Question

When a mixture containing phosphate is heated with $conc.HN{O}_3$ and ammonium molybdate solution, a canary yellow precipitate is formed.

The formula of the yellow precipitate is :

• A. ${\left(N{H}_4\right)}_{3}P{O}_4$
• B. ${\left(N{H}_4\right)}_{3}P{O}_4\cdot 12Mo{O}_4$
• C. ${\left(N{H}_4\right)}_{3}P{O}_4\cdot 12Mo{O}_3$
• D. ${\left(N{H}_4\right)}_{3}P{O}_4\cdot {\left(N{H}_4\right)}_{2}Mo{O}_4$

Answer 1

Answer: (C)

${\left(N{H}_4\right)}_{3}P{O}_4\cdot 12Mo{O}_3$

When ammonium molybdate is treated with a solution containing phosphate a canary yellow crystalline precipitate of ammonium phosphomolybdate (ammonium dodecamolybdatophosphate), $\left(N{H}_4{)}_3\right[PM{o}_{12}{O}_{40}]$ or $\left(N{H}_4{)}_3\right[P\left(M{o}_3{O}_{10}{)}_4\right]$ is formed.

In the compound formed, the $M{o}_3{O}_{10}$ group replaces each oxygen atom in phosphate.

$P{O}_4^{3-}+3N{H}_4^{+}+12Mo{O}_4^{2-}+24H^{+}\to \underset{Canary-Yellowppt}{\left(N{H}_4{)}_3\right[P\left(M{o}_3{O}_{10}{)}_4\right]\downarrow +12H_{2}O}$