Qualitative Analysis of Common Anions 1. In the test for Cl and so, explain why

Question

Qualitative Analysis of Common Anions 1. In the test for Cl and so, explain why the addition of acid will dissolve or decompose other silver salts and barium salts, which otherwise might interfere 2. Suppose a white precipitate is obtained on adding BaCl2 reagent to a neutral unknown solution. What might the precipitate be? Give as many formulas as you can of possible substances, give at least 4, assuming that barium is the counterion present in the salts. 3. A moderate test for so, is to be expected any time that you have so, or 2- unknown. Explain.

Explanation / Answer

Qualitative analysis

1. Addition of acid destroys or decomposes salts of silver and barium made with CO3^2-. PO4^3-, and S^2- ions. The carbonates results into the formation of CO2 gas, the sulfides gives H2S gas and phosphates forms phosphoric acid in solution.

2. A white precipitate is formed upon addition of BaCl2 to the unknown solution,

the most probable identity of the anion would be a sulfate ion, chloride ion, phosphate ion, or carbonate ion.

The chemical reaction occuring is,

BaCl2 + SO4^2- ---> BaSO4 (white precipitate) + 2Cl-

BaCl2 + PO4^3- --> Ba3(PO4)2 (white precipitate) + 2Cl-

BaCl2 + C2O4^2- ---> BaC2O4 (white precipitate) + 2Cl-

BaCl2 + CO3^2- --> BaCO3 (white precipitate) + 2Cl-

Possible salts, BaSO4, Ba3(PO4)2, BaC2O4, BaCO3

3. SO4^2- gets easily oxidized and reduced in solution medium to give the species SO3^2- and S^2-, therefore it is safe to assume that at any given time we have SO3^2- and S^2- in an unknown solution having SO4^2-.

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