Calculate the molar solubility of Ag2SO4 (Ksp = 1.5 x 10 ^-5)
a) in pure water
b) in 0.22 M Na2SO4
 

Answer:

a)

At equilibrium:

Ag2SO4 <----> 2 Ag+ + SO42-

   2s s

Ksp = [Ag+]^2[SO42-]

1.5*10^-5=(2s)^2*(s)

1.5*10^-5= 4(s)^3

s = 1.554*10^-2 M

Answer: 1.55*10^-2 M

b)

Na2SO4 here is Strong electrolyte

It will dissociate completely to give [SO42-] = 0.22 M

At equilibrium:

Ag2SO4 <----> 2 Ag+ + SO42-

   2s 0.22 + s

Ksp = [Ag+]^2[SO42-]

1.5*10^-5=(2s)^2*(0.22+ s)

Since Ksp is small, s can be ignored as compared to 0.22

Above expression thus becomes:

1.5*10^-5=(2s)^2*(0.22)

1.5*10^-5= 4(s)^2 * 0.22

s = 4.129*10^-3 M

Answer: 4.13*10^-3 M