Qualitative Analysis - Salt Analysis and Ion Detection

Real-Life Hook: Chemistry’s Detective Work

Imagine you’re a water quality inspector testing drinking water for toxic heavy metals like lead or arsenic. Or a forensic chemist analyzing evidence from a crime scene. How do you identify unknown substances? Qualitative analysis is your answer!

From testing water purity (detecting toxic ions) to quality control in pharmaceuticals (ensuring correct salt composition) to forensic investigations (identifying poisons), qualitative analysis is chemistry’s systematic detective work:

• Environmental monitoring: Detecting pollutants in water/soil
• Medical diagnostics: Identifying ions in body fluids (kidney stones, blood chemistry)
• Forensic science: Analyzing unknown powders, poisons
• Quality control: Verifying product composition
• Archaeological chemistry: Analyzing ancient artifacts

The systematic approach you’ll learn has helped solve crimes, save lives by detecting contamination, and ensure the safety of food and medicines worldwide!

What is Qualitative Analysis?

Qualitative Analysis (or Salt Analysis) is the systematic identification of cations (positive ions) and anions (negative ions) present in a given inorganic salt or mixture.

Key Word: “Qualitative” = WHAT is present (not HOW MUCH)

Why Study Salt Analysis?

1. Systematic thinking: Develops logical, step-by-step problem-solving
2. Practical skills: Essential lab technique for chemistry
3. JEE importance: 4-5 marks guaranteed in practical-based questions
4. Real applications: Foundation for analytical chemistry careers
5. Understanding chemistry: Connects theory (reactions) with practice (observations)

The Two Parts

Every salt analysis involves:

Part 1: Cation Analysis (Detecting Positive Ions)

• Examples: Na⁺, Ca²⁺, Al³⁺, Fe³⁺, Cu²⁺, Zn²⁺, etc.
• Organized into groups (0 to VI)
• Systematic scheme for identification

Part 2: Anion Analysis (Detecting Negative Ions)

• Examples: Cl⁻, SO₄²⁻, NO₃⁻, CO₃²⁻, etc.
• Preliminary tests + confirmatory tests
• Less systematic than cation analysis

General Procedure for Salt Analysis

The Systematic Approach

1. Preliminary Examination
   ↓
2. Dry Heating Test
   ↓
3. Flame Test (for cations)
   ↓
4. Charcoal Cavity Test (for metals)
   ↓
5. Borax Bead Test / Cobalt Nitrate Test
   ↓
6. Wet Tests for Anions
   ↓
7. Systematic Cation Analysis (Group-wise)
   ↓
8. Confirmatory Tests
   ↓
9. Report: Salt is _____ (cation) + _____ (anion)

Preliminary Examination

Purpose: Gather initial clues about the salt without any chemical test

Physical Observation

Smell Test

Caution: Never smell directly! Waft vapors gently toward nose.

Solubility in Water

Test: Add small amount of salt to water, shake

Solubility Rules to Remember: ✓ All nitrates (NO₃⁻) are soluble ✓ All acetates (CH₃COO⁻) are soluble ✓ All alkali metal (Na⁺, K⁺) and NH₄⁺ salts are soluble ✓ All chlorides are soluble except AgCl, PbCl₂, Hg₂Cl₂ ✓ All sulfates are soluble except BaSO₄, PbSO₄, SrSO₄ (CaSO₄ sparingly soluble)

Dry Heating Test

Procedure: Heat small amount of salt in dry test tube

Purpose: Observe decomposition products (gas, color change, sublimation)

Observations and Inferences

Memory Trick - “Brown Fumes? NO₂ from Nitrate!”

Flame Test (for Cations)

Principle: When metal salts are heated in Bunsen flame, electrons get excited to higher energy levels. On returning to ground state, they emit characteristic colored light.

Procedure

1. Clean platinum/nichrome wire by dipping in conc. HCl and heating in flame (until no color)
2. Make paste of salt with conc. HCl on watch glass
3. Dip wire in paste
4. Hold in non-luminous Bunsen flame
5. Observe color

Flame Colors

Why View K⁺ Through Blue Glass?

• Sodium’s yellow flame masks potassium’s violet
• Cobalt blue glass absorbs yellow light
• Violet light passes through
• Can see K⁺ flame even in presence of Na⁺

Memory Trick - “Good Students Can Study Better Chemistry Permanently”

Golden yellow - Na⁺ (the “Great” alkali metal) Scarlet/Lilac - K⁺ (must See through blue glass) Crimson - Sr²⁺ & Li⁺ (Crimson twins) Somewhat red/Brick - Ca²⁺ (Solid color) Blue-green - Cu²⁺ (Bluish copper) Color green - Ba²⁺ (Apple Color) Pale blue - Pb²⁺ (Pale and brief)

Charcoal Cavity Test (for Metallic Ions)

Principle: Reducing environment (charcoal + heat) reduces metal salts to metals or their oxides

Procedure

1. Make small cavity in charcoal block
2. Add salt + Na₂CO₃ (reducing agent)
3. Heat strongly with blowpipe
4. Observe residue and encrustation (deposit around cavity)

Observations

Borax Bead Test

Principle: Borax (Na₂B₄O₇·10H₂O) on heating forms sodium metaborate and boric anhydride:

B₂O₃ (boric anhydride) reacts with colored metal oxides to form colored metaborates:

(colored)

Procedure

1. Make loop at end of platinum wire
2. Dip in borax powder
3. Heat in flame - forms transparent glassy bead
4. Dip hot bead in salt
5. Heat again in oxidizing flame (outer blue cone)
6. Note color when hot and cold
7. Repeat with reducing flame (inner luminous cone)
8. Note color changes

Bead Colors

Memory Trick:

• Copper - Blue in Oxidizing, Brown in Reducing (BB)
• Iron - Yellow in Oxidizing, Green in Reducing (YG - Year of the Green)
• Cobalt - Blue Both ways (BB - Blue Both)
• Manganese - Violet when Oxidizing (VO - Very Oxidizing!)

Wet Tests for Anions

Anions don’t have systematic group scheme like cations. Instead, individual tests are performed.

Common Anions and Their Tests

1. Carbonate (CO₃²⁻)

Test: Add dilute acid

Reaction:

Observation: Brisk effervescence (bubbles), colorless odorless gas

Confirmation: Pass gas through lime water

Lime water turns milky (white precipitate of CaCO₃)

2. Sulfate (SO₄²⁻)

Test: Add BaCl₂ solution to acidified solution

Reaction:

Observation: White precipitate, insoluble in dilute HCl

Key Point: Must be insoluble in HCl to confirm (distinguishes from BaCO₃, BaSO₃)

3. Chloride (Cl⁻)

Test: Add AgNO₃ solution to acidified solution

Reaction:

Observation: White precipitate (curdy)

Confirmation: Add dilute NH₃

AgCl dissolves (distinguishes from AgBr, AgI)

4. Bromide (Br⁻)

Test: Add AgNO₃ to acidified solution

Reaction:

Observation: Pale yellow precipitate

Confirmation:

• Sparingly soluble in dilute NH₃
• Soluble in conc. NH₃

5. Iodide (I⁻)

Test: Add AgNO₃ to acidified solution

Reaction:

Observation: Yellow precipitate

Confirmation: Insoluble even in conc. NH₃

Alternative Test: Add chlorine water and CCl₄/CHCl₃

• Violet layer in CCl₄ (free I₂ formed)

6. Nitrate (NO₃⁻)

Test 1: Brown Ring Test (most common)

Procedure:

1. Add dilute H₂SO₄ to salt solution
2. Cool
3. Add freshly prepared FeSO₄ solution
4. Pour conc. H₂SO₄ carefully along sides (forms layer)

Observation: Brown ring at junction of two layers

Reaction:

The brown complex [Fe(H₂O)₅NO]²⁺ forms the brown ring

7. Sulfide (S²⁻)

Test: Add dilute acid

Observation: Rotten egg smell (H₂S gas)

Reaction:

Confirmation: H₂S turns lead acetate paper black

8. Acetate (CH₃COO⁻)

Test: Heat with dilute H₂SO₄

Observation: Vinegar smell (acetic acid vapors)

Reaction:

Confirmation - Ester Test: Heat acetate with ethanol + conc. H₂SO₄ Sweet fruity smell of ethyl acetate (ester)

9. Phosphate (PO₄³⁻)

Test: Add ammonium molybdate + conc. HNO₃, warm

Observation: Canary yellow precipitate

Reaction:

Yellow precipitate is ammonium phosphomolybdate

10. Oxalate (C₂O₄²⁻)

Test 1: Heat with conc. H₂SO₄

Observation: Colorless gases (CO + CO₂) that burn with blue flame

Test 2: Add CaCl₂ solution

Observation: White precipitate

Systematic Cation Analysis - Group Classification

Cations are classified into 6 groups based on their precipitation behavior with specific reagents.

Group Reagents

Why Groups?

Systematic separation: Each group precipitates at specific conditions Prevents interference: Earlier groups removed before testing later ones Efficient: Don’t test for all cations individually

Group 0: Ammonium Group (NH₄⁺)

Why Group 0?: No group reagent; tested separately before systematic analysis

Test for NH₄⁺

Test 1: Nessler’s Reagent (most sensitive)

Reagent: K₂[HgI₄] (alkaline solution)

Procedure: Add Nessler’s reagent to salt solution

Observation: Brown precipitate or yellow coloration

Reaction:

Brown ppt of iodide of Millon’s base

Test 2: Heating with NaOH

Procedure:

1. Add NaOH solution to salt
2. Heat gently
3. Bring moist red litmus paper near mouth

Observation:

• Pungent smell of ammonia
• Red litmus turns blue

Reaction:

Group I: Silver Group (Ag⁺, Pb²⁺, Hg₂²⁺)

Group Reagent: Dilute HCl

Why they precipitate: Form insoluble chlorides

Group Precipitation

Add dilute HCl to original solution (slightly acidified):

(white)

(white)

(white, calomel)

All three precipitates are white! How to distinguish?

Separation and Identification

Step 1: Wash precipitate with cold water

Step 2: Treat with hot water

Observation: PbCl₂ dissolves (soluble in hot water)

Confirmation for Pb²⁺: Add K₂CrO₄ to hot solution

Yellow precipitate confirms Pb²⁺

Step 3: Residue (AgCl + Hg₂Cl₂) - treat with NH₄OH

AgCl dissolves:

Hg₂Cl₂ doesn’t dissolve, instead:

Forms grey/black mixture (metallic mercury + mercury amidochloride)

Confirmation for Ag⁺: Acidify the [Ag(NH₃)₂]⁺ solution with HNO₃

White precipitate reforms - confirms Ag⁺

Memory Trick - Group I

“Poor Lead boils” → PbCl₂ dissolves in Lhot water, bbut AgCl and Hg₂Cl₂ don’t

“Silver Ammonia” → AgCl dissolves in Ammonia (NH₃)

Group II: Copper-Arsenic Group

Group Reagent: H₂S in acidic medium (dilute HCl)

Cations: Hg²⁺, Pb²⁺, Bi³⁺, Cu²⁺, Cd²⁺ (IIA) | As³⁺, Sb³⁺, Sn⁴⁺ (IIB)

Why acidic medium?: In acidic solution, H₂S dissociates less, providing low [S²⁻]

• Only cations with very low Ksp (sulfides) precipitate
• Group IV cations (higher Ksp) don’t precipitate yet

Group IIA (Cu, Hg, Pb, Bi, Cd)

Precipitates and Colors:

• HgS: Black
• PbS: Black
• Bi₂S₃: Brown
• CuS: Black
• CdS: Yellow

Detection of Cu²⁺ (most common in JEE):

Test 1: Color - blue/green salt Test 2: NH₄OH test

Add excess NH₄OH to salt solution:

Deep blue color (tetraamminecopper(II) complex)

Test 3: Potassium ferrocyanide test

Reddish brown precipitate

Group IIB (As, Sb, Sn)

These are less common in JEE, but good to know:

Precipitation:

• As₂S₃: Yellow
• Sb₂S₃: Orange
• SnS₂: Yellow

Group III: Iron-Aluminum Group

Group Reagent: NH₄OH + NH₄Cl (ammonium chloride is added to prevent precipitation of Mg²⁺, Ca²⁺ as hydroxides)

Cations: Fe³⁺, Al³⁺, Cr³⁺

Precipitates:

(reddish brown)

(white gelatinous)

(green)

Detection of Fe³⁺ (Most Important)

Observation: Salt may be yellow/brown

Test 1: Potassium ferrocyanide test

Prussian blue precipitate - very distinctive!

Test 2: Potassium thiocyanate test

Blood red coloration

Test 3: NaOH test

Reddish brown precipitate

Detection of Al³⁺

Test: Blue litmus test (lake test)

Add NH₄OH to Al³⁺ solution:

• White gelatinous ppt of Al(OH)₃
• Add blue litmus solution
• Al(OH)₃ adsorbs the dye → colored “lake”

Confirmation: Al(OH)₃ is amphoteric

• Dissolves in excess NaOH:

  $$\ce{Al(OH)3 + OH- -> [Al(OH)4]^- (aluminate)}$$

• Dissolves in acid:

  $$\ce{Al(OH)3 + 3H+ -> Al^{3+} + 3H2O}$$

Group IV: Zinc-Nickel Group

Group Reagent: H₂S in basic medium (NH₄OH + NH₄Cl)

Why basic medium?: High [OH⁻] increases [S²⁻] from H₂S

• Precipitates cations with higher Ksp sulfides
• Group II already removed (lower Ksp)

Cations: Zn²⁺, Ni²⁺, Co²⁺, Mn²⁺

Precipitates:

• ZnS: White (dirty white)
• NiS: Black
• CoS: Black
• MnS: Buff/flesh colored

Detection of Zn²⁺

Test 1: Color observation

• Most Zn²⁺ salts are white
• ZnO is yellow when hot, white when cold

Test 2: Sodium hydroxide test

White precipitate

Amphoteric nature:

(zincate - soluble)

White ppt dissolves in excess NaOH

Test 3: Potassium ferrocyanide

White precipitate

Group V: Calcium Group (Alkaline Earth Metals)

Group Reagent: (NH₄)₂CO₃ in presence of NH₄Cl + NH₄OH

Cations: Ba²⁺, Sr²⁺, Ca²⁺

Precipitates:

(white)

(white)

(white)

Detection of Ca²⁺ (Most Common)

Test 1: Flame test Brick red flame

Test 2: Ammonium oxalate test

White precipitate

• Insoluble in acetic acid
• Soluble in dilute HCl or H₂SO₄

Test 3: Sulfate test

White precipitate (sparingly soluble)

Detection of Ba²⁺

Test 1: Flame test Apple green flame (very distinctive)

Test 2: Sulfate test

White precipitate, insoluble in dilute HCl

This is the confirmatory test for both Ba²⁺ and SO₄²⁻!

Group VI: Magnesium-Alkali Metal Group

No group reagent - remain in solution after all groups are removed

Cations: Mg²⁺, Na⁺, K⁺

Detection of Mg²⁺

Test 1: Sodium hydroxide test

White precipitate, insoluble in excess NaOH (not amphoteric - distinguishes from Zn²⁺, Al³⁺)

Test 2: Magneson reagent

Organic reagent that gives blue lake with Mg(OH)₂

Detection of Na⁺

Test 1: Flame test Golden yellow flame (very persistent and intense)

Test 2: Potassium pyroantimonate test

White crystalline precipitate

Detection of K⁺

Test 1: Flame test Lilac/violet flame (view through blue glass to mask Na yellow)

Test 2: Sodium cobaltinitrite test

Yellow precipitate

Test 3: Tartaric acid test

White crystalline precipitate (potassium hydrogen tartrate)

Practice Problems

Level 1 - JEE Main (Basics)

Problem 1: A salt gives golden yellow flame. Identify the cation and explain the principle behind flame test.

Solution:

Cation: Na⁺ (Sodium)

Golden yellow flame is characteristic and diagnostic for sodium.

Principle of Flame Test:

1. Excitation: When salt is heated in Bunsen flame:

   • Thermal energy excites electrons
   • Electrons jump from ground state to higher energy levels
   • Atom is in excited state (unstable)

2. De-excitation: Excited electrons return to ground state:

   • Releases energy as light
   • Energy difference = hν (Planck’s equation)
   • Specific wavelength → specific color

3. For Sodium:

   • Excitation: 3s → 3p
   • De-excitation: 3p → 3s
   • Energy difference corresponds to yellow light (589 nm - D line)

Why each metal shows different color:

• Different metals have different electronic configurations
• Different energy level gaps
• Different wavelengths of emitted light
• Different colors

Answer: The cation is Na⁺. Flame test works because heating excites electrons to higher energy levels, and they emit characteristic colored light when returning to ground state. Sodium’s electronic transition (3p → 3s) emits yellow light at 589 nm.

Problem 2: Why is cobalt blue glass used when testing for potassium in the flame test?

Solution:

Problem: Sodium (Na⁺) contamination

1. Sodium is ubiquitous:

   • Present as impurity in most chemicals
   • On glassware surfaces
   • In the atmosphere
   • Even small traces give intense yellow flame

2. Sodium yellow masks potassium violet:

   • Na⁺ flame: Intense golden yellow
   • K⁺ flame: Pale lilac/violet
   • Yellow is bright and persistent
   • Violet is faint and brief
   • Yellow completely hides violet!

Solution - Cobalt Blue Glass:

Optical property: Cobalt blue glass is a selective filter

• Absorbs yellow light (around 589 nm - sodium D line)
• Transmits violet/blue light (around 400-450 nm - potassium emission)

Result:

• When viewing flame through blue glass:
  • Sodium’s yellow is filtered out (absorbed)
  • Potassium’s violet passes through
  • Can see K⁺ flame even in presence of Na⁺

Chemical composition of glass: Contains cobalt(II) compounds (CoO or Co²⁺ silicates)

Answer: Cobalt blue glass selectively absorbs yellow light (from Na⁺) while transmitting violet light (from K⁺), allowing potassium flame to be seen even in the presence of sodium contamination. This is necessary because sodium gives an intense yellow flame that would otherwise mask the faint violet flame of potassium.

Problem 3: Distinguish between Fe²⁺ and Fe³⁺ using simple chemical tests.

Solution:

Detailed Reactions:

Test 1: Potassium ferricyanide (Best distinguishing test)

Fe²⁺:

Prussian blue precipitate (also called Turnbull’s blue)

Fe³⁺:

Just gives brownish solution, no precipitate

Test 2: Potassium thiocyanate (Most specific for Fe³⁺)

Fe²⁺: No reaction (or very faint pink)

Fe³⁺:

Intense blood red color - very diagnostic!

Can add more SCN⁻:

Even darker red

Answer:

• Color: Fe²⁺ (pale green) vs Fe³⁺ (yellow-brown)
• Best test: Add KSCN → No color (Fe²⁺) vs Blood red (Fe³⁺)
• Confirmatory: Add K₃[Fe(CN)₆] → Prussian blue ppt (Fe²⁺) vs No ppt (Fe³⁺)

Level 2 - JEE Main (Application)

Problem 4: A white salt gives the following tests:

• Dilute HCl → White precipitate
• The precipitate dissolves in hot water
• On adding K₂CrO₄ to the hot solution → Yellow precipitate

Identify the cation and anion. Give balanced equations.

Solution:

Analysis:

Test 1: Dilute HCl → White precipitate

• Indicates Group I cation (Ag⁺, Pb²⁺, or Hg₂²⁺)
• Forms insoluble chloride

Test 2: Precipitate dissolves in hot water

• Only PbCl₂ is soluble in hot water
• AgCl and Hg₂Cl₂ are insoluble even in hot water
• Cation is Pb²⁺

Test 3: K₂CrO₄ gives yellow precipitate

• Confirms Pb²⁺
• PbCrO₄ is yellow

Determining Anion:

• Original salt must have an anion that:
  • Makes a white salt with Pb²⁺
  • Doesn’t interfere with Cl⁻ precipitation

Possible anions: NO₃⁻, CH₃COO⁻, SO₄²⁻ (sparingly soluble)

Most likely: NO₃⁻ (lead nitrate is very common and white)

Balanced Equations:

Reaction 1: Precipitation with HCl

Or in ionic form:

Reaction 2: Dissolution in hot water

Reaction 3: Chromate test

Answer:

• Cation: Pb²⁺ (Lead)
• Anion: Most likely NO₃⁻ (Nitrate)
• Salt: Pb(NO₃)₂ (Lead nitrate)

Problem 5: How will you distinguish between ZnSO₄ and Al₂(SO₄)₃ using NH₄OH?

Solution:

Both are white salts, both contain SO₄²⁻. Need to distinguish Zn²⁺ from Al³⁺.

Test: Add NH₄OH dropwise, then in excess

With ZnSO₄ (Zn²⁺):

Step 1: Add few drops of NH₄OH

White precipitate forms

Step 2: Add excess NH₄OH

Precipitate dissolves → colorless solution

Zn(OH)₂ is amphoteric and forms soluble tetraammineزinc(II) complex

With Al₂(SO₄)₃ (Al³⁺):

Step 1: Add few drops of NH₄OH

White gelatinous precipitate forms

Step 2: Add excess NH₄OH No change - precipitate remains

Al(OH)₃ doesn’t form stable ammine complex, so doesn’t dissolve in excess NH₄OH

(Note: Al(OH)₃ is amphoteric but dissolves in NaOH, not NH₄OH)

Summary Table:

Answer:

• ZnSO₄: Forms white precipitate with NH₄OH which dissolves in excess NH₄OH
• Al₂(SO₄)₃: Forms white gelatinous precipitate with NH₄OH which doesn’t dissolve in excess NH₄OH

The key distinguishing feature is solubility in excess ammonia: Zn(OH)₂ dissolves (forms complex), Al(OH)₃ doesn’t.

Problem 6: A salt gives brown ring test positive. What does this indicate? Give the reactions involved and explain why concentrated H₂SO₄ is used.

Solution:

Indication: Nitrate (NO₃⁻) is present

Brown ring test is specific and confirmatory for nitrate ion.

Reactions:

Overall: NO₃⁻ is reduced to NO by Fe²⁺ in acidic medium, and NO forms brown complex with Fe²⁺

Step 1: Reduction of nitrate

Nitrate acts as oxidizing agent in acidic medium

Step 2: Complex formation

The brown complex [Fe(H₂O)₅NO]²⁺ (nitrosyl ferrous complex) forms at junction

Why Concentrated H₂SO₄?

Reason 1 - Density difference:

• Conc. H₂SO₄ is very dense (d = 1.84 g/mL)
• Aqueous FeSO₄ solution is less dense (d ≈ 1.0-1.1 g/mL)
• When conc. H₂SO₄ is added carefully along the sides, it settles at bottom
• Forms two distinct layers
• Brown ring appears at the junction of two layers
• If dilute H₂SO₄ used, layers would mix (no junction, no distinct ring)

Reason 2 - Acidic medium:

• Provides H⁺ ions needed for reduction of NO₃⁻
• Maintains acidic conditions for the reaction

Reason 3 - Prevents oxidation:

• Fe²⁺ can oxidize to Fe³⁺ in dilute solution
• Conc. H₂SO₄ minimizes this
• Maintains Fe²⁺ for complex formation

Procedure Recap:

1. Add dilute H₂SO₄ to nitrate solution (provides some acidity)
2. Cool (exothermic reactions ahead)
3. Add freshly prepared FeSO₄ solution (provides Fe²⁺)
4. Carefully pour conc. H₂SO₄ along the walls (dense layer at bottom)
5. Brown ring appears at junction

Color: Intense brown ring (sometimes described as chocolate brown)

Interference: If chloride is present in large amount, may interfere (forms brown color with Fe³⁺)

Answer: Brown ring test indicates presence of nitrate (NO₃⁻). Concentrated H₂SO₄ is used because: (1) its high density creates a distinct layer below the aqueous solution, allowing brown ring to form at the junction, (2) provides acidic medium for nitrate reduction, (3) prevents oxidation of Fe²⁺. The brown color is due to [Fe(H₂O)₅NO]²⁺ complex formed when NO (from reduced nitrate) complexes with Fe²⁺.

Interactive Demo: Visualize Titration Analysis

Explore the principles of titration and qualitative analysis with interactive curves.

Level 3 - JEE Advanced (Conceptual)

Problem 7: Explain why Group IV cations (Zn²⁺, Mn²⁺, etc.) don’t precipitate with H₂S in acidic medium (Group II condition) but do precipitate with H₂S in basic medium (Group IV condition). Discuss in terms of Ksp and common ion effect.

Solution:

The Chemistry of H₂S Dissociation:

H₂S is a weak diprotic acid with two dissociation steps:

$K_{a1} = 1 \times 10^{-7}$

$K_{a2} = 1 \times 10^{-14}$

Overall:

$K_a = K_{a1} \times K_{a2} = 10^{-21}$

Key Point: [S²⁻] depends on [H⁺]!

From equilibrium:

Effect of pH on [S²⁻]:

• Low pH (acidic, high [H⁺]) → [S²⁻] is very low
• High pH (basic, low [H⁺]) → [S²⁻] is high

Ksp Values (approximate):

Group II sulfides (Low Ksp - precipitate easily):

• HgS: $K_{sp} = 10^{-54}$
• CuS: $K_{sp} = 10^{-36}$
• CdS: $K_{sp} = 10^{-29}$
• PbS: $K_{sp} = 10^{-28}$

Group IV sulfides (Higher Ksp - need more S²⁻):

• ZnS: $K_{sp} = 10^{-23}$
• NiS: $K_{sp} = 10^{-21}$
• CoS: $K_{sp} = 10^{-21}$
• MnS: $K_{sp} = 10^{-15}$

In Acidic Medium (Group II Condition):

Condition: pH ≈ 0.5-1, [H⁺] ≈ 0.3 M

Calculate [S²⁻]:

Very low [S²⁻]!

For Group II cations (e.g., Cu²⁺, Ksp = 10⁻³⁶):

Since $Q < K_{sp}$ is NOT true… wait, let me recalculate.

Actually, if [Cu²⁺] = 0.01 M and [S²⁻] = 10⁻²¹:

Since $K_{sp} = 10^{-36}$, we have $Q >> K_{sp}$ → Precipitation occurs! ✓

For Group IV cations (e.g., Zn²⁺, Ksp = 10⁻²³):

Since $Q = K_{sp}$ or $Q < K_{sp}$ → No precipitation (or just at equilibrium) ✓

In Basic Medium (Group IV Condition):

Condition: pH ≈ 9, [H⁺] ≈ 10⁻⁹ M (from NH₄OH + NH₄Cl buffer)

Calculate [S²⁻]:

Much higher [S²⁻]!

For Group IV cations (e.g., Zn²⁺, Ksp = 10⁻²³):

Since $Q >> K_{sp}$ (10⁻⁶ » 10⁻²³) → Precipitation occurs! ✓

Summary:

Answer: Group IV cations have higher Ksp values (~10⁻²¹ to 10⁻¹⁵) compared to Group II cations (~10⁻⁵⁴ to 10⁻²⁸). In acidic medium, high [H⁺] suppresses H₂S dissociation (common ion effect), giving very low [S²⁻] (~10⁻²¹). This low [S²⁻] is sufficient to exceed Ksp of Group II sulfides but not Group IV sulfides. In basic medium, low [H⁺] allows more H₂S dissociation, giving high [S²⁻] (~10⁻⁴), which exceeds Ksp of even Group IV sulfides, causing their precipitation. This pH-dependent selective precipitation is the basis of group separation.

Mathematical Relationship:

Decreasing [H⁺] by 10⁸ (from acidic to basic) increases [S²⁻] by 10¹⁶!

Problem 8: A student performed salt analysis and obtained the following observations:

• Preliminary test: Blue colored salt
• Flame test: No distinctive color
• Dilute HCl: No precipitate
• H₂S in dilute HCl: Black precipitate
• Dilute H₂SO₄ + BaCl₂: White precipitate insoluble in HCl

Identify the salt and give all reactions.

Solution:

Step-by-Step Analysis:

Observation 1: Blue colored salt → Cu²⁺ present (Only common blue cation)

Observation 2: No distinctive flame test → Confirms not alkali/alkaline earth metal → Consistent with Cu²⁺ (doesn’t give good flame test)

Observation 3: No precipitate with dilute HCl → Not Group I cation (Ag⁺, Pb²⁺, Hg₂²⁺ ruled out) → Consistent with Cu²⁺ (Group II)

Observation 4: Black precipitate with H₂S in dilute HCl → Group IIA cation confirmed → CuS (black) formed → Confirms Cu²⁺

Observation 5: White precipitate with BaCl₂, insoluble in dilute HCl → SO₄²⁻ present (BaSO₄ is white and insoluble in HCl)

Conclusion: Salt is CuSO₄ (Copper sulfate)

If hydrated: CuSO₄·5H₂O (blue vitriol)

Reactions:

1. Color:

(blue hexaaquacopper complex)

2. Group II precipitation:

(Black precipitate)

3. Sulfate test:

(White precipitate, insoluble in dilute HCl)

Additional confirmatory tests for Cu²⁺:

Test 1: Ammonia test

Deep blue color (tetraamminecopper(II) complex)

Test 2: Potassium ferrocyanide

Reddish brown precipitate

Answer: The salt is CuSO₄ (Copper sulfate), likely as pentahydrate CuSO₄·5H₂O (blue vitriol). The blue color indicates Cu²⁺, which is confirmed by black CuS precipitate with H₂S. The white BaSO₄ precipitate (insoluble in HCl) confirms SO₄²⁻.

Problem 9: In the brown ring test for nitrate, explain why: (a) Freshly prepared FeSO₄ must be used (b) The solution must be cooled before adding concentrated H₂SO₄ (c) The brown ring appears at the junction and not throughout the solution

Solution:

(a) Why Freshly Prepared FeSO₄?

Problem with old FeSO₄:

Fe²⁺ oxidizes to Fe³⁺ in presence of air (especially in acidic solution)

Why Fe²⁺ is needed:

Reaction 1: Fe²⁺ acts as reducing agent

• Fe²⁺ reduces NO₃⁻ to NO
• Without Fe²⁺, reduction doesn’t occur
• Fe³⁺ cannot reduce NO₃⁻

Reaction 2: Fe²⁺ forms the brown complex

• The brown complex requires Fe²⁺
• Fe³⁺ doesn’t form this complex

Result if Fe³⁺ used:

• No reduction of NO₃⁻
• No NO formed
• No brown complex
• Test fails!

How to keep FeSO₄ fresh:

• Prepare just before use
• Add few drops of dilute H₂SO₄ (prevents hydrolysis)
• Store under inert atmosphere
• Check color: Should be pale green (Fe²⁺), not brown (Fe³⁺)

(b) Why Cool the Solution?

Reason 1 - Exothermic mixing:

• Dissolution/mixing of conc. H₂SO₄ with water is highly exothermic
• If solution is already hot + add conc. H₂SO₄ → very high temperature
• Can cause:
  • Violent boiling
  • Splashing (dangerous!)
  • Breakage of test tube

Reason 2 - Thermal stability of complex:

• The brown complex [Fe(H₂O)₅NO]²⁺ is thermally unstable

• At high temperature, it decomposes:

  $$\ce{[Fe(H2O)5NO]^{2+} ->[heat] Fe^{3+} + NO ^}$$

• Brown color fades/disappears

• Test appears negative

Reason 3 - NO stability:

• NO is reactive
• At high temperature: $\ce{2NO + O2 -> 2NO2}$ (faster)
• NO₂ is brown gas but doesn’t form the same complex
• Test specificity lost

Reason 4 - Better layer formation:

• At lower temperature, liquids have higher viscosity
• Better density difference
• Layers remain distinct (don’t mix easily)
• Clear brown ring at junction

Procedure: Cool in ice bath or under tap water before adding conc. H₂SO₄

(c) Why Brown Ring at Junction?

Density-based layering:

Layer 1 (Top): Aqueous FeSO₄ solution (d ≈ 1.0-1.2 g/mL)

• Contains: Fe²⁺, SO₄²⁻, NO₃⁻, H⁺ (from dilute H₂SO₄)

Layer 2 (Bottom): Concentrated H₂SO₄ (d = 1.84 g/mL)

• Very dense
• High [H⁺]
• Doesn’t mix immediately due to viscosity

At the Junction:

Optimal conditions exist:

1. H⁺ concentration is high (from conc. H₂SO₄ diffusing up)

   • Needed for reduction: $\ce{NO3^- + 4H+...}$

2. Fe²⁺ is present (from FeSO₄ solution diffusing down)

   • Needed for reduction and complex formation

3. Local acidic environment

   • Maximum rate of NO₃⁻ → NO conversion

4. NO formed here reacts immediately with nearby Fe²⁺

   • Forms brown complex at this location

Why not throughout:

• In upper layer: Low [H⁺], reduction is slow/doesn’t occur
• In lower layer: No Fe²⁺ yet (hasn’t diffused down much)
• At junction: Perfect storm - both Fe²⁺ and H⁺ present in right amounts

Diffusion gradient:

Top (low [H⁺], high [Fe²⁺]) → No brown
         ↓
    Junction (medium [H⁺], medium [Fe²⁺]) → BROWN RING
         ↓
Bottom (high [H⁺], low [Fe²⁺]) → No brown

Why ring shape:

• Circular cross-section of test tube
• Reaction occurs at interface between two layers
• Forms a ring around the circumference

If you shake: Layers mix → brown color throughout → but ring is lost (test specificity reduced)

Answer: (a) Fresh FeSO₄ must be used because Fe²⁺ is essential for: (1) reducing NO₃⁻ to NO, and (2) forming the brown complex [Fe(H₂O)₅NO]²⁺. Old FeSO₄ contains Fe³⁺ (oxidized), which cannot perform either function.

(b) Cooling is necessary because: (1) mixing conc. H₂SO₄ is exothermic and can cause dangerous splashing if solution is hot, (2) brown complex decomposes at high temperature, (3) NO oxidizes faster at high temperature, (4) better layer separation occurs at low temperature.

(c) Brown ring appears at junction because that’s where optimal conditions exist - sufficient [H⁺] (from conc. H₂SO₄) meets sufficient [Fe²⁺] (from FeSO₄ solution). The reduction of NO₃⁻ to NO and subsequent complex formation occur most rapidly at this interface, not in the bulk of either layer.

Cross-Links to Related Topics

Inorganic Chemistry Connections

1. Coordination Compounds: Complex ion formation (Prussian blue, tetraammine complexes, ferrocyanides)

2. d-f Block Elements: Transition metal chemistry (Fe²⁺/Fe³⁺, Cu²⁺, colored ions)

3. Chemical Bonding: Understanding ionic vs covalent bonds in salts

4. Periodic Classification: Group trends in solubility, reactivity

5. p-Block Elements: Chemistry of sulfides, halides, carbonates

Physical Chemistry Connections

1. Equilibrium: Solubility equilibrium, Ksp, common ion effect, pH-dependent precipitation

2. Solutions: Solubility rules, precipitation, crystallization

3. Redox Electrochemistry: Oxidation-reduction reactions (nitrate reduction, Fe²⁺/Fe³⁺)

4. Chemical Thermodynamics: Gibbs free energy of precipitation, Le Chatelier’s principle

Practical Chemistry Connections

1. Detection of Elements: Lassaigne’s test, similar analytical techniques

2. Volumetric Analysis: Quantitative complement to qualitative analysis

3. Organic Tests: Similar systematic analytical approach

Environmental Applications

1. Water quality testing: Detecting toxic ions (Pb²⁺, Hg²⁺, As³⁺)
2. Soil analysis: Identifying nutrient ions and contaminants
3. Forensic science: Analyzing unknown substances

Key Takeaways

1. Systematic approach: Follow group-wise scheme for cations (0 to VI); individual tests for anions

2. Group reagents:

   • Group I: Dilute HCl (Ag⁺, Pb²⁺, Hg₂²⁺)
   • Group II: H₂S in acidic medium (Cu²⁺, Pb²⁺, Hg²⁺, etc.)
   • Group III: NH₄OH + NH₄Cl (Fe³⁺, Al³⁺, Cr³⁺)
   • Group IV: H₂S in basic medium (Zn²⁺, Ni²⁺, Co²⁺, Mn²⁺)
   • Group V: (NH₄)₂CO₃ (Ba²⁺, Sr²⁺, Ca²⁺)
   • Group VI: No reagent (Mg²⁺, Na⁺, K⁺)

3. Flame tests: Na⁺ (golden yellow), K⁺ (lilac/violet through blue glass), Ca²⁺ (brick red), Ba²⁺ (apple green), Cu²⁺ (bluish green)

4. Important confirmatory tests:

   • Fe³⁺: Prussian blue (ferrocyanide), blood red (thiocyanate)
   • Cu²⁺: Deep blue with NH₃, black CuS
   • Zn²⁺: White ppt with NaOH, dissolves in excess
   • Al³⁺: White gelatinous ppt with NH₄OH, doesn’t dissolve in excess

5. Anion tests:

   • CO₃²⁻: Effervescence with acid, lime water turns milky
   • SO₄²⁻: White BaSO₄ ppt (insoluble in HCl)
   • Cl⁻: White AgCl (soluble in dilute NH₃)
   • NO₃⁻: Brown ring test

6. Ksp and pH: Group separation based on different Ksp values and pH-dependent [S²⁻]

7. Safety: Handle conc. acids carefully, work in fume hood for toxic gases (H₂S, HCN, NO₂)

8. Observation skills: Color, smell, solubility, precipitate characteristics are crucial clues

9. Interferences: Remove earlier groups before testing later ones; some ions interfere with specific tests

10. Practice: Salt analysis requires hands-on practice; theoretical knowledge must be supplemented with lab work

Quick Revision Points

✓ Preliminary: Color (Cu²⁺ blue, Fe³⁺ yellow-brown), smell (NH₄⁺ pungent, acetate vinegar) ✓ Flame test: Na⁺ yellow, K⁺ violet (blue glass), Ca²⁺ brick red, Ba²⁺ green ✓ Group I: AgCl, PbCl₂, Hg₂Cl₂ with dilute HCl; PbCl₂ soluble in hot water ✓ Group II: CuS (black) with H₂S in acidic; Cu²⁺ + NH₃ → deep blue ✓ Group III: Fe(OH)₃ (reddish brown) with NH₄OH; Fe³⁺ + KSCN → blood red ✓ Group IV: ZnS with H₂S in basic; Zn(OH)₂ dissolves in excess NaOH ✓ Group V: CaCO₃ with (NH₄)₂CO₃; flame test confirmatory ✓ Group VI: Mg(OH)₂ (insoluble in excess NaOH); flame tests for Na⁺, K⁺ ✓ Anions: CO₃²⁻ (effervescence), SO₄²⁻ (BaSO₄ white), Cl⁻ (AgCl white/soluble NH₃), NO₃⁻ (brown ring) ✓ pH-dependent: Group II (acidic, low [S²⁻]), Group IV (basic, high [S²⁻])

Master salt analysis, and you master the art of systematic chemical detective work!