Question

calculate the molar solubility of silver chloride (ksp= 1.6x10^-6) in 0.10M FeCl3

Answer 1

FeCl3 here is Strong electrolyte

It will dissociate completely to give [Cl-] = 0.3 M

At equilibrium:

AgCl <----> Ag+ + Cl-

   s 0.3 + s

Ksp = [Ag+][Cl-]

1.6*10^-6=(s)*(0.3+ s)

Since Ksp is small, s can be ignored as compared to 0.3

Above expression thus becomes:

1.6*10^-6=(s)*(0.3)

1.6*10^-6= (s) * 0.3

s = 5.333*10^-6 M

Answer: 5.3*10^-6 M