Questions & explanations
1. A mixture gives no precipitate with dilute HCl, a black precipitate with H₂S in dilute HCl (soluble in hot dilute HNO₃, blue with NH₃), a reddish-brown gelatinous precipitate with NH₄Cl+NH₄OH (blood-red with KSCN), and a white precipitate with (NH₄)₂CO₃ (apple-green flame). The SCE gives CO₂ with dilute HCl and a white precipitate with BaCl₂ insoluble in conc HCl. Which ions are present?
- Cu²⁺, Al³⁺, Ba²⁺, CO₃²⁻, SO₄²⁻
- Cu²⁺, Fe³⁺, Ba²⁺, CO₃²⁻, SO₄²⁻
- Cu²⁺, Fe³⁺, Ca²⁺, CO₃²⁻, SO₄²⁻
- Ni²⁺, Fe³⁺, Ba²⁺, CO₃²⁻, SO₄²⁻
Answer: Cu²⁺, Fe³⁺, Ba²⁺, CO₃²⁻, SO₄²⁻
No ppt with dil HCl excludes Group I. Black ppt with H₂S in dil HCl soluble in hot HNO₃ and blue with NH₃ indicates Cu²⁺ (Group II). Reddish-brown gelatinous ppt with NH₄Cl+NH₄OH giving blood-red with KSCN indicates Fe³⁺ (Group III). White ppt with (NH₄)₂CO₃ giving apple-green flame indicates Ba²⁺ (Group V). SCE gives CO₂ with dil HCl (CO₃²⁻) and white ppt with BaCl₂ insoluble in conc HCl (SO₄²⁻). Thus ions are Cu²⁺, Fe³⁺, Ba²⁺, CO₃²⁻, SO₄²⁻.
2. A blue hydrated salt on heating first turns white, then black. Its aqueous solution gives a black precipitate with H₂S in dilute HCl, which dissolves in hot dilute HNO₃ to give a blue solution. Excess NH₄OH turns the solution deep blue. Identify the cation.
- Cu²⁺
- Ni²⁺
- Co²⁺
- Fe²⁺
Answer: Cu²⁺
The blue hydrated salt is CuSO₄·5H₂O. On heating, it loses water to form white anhydrous CuSO₄, then decomposes to black CuO. In the systematic scheme, Cu²⁺ belongs to Group II (H₂S in dil HCl gives black CuS). CuS dissolves in hot dilute HNO₃ to give Cu²⁺, which forms deep blue [Cu(NH₃)₄]²⁺ with excess NH₄OH. These observations uniquely identify Cu²⁺.
3. Which solvent is tried first when preparing the original solution for cation analysis?
- Dilute HCl
- Concentrated HCl
- Cold water
- Aqua regia
Answer: Cold water
In qualitative analysis, the solvent ladder starts with the least aggressive solvent: cold water. Only if the salt does not dissolve in cold water are stronger solvents like hot water, dilute HCl, concentrated HCl, dilute HNO3, and finally aqua regia tried in order. This ensures minimal interference and preserves the sample for subsequent tests.
4. A white salt gives a canary-yellow ppt with ammonium molybdate in conc HNO₃. OS+NH₄Cl+NH₄OH gives a white gelatinous ppt. After adding Fe(NO₃)₃ and filtering, the filtrate again gives a white gelatinous ppt with NH₄Cl+NH₄OH, soluble in excess NaOH. Cobalt nitrate on charcoal gives Thenard's blue. Identify the cation and anion.
- Fe³⁺ and PO₄³⁻
- Mg²⁺ and PO₄³⁻
- Al³⁺ and SO₄²⁻
- Al³⁺ and PO₄³⁻
Answer: Al³⁺ and PO₄³⁻
Canary-yellow ppt with ammonium molybdate confirms PO₄³⁻. White gelatinous ppt with NH₄OH soluble in excess NaOH indicates Al³⁺. Phosphate interference is removed by adding Fe(NO₃)₃ which precipitates FePO₄, leaving Al³⁺ in solution. Thenard's blue (CoAl₂O₄) from cobalt nitrate test confirms Al³⁺. Hence cation Al³⁺, anion PO₄³⁻.
5. A white salt gives a blue bead in the oxidising flame of the borax bead test. In the microcosmic-salt bead test, a translucent skeleton is observed. Which anion is present?
- Carbonate
- Sulfate
- Phosphate
- Silicate
Answer: Phosphate
The borax bead test gives a blue colour in oxidising flame due to copper(II) ions. In the microcosmic-salt bead test, copper phosphate forms a translucent skeleton because phosphate reacts with sodium metaphosphate to give a glassy bead that does not dissolve the copper salt, leaving a skeleton. Thus, the anion is phosphate.
6. A green salt on heating gives a green residue and CO₂. The OS gives a green precipitate with NH₄Cl+NH₄OH, soluble in excess NaOH. Fusion with NaOH+Na₂O₂ gives a yellow solution; acidification+H₂O₂+ether gives a blue ether layer. Borax-bead is green in both flames. Identify the cation.
- Fe²⁺
- Cr³⁺
- Ni²⁺
- Cu²⁺
Answer: Cr³⁺
The green salt (e.g., Cr₂O₃) gives green residue on heating. Cr(OH)₃ is amphoteric, dissolving in excess NaOH. Fusion with NaOH+Na₂O₂ oxidises Cr³⁺ to yellow CrO₄²⁻. Acidification with H₂O₂ forms blue CrO₅ (perchromic acid) in ether. Borax bead test: Cr³⁺ gives green in both oxidising and reducing flames. These confirm Cr³⁺.
7. A white salt gives no gas with dil H₂SO₄ cold, but with conc H₂SO₄ evolves reddish-brown fumes that intensify with Cu turnings. SCE+AgNO₃+dil HNO₃ gives a pale-yellow ppt partially soluble in conc NH₃. Chromyl chloride test is negative. Layer test with Cl₂ water+CCl₄ gives orange-brown organic layer. Brown-ring test is positive. Identify the two anions present.
- Br⁻ and I⁻
- Cl⁻ and NO₃⁻
- Br⁻ and NO₃⁻
- I⁻ and NO₃⁻
Answer: Br⁻ and NO₃⁻
Reddish-brown fumes with conc H₂SO₄ (Br₂ or NO₂) that intensify with Cu (NO₃⁻ confirmed). Pale-yellow ppt with AgNO₃ partially soluble in conc NH₃ indicates Br⁻. Negative chromyl chloride rules out Cl⁻. Orange-brown layer with Cl₂ water confirms Br⁻ (Br₂ in CCl₄). Positive brown-ring confirms NO₃⁻. Hence Br⁻ and NO₃⁻.
8. During anion analysis, the sodium carbonate extract is prepared by boiling the salt with Na2CO3 solution. What is the primary purpose of this step?
- To oxidize all anions to their highest oxidation state
- To convert anions into their sodium salts and remove cations as carbonates
- To precipitate all anions as sodium salts
- To convert cations into soluble sodium complexes
Answer: To convert anions into their sodium salts and remove cations as carbonates
Boiling the salt with Na2CO3 solution converts cations into insoluble carbonates (e.g., CaCO3, FeCO3) which are filtered off. The filtrate contains the anions as their sodium salts (e.g., Na2SO4, NaCl), making them available for testing. This is the standard procedure for preparing the solution for anion analysis.
9. A white gelatinous precipitate from Group III dissolves in excess NaOH but not in excess NH₄OH. Which cation is confirmed?
- Zn²⁺
- Al³⁺
- Cr³⁺
- Fe³⁺
Answer: Al³⁺
Al³⁺ forms white gelatinous Al(OH)₃ in Group III. Al(OH)₃ is amphoteric and dissolves in excess NaOH forming [Al(OH)₄]⁻, but does not dissolve in excess NH₄OH because Al³⁺ does not form stable ammine complexes. This distinguishes Al³⁺ from Zn²⁺ (dissolves in both) and Cr³⁺ (partially dissolves in NH₄OH).
10. A white salt gives a white incrustation on charcoal cavity. On adding cobalt nitrate and heating, the incrustation turns green. Which cation is present?
- Aluminium
- Zinc
- Magnesium
- Lead
Answer: Zinc
In the charcoal-cavity test, zinc gives a white incrustation of ZnO. The cobalt-nitrate test on this residue produces a green colour due to formation of Rinmann's green (CoZnO2). Aluminium gives blue (Thenard's blue), magnesium gives pink, and lead gives a grey metallic bead with yellow incrustation.
11. A green salt gives a green precipitate with NH4OH, which dissolves in excess NaOH. The solution turns yellow on heating with H2O2. Acidifying and adding H2O2 and ether gives a blue ether layer. The cation is:
- Ni^2+
- Fe^2+
- Cr^3+
- Mn^2+
Answer: Cr^3+
Cr^3+ forms green Cr(OH)3 precipitate with NH4OH, which dissolves in excess NaOH to give green [Cr(OH)4]^- (amphoteric). Heating with H2O2 oxidises Cr(III) to yellow CrO4^2-. Acidification and H2O2 form blue CrO5 (chromium peroxide) that extracts into ether. This sequence is unique to Cr^3+.
12. A salt gives a canary-yellow precipitate with ammonium molybdate in conc HNO3. Which anion is confirmed?
- Chromate
- Arsenate
- Silicate
- Phosphate
Answer: Phosphate
Phosphate reacts with ammonium molybdate in conc HNO3 to form a canary-yellow precipitate of ammonium phosphomolybdate, (NH4)3PO4·12MoO3. This is a specific test for phosphate. Arsenate also gives a yellow precipitate but under different conditions (e.g., heating) and is not canary-yellow.