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Enthalpy of neutralization of $H_{3} {PO}_{3}$ with $NaOH$ is $- 106.68 kJ / mol$ . If enthalpy of neutralization of $HCl$ with $NaOH$ is $- 55.84$ $kJ/mole$ , then calculate enthalpy of ionization of $H_{3} {PO}_{3}$ into its ions in kJ..

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answer is 5.

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Detailed Solution

$Δ_{neut.} H = Δ_{ion} H - 2 \times 55.84$

$Δ_{ion} H = 5 kJ$

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