The enthalpy of neutralization of `NH_(4)OH and CH_(3)COOH` is `-"10.5 kcal mol"^(-1)` and enthalpy of neutralisation of `NH_(4)OH` with a strong acid is `-"11.7 kcal mol"^(-1)`. The enthalpy of ionization of `CH_(3)COOH` will be

Options

A. `"4.0 kcal mol"^(-1)`
B. `"3.0 kcal mol"^(-1)`
C. `"1.2 kcal mol"^(-1)`
D. `"3.2 kcal mol"^(-1)`

Answer

To find the enthalpy of ionization of acetic acid (CH₃COOH), we can use the information provided about the enthalpy of neutralization of NH₄OH with CH₃COOH and with a strong acid.

### Step-by-Step Solution:

1. **Understanding Enthalpy of Neutralization**:
- The enthalpy of neutralization (ΔH_neutralization) is the heat change that occurs when an acid and a base react to form water and a salt. For strong acids and bases, this value is typically constant at around -57.1 kJ/mol (or -13.7 kcal/mol).
- For weak acids and bases, the enthalpy of neutralization will also include the energy required to ionize the weak acid and weak base.

2. **Given Data**:
- Enthalpy of neutralization of NH₄OH with CH₃COOH: ΔH_neutralization (NH₄OH + CH₃COOH) = -10.5 kcal/mol
- Enthalpy of neutralization of NH₄OH with a strong acid (e.g., HCl): ΔH_neutralization (NH₄OH + HCl) = -11.7 kcal/mol

3. **Setting Up the Equations**:
- For the neutralization of NH₄OH with CH₃COOH, we can express it as:
\[ ΔH_{neutralization} = ΔH_{ionization (NH₄OH)} + ΔH_{ionization (CH₃COOH)} + ΔH_{neutralization (water)} \]
- For the neutralization of NH₄OH with a strong acid:
\[ ΔH_{neutralization} = ΔH_{ionization (NH₄OH)} + 0 + ΔH_{neutralization (water)} \]

4. **Calculating the Ionization Energy**:
- From the second equation (NH₄OH + HCl):
\[ ΔH_{neutralization (NH₄OH + HCl)} = ΔH_{ionization (NH₄OH)} + ΔH_{neutralization (water)} \]
Here, we can assume ΔH_{ionization (NH₄OH)} is negligible for strong acids.
- Therefore, we can set:
\[ -11.7 = 0 + ΔH_{neutralization (water)} \]
- From the first equation (NH₄OH + CH₃COOH):
\[ -10.5 = ΔH_{ionization (NH₄OH)} + ΔH_{ionization (CH₃COOH)} + ΔH_{neutralization (water)} \]
- Since ΔH_{ionization (NH₄OH)} is negligible, we can simplify this to:
\[ -10.5 = ΔH_{ionization (CH₃COOH)} - 11.7 \]

5. **Solving for ΔH_{ionization (CH₃COOH)}**:
- Rearranging gives:
\[ ΔH_{ionization (CH₃COOH)} = -10.5 + 11.7 \]
\[ ΔH_{ionization (CH₃COOH)} = 1.2 \text{ kcal/mol} \]

### Final Answer:
The enthalpy of ionization of CH₃COOH is **1.2 kcal/mol**.