Question

The emf of the cell
Ag(s) | AgCI(satd), KCI(0.05 mol dm-3)/AgNO3, (01 mol dm3) I Ag(s)
is 0. 31 V at 298.15 K. The mean activity coefficient of KCl is 0.817 and that ofAgNO3, is 0.723. Calculate
the solubility product ofAgCl at 25℃.

Answer 1

Answer on Question #83960 - Chemistry - Physical Chemistry

Question:

The emf of the cell

Ag(s) | AgCl(satd), KCl(0.05 mol dm-3)/AgNO3, (01 mol dm3) | Ag(s)

is 0.31 V at 298.15 K. The mean activity coefficient of KCl is 0.817 and that of AgNO3, is 0.723. Calculate

the solubility product of AgCl at 25°C.

Solution:

Ksp = [Ag+] = [Cl-]

a = α*c;

c = a/α;

c(KCl) = 0.05/0.817 = 0,041 mol/L;

c(AgNO3) = 0.01/0.723 = 0,007 mol/L;

Ksp = 0.007.

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