JEE Challenger
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ChemMedMain 2026 • 2nd Apr - Shift 1

Identification of Cation and Anion using Qualitative Salt Analysis

A salt with few drops of conc. HCl\text{HCl} gives apple green colour in flame test. The group precipitate of the salt is dissolved in acetic acid and treated with K2CrO4\text{K}_2\text{CrO}_4 to give yellow precipitate. When the sodium carbonate extract of the salt solution is heated with conc. HNO3\text{HNO}_3 and ammonium molybdate, it resulted a canary yellow precipitate. The cation and anion present in the salt are respectively,

Options

A

Ca2+\text{Ca}^{2+} and SO42−\text{SO}_4^{2-}

B

Ba2+\text{Ba}^{2+} and PO43−\text{PO}_4^{3-}

Correct
C

Mn2+\text{Mn}^{2+} and PO43−\text{PO}_4^{3-}

D

Ba2+\text{Ba}^{2+} and SO42−\text{SO}_4^{2-}

Topics & Concepts

p-Block Elementsp-Block Elementss-Block Elements

Step-by-Step Solution

To identify the cation and anion present in the salt, we analyze the given experimental observations step-by-step:

1. Identification of the Cation:

  • Flame Test: When the salt is mixed with a few drops of concentrated HCl\text{HCl} and introduced to a non-luminous flame, it imparts an apple green colour to the flame.

    • Ba2+\text{Ba}^{2+} ions give an apple green flame.
    • Ca2+\text{Ca}^{2+} ions give a brick red flame.

    This indicates that the cation present in the salt is Ba2+\text{Ba}^{2+}.

  • Confirmatory Reaction: The Group V carbonate precipitate (barium carbonate, BaCO3\text{BaCO}_3) dissolves in acetic acid (CH3COOH\text{CH}_3\text{COOH}) to yield soluble barium acetate: BaCO3(s)+2CH3COOH(aq)→(CH3COO)2Ba(aq)+H2O(l)+CO2(g)↑\text{BaCO}_3(\text{s}) + 2\text{CH}_3\text{COOH}(\text{aq}) \rightarrow (\text{CH}_3\text{COO})_2\text{Ba}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\uparrow

    When potassium chromate (K2CrO4\text{K}_2\text{CrO}_4) is added to this solution, a yellow precipitate of barium chromate (BaCrO4\text{BaCrO}_4) is formed: Ba2+(aq)+CrO42−(aq)→CH3COOHBaCrO4(s)↓(Yellow precipitate)\text{Ba}^{2+}(\text{aq}) + \text{CrO}_4^{2-}(\text{aq}) \xrightarrow{\text{CH}_3\text{COOH}} \text{BaCrO}_4(\text{s})\downarrow \quad (\text{Yellow precipitate})

    This confirms the presence of Ba2+\text{Ba}^{2+} as the cation.

2. Identification of the Anion:

  • Ammonium Molybdate Test: When the sodium carbonate extract of the salt containing phosphate ions (PO43−\text{PO}_4^{3-}) is heated with concentrated nitric acid (HNO3\text{HNO}_3) and ammonium molybdate solution ((NH4)2MoO4)\left((\text{NH}_4)_2\text{MoO}_4\right), a canary yellow precipitate of ammonium phosphomolybdate is formed: PO43−+3NH4++12MoO42−+24H+→Δ(NH4)3PO4⋅12MoO3↓⏟Canary yellow precipitate+12H2O\text{PO}_4^{3-} + 3\text{NH}_4^+ + 12\text{MoO}_4^{2-} + 24\text{H}^+ \xrightarrow{\Delta} \underbrace{(\text{NH}_4)_3\text{PO}_4 \cdot 12\text{MoO}_3 \downarrow}_{\text{Canary yellow precipitate}} + 12\text{H}_2\text{O}

    This confirms the presence of PO43−\text{PO}_4^{3-} as the anion.

Conclusion:

  • Cation: Ba2+\text{Ba}^{2+}
  • Anion: PO43−\text{PO}_4^{3-}

Therefore, the correct option is B.

Identification of Cation and Anion using Qualitative Salt Analysis | Chemistry PYQ Solution - JEE Challenger