Question

A solution contains some or all of the ions Cu2+,Al3+, K+,Ca2+, Ba2+,Pb2+, and NH4+. The following tests were performed, in order, on the solution.1. Addition of 6 M HCl produced no reaction.2. Addition of H2S with 0.2 M HCl produced a black solid.3. Addition of (NH4)2S produced a white solid.4. Addition of (NH4)2HPO4 in NH3 produced no reaction.5. The final supernatant when heated produced a purple flame.Identify which of these ions were present in the solution (P), which were absent (A), and which were inconclusive (I)a. Cu2+b. Al3+c. K+d. Ca2+e. Ba2+f. Pb2+g. NH4+

Answer 1

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)

Answer 2

The analysis is inconclusive for the NH4+ based on this procedure.

What is qualitative analysis?

Qualitative analysis has to do with the identification of the species present in a sample.

We can see that that upon addition of H2S with 0.2 M HCl a black precipitate is obtained showing the presence of group II Cu2. The sample does not react with HCl or (NH₄)₂HPO₄ in NH₃  showing the absence if group 1 and group IV cations so Ca⁺², Pb2+, Ba⁺² and Al⁺³are absent.

The production of a purple flame shows the presence of K^+. Lastly, the analysis is inconclusive for the NH4+ based on this procedure.

Learn more about qualitative analysis: https://brainly.com/question/867804