Solid Na2S is divided into 3 solutions with 0.10M Mn2+ 0.10 M Pb2+ and 0.10 M Fe2+ each.
Ksp:
MnS = 3.0 x 10 -14
PbS = 3.4 x 10 -28
FeS = 6.0 x 10-19.
Which will precipitate first?
The wording of this problem leaves much to be desired. As worded it makes absolutely no sense what-so-ever. I know what you mean but that's not what you are saying. IF you mean that you will be adding very very small granules of Na2S to each, one at a time, and you want to know which will ppt with the lowest amount of Na2S, then the answer is the one with the smallest Ksp. That will be PbS.
Thanks.
To determine which compound will precipitate first, we need to compare the solubility product constants (Ksp) of each compound. The compound with the lowest Ksp value will precipitate first because it has the lowest solubility.
Given the following Ksp values:
MnS: Ksp = 3.0 x 10^(-14)
PbS: Ksp = 3.4 x 10^(-28)
FeS: Ksp = 6.0 x 10^(-19)
Comparing the Ksp values, we can see that PbS has the lowest solubility and will precipitate first. Hence, PbS will be the first compound to form a solid precipitate in the solution.
In summary, PbS will precipitate first in the given solutions with Mn2+, Pb2+, and Fe2+ ions.