Question

The heat of neutralization of oxalic acid is $-25.4 \, kcal \, mol ^{-1}$ using strong base, $NaOH$. Hence, the enthalpy change of the process is $H _2 C _2 O _4 \rightleftharpoons 2 H ^{\oplus}+ C _2 O _4{ }^{2-}$ is

• A. $2.0 \, kcal$
• B. $-11.8\, kcal$
• C. $1.0 \, kcal$
• D. $-1.0\, kcal$

Answer 1

Answer: (A)

Oxalic acid has two ionisableH ${ }^{\oplus}$. Hence, expected heat of neutralization, if it behaves as a strong acid would have been
$=-13.7 \times 2=-27.4\, kcal \,mol ^{-1}$
But experimental value $=-25.4 \,kcal\,mol ^{-1}$
$\therefore$ Heat of ionization $=2.0\, kcal\, mol -1$