Q. 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 $\left(\left(\text{pK}\right)_{\text{a}}\right)_{1}$ of $H_{2}S$ is
Solution:
$H_{2}S\rightleftharpoonsH^{+}+HS^{-}$
$\Delta H_{i o n}=23.6$ $kJ$
$\therefore \Delta G^\circ =23.6\times 10^{3}-T\Delta S^\circ =-2.303RTlogK_{a_{1}}$
$\therefore P_{K a_{1}}=\frac{23.6 \times 1 0^{3} - T \Delta S ^\circ }{2.303 R T}$
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