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I have been reading VOGEL'S qualitative analysis and I came across the reagent Yellow ammonium sulphide. It's written here that it is used to distinguish Group 2B elements. But my question is that how will addition of H2so4 affect YAS itself?

Answer given is that it turns the cloudless solution into white /yellow but how does it exactly do this ?what is the reaction?

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Ammonium sulphide is $\ce{(NH_4)_2S}$; it is colorless. Yellow ammonium sulphide contains some sulfur $\ce{S}$ dissolved in the preceding solution. So the anion $\ce{ S^{2-}}$ is mixed with some disulphide $\ce{S_2^{2-}}$, which is yellow. If you add an acid to this solution, you produce the reaction $$\ce{S_2^{2-} + 2 H^+ -> H_2S_2}$$ or$$\ce{S_2^{2-} + 2 H_3O^+ -> H_2S_2 + H_2O}$$ But $\ce{H_2S_2}$ is thermodynamically unstable with respect to disproportionation. It gets slowly decomposed according to $$\ce{H_2S_2 -> H_2S + S}$$The sulfur S produced here is not soluble into water, and will make the solution cloudy. If there is enough disulphide, the sulphur will make a precipitate of elementary sulphur, which may be pale yellow or white.

Ref.: N.N. Greenwood A. Earnshaw, Chemistry of the Elements, A. Wheaton & Co, Exeter, 1986, p.805 - 807.

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  • $\begingroup$ Thanks a lot for the answer $\endgroup$ – Alex Apr 16 at 12:01

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