Question

A divalent metal ion dissolved in dilute hydrochloric acid forms a precipitate when h2s is bubbled through the solution. which ion is it?

Answer 1

$\boxed{{\text{C}}{{\text{d}}^{2 + }}}$ is a divalent metal ion that forms a precipitate when dissolved in dilute hydrochloric acid, followed by the bubbling of ${{\text{H}}_{\text{2}}}{\text{S}}$.

Further Explanation:

Qualitative analysis is a technique that is used for identification and separation of cations and anions in the substance. It has nothing to do with the amount of substance.

Following are the steps of qualitative analysis:

1. The shape and color of any solid substance have to be determined.

2. The cations of different groups are separated with the help of reagents.

3. The ions of the same group are also separated from each other by performing some confirmation tests.

Initially, ions are to be separated from the original solutions. Once this is done, the individual ions are separated from each other.

The grouping of cations is done in the following manner:

1. Group I

This group includes ions like ${\text{A}}{{\text{g}}^ + }$, ${\text{P}}{{\text{b}}^{2 + }}$ and ${\text{Hg}}_2^{2 + }$. These are precipitated in the HCl solution.

2. Group II

This group includes ions like ${\text{B}}{{\text{i}}^{3 + }}$, ${\text{C}}{{\text{d}}^{2 + }}$, ${\text{S}}{{\text{n}}^{2 + }}$, ${\text{C}}{{\text{u}}^{2 + }}$ etc. These are precipitated with the help of   under acidic conditions.

3. Group III

This group includes ions like ${\text{M}}{{\text{n}}^{2 + }}$, ${\text{F}}{{\text{e}}^{3 + }}$, ${\text{A}}{{\text{l}}^{3 + }}$, ${\text{C}}{{\text{r}}^{3 + }}$, ${\text{Z}}{{\text{n}}^{2 + }}$ etc. These are precipitated with the help of ${{\text{H}}_{\text{2}}}{\text{S}}$ under alkaline conditions.

4. Group IV

This group includes ions like ${\text{M}}{{\text{g}}^{2 + }}$,${\text{C}}{{\text{a}}^{2 + }}$, ${{\text{K}}^ + }$, ${\text{B}}{{\text{a}}^{2 + }}$ etc. These are precipitated with the help of ${\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ at pH 10.

The given divalent metal ion is first dissolved in HCl and then ${{\text{H}}_{\text{2}}}{\text{S}}$ is bubbled through the solution and precipitates are formed. So this ion must belong to group II. Among the given ions, only ${\text{C}}{{\text{d}}^{2 + }}$ belong to group II. Therefore the given divalent ion is ${\text{C}}{{\text{d}}^{2 + }}$.

Learn more:

1. Balanced chemical equation: brainly.com/question/1405182
2. Identify the precipitate in the reaction: brainly.com/question/8896163

Answer Details:

Grade: College

Chapter: Qualitative Analysis

Subject: Chemistry

Keywords: ions, cations, anions, group I, group II, Cd2+, group III, group IV, Bi3+, Mn2+, Fe3+, HCl, H2S, hydrochloric acid.

Answer 2

Due to the cation forms a precipitate by adding H₂S in presence of HCl so this cation must be from group II cations 

The choices provided for this question are: Ca²⁺, Mn²⁺, Zn²⁺ and Cd²⁺

So the correct answer will be cadmium since it is the only cation from choices that present in group II cations

If you have other choices you can choose any member of group II cations and divalent like Cu²⁺ , Hg²⁺