Question

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 1.62×10−2 M . Calculate Ksp for PbCl2.

Answer 1

Answer:

Ksp for PbCl2 is 1.7 *10^-5

Explanation:

Step 1: Data given

The concentration of Pb2+ ion in the solution was found to be 1.62 *10^−2 M

Step 2: The balanced equation

PbCl2 ⇔Pb^2+ + 2Cl -

Step 2: ICE chart

The initial concentration of Pb2+ is 0 M

There will react X M  and 2X of Cl-

At the equillibrium there is X M  of Pb^2+ and 2X M of Cl-

The concentration of Pb2+ ion in the solution was found to be  1.62 *10^−2  M

Step 3: Calculate Ksp

Since PbCl2 is solid, it doesn't aply for Ksp

Ksp = [Pb^2+][Cl-]²

Ksp = X*(2X)²

Ksp = 4X³

  ⇒ X = 1.62 *10^-2 M

Ksp = 4*( 1.62 *10^-2)³

Ksp =1.7 *10^-5

Ksp for PbCl2 is 1.7 *10^-5