Question

± Introduction to Solubility and the Solubility Product Constant Learning Goal: To learn how to calculate the solubility from Kspand vice versa. Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated solution: CaF2(s)⇌Ca2+(aq)+2F−(aq) At equilibrium, the ion concentrations remain constant because the rate of dissolution of solid CaF2 equals the rate of the ion crystallization. The equilibrium constant for the dissolution reaction is Ksp=[Ca2+][F−]2 Ksp is called the solubility product and can be determined experimentally by measuring thesolubility, which is the amount of compound that dissolves per unit volume of saturated solution.

Part A A saturated solution of lead(II) chloride, PbCl2, was prepared by dissolving solid PbCl2 in water. The concentration of Pb2+ ion in the solution was found to be 1.62×10−2M . Calculate Ksp for PbCl2. Express your answer numerically. Ksp = SubmitHintsMy AnswersGive UpReview Part Part B The value of Ksp for silver sulfide, Ag2S, is 8.00×10−51. Calculate the solubility of Ag2S in grams per liter. Express your answer numerically in grams per liter.

Answer 1

Part A

PbCl₂ dissociate as

PbCl₂ Pb²⁺ + 2Cl^-

Ksp = [Pb²⁺] [Cl^-]²

[Pb²⁺] = 1.62 10^-2 then [Cl^-] = 1.62 10^-2 2 = 0.0324

Ksp = [Pb²⁺] [Cl^-]²

Ksp = (1.62 10^-2) (0.0324)² = 1.7 10^-5 = 0.000017

Part B

Ksp = [Ag+]²[S^--]

8 10^-51  =  [Ag+]²[S^--]

let, x = molar solubility of Ag₂S

8 10^-51  = 2(x)² (x)

8 10^-51  = 4x³

x³ = 8 10^-51 / 4

x³ = 2 10^-51

x = 1.2599 10^-17 = solubility in molarity

molar mass of Ag₂S =247.796 gm/mol

so, solubility in gm per liter = (1.2599 10^-17 ) (247.796) = 3.121 10^-15