In the Group analysis of cations, Ba²⁺ & Ca²⁺ are precipitated respectively as

Solution & Explanation

Core Logic

In qualitative inorganic analysis, both Ba²⁺ and Ca²⁺ belong to Group V cations. To identify Ba²⁺ & Ca²⁺, the group reagent (NH₄)₂CO₃ + NH₄Cl (with NH₄OH) is used. As a result, they precipitate out as insoluble carbonates: BaCO₃ & CaCO₃.

Pattern Recognition

Group V cations (Ba²⁺, Sr²⁺, Ca²⁺) always precipitate as Carbonates upon adding (NH₄)₂CO₃ in an ammoniacal medium.

Chapter Mix

Class 11 Chemistry: Salt Analysis

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Salt Analysis

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Q64 jee_main_2024_31_jan_morning Precipitate Colors
The compound that is white in color is
  • A. ammonium sulphide
  • B. lead sulphate
  • C. lead iodide
  • D. ammonium arsinomolybdate

Solution

Core Logic

Analyzing the colors of the given salts: Lead sulphate (PbSO₄) is white. Ammonium sulphide ((NH₄)₂S) is soluble/colorless in solution. Lead iodide (PbI₂) is bright yellow. Ammonium arsinomolybdate is yellow.

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Class 12 Chemistry: Salt Analysis

Q81 jee_main_2024_31_jan_morning Anion Identification
Molar mass of the salt from NaBr, NaNO₃, KI and CaF₂ which does not evolve coloured vapours on heating with concentrated H₂SO₄ is ________ g mol⁻¹. (Molar mass in g mol⁻¹: Na: 23, N: 14, K: 39, O: 16, Br: 80, I: 127, F: 19, Ca: 40)
Numerical Answer. Answer: 78 to 78

Solution

Step 1: Analyzing the reactions with Concentrated H2SO4

NaBr conc. H₂SO₄ Br₂ (Reddish-brown vapour evolves) NaNO₃ conc. H₂SO₄ NO₂ (Brown vapour evolves) KI conc. H₂SO₄ I₂ (Violet vapour evolves) CaF₂ conc. H₂SO₄ HF (Colorless vapour evolves)

Thus, CaF₂ does not evolve colored vapours.

Step 2: Calculating Molar Mass

Molar mass of CaF₂: Ca = 40

F₂ = 2 × 19 = 38 Molar mass = 40 + 38 = 78 g/mol
Chapter Mix

Class 12 Chemistry: Salt Analysis

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