Enthalpy of neutralization of H₃PO₃ acid is -106.68kj/mol using NaOH. If enthalpy of neutralization of HCl by NaOH is-55.84kj/mol. Calculate $∆_{ionixzation}$ of H₃PO₃ into its ions

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50.84 kJ/mol

5000 J/mol

2.5 kJ/mol

5 kJ/mol

answer is B, D.

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Detailed Solution

$H_{3} {PO}_{3} + 2 NaOH \to {Na}_{2} {HPO}_{3} + 2 H_{2} O$

$∆ H = - 106.68 kJ$

$For 2 moles of water,$

$∆ H_{Neutral} = 2 x (- 55.84 kJ) = - 111.68 kJ$

$∆ H_{Ionization}, H_{3}, {PO}_{3} = - 111.68 kJ - (- 106.68 Kj) = 5 kJ$

= 5000 J/mol

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