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)`
under same conditions ,how many mL of 0.1 m NaOH and 0.05 M `H_(2)A` (strong diprotic acid ) solution should be mixed for a total volume of 100mL to producce the hight rise in temperature ?

Options

A. `25 : 75`
B. `50 : 50`
C. `75 : 25`
D. `66.66 : 33.33`

Explanation

To solve the problem, we need to determine how many mL of 0.1 M NaOH and 0.05 M H₂A (a strong diprotic acid) should be mixed to achieve a total volume of 100 mL for maximum temperature rise during neutralization.

### Step-by-Step Solution:

1. **Understand the Neutralization Reaction:**
- For a diprotic acid (H₂A), it can donate two protons (H⁺) when it reacts with a base. The balanced reaction can be represented as:
\[ H_2A + 2OH^- \rightarrow A^{2-} + 2H_2O \]
- This means that 1 mole of H₂A reacts with 2 moles of OH⁻ from NaOH.

2. **Define Variables:**
- Let \( V \) be the volume of NaOH solution (in mL).
- The volume of H₂A solution will then be \( 100 - V \) mL since the total volume must be 100 mL.

3. **Calculate Moles of Reactants:**
- Moles of NaOH:
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times \frac{V}{1000} \, \text{L} = 0.1 \times \frac{V}{1000} \]
- Moles of H₂A:
\[ \text{Moles of H₂A} = 0.05 \, \text{mol/L} \times \frac{100 - V}{1000} \, \text{L} = 0.05 \times \frac{100 - V}{1000} \]

4. **Set Up the Neutralization Equation:**
- From the stoichiometry of the reaction, we know:
\[ \text{Moles of OH}^- = 2 \times \text{Moles of H₂A} \]
- Therefore, we can write:
\[ 0.1 \times \frac{V}{1000} = 2 \times \left(0.05 \times \frac{100 - V}{1000}\right) \]

5. **Simplify the Equation:**
- Cancel out the common factor of \( \frac{1}{1000} \):
\[ 0.1V = 2 \times 0.05(100 - V) \]
- Simplifying further gives:
\[ 0.1V = 0.1(100 - V) \]
- Expanding the right side:
\[ 0.1V = 10 - 0.1V \]

6. **Combine Like Terms:**
- Add \( 0.1V \) to both sides:
\[ 0.1V + 0.1V = 10 \]
\[ 0.2V = 10 \]

7. **Solve for V:**
- Divide both sides by 0.2:
\[ V = \frac{10}{0.2} = 50 \, \text{mL} \]

8. **Determine Volume of H₂A:**
- Since the total volume is 100 mL:
\[ \text{Volume of H₂A} = 100 - V = 100 - 50 = 50 \, \text{mL} \]

### Final Answer:
- The volumes required are:
- Volume of NaOH = 50 mL
- Volume of H₂A = 50 mL