Solubility Equilibria

56) (a) [Ba²⁺] = 0.0150 M          K_sp(BaSO₄) = 1.1 x 10^-10

            [Sr²⁺] = 0.0150 M          K_sp(SrSO₄) = 3.2 x 10^-7

            BaSO₄(s) Ba²⁺(aq) + SO₄^2-(aq)

            K_sp = [Ba²⁺] [SO₄^2-]

            1.1 x 10^-10 = 0.0150 x [SO₄^2-]

            [SO₄^2-] = 7.3 x 10^-9 M

            SrSO₄(s) Sr²⁺(aq) + SO₄^2-(aq)

            K_sp = [Sr²⁺] [SO₄^2-]

            3.2 x 10^-7 = 0.0150 x [SO₄^2-]

            [SO₄^2-] = 2.1 x 10^-5 M

            Precipitation begins when Q = K_sp and when [SO₄^2-] = 7.3 x 10^-9 M because it is the smaller of the two values.

      (b) Ba²⁺ precipitates first because it requires the least SO₄^2-.

      (c) Sr²⁺ begins to precipitate when [SO₄^2-] = 2.1 x 10^-5 M.