# Treating a sample with water, how to deduce present compounds given precipitate and solution parameters?

A sample consists of one or more of the following salts: $\ce{Cu(CH3CO2)2}$ (acetate), $\ce{BaCl2}$, $\ce{(NH4)2SO4}$, $\ce{SrCO3}$, $\ce{KNO3}$. The sample, when treated with water, gives a white precipitate that is filtered off and a blue solution. The precipitate is a white solid that is completely insoluble, both in water and in $\pu{6M HCl}$ solution.

Which compounds MUST be present? Explain why.

But $\ce{Cl}$ is soluble with all the given metals! Another question gave the parameter that the precipitate is insoluble in water and $\ce{HI}$, but that was easy because there was a salt with Hg given, so it was obvious that $\ce{HgI}$ must be the precipitate. Please help. Thank you very much.

• $\ce{BaSO4}$, obviously. – Ivan Neretin Oct 23 '17 at 21:26
• Well, what else could it be? SrCO3 is soluble in acids, and the rest are just soluble in water. – Ivan Neretin Oct 23 '17 at 22:00
• But Cl is soluble with Ba – Arthur Alex Karapetov Oct 23 '17 at 22:02
• True, but SO4 is not. – Ivan Neretin Oct 23 '17 at 22:22

You correctly point out that Cl "is soluble with all the given metals". This statement is a little unclear, though, which might be why you're confused. What this means, exactly, is that no precipitate that incorporates Cl can form. For example, the solid $\ce{KCl}$ cannot precipitate, because the combination of $\ce{K}$ and $\ce{Cl}$ is soluble.
This fact alone doesn't forbid precipitates of other species. For example, the solid $\ce{BaSO4}$ is insoluble in water. We know that $\ce{BaCl2}$ alone would be soluble in water, but the barium atoms in $\ce{BaSO4}$ don't care that there are $\ce{Cl-}$ ions in the solution; all that matters is that the combination of $\ce{Ba}$ and $\ce{SO4}$ prevents the two from separating. As a consequence, when $\ce{Ba^2+}$ and $\ce{SO4^2-}$ meet in solution, they merrily form the insoluble product $\ce{BaSO4}$, regardless of the other ions present in the solution.