The K_sp value for PbSO₄(s) is 1.3 × 10^-8. Calculate the solubility, in moles per liter, of PbSO₄(s) in a 0.0010 M solution of Na₂SO₄. A) 4.5 × 10^-6 M B) 1.3 × 10^-11 M C) 1.3 × 10^-5 M D) 1.3 × 10^-8 M E) 1.4 × 10^-4 M

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Quizplus · Accepted Answer

The equilibrium equation for the reaction is: PbSO₄(s) ⇌ Pb₂SO₄(s) Ksp = [Pb2+][S2-]^2 Let x be the solubility of PbSO₄(s), then the concentration of Pb2+ and S2- in the solution will be equal to x. Thus, Ksp = x * x^2 = x^3 Substitute the given Ksp value, Ksp = 1.3 × 10^-8, into the expression above and solve for x: 1.3 × 10^-8 = x^3 x = (1.3 × 10^-8)^(1/3) = 1.1 × 10^-5 We need to calculate the solubility of PbSO₄(s) in a 0.0010 M solution of Na₂SO₄, which means the concentration of S2- in the solution is 0.0010 M. Since the solubility of PbSO₄(s) is much smaller than 0.0010 M, we can assume that the concentration of S2- in the solution is still 0.0010 M after the precipitation of PbSO₄(s). Thus, the reaction quotient Q is: Q = [Pb2+][S2-]^2 = x * (0.0010)^2 = (1.1 × 10^-5) * (0.0010)^2 = 1.1 × 10^-11 Since Q < Ksp, PbSO₄(s) will precipitate until the concentration of Pb2+ and S2- in the solution reach the solubility product. Therefore, the solubility of PbSO₄(s) in the 0.0010 M Na₂SO₄ solution is 1.1 × 10^-5 M. The closest option to this answer is C.

The Ksp Value for PbSO4(s) Is 1

The K_sp Value for PbSO₄(s) Is 1