
In the detection of cations in 1st group concentrated solution of HCl is not used because:
A) It may precipitate AgCl first
B) Lead may dissolve in the solution
C) Testing may be done with dilute acid so to conserve acid
D) None of the above
Answer
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Hint: In order to know why the concentrated solution of HCl is not used for the detection of cations of the first group, we must first know what HCl is. HCl is the chemical formula for Hydrochloric acid and it is a very strong acid.
Complete step by step answer:
- Ions present in the first analytical group can form insoluble chlorides.
- The group reagent which is used to separate the insoluble chlorides is hydrochloric acid, which is used at a concentration of 1–2 M.
- We should not use Concentrated HCl, because it forms a soluble complex \[{[PbC{l_4}]^{2 - }}\] with \[P{b^{2 + }}\].Because of this, the \[P{b^{2 + }}\] ion would go undetected.
-Thus, dilute HCl is used in the detection of \[P{b^{2 + }}\] ion.
\[P{b^{2 + }} + 2HCl \to PbC{l_2} \downarrow \] (white ppt)
-When the \[PbC{l_2}\] is heated with \[{H_2}O\], the precipitate completely dissolves.
-Likewise, then the confirmatory test of lead ion is done.
The confirmatory tests of \[P{b^{2 + }}\] is:
\[P{b^{2 + }} + 2KI \to Pb{I_2} \downarrow \] (yellow ppt)
So, the correct answer is “Option B”.
Note: The solubility product constant (\[{K_{sp}}\]) describes the equilibrium between a solid and its constituent ions in a solution. Thus, the \[P{b^{2 + }}\] dissolves completely in concentrated HCl. Be very careful while working with concentrated mineral acids as they might cause serious dangers if exposed to skin.
Complete step by step answer:
- Ions present in the first analytical group can form insoluble chlorides.
- The group reagent which is used to separate the insoluble chlorides is hydrochloric acid, which is used at a concentration of 1–2 M.
- We should not use Concentrated HCl, because it forms a soluble complex \[{[PbC{l_4}]^{2 - }}\] with \[P{b^{2 + }}\].Because of this, the \[P{b^{2 + }}\] ion would go undetected.
-Thus, dilute HCl is used in the detection of \[P{b^{2 + }}\] ion.
\[P{b^{2 + }} + 2HCl \to PbC{l_2} \downarrow \] (white ppt)
-When the \[PbC{l_2}\] is heated with \[{H_2}O\], the precipitate completely dissolves.
-Likewise, then the confirmatory test of lead ion is done.
The confirmatory tests of \[P{b^{2 + }}\] is:
\[P{b^{2 + }} + 2KI \to Pb{I_2} \downarrow \] (yellow ppt)
So, the correct answer is “Option B”.
Note: The solubility product constant (\[{K_{sp}}\]) describes the equilibrium between a solid and its constituent ions in a solution. Thus, the \[P{b^{2 + }}\] dissolves completely in concentrated HCl. Be very careful while working with concentrated mineral acids as they might cause serious dangers if exposed to skin.
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