Assuming the final concentration of chloride anion after the addition of HCl (precipitation step) was 0.1M, what is the remaining concentration of Ag+ in the solution? (pKsp for AgCl = 9.74)

Answer 1

pKsp = 9.74 = -log Ksp
Ksp = 1.82E-10
.............AgCl ==> Ag^+ + Cl^-
I............solid....0.......0
C............solid....x.......x
E............solid....x...... 0.1

The 0.1 for Cl^- comes from the problem

Ksp AgCl =1.82E-10 = (Ag^+)(Cl^-)
1.82E-10 = (x)(0.1)
Solve for x = (Ag^+)