A solution containing a mixture of metal cations was treated with dilute HCl and

Question

A solution containing a mixture of metal cations was treated with dilute HCl and no precipitate formed. Next, H2S was bubbled through the acidic solution. A precipitate formed and was filtered off. Then, the pH was raised to about 8 and H2S was again bubbled through the solution. A precipitate again formed and was filtered off. Finally, the solution was treated with a sodium carbonate solution, which resulted in no precipitation. Which metal ions were definitely present, which were definitely absent, and which may or may not have been present in the original mixture?

Explanation / Answer

Step 1 : Addition of dilute HCl.

HCl provides Cl- ions. The metal ions which form insoluble precipitate with chloride ions are

Hg2^2+ , Hg^2+ , Ag+ and Pb^2+. Since in this step we are not getting any precipitate, we can say that

Hg2^2+ , Hg^2+ , Ag+ and Pb^2+ are definitely absent

________________________________________________________________________________

Step 2: Treatment with H2S under acidic conditions.

The insoluble sulfides under acidic conditions are HgS, PbS, CuS, Bi2S3, CdS, As2S3, Sb2S3, SnS2

Fe^2+ also forms a black precipitate with H2S( FeS)

Since there is a precipitate being formed at this step, the possible ions at this step could be

Sn^4+, Cd^2+, Pb^2+, Sb^3+ , Fe^2+

________________________________________________________________________________________

Step 3: Treatment with H2S under basic conditions.

The ions that form insoluble sulfides under basic conditions are Al^3+, Fe^3+, Co^2+, Ni^2+, Cr^3+, Zn^2+, Mn^2+

We are getting a precipitate at this step. Therefore 1 or more ions are present

Out of these, in our case possible ions are Cr^2+, Mn^2+

______________________________________________________________________________________

Step 4 : Treatment with sodium carbonate solution.

Sodium carbonate provides carbonate CO3^2- ions. Ions that could form insoluble precipitate with carbonate are at this step are Mg^2+ & Ba^2+ . But since no precipitate is fomred at this step, We can say that Mg^2+ and Ba^2+ are definitely absent

Li+ is a soluble ion, and could not be detected by all the steps we performed so far. Therefore there is a possibility that it may be present

Denitely absent : Hg2^2+ , Hg^2+ , Ag+ , Pb^2+ , Mg^2+ and Ba^2+

Possibly present : Sn^4+, Cd^2+, Pb^2+, Sb^3+ , Fe^2+, Cr^2+, Mn^2+ and Li+

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.