The answer I got was copper ion will precipitate first because its Ksp is lower

Question

The answer I got was copper ion will precipitate first because its Ksp is lower than the Ksp for MnS. I then got the concentration of [S2-] to be 3 x 10^-36M but the answer key says "(Answer: The copper ion will start to precipitate when [S2-] > 4.0 x 10-33 M)." Please help!

You decide to use sulfide precipitates to separate the copper(II) ions from the manganese(II) ions in a solution that is 0.20 M Cu2+ and 0.20 M Mn2+. Determine the minimum sulfide ion concentration that will result in the precipitation of one cation but not the other (identify the cation that precipitates).

Explanation / Answer

Ksp of MnS = 3 * 10-11 and CuS = 8 * 10-37

Since Ksp of CuS is too small than Ksp of MnS, hence on addition of sulphide ions CuS gets precipitated first.

CuS (s) = Cu2+ (aq.) + S2- (aq.)

Ksp = [Cu2+][S2-]

8 * 10-37 = 0.20 [S2-]

[S2-] = 4.0 * 10-36 M

So, CuS will get precipitated when [S2-] > 4.0 * 10-36 M only but not at > 4.0 * 10-33 M

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