Calculate the hydrolysis constant of the salt containing `NO_(2)^(-)`. Given the `K_(a) " for " HNO_(2) = 4.5 xx 10^(-10)`

Options

A. `2.22 xx 10^(-5)`
B. `2.02 xx 10^(5)`
C. `4.33 xx 10^(4)`
D. `3.03 xx 10^(-5)`

Answer

To calculate the hydrolysis constant of the salt containing \( \text{NO}_2^- \), we can use the relationship between the hydrolysis constant (\( K_h \)), the ionization constant of water (\( K_w \)), and the acid dissociation constant (\( K_a \)) of the weak acid \( \text{HNO}_2 \).

### Step-by-Step Solution:

1. **Identify the given values:**
- \( K_a \) for \( \text{HNO}_2 = 4.5 \times 10^{-10} \)
- \( K_w = 1.0 \times 10^{-14} \) (the ionization constant of water at 25°C)

2. **Use the formula for hydrolysis constant:**
The hydrolysis constant \( K_h \) can be calculated using the formula:
\[ K_h = \frac{K_w}{K_a} \]

3. **Substitute the values into the formula:**
\[ K_h = \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-10}} \]

4. **Perform the calculation:**
\[ K_h = \frac{1.0}{4.5} \times 10^{-14 + 10} = \frac{1.0}{4.5} \times 10^{-4} \]
\[ K_h = 0.2222 \times 10^{-4} \]
\[ K_h = 2.222 \times 10^{-5} \]

5. **Final result:**
The hydrolysis constant \( K_h \) for the salt containing \( \text{NO}_2^- \) is:
\[ K_h \approx 2.22 \times 10^{-5} \]

### Conclusion:
The hydrolysis constant for the salt containing \( \text{NO}_2^- \) is \( 2.22 \times 10^{-5} \).