Question

For the following reaction
$Ag_{\left(a q\right)}^{+}+Cl_{\left(a q\right)}^{-} \rightarrow AgCl_{\left(s\right)}$
Given: $\Delta G_{f}^{o}$ , AgCl = - 112.44 kJ/mol; $\Delta G_{f}^{o}$ $C l^{-}$ = - 130 kJ/mol ; $\Delta G_{f}^{o}Ag^{+}$ = 75 kJ/mol
Report your answer by rounding it upto nearest whole number. The $K_{s p}$ of AgCl is $n\times 10^{- 10}$ .The value of 'n' is.

Answer 1

$Ag_{\left(a q\right)}^{+}+Cl_{\left(a q\right)}^{-} \rightarrow AgCl_{\left(s\right)}$
$\Delta G_{f}^{o}$ , AgCl = - 112.44 kJ/mol; $\Delta G_{f}^{o}C l^{-}$ = - 130 kJ/mol ; $\Delta G_{f}^{o}Ag^{+}$ = 75 kJ/mol
$\Delta G_{r e a c t i o n}^{o}=\Delta G_{f}^{o}$ , AgCl- [ $\Delta G_{f}^{o}Ag^{+}+\Delta G_{f}^{o}C l^{-}$ ]
= - 112.44 kJ-[75 kJ- 130 kJ]
= -57.44 kJ
-2.303RTlog $K_{s p}=\Delta G_{r e a c t i o n}^{o}$
$log K_{s p}=\frac{57440}{2.303 \times 8.314 \times 298}\cong10$
$K_{s p}=1\times 10^{- 10}$
Hence n = 1