Question

A solution containing a mixture of metal cations was treated as outlined. Dilute HCl was added and a precipitate formed. The precipitate was filtered off. H 2 S was bubbled through the acidic solution. Again, a precipitate formed and was filtered off. The pH was raised to about 9 and H 2 S was again bubbled through the solution. No precipitate formed. Finally, sodium carbonate was added. A precipitate formed and was filtered off. What can be said about the presence of each of these groups of cations in the original solution?

Answer 1

Answer:

See below explanation

Explanation:

For this case, this method is applied in order to determine which species are present in a mixture of metal ions

Using this qualitative analysis, you can simply identify which metal (or cation) is present in this solution

This procedure consists of selectively precipitating each cation under selective conditions (or with a type of analyte)

The most common cations are splitted into different groups, with its respective analyte

- Group I : Ag⁺ , Pb⁺², Hg⁺² ; Analyte: Diluted HCL ; Precipitate: AgCL , PbCl₂, HgCl2₂

- Group II : As⁺³, Bi⁺², Cu⁺², Cd⁺², Sb⁺², Sn⁺³ ; Analyte: H₂S (in acid conditions) ; Precipitate: As₂S₃, BiS, CuS, CdS, SbS, Sn₂S3₃

- Group III : Co⁺² , Fe⁺², Fe⁺³, Mn⁺², Ni⁺², Al⁺³, Cr⁺³ : Analyte: NaOH or NH₃ and H₂S ; Precipitate: divalent metals as insoluble sulfides (CoS, FeS, MnS, NiS) and trivalent metals as hydroxides (Al(OH)₃, Fe(OH)₃, Cr(OH)₃)

- Group IV : Mg⁺² , Ca⁺², Sr⁺², Ba⁺² ; Analyte: Na₂CO3 ; Precipitate: MgCO₃, CaCO₃, SrCO₃, BaCO₃

- Group V : Li⁺ , Na⁺, K⁺, Rb⁺, Cs⁺, NH₄⁺; these metals has not analyte (remains in solution)

According to statement, we may say about is content:

1) When adding dilute HCL, the obtained precipitate contains one or more metals from group I

2) When H₂S is bubbles into the acidic solution, the obtained precipitate contains one or more metals from group II

3) When pH is raised into 9 and then bubbled again with H₂S, no precipitate is formed, hence, we may say that this solution does not contain any metal from group III

4) Sodium carbonate is added, and the formed precipitate contains one or more metals from group IV, and on the remaining solution we may or may not have one or more metal from group V