Answer:
3.2x10⁻⁶M of sulfate ion can be added
Explanation:
The solubility of SrSO4 is:
SrSO4(s) → Sr²⁺(aq) + SO₄²⁻(aq)
Where Ksp is defined as:
Ksp = 3.2x10⁻⁷ = [Sr²⁺] [SO₄²⁻]
The concentration of Sr is:
0.10mol / 1L = 0.10M = [Sr²⁺]
Replacing in Ksp:
3.2x10⁻⁷ = [0.10M] [SO₄²⁻]
[SO₄²⁻] = 3.2x10⁻⁶M of sulfate ion can be added
