
One of the oxoacid salts on heating with concentrated \[HN{O_3}\] and ammonium molybdate, gives yellow precipitate. What is this ion?
A. \[PO_3^{2 - }\]
B. \[PO_4^{3 - }\]
C. \[SO_4^{2 - }\]
D. \[C{l^ - }\]
Answer
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Hint: The process of determining the composition of a chemical substance is known as chemical analysis. In the chemical method of analysis, the constituent is generally converted into another compound of known characteristics and hence the parent constituent is determined. In this question, the qualitative method of analysis is used to determine a particular anion. Anions are divided into three groups for their identification in qualitative analysis. The first group consists of anions that evolve gases with dilute sulphuric acid, the second group consists of anions that evolve gases with concentrated sulphuric acid and the third group consists of special anions that do not react with both dilute and concentrated sulphuric acid.
Complete answer:
Out of the given options, \[PO_4^{3 - }\]and \[SO_4^{2 - }\] belong to the special group of anions in the qualitative analysis, and these anions are tested individually by confirmatory tests. So, we know that the ammonium molybdate test is used as a confirmatory test to check the presence of special phosphate (\[PO_4^{3 - }\]) anion.
The procedure of ammonium molybdate test-Sodium carbonate extract of the given salt-containing phosphate group is acidified with concentrated \[HN{O_3}\] and then ammonium molybdate solution is added. Then this solution is heated. Formation of canary yellow precipitate of \[{(N{H_4})_3}P{O_4}.12Mo{O_3}\] confirms the presence of \[PO_4^{3 - }\] anion.
The balanced chemical reaction for the above reaction is:
\[N{a_3}P{O_4} + 24HN{O_3} + 12{(N{H_4})_2}Mo{O_4} \to 3NaN{O_3} + 21N{H_4}N{O_3} + 12{H_2}O + {(N{H_4})_3}P{O_4}.12Mo{O_3} \downarrow \]
When any salt contains phosphate, the anion is heated with concentrated nitric acid and ammonium molybdate it gives a yellow precipitate of ammonium dodecamolybdatophosphate.
Hence, option B is the correct answer.
Note: Ammonium molybdate test is used as a confirmatory test to check the presence of phosphate anion. Ammonium molybdate test is done with sodium carbonate extract of the given salt which is prepared by boiling one part of the given salt with three parts of solid sodium carbonate in a beaker containing distilled water. Then this mixture is centrifuged and the centrifugate is the sodium carbonate extract.
Complete answer:
Out of the given options, \[PO_4^{3 - }\]and \[SO_4^{2 - }\] belong to the special group of anions in the qualitative analysis, and these anions are tested individually by confirmatory tests. So, we know that the ammonium molybdate test is used as a confirmatory test to check the presence of special phosphate (\[PO_4^{3 - }\]) anion.
The procedure of ammonium molybdate test-Sodium carbonate extract of the given salt-containing phosphate group is acidified with concentrated \[HN{O_3}\] and then ammonium molybdate solution is added. Then this solution is heated. Formation of canary yellow precipitate of \[{(N{H_4})_3}P{O_4}.12Mo{O_3}\] confirms the presence of \[PO_4^{3 - }\] anion.
The balanced chemical reaction for the above reaction is:
\[N{a_3}P{O_4} + 24HN{O_3} + 12{(N{H_4})_2}Mo{O_4} \to 3NaN{O_3} + 21N{H_4}N{O_3} + 12{H_2}O + {(N{H_4})_3}P{O_4}.12Mo{O_3} \downarrow \]
When any salt contains phosphate, the anion is heated with concentrated nitric acid and ammonium molybdate it gives a yellow precipitate of ammonium dodecamolybdatophosphate.
Hence, option B is the correct answer.
Note: Ammonium molybdate test is used as a confirmatory test to check the presence of phosphate anion. Ammonium molybdate test is done with sodium carbonate extract of the given salt which is prepared by boiling one part of the given salt with three parts of solid sodium carbonate in a beaker containing distilled water. Then this mixture is centrifuged and the centrifugate is the sodium carbonate extract.
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