Determine the molar solubility of lead(II) iodide, PbI2, in 0.050 mol/L NaI.

Ksp of PbI2 = 9.8 x 10^(-9)

To determine the molar solubility of lead(II) iodide, PbI2, in 0.050 mol/L NaI, we can use the concept of the common ion effect.

The common ion effect states that the solubility of a compound is reduced in the presence of a common ion from another compound, due to Le Chatelier's principle.

In this case, NaI is a source of iodide ion (I-) and would contribute to the concentration of iodide ions in solution.

To find the molar solubility of PbI2 in the presence of NaI, we need to first identify the equilibrium expression for the dissolution of PbI2.

The balanced equation for the dissolution of PbI2 is:
PbI2 (s) ⇌ Pb2+ (aq) + 2I- (aq)

The equilibrium expression for this reaction is:
Ksp = [Pb2+][I-]^2

Given that the value of Ksp for PbI2 is 9.8 x 10^(-9), we can set up an ICE (Initial, Change, and Equilibrium) table to determine the concentration of Pb2+ and I- ions in solution.

Let's assume the molar solubility of PbI2 is "x" mol/L. Since the stoichiometry between PbI2 and Pb2+ is 1:1, the concentration of Pb2+ ions produced will also be "x" mol/L.

In the presence of NaI, the concentration of I- ions will be equal to the initial concentration of NaI, which is 0.050 mol/L.

Therefore, the equilibrium concentrations in the ICE table are:
[Pb2+] = x mol/L
[I-] = 0.050 mol/L

Now, using the equilibrium expression for Ksp, we can substitute the equilibrium concentrations into the expression:
Ksp = [Pb2+][I-]^2

9.8 x 10^(-9) = (x)(0.050)^2
9.8 x 10^(-9) = 0.0025x^1

To solve for "x," we rearrange the equation and take the square root of both sides:
x = sqrt((9.8 x 10^(-9))/0.0025)
x = 1.76 x 10^(-3) mol/L

Therefore, the molar solubility of lead(II) iodide, PbI2, in 0.050 mol/L NaI is 1.76 x 10^(-3) mol/L.