Question

A solution contains 0.00010 M copper (II) ions and 0.0020 M lead (II) ions.

A. What minimal concentration of iodide ions added to the above solution will cause precipitation of lead (II) iodide? Ksp, PbI2 = 9.8 x 10-9

B. What minimal concentration of iodide ions will cause precipitation of copper(II) iodide? Ksp, CuI2 = 1.0 x 10-12

C. Imagine that you are adding KI solution dropwise to the above solution, which salt will precipitate first and what will remain in solution?

D. Is KI addition a viable method for separating copper (II) ions from lead (II) ion? Explain.

Answer 1

A) PbI2 (s) ----------------> Pb+2 + 2 I-  

- 0.002 s

Ksp = 0.002(s)² = 9.8x10^-9

thus [I-] = x = 2.213 x10^-3

So the minimum iodide concentration to cause precipitation of PbI2 = 2.213x10^-3 M

B) CuI2 --------------> Cu+2 + 2I-

- 1.0x10^-4 p

Thus Ksp = 1.0x10^-4 x p ² = 1.0 x10^-12

thus the [I=] = p = 1.0x10^-4 M

C) As iodie is added into solution contining both the ions CuI2 starts precipiatating first and lead remains as [I-] to precipitate CU+2 is less than Pb+2

D) No. the mimum concentration required to precipitate both of them has very small difference, so as a few drops of KI are added, the Ksp of both the salts can be exceeded , and both can be precipitated together.