When \(1\) g equivalent of a strong acid is neutralized with a strong base heat released is \(13.5\) kcal, when \(1\) g equivalent \(H_2A\) is completely neutralized against strong base \(13\) kcal is released, when \(1\) g equivalent \(B(OH)_2\) is completely neutralized against strong acid \(10\, \ kcal\) heat is released. Calculate enthalpy change when \(1 g\) mole \(H_2A\) is completely neutralized by \(B(OH)_2\).

-19 kcal

The correct answer is option 4. 19 kcal. 

Given Data:

Heat released when 1 gram equivalent of a strong acid is neutralized by a strong base: \(  \Delta H_{\text{neutralization}} = -13.5 \, \text{kcal}\)

Heat released when 1 gram equivalent of \( H_2A \) is neutralized by a strong base: \(\Delta H_{H_2A} = -13 \, \text{kcal}\)

Heat released when 1 gram equivalent of \( B(OH)_2 \) is neutralized by a strong acid: \(\Delta H_{B(OH)_2} = -10 \, \text{kcal}\)
Heat of Ionization Calculation:

For \( H_2A \):

Expected heat of neutralization if \( H_2A \) behaved like a strong acid: \(2 \times 13.5 \, \text{kcal} = 27 \, \text{kcal}\)

Actual heat released: \(26 \, \text{kcal} \quad \text{(as 2 gram equivalents of \( H_2A \) release 13 kcal each, for a total of 26 kcal)}\)

Heat of ionization of \( H_2A \): \(\Delta H_{\text{ionization, } H_2A} = 27 - 26 = 1 \, \text{kcal}\)

For \( B(OH)_2 \):

Expected heat of neutralization if \( B(OH)_2 \) behaved like a strong base: \(2 \times 13.5 \, \text{kcal} = 27 \, \text{kcal}\)

Actual heat released: \(20 \, \text{kcal} \quad \text{(as 2 gram equivalents of \( B(OH)_2 \) release 10 kcal each, for a total of 20 kcal)}\)

Heat of ionization of \( B(OH)_2 \): \(\Delta H_{\text{ionization, } B(OH)_2} = 27 - 20 = 7 \, \text{kcal}\)

Heat of Neutralization of \( H_2A \) and \( B(OH)_2 \):

Reaction: \(H_2A + B(OH)_2 \rightarrow 2H_2O + BA\)

Total heat of neutralization:

\(\Delta H_{\text{neutralization}} = -2 \times 13.5 \, \text{kcal} + 1 \, \text{kcal} + 7 \, \text{kcal}\)

Calculating step-by-step:

Heat released by strong acid and strong base: \(-2 \times 13.5 = -27 \, \text{kcal}\)

Adding the heats of ionization: \(1 \, \text{kcal} + 7 \, \text{kcal} = 8 \, \text{kcal}\)

Therefore, the total enthalpy change: \(\Delta H_{\text{neutralization}} = -27 \, \text{kcal} + 8 \, \text{kcal} = -19 \, \text{kcal}\)

Conclusion:

The enthalpy change when 1 mole of \( H_2A \) is completely neutralized by \( B(OH)_2 \) is:\(-19 \, \text{kcal}\)