Question

Given that standard free energies of formation of Ag+(aq),Cl-(aq),and AgCl(s) are 77.1kJ/mol,-131.2kJ/mol and q-109.9kJ/mol respectively calculate solubility product Ksp for AgCl(s).

Answer 1

$ΔG=ΔG(AgCl(s))−[ΔG(Ag ^+ )+ΔG(Cl ^− )]\\=−109kJ/mol−[77kJ/mol−129kJ/mol]=−57kJ/mol$

$ΔG=−nFE$

on putting values of $\Delta G$ , n and F we get

$E _ 0 =0.222V$

$E=E_ 0 − \frac{ 0.0592}n ​ log \frac1{ [Ag + ][Cl − ] ​ }$

$0.59V=0.222V− \frac{ 0.0592}1 ​ log \frac1{ K _{ sp} ​ ​ }$

$log\frac1{ K _{sp}} ​ ​ =6.2276$

$K_ {s p} ​ =5.9×10 ^{ −7 }$