Volumetric Analysis - Acid-Base and Redox Titrations

Real-Life Hook: Precision That Saves Lives

Imagine you’re a pharmaceutical quality control analyst. A batch of medicine is ready for distribution. But is the drug concentration exactly what it should be? Too little might not cure; too much could be toxic. Volumetric analysis gives you the answer with precision!

From ensuring drug potency (pharmaceutical industry) to monitoring water hardness (environmental testing) to determining vitamin C content (nutritional analysis), volumetric analysis is chemistry’s quantitative workhorse:

• Pharmaceutical QC: Verifying drug concentrations (aspirin, antibiotics)
• Environmental monitoring: Measuring water hardness, chlorine levels in pools
• Food industry: Determining acidity in juices, milk
• Forensic chemistry: Quantifying substances in evidence
• Medical diagnostics: Blood chemistry (glucose, cholesterol)
• Industrial processes: Quality control in chemical manufacturing

The titration skills you’ll learn are used millions of times daily across the world to ensure safety, quality, and precision!

What is Volumetric Analysis?

Volumetric Analysis (or Titrimetric Analysis) is a quantitative analytical technique where the concentration of a substance is determined by measuring the volume of a standard solution required to react completely with it.

Key Principle: Stoichiometry - react known quantity with unknown to find the unknown

Terminology

Titration: Process of adding a solution of known concentration to a solution of unknown concentration until the reaction is complete

Titrant: Solution of known concentration (added from burette)

Analyte: Solution of unknown concentration (taken in conical flask)

Standard Solution: Solution whose concentration is accurately known

Equivalence Point: Point where reaction is stoichiometrically complete (moles of reactants exactly as per equation)

End Point: Point where indicator changes color (detected experimentally)

Ideally: Equivalence point = End point (indicator chosen to change color exactly at equivalence point)

Types of Titrations

Classification by Reaction Type

For JEE: Focus on Acid-Base and Redox titrations

Apparatus and Glassware

Essential Equipment

1. Burette (50 mL)

• Long graduated tube with stopcock
• Holds titrant (standard solution)
• Least count: 0.1 mL
• Reading: Note meniscus level (bottom of curved surface)

2. Pipette (10 mL or 20 mL)

• Glass tube with bulb
• Transfers accurate volume of analyte
• More accurate than measuring cylinder

3. Conical Flask (250 mL)

• Receives analyte from pipette
• Conical shape prevents spillage during swirling
• Never use volumetric flask for titration!

4. Funnel

• Fill burette with titrant
• Remove before titration!

5. Wash Bottle

• Contains distilled water
• Wash down walls of flask during titration

6. White Tile

• Place under flask
• Easier to see indicator color change

7. Dropper/Glass Rod

• Add indicator

8. Beakers

• Hold stock solutions

Glassware Care

Cleaning:

• Clean all glassware with detergent and water
• Rinse with tap water, then distilled water
• For acidic solutions: rinse with acid
• For basic solutions: rinse with base

Burette Preparation:

1. Clean and rinse with water
2. Rinse with titrant (2-3 mL, 2-3 times)
3. Fill with titrant above zero mark
4. Remove air bubbles from stopcock
5. Adjust level to zero or slightly below

Pipette Preparation:

1. Clean and rinse with water
2. Rinse with analyte solution (2-3 times)
3. Fill above mark
4. Adjust to exact mark (meniscus at mark)

Why rinse with solution?

• Removes water droplets
• Prevents dilution of solution
• Ensures accurate concentration

Acid-Base Titrations

Principle

Neutralization reaction:

Or generally:

Theory and pH Changes

During titration, pH changes gradually at first, then sharply near equivalence point, then gradually again.

Example: Strong acid (HCl) vs Strong base (NaOH)

pH
14 |                    ________
   |                  /
   |                /  ← Sharp pH change
 7 |              /      at equivalence point
   |            /
 0 |__________/
   0    Volume of NaOH added (mL)    →

Equivalence point pH:

• Strong acid + Strong base: pH = 7 (neutral)
• Weak acid + Strong base: pH > 7 (slightly basic)
• Strong acid + Weak base: pH < 7 (slightly acidic)
• Weak acid + Weak base: pH ≈ 7 (depends on Ka and Kb)

Types of Acid-Base Titrations

1. Strong Acid vs Strong Base

Example: HCl vs NaOH

Reaction:

Equivalence point: pH = 7

Indicators: Phenolphthalein (8.3-10), Methyl orange (3.1-4.4), Methyl red (4.2-6.3)

Calculation:

Where M = molarity, V = volume

2. Weak Acid vs Strong Base

Example: CH₃COOH vs NaOH

Reaction:

Equivalence point: pH > 7 (≈8-9)

• Salt (CH₃COONa) hydrolyzes to give basic solution
• $\ce{CH3COO^- + H2O <=> CH3COOH + OH^-}$

Indicator: Phenolphthalein (pink in base, colorless in acid)

NOT methyl orange (changes in acidic range, misses equivalence point)

3. Strong Acid vs Weak Base

Example: HCl vs NH₄OH

Reaction:

Equivalence point: pH < 7 (≈5-6)

• Salt (NH₄Cl) hydrolyzes to give acidic solution
• $\ce{NH4^+ + H2O <=> NH3 + H3O^+}$

Indicator: Methyl orange or Methyl red

NOT phenolphthalein (changes in basic range, misses equivalence point)

4. Weak Acid vs Weak Base

Example: CH₃COOH vs NH₄OH

Equivalence point: pH ≈ 7 (depends on Ka and Kb)

Problem: No sharp pH change at equivalence point

Indicator: Very difficult to choose; this titration is rarely done

Common Indicators

Memory Trick - “Phenol Prefers Basic”

Phenolphthalein changes in PH high (basic) range → use for weak acid + strong base

Memory Trick - “Methyl in Middle”

Methyl orange and Methyl red change in acidic-middle range → use for strong acid + weak base

Procedure for Acid-Base Titration

Example: Standardizing NaOH (unknown) using oxalic acid (standard)

Step 1: Preparation of Standard Oxalic Acid Solution

1. Weigh accurately 1.26 g of pure oxalic acid (H₂C₂O₄·2H₂O)
2. Dissolve in distilled water in beaker
3. Transfer to 250 mL volumetric flask
4. Make up to mark with distilled water
5. Shake well

Molarity: $M = \frac{1.26}{126 \times 0.25} = 0.04$ M (N/25)

Step 2: Filling the Burette

1. Rinse burette with NaOH solution
2. Fill with NaOH above zero mark
3. Remove air bubbles from nozzle and stopcock
4. Adjust to zero mark (or note initial reading)

Step 3: Preparing the Conical Flask

1. Rinse pipette with oxalic acid solution
2. Pipette out 10 mL oxalic acid into conical flask
3. Add 1-2 drops of phenolphthalein indicator
4. Solution remains colorless (acidic)

Step 4: Rough Titration

1. Add NaOH from burette rapidly with continuous swirling
2. When pink color appears, stop immediately
3. Note rough burette reading
4. This is rough titre value (e.g., 9.8 mL)

Step 5: Accurate Titrations (Concordant Values)

1. Repeat titration 3-4 times
2. Add NaOH dropwise near end point
3. End point: Permanent faint pink color (lasts 30 sec)
4. Note final burette readings

Step 6: Record Observations

Concordant values: 10.0 mL (readings within 0.1 mL)

Step 7: Calculations

Reaction:

Using: $\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}$

Where n = number of H⁺ or OH⁻

For oxalic acid: n₁ = 2 (diprotic) For NaOH: n₂ = 1 (monoprotic)

Answer: Molarity of NaOH = 0.02 M (N/50)

Common Mistakes in Acid-Base Titrations

Mistake 1: Not rinsing burette/pipette with respective solutions

Problem: Water droplets dilute solutions, affect concentration

Solution: Always rinse burette with titrant, pipette with analyte (2-3 times)

Mistake 2: Air bubbles in burette nozzle

Problem: Occupies volume, gives incorrect reading

Solution: Tap burette while filling, run out some solution to remove air

Mistake 3: Reading burette from top of meniscus

Problem: Incorrect volume measurement

Solution: Always read from bottom of meniscus at eye level (avoid parallax error)

Mistake 4: Not removing funnel from burette

Problem: Drops from funnel fall into burette, increase volume

Solution: Remove funnel after filling burette

Mistake 5: Overshooting end point

Problem: Pink color appears then disappears → added too much, keep adding more

Solution: Add dropwise (or half-drops) near end point; permanent faint pink is end point

Mistake 6: Washing conical flask with titrant/analyte

Problem: Changes amount of analyte

Solution: Rinse conical flask with distilled water only (adding water doesn’t change moles of analyte)

Mistake 7: Using wrong indicator

Problem: Indicator changes before/after equivalence point

Solution: Match indicator to titration type (phenolphthalein for weak acid + strong base, etc.)

Redox Titrations

Principle

Oxidation-Reduction reactions: Transfer of electrons

Oxidation: Loss of electrons (increase in oxidation number) Reduction: Gain of electrons (decrease in oxidation number)

In titration: Oxidizing agent (titrant) oxidizes reducing agent (analyte) or vice versa

Types of Redox Titrations

1. Permanganometry

Titrant: KMnO₄ (potassium permanganate)

Medium: Acidic (dilute H₂SO₄)

Why not HCl or HNO₃?

• HCl: Oxidized by KMnO₄ → $\ce{2Cl^- -> Cl2 + 2e^-}$
• HNO₃: Already an oxidizing agent (interferes)
• H₂SO₄: Not oxidized, doesn’t interfere

Reaction in acidic medium:

Purple (MnO₄⁻) → Colorless (Mn²⁺)

Self Indicator: KMnO₄ itself acts as indicator!

• During titration: Purple color disappears (reduced to Mn²⁺)
• At end point: Permanent faint pink color (slight excess of MnO₄⁻)
• No additional indicator needed!

n-factor of KMnO₄ (acidic): 5 (each Mn goes from +7 to +2)

2. Dichromatometry

Titrant: K₂Cr₂O₇ (potassium dichromate)

Medium: Acidic (dilute H₂SO₄)

Reaction:

Orange (Cr₂O₇²⁻) → Green (Cr³⁺)

Indicator: Diphenylamine or potassium ferricyanide

• Not self-indicator (color change not sharp)

n-factor of K₂Cr₂O₇: 6 (each Cr goes from +6 to +3, two Cr atoms)

3. Iodometry and Iodimetry

Iodimetry: I₂ is titrant (standard iodine solution)

• Direct titration
• Analyte is reducing agent

Iodometry: I₂ is liberated, then titrated (indirect)

• Analyte is oxidizing agent
• Liberates I₂ from KI
• Liberated I₂ titrated with Na₂S₂O₃

Reaction:

Indicator: Starch (gives blue-black color with I₂)

• At end point: Blue color disappears (I₂ reduced to I⁻)

n-factor of I₂: 2

Permanganate Titration - Detailed Procedure

Example: Estimating Fe²⁺ using KMnO₄

Reaction:

Ratio: 1 mole MnO₄⁻ : 5 moles Fe²⁺

Step 1: Preparation of KMnO₄ Solution

1. Dissolve KMnO₄ in distilled water (approx. 0.02 M)
2. KMnO₄ cannot be used as primary standard (impure, hygroscopic)
3. Must be standardized using oxalic acid

Step 2: Standardization of KMnO₄ using Oxalic Acid

Reaction:

Procedure:

1. Pipette 10 mL standard oxalic acid (0.05 M) into conical flask
2. Add 10 mL dilute H₂SO₄ (provides acidic medium)
3. Heat to 60-70°C (reaction is slow at room temperature)
4. Titrate with KMnO₄ from burette
5. End point: Permanent faint pink color

Calculation:

For oxalic acid: n₁ = 2 (provides 2e⁻ per molecule × 2 = total 2) Actually, oxalic acid: $\ce{H2C2O4 -> 2CO2 + 2H+ + 2e^-}$

For KMnO₄: n₂ = 5

Find M₂ (molarity of KMnO₄)

Step 3: Estimation of Fe²⁺

1. Pipette 10 mL Fe²⁺ solution (Mohr’s salt, ferrous ammonium sulfate)
2. Add 10 mL dilute H₂SO₄
3. Titrate with standardized KMnO₄
4. End point: Permanent faint pink color

Calculation:

For KMnO₄: n₁ = 5 For Fe²⁺: n₂ = 1 (Fe²⁺ → Fe³⁺ + e⁻)

Find M₂ (molarity of Fe²⁺)

Why Heat Oxalic Acid During Titration?

Reason 1: Reaction kinetics

The reaction is autocatalytic:

• Initially slow (no Mn²⁺ present)
• Mn²⁺ produced acts as catalyst
• Reaction speeds up as Mn²⁺ forms
• Heating accelerates initial slow phase

Reason 2: Complete reaction

• At room temperature, reaction incomplete
• Some oxalic acid remains unreacted
• Heating ensures complete oxidation
• More accurate results

Reason 3: Preventing overshoot

• Cold solution → slow decolorization
• May add excess KMnO₄ thinking reaction incomplete
• Hot solution → rapid decolorization
• Clear indication of end point

Temperature: 60-70°C (not boiling - oxalic acid decomposes)

Iodometric Titration

Example: Estimation of Cu²⁺ using iodometry

Principle: Cu²⁺ oxidizes I⁻ to I₂, then I₂ is titrated with Na₂S₂O₃

Step 1: Liberation of I₂

White precipitate of CuI, brown color of I₂

Step 2: Titration with Thiosulfate

Indicator: Starch (added near end point)

• Starch + I₂ → Blue-black color
• End point: Blue color just disappears

Why add starch near end point?

• Starch-iodine complex is very stable
• If added initially, difficult to decompose at end point
• Add when brown color of I₂ becomes faint yellow
• Then titrate to disappearance of blue

n-factor of Na₂S₂O₃: 1 (two S atoms, each loses 0.5 e⁻)

Calculations in Volumetric Analysis

Basic Formula

For acid-base:

(if same n-factor)

General (different n-factors):

Or using normality:

Where $N = M \times n$ (normality = molarity × n-factor)

n-Factor Determination

For Acids: Number of replaceable H⁺

• HCl: n = 1
• H₂SO₄: n = 2
• H₃PO₄: n = 3 (if fully neutralized)

For Bases: Number of replaceable OH⁻

• NaOH: n = 1
• Ca(OH)₂: n = 2
• Al(OH)₃: n = 3

For Redox: n = Change in oxidation number × Number of atoms changing

Examples:

• KMnO₄ (acidic): Mn (+7 → +2), change = 5, atoms = 1 → n = 5
• K₂Cr₂O₇: Cr (+6 → +3), change = 3, atoms = 2 → n = 6
• Fe²⁺ → Fe³⁺: change = 1, atoms = 1 → n = 1
• Oxalic acid: C (+3 → +4), change = 1, atoms = 2 → n = 2

Percentage Purity Calculations

Problem: 1.0 g impure oxalic acid sample dissolved in water and made up to 250 mL. 25 mL of this required 20 mL of 0.1 N NaOH. Calculate % purity.

Solution:

Step 1: Find normality of oxalic acid

Step 2: Normality in 250 mL = 0.08 N

Step 3: Equivalent mass of oxalic acid (H₂C₂O₄·2H₂O)

Step 4: Mass of pure oxalic acid in 250 mL

Wait, this gives >100% purity! Let me recalculate…

Actually, normality of 25 mL aliquot is 0.08 N. But this 25 mL was taken from 250 mL total solution.

So mass in 250 mL:

Hmm, still same. The issue is that 25 mL required 20 mL of 0.1 N NaOH.

Let me recalculate:

This 25 mL is from 250 mL total solution. Grams of oxalic acid in 250 mL:

But sample is 1.0 g, which is LESS than 1.26 g!

Issue: Problem statement is inconsistent. Should be >1.26 g sample for <100% purity.

Let me assume sample is 1.5 g:

% Purity = $\frac{1.26}{1.5} \times 100 = 84\%$

Correct approach:

Practice Problems

Level 1 - JEE Main (Basics)

Problem 1: Why is burette rinsed with the solution to be filled in it, and not with distilled water just before filling?

Solution:

If burette is rinsed with distilled water only:

• Water droplets remain on inner walls
• When solution is added, these droplets dilute it
• Concentration decreases
• Volume measurements correct, but concentration wrong
• Calculations give incorrect results

When rinsed with the solution itself:

• Solution washes away water droplets
• Remaining droplets are of the same solution
• No dilution occurs
• Accurate concentration maintained

Procedure:

1. First, wash burette with tap water, then distilled water
2. Just before filling, rinse 2-3 times with small portions (2-3 mL) of the actual solution
3. Discard rinsings
4. Then fill burette with the solution

Same logic for pipette: Rinse with analyte solution before use

But for conical flask: Rinse with distilled water only (not with solution)

Why?

• Adding water to analyte in flask doesn’t change number of moles
• $n = M \times V$
• If V increases by dilution, M decreases proportionally, n remains same
• Titration measures moles, not concentration
• Water doesn’t react with titrant, so doesn’t interfere

Answer: Burette is rinsed with the solution to be filled (not just water) to prevent dilution by residual water droplets, which would affect the concentration and lead to inaccurate results. Conical flask is rinsed with water only because dilution doesn’t change the number of moles of analyte being titrated.

Problem 2: In titration of oxalic acid with KMnO₄, why is dilute H₂SO₄ used and not dilute HCl?

Solution:

Problem with HCl:

Cl⁻ ions from HCl can be oxidized by KMnO₄:

Result:

• KMnO₄ oxidizes both oxalic acid AND Cl⁻
• Consumes more KMnO₄ than required for oxalic acid alone
• Gives incorrect (higher) results for oxalic acid
• Cl₂ gas evolved (greenish-yellow, pungent)

Why H₂SO₄ is used:

• SO₄²⁻ ions are NOT oxidized by KMnO₄
• Very stable, doesn’t interfere
• Provides necessary acidic medium (H⁺ ions)
• No side reactions

What about HNO₃?

Also NOT used because:

• HNO₃ itself is an oxidizing agent
• Interferes with the redox reaction
• Cannot determine which oxidant caused the oxidation

Answer: Dilute H₂SO₄ is used (not HCl) because Cl⁻ ions from HCl are oxidized by KMnO₄ to Cl₂, consuming extra KMnO₄ and giving incorrect results. SO₄²⁻ ions are stable and not oxidized, providing acidic medium without interfering with the titration. HNO₃ is also avoided as it is itself an oxidizing agent.

Problem 3: Why is KMnO₄ called a self-indicator?

Solution:

Definition of Self-Indicator: A titrant that itself indicates the end point by its own color change, without needing an external indicator.

For KMnO₄:

During titration (before end point):

• Each drop of purple MnO₄⁻ added gets reduced to colorless Mn²⁺
• Reaction: $\ce{MnO4^- + 8H+ + 5e^- -> Mn^{2+} + 4H2O}$
• Purple color disappears immediately (reducing agent still present)

At equivalence point:

• All reducing agent has reacted
• No more substance to reduce MnO₄⁻

Beyond equivalence point:

• Even one drop excess of MnO₄⁻ remains unreduced
• Purple MnO₄⁻ persists in solution
• Permanent faint pink color appears

This color is the end point indicator!

Advantages:

• No need for external indicator
• Saves cost and time
• Eliminates indicator errors
• Direct visual detection

Color transition: Colorless (before end point) → Permanent faint pink (end point)

Sensitivity: Very sensitive - even 0.01 mL excess shows pink color

Answer: KMnO₄ is called self-indicator because it changes its own color at the end point. During titration, purple MnO₄⁻ is reduced to colorless Mn²⁺. At end point, when all reducing agent is consumed, even one drop excess of MnO₄⁻ remains unreduced, giving a permanent faint pink color that indicates the end point without needing any external indicator.

Level 2 - JEE Main (Application)

Problem 4: 25 mL of H₂SO₄ solution required 30 mL of 0.1 M NaOH for complete neutralization. Calculate the molarity of H₂SO₄.

Solution:

Given:

• Volume of H₂SO₄ (V₁) = 25 mL
• Volume of NaOH (V₂) = 30 mL
• Molarity of NaOH (M₂) = 0.1 M
• Molarity of H₂SO₄ (M₁) = ?

Reaction:

Method 1: Using n-factors

n-factor of H₂SO₄ = 2 (diprotic) n-factor of NaOH = 1 (monoprotic)

Method 2: Using stoichiometry

From equation: 1 mol H₂SO₄ reacts with 2 mol NaOH

Moles of NaOH = M × V = 0.1 × 0.03 = 0.003 mol

Moles of H₂SO₄ = $\frac{0.003}{2} = 0.0015$ mol

Molarity of H₂SO₄ = $\frac{\text{moles}}{\text{volume (L)}} = \frac{0.0015}{0.025} = 0.06$ M

Wait, two different answers! Let me check…

Recalculation Method 2: Moles of NaOH = 0.1 × 30/1000 = 0.003 mol From 1 H₂SO₄ : 2 NaOH Moles of H₂SO₄ = 0.003/2 = 0.0015 mol Molarity = 0.0015/(25/1000) = 0.06 M

Recalculation Method 1:

The error in Method 1: I used wrong formula!

Correct formula:

where n is from stoichiometric coefficients, not n-factor!

Actually, the correct formula for acid-base:

No wait, let me use the standard formula:

For acids and bases:

Actually this IS correct and gives M₁ = 0.24 M.

Let me recheck stoichiometry: 1 H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O

Moles NaOH = 0.1 × 0.030 = 0.003 Moles H₂SO₄ = 0.003/2 = 0.0015 Molarity H₂SO₄ = 0.0015/0.025 = 0.06 M

So stoichiometry gives 0.06 M.

The issue with first formula:

The formula $\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}$ where n = basicity/acidity

Actually means: Milliequivalents are equal

So both methods give 0.06 M when done correctly!

My error was in Method 1 setup.

Correct Answer: 0.06 M

Let me redo Method 1:

Answer: Molarity of H₂SO₄ = 0.06 M

Problem 5: In estimation of Fe²⁺ by KMnO₄ titration, why is the solution heated to 60-70°C before titration?

Solution:

Wait, Fe²⁺ solution is NOT heated!

Only oxalic acid solution is heated during titration with KMnO₄.

Let me answer for oxalic acid (which is the correct context):

For Oxalic Acid + KMnO₄ titration:

Reason 1: Reaction kinetics

• Reaction is slow at room temperature
• Initially very slow (no Mn²⁺ catalyst present)
• Heating increases reaction rate
• Makes titration faster and more practical

Reason 2: Autocatalysis

• Mn²⁺ produced acts as catalyst for further reaction
• Initial phase slowest (no Mn²⁺ yet)
• Heating compensates for this
• Once some Mn²⁺ forms, reaction self-accelerates

Reason 3: Complete reaction

• At room temperature, reaction may be incomplete
• Some oxalic acid remains unreacted
• Heat ensures complete oxidation
• Better accuracy

Reason 4: Sharp end point

• Hot solution: purple color of MnO₄⁻ disappears rapidly
• Cold solution: color fades slowly, difficult to judge end point
• May overshoot by adding excess KMnO₄

Temperature: 60-70°C

• Not too hot (oxalic acid decomposes >70°C)
• Not boiling (losses due to evaporation)
• Optimal for reaction rate

For Fe²⁺ solution: Generally NOT heated

• Reaction with KMnO₄ is fast enough at room temperature
• Heating may cause aerial oxidation: $\ce{4Fe^{2+} + O2 + 4H+ -> 4Fe^{3+} + 2H2O}$
• This would give incorrect (lower) results

Answer: For oxalic acid titration with KMnO₄, heating to 60-70°C is necessary because: (1) the reaction is slow at room temperature, (2) heating accelerates the autocatalytic reaction until enough Mn²⁺ catalyst forms, (3) ensures complete oxidation of oxalic acid, and (4) gives sharp end point. For Fe²⁺ titration, heating is NOT required as the reaction is fast at room temperature, and heating may cause aerial oxidation of Fe²⁺ to Fe³⁺.

Interactive Demo: Visualize Titration Curves

Explore how pH changes during acid-base and redox titrations with interactive curves.

Problem 6: 0.2 g of a sample of H₂O₂ is dissolved in water and the solution is made up to 100 mL. 10 mL of this solution is titrated against M/50 KMnO₄ in acidic medium. The volume of KMnO₄ required is 25 mL. Calculate the percentage purity of H₂O₂.

Solution:

Given:

• Mass of H₂O₂ sample = 0.2 g
• Total volume of solution = 100 mL
• Volume of aliquot taken = 10 mL
• Molarity of KMnO₄ = M/50 = 1/50 = 0.02 M
• Volume of KMnO₄ used = 25 mL

Reaction:

n-factors:

• KMnO₄: Mn (+7 → +2), n = 5
• H₂O₂: O (-1 → 0), 2 atoms, change = 1 each, n = 2

Step 1: Calculate milliequivalents of KMnO₄

Step 2: Milliequivalents of H₂O₂ in 10 mL = 2.5 meq

Step 3: Milliequivalents of H₂O₂ in 100 mL = 2.5 × 10 = 25 meq

Step 4: Calculate mass of pure H₂O₂

Equivalent mass of H₂O₂ = Molar mass / n-factor = 34/2 = 17

Step 5: Calculate percentage purity

This is impossible! Purity >100%!

Let me recalculate…

Rechecking Step 3: 10 mL of the 100 mL solution was used. meq in 10 mL = 2.5 meq in 100 mL = 2.5 × (100/10) = 25 meq ✓

Rechecking Step 4: Mass = (25/1000) equiv × 17 g/equiv = 0.425 g ✓

The problem: Sample is 0.2 g but contains 0.425 g pure H₂O₂?

Two possibilities:

1. Problem statement error (sample mass should be >0.425 g)
2. I made calculation error

Let me recalculate meq of KMnO₄: Normality of KMnO₄ = M × n = 0.02 × 5 = 0.1 N meq = N × V = 0.1 × 25 = 2.5 meq ✓

meq of H₂O₂ in 10 mL = 2.5 meq ✓ meq of H₂O₂ in 100 mL = 25 meq ✓

Mass of H₂O₂ = (25 meq / 1000) × 17 = 0.425 g ✓

Conclusion: The problem as stated is impossible.

If sample mass is 0.5 g: % Purity = (0.425/0.5) × 100 = 85%

Answer (with corrected sample mass of 0.5 g): 85%

If problem is as stated, there’s an error in the given data.

Level 3 - JEE Advanced (Conceptual & Numerical)

Problem 7: Explain why phenolphthalein is suitable for titration of weak acid (CH₃COOH) vs strong base (NaOH), but not for strong acid (HCl) vs weak base (NH₄OH).

Solution:

Understanding Equivalence Point pH:

Case 1: CH₃COOH (weak acid) + NaOH (strong base)

At equivalence point:

• All CH₃COOH converted to CH₃COONa
• Solution contains CH₃COO⁻ (acetate ion)
• Acetate is a weak base (conjugate of weak acid)

Hydrolysis:

• Produces OH⁻ ions
• Solution is basic at equivalence point
• pH ≈ 8-9

Phenolphthalein:

• Color change range: pH 8.3 - 10.0
• Colorless (acidic) → Pink (basic)
• Changes color right around pH 8-9
• Matches equivalence point!

Result: Sharp, accurate end point detection

Case 2: HCl (strong acid) + NH₄OH (weak base)

At equivalence point:

• All NH₄OH converted to NH₄Cl
• Solution contains NH₄⁺ (ammonium ion)
• Ammonium is a weak acid (conjugate of weak base)

Hydrolysis:

• Produces H₃O⁺ ions
• Solution is acidic at equivalence point
• pH ≈ 5-6

Phenolphthalein:

• Changes at pH 8.3 - 10.0
• At equivalence point (pH 5-6), phenolphthalein is colorless
• No color change observed!
• If we continue adding NH₄OH to see pink:
  • We overshoot equivalence point
  • Add excess base to reach pH 8-9
  • Incorrect end point!

Correct indicator for this titration: Methyl orange (pH range 3.1-4.4) or Methyl red (pH range 4.2-6.3)

• These change in acidic range
• Match the acidic equivalence point

General Rule:

Why matching is crucial:

• Indicator should change color AT equivalence point
• If indicator changes before: Underestimate titrant volume
• If indicator changes after: Overestimate titrant volume
• Matching minimizes indicator error

Answer: Phenolphthalein (pH range 8.3-10) is suitable for weak acid + strong base because the equivalence point is in the basic range (pH 8-9) due to hydrolysis of the salt (acetate ion acts as weak base). For strong acid + weak base, equivalence point is acidic (pH 5-6) due to hydrolysis of ammonium ion, which is far below phenolphthalein’s range. Phenolphthalein would remain colorless at the equivalence point, giving no indication. Methyl orange (pH 3.1-4.4) should be used instead as it changes in the acidic range.

Problem 8: In a redox titration, 100 mL of a ferrous salt solution was titrated with 50 mL of 0.1 M KMnO₄ in acidic medium. Another 100 mL of the same ferrous salt solution was titrated with 200 mL of 0.05 M K₂Cr₂O₇ in acidic medium. Show that the results are consistent.

Solution:

Titration 1: Ferrous salt + KMnO₄

Reaction:

Given:

• Volume of Fe²⁺ solution = 100 mL
• Volume of KMnO₄ = 50 mL
• Molarity of KMnO₄ = 0.1 M

n-factors:

• KMnO₄: n = 5
• Fe²⁺: n = 1

Milliequivalents of KMnO₄:

At equivalence: meq of Fe²⁺ = meq of KMnO₄ = 25

Normality of Fe²⁺:

Molarity of Fe²⁺ (n = 1):

Titration 2: Ferrous salt + K₂Cr₂O₇

Reaction:

Given:

• Volume of Fe²⁺ solution = 100 mL
• Volume of K₂Cr₂O₇ = 200 mL
• Molarity of K₂Cr₂O₇ = 0.05 M

n-factors:

• K₂Cr₂O₇: n = 6 (two Cr atoms, each +6 → +3)
• Fe²⁺: n = 1

Milliequivalents of K₂Cr₂O₇:

At equivalence: meq of Fe²⁺ = meq of K₂Cr₂O₇ = 60

Normality of Fe²⁺:

Molarity of Fe²⁺ (n = 1):

Comparison:

From KMnO₄ titration: M(Fe²⁺) = 0.25 M From K₂Cr₂O₇ titration: M(Fe²⁺) = 0.6 M

These are NOT equal! Results are inconsistent.

Let me recheck calculations…

Titration 1 recheck:

Wait, I was calculating wrong!

Let me use correct formula: From reaction: 1 MnO₄⁻ : 5 Fe²⁺

Moles of MnO₄⁻ = 0.1 × 50/1000 = 0.005 mol Moles of Fe²⁺ = 5 × 0.005 = 0.025 mol Molarity of Fe²⁺ = 0.025 / 0.1 = 0.25 M ✓ (My first calculation was correct)

Titration 2 recheck: From reaction: 1 Cr₂O₇²⁻ : 6 Fe²⁺

Moles of Cr₂O₇²⁻ = 0.05 × 200/1000 = 0.01 mol Moles of Fe²⁺ = 6 × 0.01 = 0.06 mol Molarity of Fe²⁺ = 0.06 / 0.1 = 0.6 M ✓ (This is also correct)

So the results ARE inconsistent (0.25 M vs 0.6 M)!

Either:

1. Problem statement is wrong
2. I’m misunderstanding the reaction stoichiometry
3. There’s some error in given data

Let me check if problem says “show that results are consistent” - maybe it’s asking us to prove they SHOULD be but finding they aren’t?

Actually, let me recalculate more carefully using equivalents:

Titration 1: N₁V₁ = N₂V₂ (M × n)KMnO₄ × V = (M × n)Fe²⁺ × V 0.1 × 5 × 50 = M × 1 × 100 M(Fe²⁺) = 25/100 = 0.25 M

Titration 2: 0.05 × 6 × 200 = M × 1 × 100 M(Fe²⁺) = 60/100 = 0.6 M

Results are indeed inconsistent!

For the problem to be consistent, either:

• Volume of KMnO₄ should be 50 mL AND volume of K₂Cr₂O₇ should be 50 mL (not 200 mL)
• Or concentrations should be different

Let me assume volume of K₂Cr₂O₇ is 50 mL (not 200 mL):

Titration 2 (corrected): 0.05 × 6 × 50 = M × 1 × 100 M(Fe²⁺) = 15/100 = 0.15 M

Still not matching!

Let me try: If K₂Cr₂O₇ volume is 83.33 mL: 0.05 × 6 × 83.33 = M × 1 × 100 M(Fe²⁺) = 25/100 = 0.25 M ✓ Matches!

Answer: The problem as stated has inconsistent data. For consistency, the volume of K₂Cr₂O₇ should be 83.33 mL (not 200 mL). With this correction:

• From KMnO₄: M(Fe²⁺) = 0.25 M
• From K₂Cr₂O₇: M(Fe²⁺) = 0.25 M Results are then consistent, confirming the same concentration of ferrous ions.

Problem 9: Derive the relationship between the mass of Mohr’s salt [FeSO₄(NH₄)₂SO₄·6H₂O] and volume of 0.02 M KMnO₄ required for complete oxidation.

Solution:

Mohr’s Salt: FeSO₄(NH₄)₂SO₄·6H₂O

• Contains Fe²⁺
• Molar mass = 56 + 96 + 2(18 + 96) + 6(18) = 56 + 96 + 228 + 108 = 392 g/mol

Reaction:

Stoichiometry: 1 MnO₄⁻ : 5 Fe²⁺

Let:

• Mass of Mohr’s salt = m g
• Volume of 0.02 M KMnO₄ = V mL

Step 1: Moles of Mohr’s salt

Step 2: Moles of Fe²⁺ Since 1 molecule of Mohr’s salt contains 1 Fe²⁺:

Step 3: Moles of KMnO₄ required From reaction: 5 Fe²⁺ : 1 MnO₄⁻

Step 4: Volume of 0.02 M KMnO₄

Or:

Or more precisely:

Simplified:

For V in mL, m in g:

Answer:

or

where V is volume of 0.02 M KMnO₄ in mL and m is mass of Mohr’s salt in grams.

Verification: If m = 3.92 g, then V = 1000 × 3.92 / 39.2 = 100 mL ✓

Cross-Links to Related Topics

Physical Chemistry Connections

1. Equilibrium: Acid-base equilibria, Ka, Kb, pH calculations, hydrolysis

2. Solutions: Molarity, normality, concentration units, dilution

3. Redox Electrochemistry: Oxidation-reduction, half-reactions, electrochemical series

4. Chemical Thermodynamics: Enthalpy of neutralization, spontaneity of reactions

5. Chemical Kinetics: Reaction rates (why heat oxalic acid), autocatalysis

Inorganic Chemistry Connections

1. Qualitative Analysis: Complements volumetric (qualitative vs quantitative)

2. d-Block Elements: Permanganate, dichromate chemistry, Fe²⁺/Fe³⁺

3. p-Block Elements: Oxalic acid chemistry, iodine chemistry

4. Redox Reactions: Balancing redox equations, oxidation states

Practical Chemistry Connections

1. Qualitative Analysis: Identifying what’s present before quantifying

2. Organic Tests: Quantifying organic functional groups

3. Organic Preparations: Determining yield and purity

Applications

1. Pharmaceutical: Drug potency testing, quality control
2. Environmental: Water hardness, chlorine levels, dissolved oxygen
3. Food: Acidity (pH), vitamin C content, preservatives
4. Clinical: Blood chemistry, glucose, cholesterol
5. Industrial: Quality control, process monitoring

Key Takeaways

1. Volumetric analysis: Quantitative technique using stoichiometry and volume measurements

2. Apparatus care: Rinse burette and pipette with solution to be used, not just water

3. Acid-base titrations: Match indicator to equivalence point pH

   • Strong acid + Strong base: Any indicator (pH 7)
   • Weak acid + Strong base: Phenolphthalein (pH > 7)
   • Strong acid + Weak base: Methyl orange/red (pH < 7)

4. Redox titrations:

   • KMnO₄: Self-indicator, acidic medium (H₂SO₄), n = 5
   • K₂Cr₂O₇: External indicator, acidic medium, n = 6
   • Iodometry: Starch indicator, near end point

5. Calculations: Use $\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}$ or $N_1V_1 = N_2V_2$

6. n-factor:

   • Acids/bases: H⁺ or OH⁻ provided
   • Redox: Change in oxidation number × number of atoms

7. Common mistakes: Not rinsing properly, air bubbles, parallax error, wrong indicator, overshooting end point

8. Heating: Required for oxalic acid + KMnO₄ (autocatalytic reaction); not for Fe²⁺ + KMnO₄

9. Self-indicator: KMnO₄ changes from colorless (Mn²⁺) to faint pink (MnO₄⁻) at end point

10. Precision: Concordant values (within 0.1 mL), multiple titrations, proper technique

Quick Revision Points

✓ Burette: Rinse with titrant, read meniscus bottom, 0.1 mL least count ✓ Pipette: Rinse with analyte, transfer accurate volume ✓ Conical flask: Rinse with water only (dilution OK) ✓ Phenolphthalein: Weak acid + strong base (pH 8.3-10) ✓ Methyl orange: Strong acid + weak base (pH 3.1-4.4) ✓ KMnO₄: Self-indicator, H₂SO₄ medium, n = 5, heat oxalic acid ✓ K₂Cr₂O₇: External indicator, H₂SO₄ medium, n = 6 ✓ Iodometry: Starch indicator (add near end point), blue → colorless ✓ Formula: N₁V₁ = N₂V₂ or $\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}$ ✓ n-factor: Acids (H⁺), bases (OH⁻), redox (Δ oxidation state × atoms)

Master volumetric analysis, and you master quantitative chemistry - the foundation of all analytical work!