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Question

Given Δ Ho (HCN)= 45.2 kJ mol1
Δ Ho(CH3COOH)= 2.1 kJ mol1
Which one of the following options is true?

A
pKa (HCN)= pKa(CH3COOH)
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B
pKa (HCN)> pKa(CH3COOH)
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C
pKa (HCN)< pKa(CH3COOH)
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D
pKa (HCN)= (45.17/2.07) pKa(CH3COOH)
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Solution

The correct option is A pKa (HCN)> pKa(CH3COOH)
Delta H0 (HCN)=45.2 KJ/mol
Delta H0(CH3COOH)=2.1 KJ/mol
HCN has greater heat of ionisation, and is a weak acid. Thus, it has high pKa value.
pKa(HCN)>pKa(CH3COOH)

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