Question

Which of the following anions may be identified by their ppt. reaction in aqueous solution?
(A) \[Cr{O_4}^{2 - }\]
(B) $S{O_4}^{2 - }$
(C) $P{O_4}^{3 - }$
(D) $Mn{O_4}^ - $

Answer 1

Hint: A precipitate is an insoluble substance that does not dissolve in the solution. A precipitate usually appears when it cannot dissolve or the saturation point has reached. Some ionic compounds do not dissolve in the solution due to contributing factors like lattice energy and hydration energy. Whereas some dissolve on the principle of like dissolves like.

Complete Step by Step Solution:
Option A: \[Cr{O_4}^{2 - }\]can be precipitated by the reaction of \[PbN{O_3}\](potassium nitrate) with ${K_2}Cr{O_4}$(potassium chromate). $KN{O_3}$(Potassium nitrate) is formed and $PbCr{O_4}$(potassium chromate) separates out as a solid.

Option B: A very well-known precipitate also used in the qualitative analysis of detection of sulphate ion is $PbS{O_4}$and$BaS{O_4}$.
$BaS{O_4}$ is insoluble in water due to high lattice energy and low hydration energy even though it is an ionic compound. Lattice energy binds the particles together and prevents dissolution.

Option C: in the qualitative analysis of phosphate, precipitate \[{(N{H_4})_3}P{O_4}.12Mo{O_{4}}\;\]appears having canary yellow in colour. This name is especially given to the colour of the compound precipitated after the reaction of phosphoric acid and nitric acid with\[{(N{H_4})_2}.Mo{O_4}\]. This reaction is of extreme importance in phosphorus chemistry.
\[2B{a_3}{(P{O_4})^{^{_2}}}^ - + 6HN{O_3} \to 3Ba{(N{O_3})_{2}} + 2{H_3}P{O_4}\]​
\[{H_3}P{O_4} + 21{(N{H_4})_2}.Mo{O_4} + 21HN{O_3} \to {(N{H_4})_3}P{O_4}.12Mo{O_{4}}\; + 12{H_2}O + 21N{H_4}N{O_3}.\]

Option D: there is as such no precipitate of $Mn{O_4}^ - $.
Hence correct options are A, B, C.

Note: The precipitation concept can be understood by the concept of solubility product and ionic product. When the value of an ionic product exceeds the value of a solubility product, precipitation occurs. Solubility is favoured by high hydration and low lattice energy.