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Question: Given \(\Delta\)ioniz H<sup>0</sup> (HCN) = 45.2 kJ mol–1 and \(\Delta\)ioniz H<sup>0</sup> (CH3COOH...

Given Δ\Deltaioniz H0 (HCN) = 45.2 kJ mol–1 and Δ\Deltaioniz H0 (CH3COOH) = 2.1 kJ mol–1. Which one of the following facts is true ?

A

pKa (HCN) = pKa (CH3COOH)

B

pKa (HCN) >pKa (CH3COOH)

C

pKa (HCN) <pKa (CH3COOH)

D

pKa (HCN) = (45.17/2.07) pKa (CH3COOH)

Answer

pKa (HCN) >pKa (CH3COOH)

Explanation

Solution

Larger the value of Δ\DeltaionizH0, weaker the acid and larger the value of pKa. Hence, pKa (HCN) > pKa