Answer:
[Ca²⁺] = 0,00365M
[SO₄²⁻] = 0,00865M
Explanation:
The reaction:
CaSO₄(s) ⇄ Ca²⁺ + SO₄²⁻ has a Ksp of 10^-4.5 = 3,16x10⁻⁵ = [Ca²⁺] [SO₄²⁻] (1)
Concentrations are
[Ca²⁺] = X
[SO₄²⁻] = X
The addition of 5.00x10⁻³M of ions of SO₄⁻ gives a concentration of:
[Ca²⁺] = X
[SO₄²⁻] = (X + 0,005M)
Replacing in (1):
3,16x10⁻⁵ = X (X + 0,005M)
3,16x10⁻⁵ = X² + 0,005X
0 = X² + 0,005X - 3,16x10⁻⁵
Solving for X:
X = -0,00865M
X = 0,00365M
Thus, concentrations in equilibrium are:
[Ca²⁺] = 0,00365M
[SO₄²⁻] = 0,00365M + 0,005M = 0,00865M
I hope it helps!