[NYM]

Four different solutions in four different test tubes: Na₂SO₄, K₂CO₃, Ni(NO₃)₂ and Na₃PO₄.

Solution B and solution C each have pH > 7. Solution C and solution D forms a white precipitate with BaCl₂. Solution A forms a black precipitate with H₂S.
Identify each of the four solutions.


My guesses:
A is Ni(NO₃)₂: NiS is formed.
B and C are K₂CO₃ and Na₃PO₄: pH > 7 (looked at pK_B values).
Then D is Na₂SO₄: BaSO₄ is formed.

How do I distinguish B and C? According to my solubility table, both carbonate and phosphate form insoluble salts with Ba²⁺.

[Borek]

You are right, and IMHO there is not enough information to answer that question.

[ARGOS++]

Dear NYM;

Sorry!,  - But IMHO:  After having identified A as Ni(NO₃)₂ I would use the build HNO₃ to release CO₂ from K₂CO₃.
(Or would such not be allowed?)


Good Luck!
                    ARGOS⁺⁺

[NYM]

You are right, and IMHO there is not enough information to answer that question.

Good to hear! I'll ask my teacher then.

Dear NYM;

Sorry!,  - But IMHO:  After having identified A as Ni(NO₃)₂ I would use the build HNO₃ to release CO₂ from K₂CO₃.
(Or would such not be allowed?)


Good Luck!
                    ARGOS⁺⁺

Unfortunately, this is a purely theoretical "lab test." I should be able to solve this without any additional information, but it looks like *somebody* screwed up (could easily be me, though, I just copied this assignment from the black board in the last minute.)

[ARGOS++]

Dear NYM;

Unfortunately! there is absolute no additional or special knowledge required, not at all!

If theoretically or not:
Everybody who uses H₂S in this situation your way should/must know:

• As soon as with H₂S any Precipitate is formed from a water-soluble metal-salt, then the corresponding Acid will be build.
• And there is not many, if any, build Acid that is not easy able to release CO₂ from a present “Carbonate”, expect the H₂S (pKs ~7.0).

A simple rxn is enough to tell it.

Good Luck!
                    ARGOS⁺⁺

[kevins]

Given: Both C and D can form a white ppt with Ba and the pH of C is >7.
So B is K2CO3 because it form a BaCO3 ppt.