_H = \frac{K_w}{K_a} \]