Enthalpy of neutralization of $NaOH$ with $H_{2} {SO}_{4}$ is $- 57.3 kJ {eq}^{- 1}$ and with ethanoic acid is $- 55.2 kJ {eq}^{- 1}$ . Which of the following is the best explanation of this difference?

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Ethanoic acid is monobasic while $H_{2} {SO}_{4}$ is dibasic.

Ethanoic acid is a weak acid and thus requires less $NaOH$ for neutralization.

Ethanoic acid is only partly ionized I neutralization is therefore incomplete

Some heat is used to ionize ethanoic acid completely

answer is D.

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Detailed Solution

Enthalpy of neutralization for $NaOH = - 57.3 kJ {eq}^{- 1}$

Enthalpy of neutralization for $H_{2} {SO}_{4} = - 55.2 kJ {eq}^{- 1}$

Some heat is used to ionize ethanoic acid completely .

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