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 (K_h) of the salt containing the nitrite ion (NO₂⁻), we will use the relationship between the hydrolysis constant, the dissociation constant of the weak acid (K_a), and the ion product of water (K_w).

### Step-by-Step Solution:

1. **Identify the given values**:
- The dissociation constant of HNO₂ (weak acid) is given as:
\[ K_a = 4.5 \times 10^{-10} \]
- The ion product of water at 25°C is:
\[ K_w = 1.0 \times 10^{-14} \]

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

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

4. **Perform the division**:
\[ K_h = \frac{1.0}{4.5} \times 10^{-14 + 10} = \frac{1.0}{4.5} \times 10^{-4} \]
\[ K_h \approx 0.222 \times 10^{-4} = 2.22 \times 10^{-5} \]

5. **Final result**:
The hydrolysis constant (K_h) for the salt containing NO₂⁻ is:
\[ K_h \approx 2.22 \times 10^{-5} \]