The Ksp of SrSO4 is 3.2 × 10–7. What is the equilibrium concentration of sulfate ion in a 1.0-L solution of strontium sulfate to which 0.10 mol of Sr(CH3CO2)2 has been added?

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

c (diamond),

sio2 (quartz)