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Question

The heat of neutralization of LiOH and HCl at 25C is 34.868KJmol1. The heat ionization of LiOH will be:

A
44.674kJ
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B
22.232kJ
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C
32.684kJ
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D
96.464kJ
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Solution

The correct option is B 22.232kJ
Solution:- (B) 22.232kJ/mol
NaOH+HClNaCl+H2OΔH=57.1kJ/mol

As we know that heat of neutralization of a strong acid by a strong base is heat of formation of water.
Therefore,
H++OHH2OΔH=57.1kJ/mol
H2OH++OHΔH=57.1kJ/mol.....(1)

Given that heat of neutralisation of LiOH and HCl is 34.868kJ/mol
Therefore,
H++LiOHLi++H2OΔH=34.868kJ/mol.....(2)

Adding eqn(1)&(2), we have
H2O+LiOH+H+H++OH+Li++H2O;ΔH=34.868+57.1
LiOHLi++OH;ΔH=22.232kJ/mol

Hence the heat of ionisation of LiOH is 22.232kJ/mol.

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