Enthalpy of neutralzation 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)`
What is `DeltaH^(@)` for complate neutralization of strong diacidic base `A(OH)_(2)by HNO_(3)`?

Options

A. `-55.84` kJ/mol
B. `111.68` kJ/mol
C. 55.84 kJ/mol
D. None of these

Explanation

To find the enthalpy change (ΔH°) for the complete neutralization of a strong diacidic base \( A(OH)_2 \) by \( HNO_3 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between the strong diacidic base \( A(OH)_2 \) and nitric acid \( HNO_3 \) can be written as:
\[ A(OH)_2 + 2 HNO_3 \rightarrow A(NO_3)_2 + 2 H_2O \]
This equation shows that one mole of \( A(OH)_2 \) reacts with two moles of \( HNO_3 \) to produce one mole of \( A(NO_3)_2 \) and two moles of water.

### Step 2: Identify the net ionic reaction
The net ionic reaction for the neutralization can be simplified to:
\[ 2 H^+ (aq) + 2 OH^- (aq) \rightarrow 2 H_2O (l) \]
This indicates that two moles of \( H^+ \) ions from the acid react with two moles of \( OH^- \) ions from the base to form water.

### Step 3: Use the enthalpy of neutralization for strong acid and strong base
For the neutralization of a strong acid with a strong base, the enthalpy change (ΔH°) is given as:
\[ \Delta H_{neutralization} = -55.84 \, \text{kJ/mol} \]
This value represents the enthalpy change for the neutralization of one mole of \( H^+ \) with one mole of \( OH^- \).

### Step 4: Calculate the total enthalpy change for the reaction
Since the reaction involves two moles of \( H^+ \) and two moles of \( OH^- \), the total enthalpy change for the complete neutralization of the diacidic base will be:
\[ \Delta H° = 2 \times (-55.84 \, \text{kJ/mol}) = -111.68 \, \text{kJ} \]

### Conclusion
Thus, the enthalpy change for the complete neutralization of the strong diacidic base \( A(OH)_2 \) by \( HNO_3 \) is:
\[ \Delta H° = -111.68 \, \text{kJ} \]

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