The heat of neutralization of NaOH with HCl is 57.3 KJ and with HCN is 12.1 KJ. The heat of ionization of HCN is

Options

A. `+69.4 ` KJ
B. `+45.2` KJ
C. `-45.2` KJ
D. `-69.4` KJ

Answer

To solve the problem, we need to find the heat of ionization of HCN given the heat of neutralization of NaOH with HCl and HCN.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Heat of neutralization of NaOH with HCl (ΔH_NaOH + HCl) = 57.3 kJ
- Heat of neutralization of NaOH with HCN (ΔH_NaOH + HCN) = 12.1 kJ

2. **Understanding the Reaction:**
- When NaOH neutralizes HCl, it releases a certain amount of heat (57.3 kJ). This is the complete neutralization of a strong acid with a strong base.
- When NaOH neutralizes HCN, the heat released is less (12.1 kJ) because HCN is a weak acid and partially ionizes in solution.

3. **Using the Heat of Neutralization Equation:**
- The heat of ionization of HCN can be expressed as:
\[ \text{Heat of Ionization of HCN} + \text{Heat of Neutralization of NaOH with HCN} = \text{Heat of Neutralization of NaOH with HCl} \]

4. **Rearranging the Equation:**
- We can rearrange the equation to solve for the heat of ionization of HCN:
\[ \text{Heat of Ionization of HCN} = \text{Heat of Neutralization of NaOH with HCl} - \text{Heat of Neutralization of NaOH with HCN} \]

5. **Substituting the Values:**
- Now, substitute the known values into the equation:
\[ \text{Heat of Ionization of HCN} = 57.3 \, \text{kJ} - 12.1 \, \text{kJ} \]

6. **Calculating the Result:**
- Perform the subtraction:
\[ \text{Heat of Ionization of HCN} = 57.3 - 12.1 = 45.2 \, \text{kJ} \]

### Final Answer:
The heat of ionization of HCN is **45.2 kJ**.