Question

Describe how you could use pH control to separate lead and iron if they were both precipitated as their hydroxides. (Hint: how could pH be used to dissolve one of the hydroxide precipitates but not the other?)

Answer 1

We know about the solubility product (Ksp) values of lead and iron hydroxide.

Ksp for Pb(OH)2 = 1.43 x 10^(-20)

Ksp for Fe(OH)2 = 4.87 x 10^(-17)

Lets suppose, we have 1 M Fe2+ and Pb2+ each. We know precipitation will occur when ionic product (Q) is greater than solubility product (Ksp).

Ksp(Pb(OH)2) =[Pb2+][OH-]^2

[OH-] = [1.43x10^(-20)/1 M]^(1/2) = 1.19x10^(-10) M

Similarily,

Ksp (Fe(OH)2) = [Fe2+][OH-]^2

[OH-]= [4.87 x 10^(-17)/1M]^(1/2) = 6.98x10^(-9) M

From the above calcutions, we know that the precipitate of Pb(OH)2 require low hydoxide concentration while on the other hand Fe(OH)2 require high hydroxide concentration. That mean to precipitate out Fe(OH)2, we need high pH compared to precipitate out Pb(OH)2.

To separate out Fe(OH)2 and Pb(OH)2, we have to increase the pH slowly slowly. First ppt formed is of Pb(OH)2 separate that. Next again increase the pH will separate Fe(OH)2.

This is the method to separate lead hydroxide and iron hydroxide.