Question

Consider the following reactions.
\(PbCl_2 + K_2CrO_4 \rightarrow A + 2KCl\) (Hot solution)
\(A + NaOH \rightarrow B + Na_2CrO_4\)
\(PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2SO_4 + X\)
In the above reactions, A, B and X are respectively:

• A. \(Na_2[Pb(OH)_2]\), \(PbCrO_4\) and \((NH_4)_2[Pb(CH_3COO)_4]\)
• B. \(Na_2[Pb(OH)_2]\), \(PbCrO_4\) and \([Pb(NH_3)_4]SO_4\)
• C. \(PbCrO_4\), \(Na_2[Pb(OH)_4]\) and \([Pb(NH_3)_4]SO_4\)
• D. \(PbCrO_4\), \(Na_2[Pb(OH)_4]\) and \((NH_4)_2[Pb(CH_3COO)_4]\)

Hint:

PbSO4 is soluble in ammonium acetate. This is a distinguishing test for Pb in qualitative analysis.

Answer 1

The Correct Option is D

Step 1: Reaction 1:
Lead chloride reacts with chromate to form Lead Chromate (Yellow ppt). \[ PbCl_2 + K_2CrO_4 \rightarrow PbCrO_4 \downarrow (\text{A}) + 2KCl \] Step 2: Reaction 2:
Lead Chromate reacts with excess NaOH to form a soluble plumbite complex. \[ PbCrO_4 + 4NaOH \rightarrow Na_2[Pb(OH)_4] (\text{B}) + Na_2CrO_4 \] Step 3: Reaction 3:
Lead sulfate dissolves in ammonium acetate forming a soluble lead acetate complex. \[ PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2[Pb(CH_3COO)_4] (\text{X}) + (NH_4)_2SO_4 \] Step 4: Final Answer:
A = \(PbCrO_4\), B = \(Na_2[Pb(OH)_4]\), X = \((NH_4)_2[Pb(CH_3COO)_4]\). Matches Option (D).