Calculate the solubility product:
Problem:
Calculate the solubility product of AgCl from the data
Ag++ le → Ag(s) E0Ag+,Ag = +0.799V
AgCl(s) + le → Ag (s) + Cl- E0AgCl,Ag = +0.222V
Answer:
cell reaction can be written as:
AgCl(s) +le→ Ag(s) +Cl
Ag(s) → Ag (s) +le
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AgCl (s) =→ Ag+ +Cl-
Log Ksp = 1[0.222-(+0.799)]/ 0.0591
Ksp = 1.82× 10-10