Which relation is correct for `NH_4` CI ?

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_w.K_a`

Answer

To solve the question regarding the correct relation for `NH4Cl`, we will analyze the hydrolysis of ammonium chloride and derive the appropriate expression step by step.

### Step-by-Step Solution:

1. **Understanding Hydrolysis of NH4Cl**:
- Ammonium chloride (NH4Cl) dissociates in water to form ammonium ions (NH4+) and chloride ions (Cl-).
- The relevant reaction is:
\[ NH_4Cl \rightarrow NH_4^+ + Cl^- \]

2. **Behavior of NH4+ in Water**:
- The ammonium ion (NH4+) is a weak acid and can react with water:
\[ NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+ \]
- This reaction indicates that NH4+ donates a proton (H+) to water, forming ammonia (NH3) and hydronium ions (H3O+).

3. **Equilibrium Constant for the Reaction**:
- The equilibrium constant for the above reaction (acid dissociation constant) can be expressed as:
\[ K_a = \frac{[NH_3][H_3O^+]}{[NH_4^+]} \]

4. **Hydrolysis Constant (Kh)**:
- The hydrolysis constant (Kh) for the reaction can be defined as:
\[ K_h = \frac{[NH_3][H_3O^+]}{[NH_4^+]} \]
- Since H3O+ can be represented by H+ in dilute solutions, we can write:
\[ K_h = \frac{[NH_3][H^+]}{[NH_4^+]} \]

5. **Relation Between Kw, Ka, and Kb**:
- We know that the product of the ionization constants of water (Kw), the acid dissociation constant (Ka), and the base dissociation constant (Kb) is given by:
\[ K_w = K_a \times K_b \]
- Rearranging this gives:
\[ K_b = \frac{K_w}{K_a} \]

6. **Substituting Kb in Kh Expression**:
- From the hydrolysis constant expression, we can substitute Kb:
\[ K_h = \frac{K_w}{K_b} \]
- Therefore, we can express Kh in terms of Kw and Ka:
\[ K_h = \frac{K_w}{K_a} \]

7. **Conclusion**:
- Thus, the correct relation for NH4Cl is:
\[ K_h = \frac{K_w}{K_a} \]
- This corresponds to option B.

### Final Answer:
The correct relation for NH4Cl is:
\[ K_h = \frac{K_w}{K_a} \]