Easy Practice — Set 1

Quick Recap — Detection & Analysis

• Functional-group tests: $-COOH$ (effervescence with $NaHCO_3$), $-OH$ (Na $\to H_2$; neutral $FeCl_3$ violet for phenol), $-CHO$ (Tollens'/Fehling's), C=C (bromine water / Baeyer's), 1$^\circ$ amine (carbylamine), methyl ketone/ethanol (iodoform).
• Anion tests: carbonate (effervescence), chloride (white AgCl, soluble in $NH_3$), sulphate (white $BaSO_4$, acid-insoluble), nitrate (brown ring).
• Cation groups (systematic analysis): I dilute HCl; II $H_2S$/acid; III $NH_4OH$; IV $H_2S$/basic; V $(NH_4)_2CO_3$; VI Mg.
• Titrations: acid-base (phenolphthalein, methyl orange), redox ($KMnO_4$ is self-indicating); $N_1V_1=N_2V_2$.

Beyond-NCERT JEE Formulae

Practical Chemistry in JEE Main rewards two skills: fast volumetric (titration) arithmetic and quick recall of salt-analysis facts. Work in equivalents to avoid balancing every equation, keep chemical species and units as plain text, and reserve symbols for the numbers.

1. Volumetric analysis — master relations

When to use: whenever one solution exactly reacts with another at an end point.

• Equivalents are equal at the equivalence point: $N_1 V_1 = N_2 V_2$ (both volumes in the same unit).
• In milliequivalents: $\text{meq} = N \times V(\text{mL}) = M \times n \times V(\text{mL})$.
• Molarity to normality: $N = M \times n$, with $n$ the n-factor.
• Percentage purity of a sample:

$$\text{purity} = \dfrac{\text{mass of pure substance}}{\text{mass of sample}} \times 100$$

• Strength in g/L: $\text{strength} = N \times E = M \times M_m$, where $E$ is the equivalent weight and $M_m$ the molar mass.

[JEE Tip] $N_1 V_1 = N_2 V_2$ holds ONLY in equivalents. Given molarities across a non 1:1 reaction, either switch to normality first or use $M_1 n_1 V_1 = M_2 n_2 V_2$; never force $M_1 V_1 = M_2 V_2$ onto a reaction whose mole ratio is not one.

2. n-factor and equivalent weight of common titrants

Here $E = \dfrac{M_m}{n}$, and the n-factor is electrons exchanged (redox) or protons exchanged (acid-base) per formula unit.

• KMnO4 (Mm 158): acidic $n = 5$ (Mn goes +7 to +2), $E = 31.6$; neutral or faintly alkaline $n = 3$ (to MnO2), $E = 52.7$; strongly alkaline $n = 1$.
• K2Cr2O7 (Mm 294): acidic $n = 6$ (two Cr, each +6 to +3), $E = 49$.
• Oxalic acid dihydrate (Mm 126): $n = 2$, $E = 63$ (the anhydrous acid, Mm 90, gives $E = 45$).
• FeSO4 and Mohr's salt (Mm 392): $n = 1$ (Fe2+ to Fe3+), so $E$ equals the molar mass.
• Sodium thiosulphate, hypo (Mm 158): iodometric $n = 1$, so $E = 158$.
• Iodine (Mm 254): as an oxidant $n = 2$, $E = 127$; one I2 consumes two hypo.

[JEE Tip] KMnO4's n-factor follows the medium: 5 in acid, 3 in neutral, 1 in strong alkali. A stem that quietly says "neutral" wants $n = 3$, not 5 — a favourite trap.

3. Redox titration stoichiometry

• KMnO4 vs oxalate (self-indicating; warm to about 60 C since Mn2+ autocatalyses the reaction):

2 KMnO4 + 5 H2C2O4 + 3 H2SO4 -> K2SO4 + 2 MnSO4 + 10 CO2 + 8 H2O

so one KMnO4 is equivalent to five Fe2+ (or oxidises five-halves of an oxalate).

• Iodometric copper: Cu2+ sets free iodine from excess KI, and that iodine is titrated with hypo (starch added late):

2 CuSO4 + 4 KI -> Cu2I2 (white) + I2 + 2 K2SO4, then I2 + 2 Na2S2O3 -> 2 NaI + Na2S4O6

Net one Cu is equivalent to one hypo, so millimoles of copper equal millimoles of thiosulphate.

[JEE Tip] In every iodometric problem the analyte first releases iodine, which is what you actually titrate. Add starch only at the pale straw-yellow stage; early starch traps iodine and the end point overshoots.

4. Double-indicator titration (NaOH / Na2CO3 / NaHCO3 vs HCl)

Phenolphthalein turns near pH 8.3 and registers all OH- plus only the first proton of carbonate (CO3 2- to HCO3-); methyl orange turns near pH 3.7 and registers the rest (HCO3- to H2CO3). Let $V_p$ be the acid to the phenolphthalein end point, $V_m$ the FURTHER acid on to the methyl-orange end point, and $M$ the acid molarity.

• $V_m = 0$: NaOH only, moles $= M V_p$.
• $V_p = V_m$: Na2CO3 only, moles $= M V_p$.
• $V_p = 0$: NaHCO3 only, moles $= M V_m$.
• $V_p > V_m$: NaOH with Na2CO3 — NaOH $= M(V_p - V_m)$, Na2CO3 $= M V_m$.
• $V_p < V_m$: Na2CO3 with NaHCO3 — Na2CO3 $= M V_p$, NaHCO3 $= M(V_m - V_p)$.

[JEE Tip] Compare $V_p$ and $V_m$ first to name the mixture, then read the amounts off the last two cases. NaOH and NaHCO3 can never coexist (they react), so that pairing is always a distractor.

5. Back titration, standardisation, primary standards

• Back titration: for an analyte that is insoluble, slow, or lacks a sharp end point. Add a known EXCESS of standard reagent A, let it react, then titrate the UNREACTED A with standard reagent B. Then equivalents of analyte $=$ equivalents of A taken $-$ equivalents of A left.
• Standardisation fixes the exact concentration of a secondary standard (KMnO4, NaOH, HCl, hypo) against a primary standard.
• Primary standards are pure, stable, non-hygroscopic solids that can be weighed directly: oxalic acid dihydrate, K2Cr2O7, Na2CO3, AgNO3. KMnO4 and hypo are NOT primary standards.

[JEE Tip] In a back titration you titrate the LEFTOVER reagent, so the analyte is always a difference of equivalents. Forgetting to subtract the leftover is the classic error.

6. Qualitative analysis — quick reference (facts)

Cation groups (group reagent): I — Pb2+, Ag+, Hg2 2+ (dilute HCl); II — Cu2+, Cd2+, Hg2+, Bi3+, As, Sb, Sn (H2S in dilute HCl); III — Fe3+, Al3+, Cr3+ (NH4Cl + NH4OH); IV — Co2+, Ni2+, Mn2+, Zn2+ (H2S with NH4OH); V — Ba2+, Sr2+, Ca2+ ((NH4)2CO3 with NH4OH); VI — Mg2+ (no group reagent).

Confirmatory tests: Cu2+ gives deep-blue [Cu(NH3)4]2+ with excess NH3 and a chocolate ppt with K4[Fe(CN)6]; Fe3+ gives blood-red colour with KSCN and Prussian blue with K4[Fe(CN)6]; Ni2+ gives a rosy-red ppt with DMG; Ba2+ an apple-green flame and yellow BaCrO4; Ca2+ a brick-red flame and white ppt with ammonium oxalate.

Anion tests: CO3 2- gives effervescence of CO2 (lime water milky); Cl- a white AgCl soluble in NH3 and red chromyl-chloride vapours; Br- / I- an orange / violet layer in CCl4 with chlorine water; SO4 2- a white BaSO4 insoluble in acid; NO3- the brown ring; S2- black PbS and a violet colour with sodium nitroprusside.

[JEE Tip] Only CO3 2-, SO3 2-, S2- and NO2- (plus S2O3 2-) evolve a gas with DILUTE acid; Cl-, Br-, I- and NO3- need CONCENTRATED acid. Assertion-reason questions live on this dilute-versus-concentrated split.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Percentage purity from a KMnO4 titration. A 1.96 g sample of impure Mohr's salt (FeSO4.(NH4)2SO4.6H2O, Mm 392) is dissolved in dilute H2SO4 and titrated with 0.02 M KMnO4, needing 40.0 mL. One MnO4- oxidises five Fe2+. Find the percentage purity.

• Moles of KMnO4 $= 0.02 \times 0.0400 = 8 \times 10^{-4}$.
• Moles of Fe2+ $= 5 \times 8 \times 10^{-4} = 4 \times 10^{-3}$; one Fe2+ per formula unit, so moles of pure salt $= 4 \times 10^{-3}$.
• Mass of pure salt $= 4 \times 10^{-3} \times 392 = 1.568$ g.
• Purity $= \dfrac{1.568}{1.96} \times 100 = 80$.

Answer: 80% pure.

Example 2 — Iodometric estimation of copper. 0.635 g of a copper alloy is dissolved; the Cu2+ liberates iodine from excess KI, and that iodine needs 15.0 mL of 0.5 M Na2S2O3 (starch near the end). Find the percentage of copper (atomic mass 63.5). Because 2 CuSO4 + 4 KI -> Cu2I2 + I2 + 2 K2SO4 and I2 + 2 Na2S2O3 -> 2 NaI + Na2S4O6, one copper is equivalent to one thiosulphate.

• Moles of hypo $= 0.5 \times 0.0150 = 7.5 \times 10^{-3}$, so moles of Cu $= 7.5 \times 10^{-3}$.
• Mass of Cu $= 7.5 \times 10^{-3} \times 63.5 = 0.47625$ g.
• Percentage $= \dfrac{0.47625}{0.635} \times 100 = 75$.

Answer: 75% copper.

Example 3 — Unknown concentration by N1V1 = N2V2 (standardisation). 25 mL of 0.1 N oxalic acid exactly oxidises 20 mL of a KMnO4 solution in acid. Find the normality and molarity of the KMnO4.

• Equivalents balance, so the normality of KMnO4 is $\dfrac{0.1 \times 25}{20} = 0.125$ N.
• In acid the n-factor is 5, hence the molarity is $\dfrac{0.125}{5} = 0.025$ M.

Answer: 0.125 N, i.e. 0.025 M.

Example 4 — Double indicator: carbonate + bicarbonate. A solution of Na2CO3 and NaHCO3 is titrated with 0.1 M HCl. Phenolphthalein gives the end point at $V_p = 15$ mL; methyl orange then needs a further $V_m = 20$ mL. Find the mass of each salt.

• Since $V_p < V_m$, the mixture is Na2CO3 plus NaHCO3.
• Na2CO3: moles $= M V_p = 0.1 \times 0.015 = 1.5 \times 10^{-3}$, so mass $= 1.5 \times 10^{-3} \times 106 = 0.159$ g.
• NaHCO3: moles $= M(V_m - V_p) = 0.1 \times 0.005 = 5 \times 10^{-4}$, so mass $= 5 \times 10^{-4} \times 84 = 0.042$ g.

Answer: 0.159 g Na2CO3 and 0.042 g NaHCO3.

Example 5 — Double indicator: caustic soda contaminated with carbonate (strength in g/L). 25 mL of a NaOH solution contaminated with Na2CO3 is titrated with 0.1 M HCl: $V_p = 20$ mL to phenolphthalein and a further $V_m = 5$ mL to methyl orange. Find the strength (g/L) of each.

• Since $V_p > V_m$, the mixture is NaOH plus Na2CO3.
• Na2CO3: moles $= M V_m = 0.1 \times 0.005 = 5 \times 10^{-4}$, so mass $= 5 \times 10^{-4} \times 106 = 0.053$ g in 25 mL, giving $\dfrac{0.053}{0.025} = 2.12$ g/L.
• NaOH: moles $= M(V_p - V_m) = 0.1 \times 0.015 = 1.5 \times 10^{-3}$, so mass $= 1.5 \times 10^{-3} \times 40 = 0.06$ g in 25 mL, giving $\dfrac{0.06}{0.025} = 2.4$ g/L.

Answer: 2.4 g/L NaOH and 2.12 g/L Na2CO3.

Example 6 — Back titration for CaCO3 purity. 1.25 g of impure limestone (CaCO3, equivalent weight 50) is treated with 30 mL of 1 N HCl (an excess). The unreacted acid then needs 10 mL of 1 N NaOH. Find the percentage purity.

• Equivalents of HCl taken $= 1 \times 0.030 = 0.030$; equivalents left over (found from the NaOH) $= 1 \times 0.010 = 0.010$.
• Equivalents used up by CaCO3 $= 0.030 - 0.010 = 0.020$, so mass $= 0.020 \times 50 = 1.0$ g.
• Purity $= \dfrac{1.0}{1.25} \times 100 = 80$.

Answer: 80% pure.

Example 7 — Strength in g/L and molarity from normality. A KMnO4 solution used in acidic medium is 0.1 N. Taking its equivalent weight as 31.6 (n-factor 5, Mm 158), find its strength in g/L and its molarity.

• Strength $= N \times E = 0.1 \times 31.6 = 3.16$ g/L.
• Molarity $= \dfrac{N}{n} = \dfrac{0.1}{5} = 0.02$ M; as a check, $0.02 \times 158 = 3.16$ g/L.

Answer: 3.16 g/L, i.e. 0.02 M.