While performing a thermodynamics experiment, a student made the following observations:
$$HCl + NaOH \rightarrow NaCl + H_2O \quad \Delta H = -57.3 \text{ kJ mol}^{-1}$$
$$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \quad \Delta H = -55.3 \text{ kJ mol}^{-1}$$
The enthalpy of ionization of $$CH_3COOH$$ as calculated by the student is ______ $$\text{kJ mol}^{-1}$$.

We need to find the enthalpy of ionization of acetic acid using the given data.

$$HCl + NaOH \rightarrow NaCl + H_2O \quad \Delta H_1 = -57.3 \text{ kJ/mol}$$

$$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \quad \Delta H_2 = -55.3 \text{ kJ/mol}$$

Understand the reactions

The first reaction is the neutralization of a strong acid (HCl) with a strong base (NaOH). The enthalpy of this reaction represents the enthalpy of neutralization of a strong acid-strong base pair, which is essentially the enthalpy of formation of water from $$H^+$$ and $$OH^-$$ ions.

The second reaction involves a weak acid ($$CH_3COOH$$) with a strong base. In this case, the weak acid must first ionize before neutralization can occur.

Apply Hess's law

The neutralization of a weak acid can be broken into two steps:

(i) Ionization of $$CH_3COOH$$: $$\Delta H_{ionization}$$

(ii) Neutralization of $$H^+$$ and $$OH^-$$: $$\Delta H_{neutralization} = -57.3 \text{ kJ/mol}$$

$$\Delta H_2 = \Delta H_{ionization} + \Delta H_{neutralization}$$

Calculate enthalpy of ionization

$$\Delta H_{ionization} = \Delta H_2 - \Delta H_{neutralization}$$

$$\Delta H_{ionization} = -55.3 - (-57.3)$$

$$\Delta H_{ionization} = -55.3 + 57.3 = 2.0 \text{ kJ/mol}$$

The answer is 2.