The solubility of lead(II) chloride is 0.45 g/100 mL of solution. What is the K_sp of PbCl₂? A)4.9 × 10 ⁻² B)1.7 × 10 ⁻⁵ C)8.5 × 10 ⁻⁶ D)4.2 × 10 ⁻⁶ E)

Question

The solubility of lead(II) chloride is 0.45 g/100 mL of solution. What is the K_sp of PbCl₂? A)4.9 × 10 ⁻² B)1.7 × 10 ⁻⁵ C)8.5 × 10 ⁻⁶ D)4.2 × 10 ⁻⁶ E)< 1.0 × 10 ⁻⁶

Quizplus · Accepted Answer

The solubility product constant (Ksp) expression for PbCl2 is:

Ksp = [Pb2+][Cl-]^2

To find the value of Ksp, we need to know the equilibrium concentrations of Pb2+ and Cl-. Since PbCl2 dissociates completely, the concentration of Pb2+ will be equal to the solubility of PbCl2, which is given as 0.45 g/100 mL.

First, we need to convert grams of PbCl2 to moles:

0.45 g PbCl2 x (1 mol PbCl2/278.1 g PbCl2) = 0.00162 mol PbCl2

Since the solubility is in 100 mL of solution, the volume is 0.1 L.

Now we can calculate the concentration of Pb2+:

[Pb2+] = 0.00162 mol/0.1 L = 0.0162 M

Since PbCl2 dissociates into 1 Pb2+ ion and 2 Cl- ions, the concentration of Cl- ions will be twice the concentration of Pb2+ ions:

[Cl-] = 2 x [Pb2+] = 2 x 0.0162 M = 0.0324 M

Now we can plug in the values of [Pb2+] and [Cl-] into the Ksp expression:

Ksp = [Pb2+][Cl-]^2 = (0.0162 M)(0.0324 M)^2 = 1.7 x 10^-5

Therefore, the answer is B (1.7 x 10^-5).

The Solubility of Lead(II) Chloride Is 0