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.

Question

Quizplus · Accepted Answer

To solve for the solubility (S) of PbS, we use the expression:

KS = [Pb2+][S2-]

where KS is the solubility product constant, [Pb2+] is the molar concentration of Pb2+ ions, and [S2-] is the molar concentration of S2- ions. We can express [S2-] in terms of the concentration of Na2S solution:

[S2-] = [Na2S] - [S2O3 2-]

where [S2O3 2-] is the molar concentration of thiosulfate ions in solution. We can assume that the concentration of thiosulfate ions is negligible, since thiosulfate is not part of the solubility equilibrium equation for PbS. Therefore:

[S2-] = [Na2S]

Substituting into the solubility product expression and solving for S, we get:

KS = [Pb2+][Na2S]

S = KS / [Na2S]

Since the solubility product constant (KS) and the ionic strength of the solution ([Na2S]) are given, we can plug in the values and solve for S:

S = 1.3E-14 / (0.0010 M)

S = 1.3E-11 M

Therefore, the solubility of PbS in a 0.0010 M solution of Na2S is 1.3E-11 M, which is equivalent to choice C.

The Ksp Value for PbSO4(s) is 1

The K_sp Value for PbSO₄_(s)is 1