Question

The standard enthalpy of neutralization of strong acid and strong base is $-57.3$ $kJ$ ${\text{equiv}}^{-1}.$ If the enthalpy of neutralization of the first proton of aqueous $H_{2}S$ is $-33.7$ $kJmo{l}^{-1}$ then the ${\left({\mathrm{pK}}_{\mathrm{a}}\right)}_1$ of ${\mathrm{H}}_2\mathrm{S}$ is

• A. $\left(\frac{23.6\times 10^3-T\Delta S^{\circ }}{2.303RT}\right)$
• B. $\left(\frac{2.303RT}{23.6-T\Delta S^{\circ }}\right)$
• C. $\left(\frac{T\Delta S^{\circ }-23.6}{RT}\right)$
• D. $2.303\left(\frac{T\Delta S^{\circ }-23.6}{RT}\right)$

Answer 1

Answer: (A)

$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}$

$\Delta H_{ion}=23.6$ $kJ$

$\therefore \Delta G^{\circ }=23.6\times 10^3-T\Delta S^{\circ }=-2.303RTlog{K}_{a_1}$

$\therefore P_{K{a}_1}=\frac{23.6\times 10^3-T\Delta S^{\circ }}{2.303RT}$