The solubility of PbCl2 in water is 0.99 g PbCl2 per litre of solution.?

Question

The solubility of PbCl2 in water is 0.99 g PbCl2 per litre of solution.
(a) What is the Ksp of PbCl2?
(b) What is the solubility of PbCl2 in 0.5 M lead nitrate solution, expressed in g L-1?
please gie long answer

Answer 1

Ksp = [Pb+2]*[Cl-]^2/[PbCl2]

by definition, [PbCl2] = 1

Also, based on the reaction:

PbCl2 =====> Pb+2 + 2 Cl-

[Cl-] = 2 [Pb+2]

Substitute this into the equation, and you get:

Ksp = [Pb+2] * [Cl-]^2 = [Pb+2]*(2[Pb+2])^2 = 4[Pb+2]^3

Now, since you know that 0.99 grams of PbCl2 is dissolved in your liter of solution, you can calculate the molarity from:

(0.99 g)/ (278.1 grams/mole) = 3.56x10^-3 moles = [Pb+2]

Ksp = 4*(3.56x10^-3)^3 = 1.8 x 10^-7

The text value for the Ksp of lead (II) chloride is 1.7 x 10^-5
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Okay, now for the second part of your question:

If you have 0.5M lead nitrate, which is soluble, you will have 0.5M Pb+2

So, let's plug 'n chug (we'll use the text value for Ksp to show the process):

Ksp = 1.7x10^-5 = [Pb+2]*[Cl-]^2

but, with the lead nitrate in solution, [Pb+2] = 0.5 moles/liter (the amount you'd get from the dissolution of lead chloride will be so small as to be negligible), so:

Ksp = 1.7x10^-5 = [0.5]*[Cl-]^2

[Cl-]^2 = (1.7x10^-5)/(0.5) = 3.4x10^-5

so [Cl-] = sqrt(3.4x10^-5) = 5.83 x 10^-3 and since [Cl-] = 2 [Pb+2], we halve that value to get:

solubility = 5.83x10^-3/2 = 2.92x10^-3 moles per liter

(2.92x10^-3 moles/liter) * 278.1 grams/mole = 0.811 grams/liter

If you use the value of the Ksp calculated in the first half of the question, then the solubility of PbCl2 in your solution is 3x10^-4 moles/liter, or 0.083 grams/liter