Enthalpy of neutralization is defined as the enthalpy change when 1 mole of acid`/ /`base is completely neutralized by base `//`acid in dilute solution .
For Strong acid and strong base neutralization net chemical change is
`H^(+) (aq)+OH^(-)(aq)to H_(2)O(l)`
`Delta_(r)H^(@)=-55.84KJ//mol`
`DeltaH_("ionization")^(@)` of aqueous solution of strong acid and strong base is zero .
when a dilute solution of weak acid or base is neutralized, the enthalpy of neutralization is somewhat less because of the absorption of heat in the ionzation of the because of the absorotion of heat in the ionization of the weak acid or base ,for weak acid /base
`DeltaH_("neutrlzation")^(@)=DeltaH_("ionization")^(@)+ Delta _(r)H^(@)(H^(+)+OH^(-)to H_(2)O)`
If enthalpy of neutralization of `CH_(3)COOH` by NaOH is -49.86KJ`//`mol then enthalpy of ionization of `CH_(3)COOH` is:
(a)5.98 kJ/mol
(b)`-5.98` kJ/mol
(c)105.7 kJ/mol
(d)None of these

Options

A. 5.98 kJ/mol
B. `-5.98` kJ/mol
C. 105.7 kJ/mol
D. None of these

Explanation

To find the enthalpy of ionization of acetic acid (CH₃COOH) when neutralized by sodium hydroxide (NaOH), we can use the given relationship for the enthalpy of neutralization:

\[ \Delta H_{\text{neutralization}}^{\circ} = \Delta H_{\text{ionization}}^{\circ} + \Delta H_{r}^{\circ} \]

Where:
- \(\Delta H_{\text{neutralization}}^{\circ}\) is the enthalpy of neutralization,
- \(\Delta H_{\text{ionization}}^{\circ}\) is the enthalpy of ionization of the weak acid,
- \(\Delta H_{r}^{\circ}\) is the enthalpy change for the reaction of a strong acid with a strong base.

### Step 1: Identify the given values
From the question, we know:
- \(\Delta H_{\text{neutralization}}^{\circ} = -49.86 \, \text{kJ/mol}\)
- \(\Delta H_{r}^{\circ} = -55.84 \, \text{kJ/mol}\)

### Step 2: Substitute the values into the equation
We can rearrange the equation to solve for \(\Delta H_{\text{ionization}}^{\circ}\):

\[ \Delta H_{\text{ionization}}^{\circ} = \Delta H_{\text{neutralization}}^{\circ} - \Delta H_{r}^{\circ} \]

Substituting the known values:

\[ \Delta H_{\text{ionization}}^{\circ} = -49.86 \, \text{kJ/mol} - (-55.84 \, \text{kJ/mol}) \]

### Step 3: Calculate \(\Delta H_{\text{ionization}}^{\circ}\)
Now, perform the calculation:

\[ \Delta H_{\text{ionization}}^{\circ} = -49.86 + 55.84 \]

\[ \Delta H_{\text{ionization}}^{\circ} = 5.98 \, \text{kJ/mol} \]

### Conclusion
The enthalpy of ionization of acetic acid (CH₃COOH) is:

\[ \Delta H_{\text{ionization}}^{\circ} = 5.98 \, \text{kJ/mol} \]

### Answer
(a) 5.98 kJ/mol

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