Chapter 22: Principles Related to Practical Chemistry

Lassaigne's Test

Preparation of Lassaigne's Extract

• Heat a small piece of sodium in an ignition tube until it melts.

• Add 50-60 mg of organic compound; heat strongly for 2-3 minutes to fuse.

• Cool and break tube into distilled water; crush and boil the mixture.

• Formed sodium salts (NaCN, Na2S, NaX, NaSCN) dissolve; excess sodium forms NaOH.

• This alkaline solution is Lassaigne's extract.

• Lithium and potassium are not used due to slow reaction and vigorous reaction respectively.

Detection of Nitrogen

• Nitrogen and carbon form sodium cyanide (NaCN) in extract.

• NaCN reacts with FeSO4 to form sodium ferrocyanide, which with Fe3+ forms Prussian blue (ferric ferrocyanide).

• FeSO4 + 2NaOH → Fe(OH)2 + Na2SO4

• Fe(OH)2 + 6NaCN → Na4[Fe(CN)6] + 2NaOH

• 4Fe3+ + 3Na4[Fe(CN)6] → Fe4[Fe(CN)6]3 (Prussian blue)

• Test fails if nitrogen present without carbon (e.g., hydrazine).

• If nitrogen and sulphur both present, blood red colour due to ferric thiocyanate forms.

• Na + C + N + S → NaCNS; 3NaCNS + Fe3+ → Fe(CNS)3 (blood red colour)

• Excess sodium decomposes sodium thiocyanate, no red colour appears.

Detection of Sulphur

• Sulphur converts to sodium sulphide (Na2S) in extract.

• Na + S → Na2S (on heating)

• Na2S reacts with sodium nitroprusside to give violet colour.

• Na2S + Na2[Fe(CN)5(NO)] → Na4[Fe(CN)5NOS] (violet)

• Na2S + Pb(CH3COO)2 → PbS (black precipitate) + 2CH3COONa

Detection of Halogens

• Beilstein test: Burning organic compound on copper wire gives coloured flame if halogens present (except fluorine).

• In Lassaigne's extract, halogens form sodium halides.

• Boil extract with dilute HNO3 to remove interference.

• React with AgNO3 to get precipitates:

  • AgCl: white, soluble in NH4OH (chlorine)

  • AgBr: pale yellow, sparingly soluble (bromine)

  • AgI: yellow, insoluble (iodine)

• Chlorine water test: Boil extract with dilute H2SO4, add CS2 and chlorine water.

  • Orange colour in CS2 layer: bromine

  • Violet colour in CS2 layer: iodine

Solved Examples

Ex 1: Identify nitrogen in an organic compound using Lassaigne's test.

• Fuse compound with sodium to form NaCN.

• Boil extract with FeSO4; observe Prussian blue colour.

Ex 2: Detect halogen in compound.

• Prepare Lassaigne's extract.

• Boil with dilute HNO3, add AgNO3.

• Observe precipitate colour and solubility in NH4OH.

Practice Set

Q1: Which metal is used in Lassaigne's test?

• A. Lithium

• B. Sodium

• C. Potassium

• D. Calcium

Q2: What colour indicates nitrogen presence in Lassaigne's test?

• A. Blood red

• B. Prussian blue

• C. Violet

• D. Black

Q3: Which precipitate is formed for bromine detection in Lassaigne's test?

• A. White AgCl

• B. Pale yellow AgBr

• C. Yellow AgI

• D. Black PbS

Answer Key

Q1: B. Sodium - Reacts efficiently to form sodium salts.

Q2: B. Prussian blue - Ferric ferrocyanide indicates nitrogen.

Q3: B. Pale yellow AgBr - Characteristic for bromine.

Memory Aids

• Na used in Lassaigne's test; Li slow, K too reactive.

• Prussian blue = nitrogen; blood red = N + S.

• AgCl white soluble, AgBr pale yellow sparingly soluble, AgI yellow insoluble.

Alcohols

Detection Tests

• Sodium metal test: Alcohol + Na → Sodium alkoxide + H2 (effervescence)

• 

  Ester test: Alcohol + Carboxylic acid + conc. H2SO4 → Ester (fruity smell)

• 

  Ceric ammonium nitrate test: Alcohol + ceric ammonium nitrate → Pink/red colour

• 

• Acetyl chloride test: Alcohol + acetyl chloride → Ester + HCl (white fumes with NH3)

Distinction of Alcohols

• Lucas test: Alcohol + Lucas reagent (HCl + ZnCl2) → Alkyl halide (cloudiness)

  • Tertiary: immediate cloudiness

  • Secondary: cloudiness in 5-10 min

  • Primary: no cloudiness at room temp

• 

  Victor Meyer test: Series of reactions producing colour:

  • Primary: blood red

  • Secondary: blue

  • Tertiary: colourless

• 

  Catalytic dehydrogenation:

  • Primary → Aldehydes

  • Secondary → Ketones

  • Tertiary → Olefins (dehydration)

Solved Examples

Ex 1: Identify type of alcohol using Lucas test.

• Mix alcohol with Lucas reagent at room temp.

• Observe time for cloudiness.

• Immediate: tertiary; 5-10 min: secondary; no cloudiness: primary.

Ex 2: Confirm alcohol presence by ceric ammonium nitrate test.

• Add ceric ammonium nitrate solution to sample.

• Pink/red colour confirms alcohol.

Practice Set

Q1: Which alcohol gives immediate cloudiness with Lucas reagent?

• A. Primary

• B. Secondary

• C. Tertiary

• D. None

Q2: Product of catalytic dehydrogenation of secondary alcohol is?

• A. Aldehyde

• B. Ketone

• C. Alkene

• D. Ester

Q3: Reaction of alcohol with acetyl chloride produces?

• A. Alcohol + HCl

• B. Ester + HCl

• C. Ether + H2O

• D. Ketone + H2O

Answer Key

Q1: C. Tertiary - fastest reaction forming alkyl halide.

Q2: B. Ketone - secondary alcohols oxidize to ketones.

Q3: B. Ester + HCl - acetyl chloride reacts with alcohol forming ester.

Memory Aids

• Lucas test: Tertiary > Secondary > Primary (rate of carbocation formation).

• Dehydrogenation: Primary → Aldehyde, Secondary → Ketone, Tertiary → Alkene.

• Acetyl chloride test: White fumes with NH3 confirm ester formation.

Phenols

Detection Tests

• Litmus test: Phenols turn blue litmus red; no effervescence with Na2CO3.

• Ferric chloride test: Phenol + FeCl3 → coloured complex (varies with phenol type).

• Liebermann’s nitroso test: Phenol + NaNO2 + conc. H2SO4 → blue colour, changes to red on dilution, blue restored in alkaline medium.

Solved Examples

Ex 1: Confirm phenol by ferric chloride test.

• Add FeCl3 solution to phenol; observe colour change.

Ex 2: Perform Liebermann’s nitroso test.

• React phenol with sodium nitrite and conc. H2SO4; observe blue colour.

Practice Set

Q1: Phenols react with ferric chloride to give?

• A. Colourless solution

• B. Coloured complex

• C. Precipitate

• D. No reaction

Q2: Phenols turn blue litmus paper?

• A. Blue

• B. Red

• C. No change

• D. Green

Q3: Liebermann’s nitroso test colour changes from blue to?

• A. Green

• B. Red

• C. Yellow

• D. Violet

Answer Key

Q1: B. Coloured complex - characteristic of phenols.

Q2: B. Red - phenols are acidic.

Q3: B. Red - blue changes to red on dilution.

Memory Aids

• Phenols: acidic, blue to red litmus.

• Ferric chloride test: coloured complexes.

• Liebermann’s test: blue → red → blue (alkaline).

Carbonyl Compounds

Detection Tests

• 2,4-Dinitrophenyl hydrazine (DNP) test: Carbonyl + DNP → yellow-orange precipitate (hydrazone).

• 

  Sodium bisulphite test: Aldehydes and methyl ketones + NaHSO3 → crystalline addition product.

• 

  Tollens’ test: Aldehydes + Tollens’ reagent → silver mirror (Ag deposition).

• 

  Fehling’s test: Aldehydes + Fehling’s solution → red Cu2O precipitate.

• 

  Benedict’s test: Similar to Fehling’s; aldehydes give red precipitate.

• Schiff’s test: Aldehydes + Schiff’s reagent → pink/magenta colour.

Distinction Between Aldehydes and Ketones

• Aldehydes give Tollens’, Fehling’s, Benedict’s, Schiff’s tests; ketones do not.

• Ketones give m-dinitrobenzene test (violet colour) and sodium nitroprusside test (red complex).

Solved Examples

Ex 1: Identify aldehyde using Tollens’ test.

• Warm compound with Tollens’ reagent; silver mirror confirms aldehyde.

Ex 2: Distinguish ketone by m-dinitrobenzene test.

• Mix compound with m-dinitrobenzene and alkali; violet colour confirms ketone.

Practice Set

Q1: Which test gives silver mirror for aldehydes?

• A. Fehling’s

• B. Tollens’

• C. Schiff’s

• D. DNP

Q2: Ketones give positive test with?

• A. Tollens’ reagent

• B. m-Dinitrobenzene

• C. Fehling’s solution

• D. Schiff’s reagent

Q3: Sodium bisulphite test forms?

• A. Red precipitate

• B. Crystalline addition product

• C. Silver mirror

• D. No reaction

Answer Key

Q1: B. Tollens’ - specific for aldehydes.

Q2: B. m-Dinitrobenzene - ketone test.

Q3: B. Crystalline addition product - aldehydes and methyl ketones.

Memory Aids

• Aldehydes: Tollens’, Fehling’s, Schiff’s positive.

• Ketones: m-Dinitrobenzene, sodium nitroprusside positive.

• DNP test: yellow-orange precipitate for carbonyls.

Carboxylic Acids

Detection Tests

• Litmus test: Turn blue litmus red (acidic).

• Ester test: Carboxylic acid + ethyl alcohol + conc. H2SO4 → ester (fruity smell).

• 

  Sodium bicarbonate test: Carboxylic acid + NaHCO3 → brisk effervescence (CO2 gas).

Solved Examples

Ex 1: Confirm carboxylic acid by sodium bicarbonate test.

• Add NaHCO3 to compound; observe effervescence.

Ex 2: Perform ester test.

• Mix compound with ethyl alcohol and conc. H2SO4; heat and detect fruity smell.

Practice Set

Q1: Carboxylic acids turn blue litmus?

• A. Blue

• B. Red

• C. No change

• D. Green

Q2: Effervescence in sodium bicarbonate test is due to?

• A. O2

• B. CO2

• C. H2

• D. N2

Q3: Fruity smell in ester test confirms presence of?

• A. Alcohol

• B. Carboxyl

• C. Ester

• D. Ketone

Answer Key

Q1: B. Red - acidic nature.

Q2: B. CO2 - from reaction with bicarbonate.

Q3: B. Carboxyl - ester formed from carboxylic acid.

Memory Aids

• Carboxylic acids: acidic, effervescence with NaHCO3.

• Ester test: fruity smell confirms carboxyl group.

Amines

Detection Tests

• Litmus test: Amines turn red litmus blue (basic).

• Solubility test: Amines dissolve in mineral acids forming salts.

• Carbylamine test: Primary amines + chloroform + alcoholic KOH → isocyanide (offensive smell).

• 

  Azo-dye test: Primary aromatic amines + nitrous acid + phenol → coloured azo dye.

• 

  Nitrous acid test:

  • Primary amines liberate N2 gas (bubbles).

  • Secondary amines form yellow oily N-nitrosoamines.

  • Tertiary amines form water-soluble salts.

• 

  Hinsberg test: Primary and secondary amines react with benzene sulphonyl chloride forming sulphonamides; primary soluble in NaOH, secondary insoluble; tertiary do not react.

Solved Examples

Ex 1: Identify primary amine by carbylamine test.

• Mix amine with chloroform and alcoholic KOH; warm and detect foul smell.

Ex 2: Differentiate amines by Hinsberg test.

• React amine with benzene sulphonyl chloride; test solubility of product in NaOH.

Practice Set

Q1: Carbylamine test is positive for?

• A. Primary amines

• B. Secondary amines

• C. Tertiary amines

• D. All amines

Q2: In Hinsberg test, product soluble in NaOH indicates?

• A. Primary amine

• B. Secondary amine

• C. Tertiary amine

• D. None

Q3: Primary amines react with nitrous acid to liberate?

• A. O2

• B. N2

• C. H2

• D. CO2

Answer Key

Q1: A. Primary amines - form isocyanides.

Q2: A. Primary amine - sulphonamide soluble in NaOH.

Q3: B. N2 - nitrogen gas bubbles.

Memory Aids

• Carbylamine test: only primary amines give foul smell.

• Hinsberg test: solubility in NaOH distinguishes primary and secondary.

• Nitrous acid: primary amines liberate N2 gas.

Acetanilide

Theory and Preparation

• Prepared by acetylation of aniline with acetic anhydride in presence of glacial acetic acid.

• Aniline + Acetic anhydride → Acetanilide + Acetic acid

• Reaction carried out by boiling mixture with reflux, then cooling to get crude acetanilide.

• Pure acetanilide obtained by recrystallisation.

• Zinc added to prevent oxidation and impurities.

• Cooling before pouring into ice water prevents hydrolysis.

Chemistry and Reactions

• Acetylation can be done with acetyl chloride using pyridine as catalyst.

• Acetanilide protects amino group in reactions.

• Nitration of acetanilide with nitrating mixture at <25°C gives mainly p-nitroacetanilide.

• 

  Direct nitration of aniline is avoided due to oxidation.

Solved Examples

Ex 1: Prepare acetanilide from aniline.

• Mix aniline, acetic anhydride, glacial acetic acid; reflux and cool.

• Filter crude product; recrystallise for pure acetanilide.

Ex 2: Nitrate acetanilide.

• Mix acetanilide with nitrating mixture below 25°C.

• Isolate p-nitroacetanilide by recrystallisation.

Practice Set

Q1: Acetanilide is prepared by acetylation of?

• A. Phenol

• B. Aniline

• C. Benzene

• D. Nitrobenzene

Q2: Catalyst used in acetyl chloride acetylation is?

• A. Glacial acetic acid

• B. Pyridine

• C. Zinc

• D. Sulfuric acid

Q3: Major product of nitration of acetanilide is?

• A. o-Nitroacetanilide

• B. p-Nitroacetanilide

• C. m-Nitroacetanilide

• D. Nitroaniline

Answer Key

Q1: B. Aniline - starting amine for acetanilide.

Q2: B. Pyridine - catalyst with acetyl chloride.

Q3: B. p-Nitroacetanilide - major nitration product.

Memory Aids

• Acetanilide = aniline + acetyl group.

• Pyridine catalyses acetyl chloride reaction.

• Nitration: para product major due to sterics.

Aniline Yellow

Theory and Preparation

• Basic azo dye with azo group (-N=N-).

• Prepared by coupling aniline with benzenediazonium chloride in acidic medium.

• Yellow solid formed; sensitive to acids, not used as dye.

Solved Examples

Ex 1: Prepare aniline yellow.

• Mix aniline with cold benzenediazonium chloride solution; shake vigorously.

• Yellow precipitate of aniline yellow forms.

Practice Set

Q1: Aniline yellow contains which characteristic group?

• A. Nitro

• B. Azo (-N=N-)

• C. Amino

• D. Carbonyl

Q2: Aniline yellow is prepared by coupling of?

• A. Aniline and nitrobenzene

• B. Aniline and benzenediazonium chloride

• C. Benzene and diazonium salt

• D. Aniline and benzene

Q3: Aniline yellow is sensitive to?

• A. Bases

• B. Acids

• C. Heat

• D. Light

Answer Key

Q1: B. Azo (-N=N-) - characteristic azo dye group.

Q2: B. Aniline and benzenediazonium chloride - coupling reaction.

Q3: B. Acids - dye is acid sensitive.

Memory Aids

• Azo dyes contain -N=N- group.

• Aniline yellow formed by diazonium coupling.

• Acid sensitive; limits dye use.

Iodoform Test

Theory and Preparation

• Iodoform (CHI3) is yellow solid with antiseptic properties.

• Prepared from acetone or ethyl alcohol (better yield from acetone).

• Reaction involves oxidation (for ethanol), iodination, and hydrolysis steps.

• Ethyl alcohol oxidized to acetaldehyde, then iodinated to triiodoacetaldehyde, hydrolyzed to iodoform and sodium formate.

• Acetone directly iodinated and hydrolyzed to iodoform and sodium acetate.

Solved Examples

Ex 1: Prepare iodoform from acetone.

• Mix acetone, KI, NaOH, and sodium hypochlorite; yellow precipitate forms.

• Filter and recrystallize iodoform.

Practice Set

Q1: Iodoform test detects which groups?

• A. -CH2OH

• B. -CH(OH)- and -CO- (methyl ketone)

• C. -OH only

• D. -COOH

Q2: Better yield of iodoform is obtained from?

• A. Ethanol

• B. Acetone

• C. Methanol

• D. Benzene

Q3: Iodoform is used as?

• A. Solvent

• B. Antiseptic

• C. Dye

• D. Fuel

Answer Key

Q1: B. -CH(OH)- and -CO- groups - methyl alcohol and methyl ketone groups.

Q2: B. Acetone - gives better yield.

Q3: B. Antiseptic - used for wound cleaning.

Memory Aids

• Iodoform test positive for methyl ketones and ethanol.

• Yellow precipitate confirms presence.

• Used as antiseptic.

Preparation of Mohr's Salt and Potash Alum

Key Concepts

• Mohr's salt: FeSO4(NH4)2SO4·6H2O, a double salt of ferrous sulfate and ammonium sulfate.

• Potash alum: K2SO4·Al2(SO4)3·24H2O, a double sulfate of potassium and aluminum.

• Double salts dissociate completely into their ions in aqueous solution.

• General formula of alums: M2SO4·M'2(SO4)3·24H2O, where M = monovalent metal, M' = trivalent metal.

• Mohr's salt preparation: equimolar hydrated ferrous sulfate + ammonium sulfate + dilute H2SO4 in boiled water to avoid oxidation.

• Potash alum preparation: equimolar hydrated aluminum sulfate + potassium sulfate + dilute H2SO4.

• Crystallization yields light green monoclinic Mohr's salt crystals and octahedral potash alum crystals.

• Mohr's salt is stable, resists oxidation and loss of water of crystallization.

• Potash alum melts and swells on heating, losing water to form burnt alum.

Solved Examples

• Ex 1: Write the reaction for preparation of Mohr's salt.
  FeSO4·7H2O + (NH4)2SO4 + H2SO4 → FeSO4(NH4)2SO4·6H2O + H2O

• Ex 2: Identify the type of crystals formed by potash alum.
  Octahedral crystals formed by crystallization.

Practice Set

• Q1: Mohr's salt is a
  A. Simple salt
  B. Double salt
  C. Complex salt
  D. Acid salt

• Q2: Which acid is added during preparation of Mohr's salt to prevent hydrolysis?
  A. Hydrochloric acid
  B. Nitric acid
  C. Sulfuric acid
  D. Acetic acid

• Q3: Potash alum contains which of the following ions?
  A. K+ and Fe3+
  B. K+ and Al3+
  C. Na+ and Al3+
  D. Na+ and Fe3+

Answer Key

• Q1: B - Mohr's salt is a double salt formed by combination of ferrous sulfate and ammonium sulfate.

• Q2: C - Sulfuric acid is added to prevent hydrolysis during preparation.

• Q3: B - Potash alum contains potassium (K+) and aluminum (Al3+) ions.

Memory Aids

• Mohr's salt = Fe + NH4 + SO4, stable ferrous ammonium sulfate.

• Alum formula: M2SO4·M'2(SO4)3·24H2O (M = monovalent, M' = trivalent metal).

• Boil water before Mohr's salt prep to remove O2 and prevent oxidation.

Qualitative Salt Analysis

Dry Tests

• Action of Heat: Heating salts causes color changes useful for identification.

• Example: PbO changes from light buff (cold) to red (hot).

• 

  Flame Test: Salt + conc. HCl heated on Pt wire; metal ions give characteristic flame colors.

• Na+: golden yellow; K+: lilac; Ca2+: brick red; Ba2+: apple green.

• Pt wire preferred as it is inert and does not color flame.

• 

  Borax Bead Test: Borax heated on Pt wire forms sodium metaborate bead; colored salts produce colored metaborates in oxidizing flame.

Wet Tests: Identification of Acid Radicals

• First Group Acid Radicals: CO32−, S2−, SO32−, NO2−; reagent: dilute H2SO4.

• Carbonate: Effervescence of CO2 with dilute acid; lime water turns milky.

• Sulphide: H2S gas with rotten egg smell; black precipitate with Pb(CH3COO)2.

• Sulphite: SO2 gas evolved; reduces K2Cr2O7 from orange to green.

• Nitrite: Brown gas (NO2) formed; blackens ferrous sulfate solution.

• 

  Second Group Acid Radicals: Cl−, Br−, I−, NO3−; reagent: concentrated H2SO4.

• Chloride: Colorless gas with sharp smell; white fumes with NH4OH; confirmed by chromyl chloride and silver nitrate tests.

• Bromide: Brown gas; pale yellow precipitate with AgNO3.

• Iodide: Violet vapors; blue starch paper; yellow precipitate with AgNO3.

• Nitrate: Brown ring test with FeSO4 and H2SO4.

• 

  Third Group Acid Radical: SO42−; confirmed by white precipitate of BaSO4 with BaCl2 in acidified solution.

Wet Tests: Identification of Basic Radicals

• Basic radicals (cations) are separated into groups I to VI and zero based on their precipitation reactions.

• Group I: Pb2+, Ag+; reagent: dilute HCl; white precipitate formed.

• Group II: Pb2+, Hg2+, Cu2+, Cd2+; reagent: H2S in dilute HCl; black or yellow precipitates.

• Group III: Al3+, Fe2+, Fe3+; reagent: NH4Cl + NH4OH; gelatinous or colored precipitates.

• Group IV: Zn2+, Mn2+, Co2+, Ni2+; reagent: NH4Cl + NH4OH + H2S; white, pink, or black precipitates.

• Group V: Ca2+, Sr2+, Ba2+; reagent: NH4Cl + NH4OH + (NH4)2CO3; white precipitates; flame tests confirm identity.

• Group VI: Mg2+; reagent: NH4OH + NH4HPO4; white precipitate.

• Zero group: NH4+; identified by evolution of NH3 gas on heating with NaOH.

• H2S in dilute HCl precipitates only group II cations due to common ion effect reducing S2− concentration.

Solved Examples

• Ex 1: Identify the cation if a white precipitate forms with dilute HCl and dissolves in hot water.
  Answer: Pb2+

• Ex 2: Which test confirms presence of NH4+ ions?
  Answer: Heating with NaOH produces NH3 gas turning moist red litmus blue; confirmed by Nessler's reagent.

Practice Set

• Q1: Which acid radical gives a white precipitate with BaCl2 in acidified solution?
  A. Carbonate
  B. Sulfate
  C. Sulfide
  D. Nitrate

• Q2: Which group of cations precipitates as sulfides in presence of H2S and dilute HCl?
  A. Group I
  B. Group II
  C. Group III
  D. Group IV

• Q3: In flame test, which ion gives lilac color?
  A. Na+
  B. K+
  C. Ca2+
  D. Ba2+

Answer Key

• Q1: B - Sulfate ions form white BaSO4 precipitate with BaCl2.

• Q2: B - Group II cations precipitate as sulfides in H2S and dilute HCl.

• Q3: B - Potassium ion gives lilac flame color.

Memory Aids

• Group I cations: Pb2+, Ag+ precipitate with dilute HCl.

• Group II cations: precipitate as sulfides in acidic H2S solution.

• Flame test colors: Na - yellow, K - lilac, Ca - brick red, Ba - apple green.

• NH4+ detected by NH3 gas on heating with NaOH.

Volumetric Analysis

Overview

• Measures concentration of a substance by volume of its solution in a chemical reaction.

• Main process: titration.

• Types: acid-base, redox, iodine, complexometric titrations.

Solved Examples

• Ex 1: Calculate normality if 25 mL of titrant neutralizes 20 mL of analyte with known normality 0.1 N.
  Solution: N1V1 = N2V2 → N2 = (N1V1)/V2 = (0.1 × 20)/25 = 0.08 N.

Practice Set

• Q1: What is the main purpose of volumetric analysis?
  A. Measure mass
  B. Measure volume
  C. Measure concentration
  D. Measure temperature

• Q2: Which of the following is NOT a type of titration?
  A. Acid-base
  B. Oxidation-reduction
  C. Precipitation
  D. Chromatographic

• Q3: In titration, the point where reaction is just complete is called:
  A. End point
  B. Equivalence point
  C. Indicator point
  D. Titration point

Answer Key

• Q1: C - Volumetric analysis measures concentration by volume.

• Q2: D - Chromatographic is not a titration type.

• Q3: B - Equivalence point is where reaction is stoichiometrically complete.

Memory Aids

• Volumetric = Volume-based concentration measurement.

• Equivalence point = Stoichiometric completion.

• Types: Acid-base, Redox, Iodine, Complexometric.

Titration Terms

Key Definitions

• Analyte (Titre): Substance being analyzed.

• Titrant: Solution added to analyte.

• Equivalence Point: Reaction just complete; stoichiometric amounts reacted.

• Indicator: Compound changing color near equivalence point.

• Standard Solution: Solution with known concentration.

• Standardization: Process to determine concentration of reagent using known standard.

• Primary Standard: Pure reagent for direct standard solution preparation (e.g., oxalic acid, Mohr’s salt).

• Secondary Standard: Reagent not pure enough for direct standard solution (e.g., KMnO4, NaOH).

Solved Examples

• Ex 1: Identify primary standard among NaOH, oxalic acid, KMnO4.
  Answer: Oxalic acid (pure, stable, non-hygroscopic).

Practice Set

• Q1: What is the role of an indicator?
  A. Reactant
  B. Change color at equivalence point
  C. Titrant
  D. Solvent

• Q2: Which is a primary standard?
  A. KMnO4
  B. NaOH
  C. Oxalic acid
  D. HCl

• Q3: Standardization is:
  A. Preparing solution
  B. Measuring volume
  C. Determining concentration of reagent
  D. Mixing solutions

Answer Key

• Q1: B - Indicator changes color at equivalence point.

• Q2: C - Oxalic acid is primary standard.

• Q3: C - Standardization finds reagent concentration.

Memory Aids

• Primary standard: Pure, stable, non-hygroscopic.

• Indicator = Color change at equivalence.

• Standardization = Concentration determination.

Law of Chemical Equivalence

Concepts and Equations

• Number of equivalents of analyte = Number of equivalents of titrant.

• Equivalents = Normality (N) × Volume (V in L).

• Normality equation: N1V1 = N2V2.

• For volume in mL: milliequivalents = Normality × Volume (mL).

• Molarity equation (for dilution): M1V1 = M2V2.

• Normality = Molarity × valency factor (n).

• Combined: M1V1n1 = M2V2n2.

Solved Examples

• Ex 1: Calculate normality if 25 mL of 0.1 M acid reacts with 30 mL base of valency 1.
  Solution: NacidVacid = NbaseVbase → (0.1×25×1) = Nbase×30×1 → Nbase = 0.0833 N.

Practice Set

• Q1: Normality is defined as:
  A. Molarity × Volume
  B. Molarity × Valency factor
  C. Volume × Valency
  D. Molarity / Valency

• Q2: If 20 mL of 0.1 N acid neutralizes 25 mL base, normality of base is:
  A. 0.08 N
  B. 0.125 N
  C. 0.1 N
  D. 0.2 N

• Q3: For diprotic acid titrated with monobasic base, relation is:
  A. M1V1 = M2V2
  B. M1V1n1 = M2V2n2
  C. N1V1 = N2V2
  D. None

Answer Key

• Q1: B - Normality = Molarity × valency.

• Q2: A - Nbase = (0.1×20)/25 = 0.08 N.

• Q3: B - Use molarity × volume × valency relation.

Memory Aids

• Normality = Molarity × n (valency).

• Use N1V1 = N2V2 for titrations.

• Valency factor = number of replaceable H+ or OH−.

Acid-Base Titration

Key Points

• Neutralization: H+ + OH− → H2O.

• Equivalence point: sudden pH change; end point detected by indicator.

• Indicators: weak acids/bases changing color near equivalence point.

• Common indicators: phenolphthalein (colorless to pink), methyl orange (red to yellow).

• Indicator color change corresponds to pH range near equivalence point.

• Strong acid + strong base: equivalence pH = 7.

• Weak acid + strong base: equivalence pH > 7 (basic salt hydrolysis).

• Strong acid + weak base: equivalence pH < 7 (acidic salt hydrolysis).

• Weak acid + weak base: no sharp end point; pH ~7.

Solved Examples

• Ex 1: Identify suitable indicator for strong acid-strong base titration.
  Answer: Phenolphthalein or methyl orange.

• Ex 2: pH at equivalence point for HCl + NaOH titration.
  Answer: 7 (neutral).

Practice Set

• Q1: Which indicator is best for weak acid-strong base titration?
  A. Methyl orange
  B. Phenolphthalein
  C. Thymol blue
  D. Bromothymol blue

• Q2: pH at equivalence point of strong acid-weak base titration is:
  A. 7
  B. >7
  C. <7
  D. Variable

• Q3: In weak acid-weak base titration, the titration curve:
  A. Has sharp end point
  B. Has no sharp end point
  C. pH at equivalence is 14
  D. pH at equivalence is 0

Answer Key

• Q1: B - Phenolphthalein suitable for weak acid-strong base.

• Q2: C - pH < 7 due to acidic salt hydrolysis.

• Q3: B - No sharp end point; pH ~7.

Memory Aids

• Strong acid + strong base → pH 7 equivalence.

• Weak acid + strong base → pH > 7 equivalence.

• Strong acid + weak base → pH < 7 equivalence.

• Phenolphthalein pink in base; methyl orange red in acid.

Redox Titrations

Essentials

• Use KMnO4 as oxidizing agent in acidic medium.

• Reducing agents: Mohr’s salt, oxalic acid, H2O2, etc.

• KMnO4 acts as self-indicator: colorless to pink at end point.

• Redox reaction example: MnO4− + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O.

• Equivalent mass of KMnO4 in acid = Molar mass / 5 = 31.6 g/equiv.

Solved Examples

• Ex 1: Calculate equivalent mass of KMnO4.
  Answer: 158/5 = 31.6 g/equiv.

• Ex 2: Write ionic equation for oxalic acid titration with KMnO4.
  Answer: 2MnO4− + 16H+ + 5C2O42− → 2Mn2+ + 8H2O + 10CO2.

Practice Set

• Q1: KMnO4 acts as:
  A. Reducing agent
  B. Oxidizing agent
  C. Indicator only
  D. None

• Q2: End point color change in KMnO4 titration is:
  A. Colorless to pink
  B. Pink to colorless
  C. Blue to colorless
  D. Yellow to red

• Q3: Number of electrons lost by KMnO4 in acidic medium:
  A. 1
  B. 3
  C. 5
  D. 7

Answer Key

• Q1: B - KMnO4 is oxidizing agent.

• Q2: A - Colorless to permanent pink.

• Q3: C - KMnO4 loses 5 electrons.

Memory Aids

• KMnO4 self-indicator: pink color at end point.

• Equivalent mass = molar mass / electrons transferred.

• Redox titrations in acidic medium.

Enthalpy of Solution

Concepts

• Enthalpy change when 1 mole of substance dissolves in excess solvent.

• Heat of solution can be endothermic (+) or exothermic (−).

• Measured using calorimeter (polythene bottle).

• Formula: ΔH = (W + m) × Δt × 4.2 × M / w (J/mol), where W = water equivalent, m = water mass, Δt = temperature change, M = molar mass, w = solute mass.

• Lattice energy: energy to dissociate ionic solid into ions.

• Hydration energy: energy released when ions hydrate.

• If lattice energy > hydration energy → endothermic dissolution.

• If lattice energy < hydration energy → exothermic dissolution.

• Example: CuSO4 anhydrous dissolution is exothermic; KCl dissolution is endothermic.

Solved Examples

• Ex 1: Calculate enthalpy if 5 g CuSO4 dissolved in 150 mL water causes 2°C rise, W=10 g, M=160 g/mol.
  Solution: ΔH = (10+150)×2×4.2×160/5 = 21504 J/mol.

Practice Set

• Q1: Enthalpy of solution is:
  A. Heat change on dissolving 1 mole
  B. Heat change on mixing
  C. Heat change on reaction
  D. Heat change on evaporation

• Q2: If lattice energy > hydration energy, dissolution is:
  A. Exothermic
  B. Endothermic
  C. Neutral
  D. Variable

• Q3: Which salt dissolution is exothermic?
  A. KCl
  B. CuSO4 anhydrous
  C. CuSO4.5H2O
  D. NaCl

Answer Key

• Q1: A - Enthalpy of solution is heat change on dissolving 1 mole.

• Q2: B - Endothermic if lattice energy > hydration energy.

• Q3: B - Anhydrous CuSO4 dissolution is exothermic.

Memory Aids

• ΔH positive = endothermic; negative = exothermic.

• Lattice energy vs hydration energy decides heat sign.

• Hydrated salts often dissolve endothermically.

Preparation of Lyophilic and Lyophobic Sols

Types and Methods

• Lyophilic sols: Dispersed phase has affinity for medium; stable; reversible; prepared by direct mixing (e.g., starch in water).

• Lyophobic sols: Little affinity; unstable; irreversible; prepared by special methods.

• Lyophobic sols prepared by condensation (chemical/physical) and dispersion methods.

• Chemical methods: double decomposition, oxidation, reduction, hydrolysis.

• Physical methods: solvent exchange, excessive cooling.

• Dispersion methods: mechanical dispersion, electrical disintegration (Bredig’s arc), peptization.

Solved Examples

• Ex 1: Prepare ferric hydroxide sol.
  Answer: Hydrolysis of FeCl3 with water.

• Ex 2: Method to prepare gold sol.
  Answer: Electrical disintegration (Bredig’s arc).

Practice Set

• Q1: Lyophilic sols are:
  A. Unstable
  B. Stable and reversible
  C. Irreversible
  D. Prepared by special methods only

• Q2: Peptization is:
  A. Breaking particles
  B. Converting precipitate to sol
  C. Mixing solvents
  D. Cooling solution

• Q3: Bredig’s arc method is used for:
  A. Lyophilic sols
  B. Lyophobic sols of metals
  C. Organic sols
  D. None

Answer Key

• Q1: B - Lyophilic sols are stable and reversible.

• Q2: B - Peptization converts precipitate to sol.

• Q3: B - Bredig’s arc prepares metal lyophobic sols.

Memory Aids

• Lyophilic = love solvent; stable, reversible.

• Lyophobic = fear solvent; unstable, special prep.

• Peptization = precipitate to sol by shaking with electrolyte.

Kinetic Study of Iodide with Hydrogen Peroxide

Reaction and Rate

• Reaction: H2O2 + 2I− + 2H+ → I2 + 2H2O (slow).

• Iodine reacts with sodium thiosulphate: I2 + 2S2O32− → 2I− + S4O62− (fast).

• When thiosulphate is consumed, iodine reacts with starch to form blue complex.

• Time for blue color appearance indicates reaction rate.

• Rate ∝ concentration of reactants.

• Rate law: Rate = k[X]a[Y]b, where a,b determined experimentally.

• Order = a + b; molecularity = sum of stoichiometric coefficients.

Solved Examples

• Ex 1: Write rate expression for A + B → products if order w.r.t A is 1 and B is 2.
  Answer: Rate = k[A][B]2.

Practice Set

• Q1: Rate of reaction is:
  A. Change in concentration per unit time
  B. Change in volume
  C. Change in mass
  D. Change in temperature

• Q2: In rate law Rate = k[X]a[Y]b, a and b are:
  A. Stoichiometric coefficients
  B. Experimental orders
  C. Rate constants
  D. Concentrations

• Q3: Which factor does NOT affect reaction rate?
  A. Concentration
  B. Temperature
  C. Catalyst
  D. Color of reactants

Answer Key

• Q1: A - Rate is change in concentration per time.

• Q2: B - a and b are experimentally determined orders.

• Q3: D - Color does not affect rate.

Memory Aids

• Rate = change in concentration/time.

• Order determined experimentally, not always stoichiometric.

• Factors affecting rate: concentration, temperature, catalyst, surface area.