Question

In qualitative test for identification of presence of phosphorous, the compound is heated with an oxidising agent. Which is further treated with nitric acid and ammonium molybdate respectively. The yellow coloured precipitate obtained is :

• A. Na$_3$PO$_4$.12MoO$_3$
• B. (NH$_4$)$_3$PO$_4$.12(NH$_4$)$_2$MoO$_4$
• C. (NH$_4$)$_3$PO$_4$.12MoO$_3$
• D. MoPO$_4$.21NH$_4$NO$_3$

Answer 1

The Correct Option is C

To identify the presence of phosphorus in a qualitative test, the given compound is initially heated with an oxidizing agent. This process helps in converting phosphorus into a more analyzable form. Here's a step-by-step explanation of the test used:

1. The compound is first oxidized to form phosphoric acid (H₃PO₄).
2. Subsequently, the oxidized product is treated with nitric acid (HNO₃). This ensures complete conversion of phosphorus to its +5 oxidation state.
3. Next, ammonium molybdate ((NH₄)₂MoO₄) is added under acidic conditions.

The reaction results in the formation of a yellow precipitate known as ammonium phosphomolybdate. The chemical formula for this precipitate is (NH₄)₃PO₄.12MoO₃. This unique yellow color and its composition confirm the presence of phosphorus in the tested compound.

Let's evaluate the options provided:

• \(Na_3PO_4.12MoO_3\): Sodium corresponds to another type of compound not typically involved in this qualitative test.
• (NH₄)₃PO₄.12(NH₄)₂MoO₄: This formula is not representing the correct ammonium phosphomolybdate complex.
• (NH₄)₃PO₄.12MoO₃: Matches the typical formulation of ammonium phosphomolybdate precipitate. Thus, this is the correct answer.
• \(MoPO_4.21NH_4NO_3\): Not relevant to the classic phosphorus analysis test.

Therefore, the correct answer is the formation of (NH₄)₃PO₄.12MoO₃, which is indicative of the presence of phosphorus.

Answer 2

In the qualitative analysis of phosphorus, the following reaction occurs:
\[ \text{PO}_4^{3-} \text{ or } \text{HPO}_4^{2-} + (\text{NH}_4)_2\text{MoO}_4 \xrightarrow{\text{H}^+} (\text{NH}_4)_3\text{PO}_4 \cdot 12\text{MoO}_3 \downarrow \]
The product, $(\text{NH}_4)_3\text{PO}_4 \cdot 12\text{MoO}_3$, is a canary yellow precipitate known as ammonium phosphomolybdate.