Question

The solubility product of calcium sulfate is 2.4x10^-5. What is the equilibrium concentration of Ca²⁺ in pure water? For the calcium sulfate in #4 above, what would the equilibrium concentration of Ca²⁺ be if the calcium sulfate was placed in a solution of 0.010 M Na₂SO₄?

Answer 1

1)

At equilibrium:
CaSO4 <----> Ca2+ + SO42-   

s s

Ksp = [Ca2+][SO42-]
2.4*10^-5=(s)*(s)
2.4*10^-5= 1(s)^2
s = 4.90*10^-3 M
Answer: 4.90*10^-3 M

2)

Na2SO4 here is Strong electrolyte
It will dissociate completely to give [SO42-] = 0.01 M = 1*10^-2 M
At equilibrium:
CaSO4 <----> Ca2+ + SO42-   
s 1*10^-2 + s
Ksp = [Ca2+][SO42-]
2.4*10^-5=(s)*(1*10^-2+ s)
Since Ksp is small, s can be ignored as compared to 1*10^-2
Above expression thus becomes:
2.4*10^-5=(s)*(1*10^-2)
2.4*10^-5= (s) * 10^-2
s = 2.4*10^-3 M
Answer: 2.4*10^-3 M