Question

At $25^\circ C$ the enthalpy change, for the ionization of trichloroacetic acid is $+6.3 \, kJ \, mol^{- 1}$ and the entropy change, is $+0.0084 \, kJ \, mol^{- 1} \, K^{- 1}$ . Then $pKa$ of tricholoro acetic acid is

• A. $1.74$
• B. $2.52$
• C. $0.66$
• D. $4.72$

Answer 1

Answer: (C)

$\Delta G^\circ =\Delta H^\circ -T\Delta S^\circ $

$\Delta G^\circ =6.3-\left(\right.298\times 0.0084\left.\right)=3.8 \, KJ.mol^{- 1}$

$\therefore\left(\Delta \mathrm{G}^O=2.303 \quad \mathrm{RT} \ln \mathrm{K}\right)$
$logKeq=logKa=-pKa=\frac{- \Delta G ^\circ }{2.303 R T}$

$-pKa=\frac{- \Delta G ^\circ }{2.303 R T}=\frac{3.8}{2.303 \times 0.008314 \times 298}$

$=0.66$