Use the following information to calculate the solubility product constant, K_sp, for PbCl₂. A saturated solution of PbCl₂ in water was prepared and filtered. From the filtrate, 1.0 L was measured out into a beaker and evaporated to dryness. The solid PbCl₂ residue recovered in the beaker amounted to 0.0162 moles. A) K_sp = 6.9 *10¯⁸ B) K_sp = 4.3 * 10¯⁶ C) K_sp = 1.7 * 10¯⁵ D) K_sp = 2.6 *10¯⁴ E) K_sp = 3.2 *10¯²

Question

Quizplus · Accepted Answer

The balanced chemical equation for the dissolution of PbCl2 in water is:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)

The equation for the solubility product constant expression is:
Ksp = [Pb2+][Cl-]^2

To determine Ksp, we need to find the concentrations of Pb2+ and Cl- in the saturated solution. From the given information, we know that the amount of PbCl2 residue recovered in the beaker is 0.0162 moles. Since PbCl2 dissociates into one Pb2+ ion and two Cl- ions, the amount of Pb2+ in solution is 0.0162 moles and the amount of Cl- is 0.0324 moles.

The volume of the solution is 1.0 L, so the concentration of Pb2+ and Cl- ions in the saturated solution is:

[Pb2+] = 0.0162 mol/L
[Cl-] = 0.0324 mol/L

Substituting the values into the Ksp expression, we get:

Ksp = [Pb2+][Cl-]^2
Ksp = (0.0162 mol/L)(0.0324 mol/L)^2
Ksp = 1.72 x 10^-15 mol^3/L^3

Rounding off to two significant figures and using scientific notation, we get:

Ksp = 1.7 x 10^-15

Therefore, the best choice is C, with the correct value of Ksp.

Use the Following Information to Calculate the Solubility Product Constant