Analytical ChemistryHard
Question
Consider the following statements,
S1 : Fe(OH)3 and Cr(OH)3 precipitates can be separated using NaOH + H2O2 but not by excess of NaOH alone.
S2 : Ag2CrO4 precipitate is soluble in dilute HNO3 and ammonia solution.
S3 : both HgI2 and BiI3 precipitates form colourless soluble complexes with excess of potassium iodide solution.
S4 : white precipitate of PbCl2 is turned black by H2S (not taken in excess) in saturated solution of KCl.
and arrange in the order of true / false.
S1 : Fe(OH)3 and Cr(OH)3 precipitates can be separated using NaOH + H2O2 but not by excess of NaOH alone.
S2 : Ag2CrO4 precipitate is soluble in dilute HNO3 and ammonia solution.
S3 : both HgI2 and BiI3 precipitates form colourless soluble complexes with excess of potassium iodide solution.
S4 : white precipitate of PbCl2 is turned black by H2S (not taken in excess) in saturated solution of KCl.
and arrange in the order of true / false.
Options
A.T F T T
B.F T F F
C.T T T T
D.T F T F
Solution
S1 : The two can be separated by NaOH and H2O2, and by excess of NaOH taken alone.
Cr(OH)3 + 2NaOH + H2O2 → Na2CrO4 (yellow colour solution) + H2O
But no reaction with Fe(OH)3 take place.
OH- + Cr(OH)3 ⇋ [Cr(OH)4]- (green solution)
Fe(OH)3 is not soluble in excess NaOH.
S2 : 2Ag2CrO4↓ (red) + 2H+ ⇋ 4Ag+ + Cr2O72- + H2O
Ag2CrO4 + 4NH3 → 2[Ag(NH3)2]+ + CrO42-
S3 : HgI2 + 2I- → [HgI4]2- (colourless soluble complex)
BiI3 + I- → [BiI4]- orange soluble complex
S4 : 2Pb2+ + H2S + 2Cl- → Pb2SCl2↓ (red) + 2H+
Pb2SCl2 + H2S (more) → 2PbS↓ (black) + 2Cl- + 2H+
Cr(OH)3 + 2NaOH + H2O2 → Na2CrO4 (yellow colour solution) + H2O
But no reaction with Fe(OH)3 take place.
OH- + Cr(OH)3 ⇋ [Cr(OH)4]- (green solution)
Fe(OH)3 is not soluble in excess NaOH.
S2 : 2Ag2CrO4↓ (red) + 2H+ ⇋ 4Ag+ + Cr2O72- + H2O
Ag2CrO4 + 4NH3 → 2[Ag(NH3)2]+ + CrO42-
S3 : HgI2 + 2I- → [HgI4]2- (colourless soluble complex)
BiI3 + I- → [BiI4]- orange soluble complex
S4 : 2Pb2+ + H2S + 2Cl- → Pb2SCl2↓ (red) + 2H+
Pb2SCl2 + H2S (more) → 2PbS↓ (black) + 2Cl- + 2H+
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