A saturated solution of lead(II)sulfate can be prepared by diluting 0.0101 g of PbSO₄ to 250 mL.What is the K_sp of lead(II)sulfate?

Question

Quizplus · Accepted Answer

To find the Ksp of lead(II) sulfate, we need to first write the balanced chemical equation and the corresponding solubility product expression:

PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq)
Ksp = [Pb2+][SO42-]

The question tells us that we have a saturated solution of lead(II) sulfate. This means that the solution is at equilibrium and the amount of solid lead(II) sulfate that has dissolved is at its maximum. We can use this information, along with the molar mass of PbSO4, to calculate the concentration of Pb2+ and SO42- ions in solution:

moles of PbSO4 = 0.0101 g / 303.26 g/mol = 3.34 × 10^-5 mol
concentration of Pb2+ = concentration of SO42- = (3.34 × 10^-5 mol) / 0.250 L = 1.34 × 10^-4 M

Now we can substitute these values into the Ksp expression to solve for Ksp:

Ksp = (1.34 × 10^-4)^2 = 1.8 × 10^-8

Therefore, the correct answer is C: 1.8 × 10^-8.

A Saturated Solution of Lead(II)sulfate Can Be Prepared by Diluting