JEE Mains Chemistry · Some Basic Concepts of Chemistry

Equivalents and Titrations

Equal milliequivalents at the end point for acid–base and redox titrations, and the rules that make a substance a primary standard.

All Some Basic Concepts of Chemistry notes JEE Mains Chemistry strategy

Why this matters

Thirty-one PYQs, twenty of them numerical and six from 2026. Twelve are acid–base neutralisations, where the number of H⁺ or OH⁻ sets the n-factor. Thirteen are redox titrations with permanganate, dichromate or thiosulphate, and six test the rules for a primary standard. One equation, equal milliequivalents, solves every calculation on the page.

Concept 1 of 3: Acid–base titration and the n-factor

At the end point the acid has given as many

\mathrm{H^+}

as the base has taken. Count each side as molarity

\times

volume

\times

the number of

\mathrm{H^+}

\mathrm{OH^-}

per formula unit, and set them equal.

Definition

Milliequivalents $=M\times n\times V(\mathrm{mL})$ . At the end point, meq of acid $=$ meq of base.
n-factor: HCl and $\mathrm{HNO_3}$ 1; $\mathrm{H_2SO_4}$ and $\mathrm{H_2C_2O_4}$ 2; $\mathrm{H_3PO_4}$ 3 when fully neutralised; NaOH 1; $\mathrm{Ca(OH)_2}$ and $\mathrm{Ba(OH)_2}$ 2.
Normality $N=M\times n$ .
Excess acid or base: subtract the meq, then divide by the total volume.
Back titration: add a known excess, titrate what is left; the amount that reacted is added $-$ left.
A compound that makes acids in water counts all of them: $\mathrm{SO_2Cl_2+2H_2O\rightarrow H_2SO_4+2HCl}$ gives 4 $\mathrm{H^+}$ .

End point

M_1n_1V_1=M_2n_2V_2

Worked example

What volume of

0.2

\mathrm{H_2SO_4}

neutralises 30 mL of

0.1

\mathrm{Ca(OH)_2}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q48Moderate

Example 1 · Some Basic Concepts of Chemistry · Equivalents and Titrations

x mg of pure HCl was used to make an aqueous solution. 25.0 mL of

0.1MBa(OH)_{2}

solution is used when the HCl solution was titrated against it. The numerical value of

x

\_\_\_\_

\times 10^{- 1}

. (Nearest integer) Given : Molar mass of HCl and

Ba(OH)_{2}

are 36.5 and

171.0\text{ }g{\text{ }mol}^{- 1}

respectively.

Forgetting n = 2

\mathrm{Ba(OH)_2}

\mathrm{Ca(OH)_2}

and

\mathrm{H_2SO_4}

each carry two. The options are built a factor of two apart for exactly this slip.

Drill 11 more on acid–base titration and the n-factor

Concept 2 of 3: Redox titrations: permanganate, dichromate and iodometry

The same equal-equivalents rule, with

n

now the electrons one formula unit gains or loses. In iodometry, iodine is only a messenger: count the thiosulphate and trace it back to what freed the iodine.

Definition

n-factors: $\mathrm{MnO_4^-}$ 5 in acid, 3 in neutral or basic solution; $\mathrm{Cr_2O_7^{2-}}$ 6; $\mathrm{Fe^{2+}}$ 1; $\mathrm{C_2O_4^{2-}}$ 2; $\mathrm{FeC_2O_4}$ 3, since both Fe and C are oxidised; $\mathrm{S_2O_3^{2-}}$ 1.
Iodometry: $\mathrm{I_2+2S_2O_3^{2-}\rightarrow2I^-+S_4O_6^{2-}}$ . With copper, $\mathrm{2Cu^{2+}+4I^-\rightarrow2CuI+I_2}$ , so one $\mathrm{Cu^{2+}}$ per thiosulphate.
$\mathrm{KMnO_4}$ against oxalic acid: warm to about 60 °C at the start; the $\mathrm{Mn^{2+}}$ formed then catalyses the reaction.
$\mathrm{KMnO_4}$ against Mohr's salt: no heating, or air oxidises the $\mathrm{Fe^{2+}}$ .
$\mathrm{KMnO_4}$ is its own indicator: the end point is the first lasting pink.

Redox end point

M_1n_1V_1=M_2n_2V_2,\qquad n=\text{electrons per formula unit}

Worked example

What volume of

0.02

\mathrm{KMnO_4}

oxidises 25 mL of

0.1

\mathrm{FeSO_4}

in acid?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q48Moderate

Example 2 · Some Basic Concepts of Chemistry · Equivalents and Titrations

200 cc of

x \times 10^{- 3}M

potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here

x =

Count every oxidisable part

\mathrm{FeC_2O_4}

both

\mathrm{Fe^{2+}}

(1 electron) and oxalate (2 electrons) are oxidised by permanganate, so

n=3

\mathrm{Fe^{3+}}

and sulphate take no oxidant at all.

Drill 12 more on redox titrations: permanganate, dichromate and iodometry

Concept 3 of 3: Primary standards

A primary standard is weighed out to make a solution of exactly known concentration. So it must weigh what it says it weighs: pure, dry, stable in air, not a water absorber, and heavy enough that small weighing errors do not matter.

Definition

Available pure and dry.
Stable in air: not oxidised, not reacting with $\mathrm{CO_2}$ .
Not hygroscopic, and not losing water either.
High molar mass, so weighing errors are small.
Soluble in water, and reacts quickly and stoichiometrically.
A fixed amount of water of crystallisation is allowed: oxalic acid dihydrate, borax and Mohr's salt are standards.

Substance	Primary standard?	Reason
Oxalic acid dihydrate, $\mathrm{H_2C_2O_4\cdot2H_2O}$	Yes	Stable crystals of fixed composition
Potassium hydrogen phthalate (KHP)	Yes	High molar mass, not hygroscopic; standardises NaOH with phenolphthalein
Mohr's salt, $\mathrm{(NH_4)_2Fe(SO_4)_2\cdot6H_2O}$	Yes	Its $\mathrm{Fe^{2+}}$ resists air oxidation, unlike ferrous sulphate
$\mathrm{K_2Cr_2O_7}$	Yes	Pure, stable and not hygroscopic
Borax, $\mathrm{Na_2B_4O_7\cdot10H_2O}$	Yes	Used to standardise acids
NaOH	No	Absorbs water and $\mathrm{CO_2}$ from air
$\mathrm{KMnO_4}$	No	Hard to get pure; slowly reduced by light and traces of organic matter
$\mathrm{Na_2Cr_2O_7}$	No	Hygroscopic, unlike the potassium salt
Ferrous sulphate hydrates	No	Air oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$

The potassium dichromate is a standard; the sodium one is not.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q106Moderate

Example 3 · Some Basic Concepts of Chemistry · Equivalents and Titrations

Compounds that should not be used as primary standards in titrimetric analysis are A.

{Na}_{2}{Cr}_{2}O_{7}

B. Oxalic acid C. NaOH D.

{FeSO}_{4} \cdot 6H_{2}O

E. Sodium tetraborate Choose the most appropriate answer from the options given below:

Hydrated salts can qualify

Water of crystallisation is fine when its amount is fixed. What rules a salt out is taking up or losing water, or being oxidised by air.

Drill 5 more on primary standards

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Acid–base titration and the n-factor
End point
$M_1n_1V_1=M_2n_2V_2$
Redox titrations: permanganate, dichromate and iodometry
Redox end point
$M_1n_1V_1=M_2n_2V_2,\qquad n=\text{electrons per formula unit}$

Reference tables (1)

Primary standards9 rows

Substance	Primary standard?	Reason
Oxalic acid dihydrate, $\mathrm{H_2C_2O_4\cdot2H_2O}$	Yes	Stable crystals of fixed composition
Potassium hydrogen phthalate (KHP)	Yes	High molar mass, not hygroscopic; standardises NaOH with phenolphthalein
Mohr's salt, $\mathrm{(NH_4)_2Fe(SO_4)_2\cdot6H_2O}$	Yes	Its $\mathrm{Fe^{2+}}$ resists air oxidation, unlike ferrous sulphate
$\mathrm{K_2Cr_2O_7}$	Yes	Pure, stable and not hygroscopic
Borax, $\mathrm{Na_2B_4O_7\cdot10H_2O}$	Yes	Used to standardise acids
NaOH	No	Absorbs water and $\mathrm{CO_2}$ from air
$\mathrm{KMnO_4}$	No	Hard to get pure; slowly reduced by light and traces of organic matter
$\mathrm{Na_2Cr_2O_7}$	No	Hygroscopic, unlike the potassium salt
Ferrous sulphate hydrates	No	Air oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$

The potassium dichromate is a standard; the sodium one is not.

Watch out for (3)

Forgetting n = 2
→ Acid–base titration and the n-factor
Count every oxidisable part
→ Redox titrations: permanganate, dichromate and iodometry
Hydrated salts can qualify
→ Primary standards

Test yourself on Some Basic Concepts of Chemistry

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Some Basic Concepts of Chemistry chapter test Drill all 31 questions