The ionization constant of ammonium hydroxide is `1.77xx10^(-5)` at `298 K`. Hydrolysis constant of ammonium chloride is

Options

A. `5.65xx10^(-10)`
B. `6.50xx10^(-12)`
C. `5.65xx10^(-13)`
D. `5.65xx10^(-12)`

Explanation

To find the hydrolysis constant (K_h) of ammonium chloride (NH4Cl), we can use the relationship between the ionization constant of ammonium hydroxide (K_b) and the hydrolysis constant (K_h) along with the ion product of water (K_w).

### Step-by-Step Solution

1. **Identify the Given Values**:
- Ionization constant of ammonium hydroxide (K_b) = \(1.77 \times 10^{-5}\)
- Ion product of water (K_w) at 298 K = \(1.0 \times 10^{-14}\)

2. **Write the Relationship Between K_h, K_b, and K_w**:
The relationship is given by:
\[ K_h = \frac{K_w}{K_b} \]

3. **Substitute the Known Values**:
Substitute the values of K_w and K_b into the equation:
\[ K_h = \frac{1.0 \times 10^{-14}}{1.77 \times 10^{-5}} \]

4. **Perform the Calculation**:
- First, calculate the division:
\[ K_h = \frac{1.0 \times 10^{-14}}{1.77 \times 10^{-5}} \approx 5.65 \times 10^{-10} \]

5. **Conclusion**:
The hydrolysis constant (K_h) of ammonium chloride is approximately \(5.65 \times 10^{-10}\).

### Final Answer:
The hydrolysis constant of ammonium chloride is \(5.65 \times 10^{-10}\).