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 $-33.7$ $kJmol^{- 1}$ then the $\left(\mathrm{pK}_{\mathrm{a}}\right)_1$ of $\mathrm{H}_2 \mathrm{~S}$ is

• A. $\left(\frac{23.6 \times 1 0^{3} - T \Delta S ^\circ }{2.303 R T}\right)$
• B. $\left(\frac{2.303 R T}{23.6 - T \Delta S ^\circ }\right)$
• C. $\left(\frac{T \Delta S ^\circ - 23.6}{R T}\right)$
• D. $2.303\left(\frac{T \Delta S ^\circ - 23.6}{R T}\right)$

Answer 1

Answer: (A)

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

$\Delta H_{i o n}=23.6$ $kJ$

$\therefore \Delta G^\circ =23.6\times 10^{3}-T\Delta S^\circ =-2.303RTlogK_{a_{1}}$

$\therefore P_{K a_{1}}=\frac{23.6 \times 1 0^{3} - T \Delta S ^\circ }{2.303 R T}$