Question

Identify A, B, and C in the given reaction sequence: \[ \text{A} \xrightarrow{\text{HNO}_3} \text{Pb(NO}_3)_2 \xrightarrow{\text{H}_2\text{SO}_4} \text{B} \rightarrow \text{C (Yellow ppt)}. \]

• A. PbCl$_2$, PbSO$_4$, PbCrO$_4$
• B. PbS, PbSO$_4$, PbCrO$_4$
• C. PbS, PbSO$_4$, Pb(CH$_3$COO)$_2$
• D. PbCl$_2$, Pb(SO$_4$)$_2$, PbCrO$_4$

Hint:

The reaction sequence involves the formation of lead nitrate, lead sulfate, and lead chromate. This is a typical sequence for testing the presence of lead in reactions with chromates.

Answer 1

The Correct Option is B

Step 1: Identify compound A

The first reaction is:

A ⟶_HNO₃ Pb(NO₃)₂

This indicates that compound A reacts with nitric acid to form lead nitrate. Hence, A must be metallic lead (Pb).

So, A = Pb

Step 2: Identify compound B

Pb(NO₃)₂ reacts with H₂SO₄ to give compound B. The reaction is: \[ \text{Pb(NO}_3)_2 + \text{H}_2\text{SO}_4 \rightarrow \text{PbSO}_4 \downarrow + 2\text{HNO}_3 \] PbSO₄ is a white precipitate formed by double displacement.

So, B = PbSO₄

Step 3: Identify compound C (Yellow precipitate)

Compound B is then treated to give a yellow precipitate C. The only classic yellow precipitate with lead is lead chromate (PbCrO₄), which is formed when Pb²⁺ reacts with chromate ions: \[ \text{Pb}^{2+} + \text{CrO}_4^{2-} \rightarrow \text{PbCrO}_4 \downarrow \, (\text{yellow ppt}) \] This confirms C is lead chromate.

So, C = PbCrO₄

Final Identifications:
A = Pb, B = PbSO₄, C = PbCrO₄