Question

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

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

Answer 1

Answer: (C)

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

$\Delta G^{\circ }=6.3-(298\times 0.0084)=3.8KJ.mo{l}^{-1}$

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

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

$=0.66$