A solution containing several metal ions is treated with dilut...

Question

A solution containing several metal ions is treated with dilute HCl; no precipitate forms. The pH is adjusted to about 1, and H₂S is bubbled through. Again, no precipitate forms. The pH of the solution is then adjusted to about 8. Again, H₂S is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with (NH₄)₂HPO₄. No precipitate forms. Which of these metal cations are either possibly present or definitely absent: Al³⁺, Na⁺, Ag⁺, Mg²⁺?

Accepted Answer

Step 1: Understand the behavior of metal ions with HCl. When dilute HCl is added and no precipitate forms, it suggests that metal ions like Ag^+ (which forms AgCl) are absent.. Step 2: Consider the effect of H2S at pH 1. At this pH, metal sulfides of Group II (like $$Cd^2$$+, $$Bi^3$$+, $$Cu^2+) $$would precipitate. Since no precipitate forms, these ions are likely absent.. Step 3: Analyze the effect of H2S at pH 8. At this pH, metal sulfides of Group III (like $$Co^2$$+, $$Ni^2$$+, $$Zn^2$$+, $$Mn^2$$+, $$Fe^2+) $$can precipitate. The formation of a precipitate suggests the presence of one or more of these ions.. Step 4: Evaluate the addition of (NH4)2HPO4. This reagent is used to precipitate Group IV cations like $$Mg^2+ as $$MgNH4PO4. Since no precipitate forms, $$Mg^2+ is $$likely absent.. Step 5: Conclude the presence or absence of the given ions. $$Al^3+ $$does not precipitate with H2S or (NH4)2HPO4, so it could be present. Na^+ does not form precipitates in these conditions, so it is likely present. Ag^+ and $$Mg^2+ $$are likely absent based on the lack of precipitates in earlier steps.

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Key Concepts

Solubility and Precipitation Reactions

pH and its Effect on Metal Ion Solubility

Complex Ion Formation