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

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? K_sp, PbI_2 = 9.8 times 10^-9 b. What minimal concentration of iodide ions will cause precipitation of copper(II) iodide? K_sp, Cul_2 = 1.0 times 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.

Explanation / Answer

(a)

PbI2 (s) = Pb2+ (aq.) + 2 I- (aq.)

Ksp = [Pb2+][I-]2

9.8 * 10-9 = (0.0020)[I-]2

[I-] = 2.21 * 10-3 M

(b)

CuI2 (s) = Cu2+(aq.) + 2 I- (aq.)

Ksp = [Cu2+][I-]2

1.0 * 10-12 = (0.00010) [I-]2

[I-] = 1.0 * 10-4 M

(c)

Since Ksp of CuI2 is less than that of PbI2,

Cu2+ will be precipitated first and Pb2+ will remain in the solution

(d)

Yes. It is suitable method to seperate Pb2+ and Cu2+ ions.

As kSp of CuI2 is low addition of KI first precipitates Cu2+. FIlter it off and later on addtion of KI it precipitates Pb2+

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