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Enthalpy of neutralization of $H 3 P O 3$ acid is -106.68 kj / mol using NaOH.Enthalpy

of neutralization of HCl by NaOH is -55.84 kj/mol. Calculate $Δ H i o n i z a t i o n$ of $H 3 PO 3$ into its ions:

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50.84 kJ/mol

5 kJ/mol

2.5 kJ/mol

10 kJ/mol

answer is B.

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Detailed Solution

H 3 P O 3 → 2 H + + H P O 3 2 −; Δ r H = ?

2 H + + 2 O H − → 2 H 2 O

Δ r H = − 55.84 × 2 = − 111.68

− 106.68 = Δ i o n H − 55.84 × 2

Δ i o n H = 5 k J / m o l

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