Answer:
See below explanation
Explanation:
When having a mixture of metals in solution, you may perform an analytical study (using selective chemical conditions), that may help you to determine whether a metal (cation) is present or not
Using selective analytes (or conditions), leads to consecutive precipitations, until most of the cations are separated in precipitates
With this technique, you may identify metals in different groups, each group will have its analyte (or condition), which will help to have a different precipitate:
- Group I: Ag⁺, Pb⁺², Hg⁺²; Analyte: HCL ; Precipitate: AgCl (white) , PbCl₂, HgCl₂
- Group II: As⁺³ , Bi⁺³, Cd⁺², Cu⁺² , Sb⁺³, Sn⁺² ; Analyte: H₂S (g) with HCL ; Precipitate: As₂S₃ , Bi₂S₃ , CdS (yellow) , CuS (black), Sb₂S₃, SnS
- Group III: Co⁺², Fe⁺², Fe⁺³, Mn⁺², Ni⁺², Zn⁺², Al⁺³, Cr⁺³; Analyte: NaOH or NH₃ with (NH₄)₂S (ac) ; Precipitate: CoS (black) , FeS, MnS , NiS (black), ZnS (white) , Al(OH)₃ (white), Cr(OH)₃
- Group IV: Mg⁺², Ca⁺², Sr⁺², Ba⁺²; Analyte: Na₂CO₃ (ac) or (NH₄)₂HPO₄ (ac); Precipitate: respective carbonate or phosphate MgCO₃/MgHPO₄, CaCO₃/CaHPO₄ , SrCO₃/SrHPO₄, BaCO₃/BaHPO₄
- Group V: Li⁺, K⁺, Na⁺, Rb⁺, Cs⁺, NH₄⁺ ; will remain all in final solution
According to the original statement:
A solution contains one or more of the following: Cu⁺², Al⁺³, K⁺, Ca⁺², Ba⁺², Pb⁺², NH₄⁺
1) Addition on HCl 6M produces no change: we can say the sample does not contain Pb⁺² (group I)
2) Addition of H₂S with 0.2 M HCL produced a black solid: we could say sample contains Cu⁺²(group II)
3) Addition of (NH₄)₂HPO₄ in NH₃ produces no reaction: we could say we don´t have Ca⁺² and /or Ba⁺² (group IV)
4) The final supernatant, when heated produced a purple flame: in the final solution, we have K⁺ (group V), which produces a purple flame (based on its characteristic emission spectrum when subjected to flame)
This analysis will be inconclusive for NH₄⁺ (according to above describe technique)
= 0.78 mol Mg₃(Si₂O₅)₂(OH)₂