Question

The Ksp for lead (II) chromate is 2 x 10 -14. A 1.0 L solution is prepared by mixing 50 mg of lead (II) nitrate with 0.20 mg of potassium chromate.

A. will a precipitate form?

B. What should the [Pb + 2] be to just start precipitation?

Answer 1

precipitaion starts only when the ionic product exceeds the solubility product

so for PbCrO4 to precipitate the

[Pb²⁺] X [CrO₄^2-] > Ksp

[Pb²⁺] = [ 50mg of Pb(NO₃)₂]

=1.51 X 10^-4 M

and that for

[CrO₄^2-] = [0.2 mg of K₂CrO₄]

=1.03 X 10^-6

hence their ionic product = 1.51 X 10^-4 X 1.03 X 10^-6

=1.56 X 10^-10

thus ionic product exceedes solubility product, hence precipitation will take place

B.)

Ksp = [Pb²⁺] [CrO₄^2-]

let the minimum conc of Pb ²⁺ to precipitate be x, so

[x] X [1.03 X10^-6] = Ksp = 2 X 10^-14

x= [Pb²⁺] = 1.94 X10^-8 mol