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Q.

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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a

50.84 kJ/mol

b

5000 J/mol

c

2.5 kJ/mol

d

5 kJ/mol

answer is B, D.

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Detailed Solution

H3PO3+2NaOH→Na2HPO3+2H2O

∆H=-106.68kJ

For 2 moles of water,

∆HNeutral=2x(-55.84kJ)=-111.68kJ

∆HIonization, H3, PO3=-111.68kJ-(-106.68Kj)=5kJ

= 5000 J/mol

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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