Question

A solution containing a mixture of metal cations was treated with dilute HCl and a precipitate formed. The solution was filtered and H2S was bubbled through the acidic solution. A precipitate again formed and was filtered off. Then, the pH was raised to about 8 and H2S was again bubbled through the solution. This time, no precipitate formed. Finally, the solution was treated with a sodium carbonate solution, which resulted in formation of a precipitate. Which metal ions were definitely present, which were definitely absent, and which may or may not have been present in the original mixture? Definitely present Definitely absent bly present Ca2+ Zn2+N+ Co2 Sn+ Lit M

Answer 1

ANSWER:

The answer can be found out easily by following the qualitative analysis flow sequences given below.

1. Solution + HCl ------> Insoluble chlorides of Hg₂²⁺, Ag⁺ and Pb²⁺ (First group cations). Take out the filtrate 1
2. Filtrate 1 + H₂S -------> Insoluble sulphides of Sn²⁺, As³⁺, Sb³⁺, Bi³⁺, Hg²⁺, Cd²⁺ and Cu²⁺ (Second group cations) in acidic conditions. Take out the filtrate 2
3. Filtrate 2 + NaOH and H₂S ------> Insoluble sulphides of Cr³⁺, Co²⁺, Fe²⁺, Al³⁺, Fe³⁺, Zn²⁺, Ni²⁺ and Mn²⁺ (Third group cations). Take out the filtrate 3
4. Filtrate 3 + Na₂CO₃ --------> Insoluble carbonates of Ba²⁺, Mg²⁺, Sr²⁺ and Ca²⁺ (Fourth group cations). Take out the filtrate 4.
5. This filtrate 4 may contain the cations of alkali metals (Li⁺ to Cs⁺) and ammonium ion (Fifth group cations).

The above sequence is followed in the qualitative analysis of a given solution.

In the first step of the analysis, it is stated that the solution formed a precipitate when HCl was added. Hence, it must contain the first group cations (Hg₂²⁺, Ag⁺ and Pb²⁺ ). As we are given with all of them, the cations Hg₂²⁺, Ag⁺ and Pb²⁺ are definitely present.

In the second step of the analysis, the solution had formed a precipitate again when H₂S was bubbled which means that the solution must contain the second group cations (Sn²⁺, As³⁺, Sb³⁺, Bi³⁺, Hg²⁺, Cd²⁺ and Cu²⁺). Since we are given with Bi³⁺, Hg²⁺ and Cd²⁺ ions, they are definitely present in the solution.

In the third step of the analysis when NaOH was added and H₂S was bubbled, no precipitate was formed. This is indicating the definite absence of third group cations (Cr³⁺, Co²⁺, Fe²⁺, Al³⁺, Fe³⁺, Zn²⁺, Ni²⁺ and Mn²⁺). Hence, the given third group cations Zn²⁺, Ni²⁺, and Co²⁺ are definitely absent.

When the solution was treated finally with Na₂CO₃, the formation of a precipitate was again observed which is indicating the definite presence of fourth group cations (Ba²⁺, Mg²⁺, Sr²⁺, and Ca²⁺). Among these fourth group cations, we are given with Mg²⁺ and Ca²⁺. Hence, they are definitely present.

The final filtrate may or may not contain Li⁺ and Sn⁴⁺. (Note: Only Sn²⁺ belongs to the second group cations. Not Sn⁴⁺ and Sn⁴⁺ does not precipitate with H₂S under acidic conditions.)

Hence, we can summarize as follows

Definitely Present: Hg₂²⁺, Ag⁺, Pb²⁺, Bi³⁺, Hg²⁺, Cd²⁺, Mg²⁺ and Ca²⁺

Definitely Absent: Zn²⁺, Ni²⁺, and Co²⁺

Possibly Present: Li⁺ and Sn⁴⁺