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GATE Analytical Chemistry — Separation Methods and Their Numericals

By Aniket Bhardwaj · 10 September 2026 · GATE Chemistry

Separation is the half of analytical chemistry that actually carries numericals. Detection and instrumentation give you concept questions; separation gives you numerical-answer-type questions where a careless line of arithmetic costs the whole mark. This guide covers the separation family that GATE Chemistry keeps returning to — solvent extraction, selective precipitation, ion exchange and distillation — with every calculation done step by step.

The distribution law and the distribution ratio

Shake a solute with two immiscible liquids and it distributes itself between them. For a single chemical species that does not dissociate, associate or complex in either phase, the Nernst distribution law applies:

KD = [A]org / [A]aq   (constant at fixed temperature)

Real solutes rarely behave that simply. A weak acid may be partly ionised in water and partly dimerised in the organic phase. So analytical chemists use the distribution ratio D, which counts all forms of the element in each phase:

D = (total analytical concentration in organic phase) / (total analytical concentration in aqueous phase)

KD is a true thermodynamic constant for one species. D is an operational quantity that depends on pH, on masking agents and on how much complexing ligand you added. Every extraction calculation in an exam is written in terms of D, not KD — and confusing the two is the single most common error in this topic.

Fraction extracted, and why many small extractions win

If Vorg and Vaq are the two volumes, the fraction of solute extracted into the organic phase in one shake is:

E = D Vorg / (D Vorg + Vaq)   →   with equal volumes, E = D/(D + 1)

The fraction remaining in the aqueous phase after n successive extractions, each with a fresh Vorg, is the single-step fraction raised to the power n:

(fraction left after n extractions) = [ Vaq / (Vaq + D Vorg) ]n

Worked example 1 — one big extraction against five small ones

A solute has D = 10 between an organic solvent and water. You have 50 mL of aqueous solution and 50 mL of organic solvent in total. Compare extracting once with all 50 mL against five extractions of 10 mL each.

Route A — one 50 mL extraction.
fraction left = 50 ÷ (50 + 10 × 50) = 50 ÷ (50 + 500) = 50 ÷ 550 = 0.09091
fraction extracted = 1 − 0.09091 = 0.90909 → 90.91 %

Route B — five 10 mL extractions.
fraction left after one 10 mL shake = 50 ÷ (50 + 10 × 10) = 50 ÷ 150 = 1/3 = 0.33333
after five shakes = (1/3)5 = 1 ÷ 243 = 0.0041152
fraction extracted = 1 − 0.0041152 = 0.9958848 → 99.59 %

Conclusion. The same total volume of solvent, split into five portions, leaves 0.41 % behind instead of 9.09 % — about 22 times less. This is the standard result GATE tests: for a fixed total solvent volume, many small extractions always beat one large extraction, and the advantage grows as n grows.

Cross-check. Route A can also be read from E = D/(D+1) = 10/11 = 0.9091, since the volumes there are equal. The two routes agree.

Separation by selective precipitation

The second big numerical family uses the solubility product. If two cations both form precipitates with the same anion, and their Ksp values differ enough, adding the anion slowly precipitates one essentially completely before the other begins.

Precipitation of MxXy begins when [M]x[X]y > Ksp

A separation is usually called quantitative when less than 0.1 % of the first ion is still in solution at the moment the second ion begins to precipitate.

Worked example 2 — separating Ag⁺ from Pb²⁺ with chloride

A solution is 0.010 M in Ag⁺ and 0.010 M in Pb²⁺. Chloride is added slowly. Take Ksp(AgCl) = 1.8 × 10⁻¹⁰ and Ksp(PbCl₂) = 1.7 × 10⁻⁵. (Tabulated Ksp values differ slightly between textbooks — always use the values printed in the question.) Can the two be separated?

Step 1 — chloride needed to start AgCl. AgCl ⇌ Ag⁺ + Cl⁻, so Ksp = [Ag⁺][Cl⁻]:
[Cl⁻] = Ksp ÷ [Ag⁺] = 1.8 × 10⁻¹⁰ ÷ 0.010 = 1.8 × 10⁻⁸ M

Step 2 — chloride needed to start PbCl₂. PbCl₂ ⇌ Pb²⁺ + 2Cl⁻, so Ksp = [Pb²⁺][Cl⁻]²:
[Cl⁻]² = 1.7 × 10⁻⁵ ÷ 0.010 = 1.7 × 10⁻³
[Cl⁻] = √(1.7 × 10⁻³) = 0.0412 M

Step 3 — which precipitates first? AgCl needs only 1.8 × 10⁻⁸ M chloride against 0.0412 M for PbCl₂, so AgCl precipitates first, by a factor of about 2.3 million in chloride concentration.

Step 4 — how much Ag⁺ is left when PbCl₂ just starts? At that instant [Cl⁻] = 0.0412 M, so
[Ag⁺] = Ksp(AgCl) ÷ [Cl⁻] = 1.8 × 10⁻¹⁰ ÷ 0.0412 = 4.37 × 10⁻⁹ M

Step 5 — fraction of silver still in solution.
4.37 × 10⁻⁹ ÷ 0.010 = 4.4 × 10⁻⁷, i.e. about 0.000044 % of the original silver.

Answer: the separation is quantitative — far better than the usual 0.1 % criterion. Note the shape of the working: the 1:1 salt uses a direct division, the 1:2 salt needs a square root. Missing that square root is the classic slip here.

Ion exchange

An ion-exchange resin is a cross-linked polymer carrying fixed charged groups. A cation exchanger carries fixed anionic groups (strongly acidic sulfonate, –SO₃⁻; weakly acidic carboxylate, –COO⁻) and exchanges mobile cations. An anion exchanger carries fixed cationic groups (strongly basic quaternary ammonium; weakly basic amine) and exchanges mobile anions.

The exchange is a genuine equilibrium, so a selectivity coefficient can be written. Two trends are worth memorising because they are asked directly:

Capacity is quoted in milliequivalents per gram of dry resin. Deionised water is made by passing water through a cation exchanger in the H⁺ form and an anion exchanger in the OH⁻ form in series; the released H⁺ and OH⁻ neutralise each other. Ion exchange is also how the lanthanides were first separated cleanly, using an eluting complexing agent that discriminates on ionic radius.

Distillation and the role of relative volatility

Distillation separates on the basis of vapour pressure. For an ideal binary mixture obeying Raoult's law, the relative volatility α controls how easy the separation is:

α = P°A / P°B   (ratio of pure-component vapour pressures)

α close to 1 means the vapour has nearly the same composition as the liquid and a simple distillation achieves almost nothing — you need a fractionating column with many theoretical plates. The key exam concepts are: azeotropes cannot be separated further by ordinary distillation because vapour and liquid have identical composition at that point; steam distillation lets a water-immiscible organic compound distil below 100 °C because the total pressure is the sum of the two independent vapour pressures; and vacuum distillation lowers the boiling point to protect thermally unstable compounds.

Where chromatography fits

Chromatography is a separation method too, but it is a large enough topic to deserve its own treatment. In one sentence: it is a repeated partitioning of the analyte between a stationary and a mobile phase, so it behaves like thousands of tiny extractions in series — which is exactly why it separates species whose single-stage D values are close. The quantities to know are the retention factor, the plate number, resolution and the van Deemter equation, and they belong together in a dedicated article rather than as a footnote here.

Choosing the right method

MethodProperty exploitedBest when
Solvent extractionDifferential solubility in two immiscible liquidsD values differ by a large factor; scale-up is easy
Selective precipitationSolubility productKsp values differ by several orders of magnitude
Ion exchangeCharge and hydrated radiusIonic species; trace enrichment; deionisation
DistillationVapour pressure / relative volatilityVolatile liquids, α clearly above 1, no azeotrope
ChromatographyRepeated partition or adsorptionVery similar species; small sample amounts
MaskingSelective complex formationAn interfering ion must be kept in solution rather than removed

Common mistakes that cost marks

  • Using KD where the question means D. The moment the solute ionises, dimerises or complexes, only D describes the real distribution.
  • Forgetting the square root for a 1:2 salt. For MX₂, [X⁻] = √(Ksp/[M²⁺]), not Ksp/[M²⁺].
  • Adding fractions extracted instead of multiplying fractions remaining. Successive extractions multiply the remaining fraction; they do not add the extracted fractions.
  • Ignoring pH in chelate extraction. Extraction of a metal as a neutral chelate depends steeply on pH because the ligand must first lose its proton. A question that gives you a pH is telling you it matters.
  • Assuming the smallest ion binds a resin most strongly. It is the hydrated radius that decides, which reverses the order for the alkali metals.
  • Trying to break an azeotrope with a longer column. No number of plates helps; you need an entrainer, pressure swing or a different method entirely.

Where this appears in GATE Chemistry

Separations sit in the analytical portion of the GATE Chemistry (CY) syllabus and overlap with physical chemistry equilibrium. Confirm the exact syllabus wording, question count and marking scheme for your year from the current official GATE information brochure — do not rely on second-hand summaries.

Question typeWhat you must do
Multiple extraction (NAT)Apply the power-n formula; report percent extracted or percent remaining
Minimum solvent volumeRearrange the extraction formula for Vorg at a target recovery
Selective precipitation (NAT)Compare threshold anion concentrations; find residual first-ion concentration
Ion-exchange order MCQRank ions by charge, then by hydrated radius
Distillation concept MCQAzeotropes, steam distillation, effect of α on plate requirement
Method-choice MCQMatch the separation problem to the property being exploited

Check the precipitation arithmetic instantly. The Ksp and solubility calculator handles 1:1, 1:2 and 1:3 salts, does the square and cube roots for you, and gives the ion concentrations at the onset of precipitation — exactly the step where the selective-precipitation numerical usually goes wrong.

Open the Ksp & Solubility Calculator →

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