Instrumental Methods of Analysis BCH6T16

Syllabus topics (RTMNU)

• Electromagnetic spectrum and various regions in it
• Atomic versus molecular spectrum
• Absorption versus emission spectrum
• Lambert's law, Beer's law, derivation, verification and deviations of Beer's law
• Concept of transmittance, absorbance and absorptivity
• Numerical problems
• Colorimetry versus spectrophotometry
• Single and double beam instrumentation
• Working of colorimeter and spectrophotometer with simple schematic diagrams
• Application of colorimeter and spectrophotometer in quantitative analysis with reference to estimation of Cu(II) as Cu-ammonia complex

Electromagnetic spectrum and its regions

Electromagnetic radiation travels as waves with wavelength λ and frequency ν (c = λν) and energy E = hν = hc/λ. Shorter wavelength means higher energy.

Atomic vs molecular spectrum; absorption vs emission spectrum

• Atomic spectrum: from free atoms; only electronic transitions → sharp lines (e.g. sodium D-lines).
• Molecular spectrum: electronic + vibrational + rotational levels → closely spaced lines merge into bands.
• Absorption spectrum: radiation passed through the sample; dark lines/bands where energy is absorbed (ground → excited state).
• Emission spectrum: excited sample (flame, arc) gives out radiation; bright lines on a dark background (excited → lower state).

Lambert's law, Beer's law and derivation

Lambert's law: when monochromatic light passes through an absorbing medium, the rate of decrease of intensity with thickness is proportional to the intensity: −dI/db = kI.

Beer's law: the rate of decrease of intensity is proportional to the intensity and to the concentration of the absorbing solution: −dI/db = k′Ic.

Derivation (combined): −dI/I = k′c db. Integrating from I₀ (b = 0) to I (b = b): ln(I₀/I) = k′cb. Converting to log₁₀: log(I₀/I) = (k′/2.303)cb = εbc.

Transmittance, absorbance and absorptivity

• Transmittance T = I/I₀; %T = 100 × I/I₀.
• Absorbance A = log(1/T) = −log T = 2 − log(%T). A has no unit.
• Absorptivity a (c in g L^–1) or molar absorptivity ε (c in mol L^–1): absorbance of a 1 M solution in a 1 cm cell; a constant for a substance at a given λ.

Verification and deviations of Beer's law

Verification: prepare standard solutions, measure A at λ_max and plot A vs c — a straight line through the origin (calibration curve) verifies the law; slope = εb.

Deviations:

1. Real: at high concentration (> 0.01 M) molecules interact and refractive index changes; the law holds only for dilute solutions.
2. Chemical: association, dissociation, complex formation or reaction with solvent change the absorbing species (e.g. Cr₂O₇^2– ⇌ CrO₄^2– with pH).
3. Instrumental: non-monochromatic light, stray light, wide slit, unmatched cells.

Numerical problems

Example 1. A solution of ε = 1.5 × 10⁴ L mol^–1 cm^–1 shows A = 0.45 in a 1 cm cell. Find c and %T.
c = A/(εb) = 0.45/(1.5 × 10⁴ × 1) = 3.0 × 10^–5 mol L^–1. T = 10^–0.45 = 0.355 → %T = 35.5%.

Example 2. A solution transmits 40% of incident light. Find absorbance.
A = −log(0.40) = 0.398 (or 2 − log 40 = 2 − 1.602 = 0.398).

Colorimetry vs spectrophotometry; single and double beam instruments

Working of a colorimeter (schematic): light source (tungsten lamp) → lens/slit → filter (colour complementary to the solution) → cuvette with sample → photocell/phototube detector → galvanometer/digital meter reading A or %T. The instrument is set to zero absorbance with the blank (solvent).

Spectrophotometer: source (tungsten for visible, deuterium/hydrogen lamp for UV) → entrance slit → monochromator → exit slit → sample cell (glass for visible, quartz for UV) → detector (photomultiplier) → amplifier and readout.

• Single beam: one light path; blank and sample measured one after the other; simple and cheap, but lamp fluctuations cause error.
• Double beam: a rotating mirror (chopper) splits the beam through reference and sample cells alternately; the ratio is read directly, cancelling source drift; faster scanning of spectra.

Application — estimation of Cu(II) as copper–ammonia complex

1. Prepare standard Cu²⁺ solutions (from CuSO₄·5H₂O) of known concentration.
2. Add excess ammonia to each → deep blue [Cu(NH₃)₄]²⁺; dilute to a fixed volume.
3. Choose λ_max (about 600–620 nm; orange/red filter in a colorimeter) and set zero with a blank (ammonia + water).
4. Measure absorbance of standards and plot the calibration curve (A vs concentration).
5. Treat the unknown the same way, read its A and find its concentration from the graph.

Likely exam questions

1. (Long, 8 marks) State and derive the Beer–Lambert law. Discuss the deviations from Beer's law.
2. (Long, 8 marks) Describe the working of a double-beam spectrophotometer with a block diagram. How is Cu(II) estimated colorimetrically?
3. (Short, 4 marks) A solution transmits 25% of light in a 1 cm cell. Calculate its absorbance.
4. (Short, 4 marks) Distinguish between colorimetry and spectrophotometry.
5. (Short, 4 marks) Distinguish between absorption and emission spectra.

Syllabus topics (RTMNU)

• Concepts of electrical resistance, conductance, resistivity and conductivity
• Specific, molar and equivalent conductance and effect of dilution on them
• Measurement of conductance
• Kohlrausch's law
• Applications of Kohlrausch law in determination of limiting equivalent and molar conductivity, degree of dissociation, dissociation constant, solubility, solubility product
• Conductometric titrations: Method, advantages and limitations
• Titration curves for single and multi-component analysis

Resistance, conductance, resistivity and conductivity

• Resistance (R): opposition to current; R = V/I (Ohm's law); unit ohm (Ω).
• Conductance (G): ease of flow of current; G = 1/R; unit siemens (S) = Ω^–1 (mho).
• Resistivity (ρ): R = ρ(l/a); ρ is the resistance of a conductor of 1 cm length and 1 cm² area; unit Ω cm.
• Conductivity / specific conductance (κ): κ = 1/ρ = G × (l/a); conductance of a solution between two electrodes 1 cm apart with 1 cm² area (1 cm³ solution); unit S cm^–1. The ratio l/a is the cell constant (cm^–1).

Specific, molar and equivalent conductance and effect of dilution

• Molar conductance: Λ_m = κ × 1000 / C (C in mol L^–1); unit S cm² mol^–1. Conductance of all ions from 1 mole of electrolyte.
• Equivalent conductance: Λ_eq = κ × 1000 / N (N = normality); unit S cm² eq^–1.

Effect of dilution: κ decreases on dilution (fewer ions per cm³), but Λ_m and Λ_eq increase (more ionisation and less inter-ionic attraction), approaching a limiting value Λ°_m at infinite dilution. For strong electrolytes Λ_m = Λ°_m − b√C (Debye–Hückel–Onsager), a straight line on extrapolation; for weak electrolytes Λ_m rises steeply only at very low concentration, so Λ° cannot be found by extrapolation.

Example: κ of 0.1 M KCl = 0.0129 S cm^–1 → Λ_m = 0.0129 × 1000/0.1 = 129 S cm² mol^–1.

Measurement of conductance

Conductance is measured with a Wheatstone bridge using AC (about 1000 Hz) to avoid electrolysis and polarisation, with a conductivity cell of platinised platinum electrodes; the null point is found by a headphone or magic-eye/galvanometer (modern conductivity meters give κ directly). The cell constant is first found using standard KCl solution of known κ: cell constant = κ_KCl / G_observed.

Kohlrausch's law and its applications

Kohlrausch's law of independent migration of ions: at infinite dilution each ion migrates independently and contributes a definite amount to the limiting conductance: Λ°_m = ν₊λ°₊ + ν_–λ°_–.

1. Λ° of a weak electrolyte: Λ°(CH₃COOH) = Λ°(HCl) + Λ°(CH₃COONa) − Λ°(NaCl) = 426.1 + 91.0 − 126.5 = 390.6 S cm² mol^–1.
2. Degree of dissociation: α = Λ_m/Λ°_m.
3. Dissociation constant: K_a = Cα²/(1 − α) (Ostwald's dilution law). Example: 0.001 M acetic acid has Λ_m = 48.15 S cm² mol^–1. α = 48.15/390.6 = 0.1233; K_a = 0.001 × (0.1233)²/(1 − 0.1233) = 1.73 × 10^–5.
4. Solubility and solubility product of sparingly soluble salts (AgCl, BaSO₄): measure κ of saturated solution, subtract κ of water; since the solution is very dilute, Λ_m ≈ Λ°_m; solubility S (mol L^–1) = 1000 κ / Λ°_m; K_sp = S² for AgCl. Example: κ(AgCl) = 1.86 × 10^–6 S cm^–1, Λ° = 138.3 → S = 1.34 × 10^–5 mol L^–1; K_sp = 1.8 × 10^–10.

Conductometric titrations — method, advantages and limitations

Principle: conductance of the solution changes during titration because ions of different mobility replace each other. Conductance is plotted against volume of titrant; the end point is the intersection of two straight lines.

Method: take the analyte in a beaker with the conductivity cell, add titrant (about 10 times more concentrated, to avoid dilution) from a burette in small portions, stir and note conductance after each addition; plot and extrapolate.

Titration curves

1. Strong acid vs strong base (HCl vs NaOH): conductance falls sharply (fast H⁺, λ° ≈ 350, replaced by slow Na⁺) until the end point, then rises (excess OH^–, λ° ≈ 198) — a V-shaped curve.
2. Weak acid vs strong base (CH₃COOH vs NaOH): small initial conductance, slight fall then gradual rise (salt formation) up to end point, then steep rise due to excess OH^–.
3. Strong acid vs weak base (HCl vs NH₄OH): sharp fall to end point, then nearly constant (excess NH₄OH hardly ionises).
4. Weak acid vs weak base: rise up to end point, then almost flat — end point clear though indicators fail.
5. Precipitation (AgNO₃ vs KCl): almost flat (Ag⁺ replaced by K⁺ of similar mobility), then rise after end point.
6. Mixture of strong and weak acid (HCl + CH₃COOH vs NaOH) — multi-component: first a fall (HCl neutralised), then a gentle rise (CH₃COOH neutralised), then a steep rise (excess NaOH) — two breaks give the amount of each acid.

Advantages: works for coloured/turbid solutions, very dilute solutions, weak acid–weak base titrations and mixtures; no indicator; end point found graphically, so accurate.

Limitations: not suitable when large amounts of foreign electrolyte are present (small relative change); redox titrations less suitable; temperature must be constant; needs several readings and graph.

Likely exam questions

1. (Long, 8 marks) State Kohlrausch's law. Explain its applications in determining the degree of dissociation, dissociation constant and solubility of a sparingly soluble salt.
2. (Long, 8 marks) What are conductometric titrations? Explain the titration curves for strong acid–strong base, weak acid–strong base and a mixture of strong and weak acids against NaOH.
3. (Short, 4 marks) Define specific conductance and molar conductance. How do they vary with dilution?
4. (Short, 4 marks) The conductivity of 0.05 M KCl is 6.67 × 10⁻³ S cm⁻¹. Calculate its molar conductance.
5. (Short, 4 marks) Give the advantages of conductometric titrations.

Syllabus topics (RTMNU)

• Theory of electrode potential: Electronation and de-electronation phenomena
• Single electrode potential and standard electrode potential
• Circuit diagram of simple potentiometer
• Indicator electrodes: hydrogen electrode, quinhydrone electrode and glass electrode
• Reference electrode: Calomel electrode and Ag/AgCl electrode
• Ionic product of water and pH scale
• Buffer solutions. Types of buffers and buffering action
• Henderson-Hasselbalch equation and its application
• Theory of potentiometric titrations
• Titration curves of monoprotic and polyprotic acids and bases
• Titrations of mixture of acids and bases

Theory of electrode potential — electronation and de-electronation

When a metal rod is dipped in a solution of its ions, two opposite tendencies operate: metal atoms may lose electrons and go into solution as ions (de-electronation = oxidation, M → Mⁿ⁺ + ne^–), or ions may gain electrons and deposit on the metal (electronation = reduction, Mⁿ⁺ + ne^– → M). Whichever is stronger leaves the metal negatively or positively charged and an electrical double layer (Helmholtz layer) forms at the interface. The potential difference so set up is the single electrode potential.

Single electrode potential and standard electrode potential

• A single electrode potential cannot be measured alone; it is measured against a reference electrode (the standard hydrogen electrode, SHE, given 0.00 V).
• Standard electrode potential E°: potential of an electrode in contact with its ions at unit activity (1 M), at 298 K and 1 atm (for gases). By IUPAC convention it is written as a reduction potential.
• Nernst equation: E = E° − (0.0591/n) log (1/[Mⁿ⁺]) at 298 K, i.e. E = E° + (0.0591/n) log[Mⁿ⁺].

Circuit diagram of a simple potentiometer

A potentiometer measures emf without drawing current (null method). A long uniform wire AB is connected to a steady battery (driver cell). The cell of unknown emf is connected with like poles at A through a galvanometer to a sliding jockey. The jockey is moved till the galvanometer shows zero deflection; the balancing length is proportional to the emf. A standard cell (Weston cadmium cell, about 1.018 V) is used to calibrate: E_x/E_s = l_x/l_s. Modern pH meters/potentiometers are electronic high-impedance voltmeters.

Indicator electrodes

1. Hydrogen electrode: platinised Pt foil in the solution with H₂ gas bubbling at 1 atm: H⁺ + e^– ⇌ ½H₂; E = −0.0591 pH. Primary standard (E° = 0) but inconvenient: needs pure H₂, poisoned by sulphides/arsenic, not used with oxidising agents.
2. Quinhydrone electrode: Pt wire in the solution saturated with quinhydrone (1:1 complex of quinone Q and hydroquinone QH₂): Q + 2H⁺ + 2e^– ⇌ QH₂; E = E°_Q − 0.0591 pH with E° ≈ 0.699 V. Simple and quick; not usable above pH about 8 or with strong oxidants/reductants.
3. Glass electrode: thin bulb of special glass containing 0.1 M HCl and an Ag/AgCl internal electrode; the potential across the glass membrane depends on H⁺ activity outside: E = E°_G − 0.0591 pH. Used in all pH meters; works in coloured, oxidising and turbid solutions; has high resistance (needs electronic meter) and shows alkaline (sodium) error at high pH.

Reference electrodes — calomel and Ag/AgCl

• Calomel electrode: Hg | Hg₂Cl₂(s) | KCl solution. Hg₂Cl₂ + 2e^– ⇌ 2Hg + 2Cl^–. Potential depends on [Cl^–]: saturated calomel electrode (SCE) ≈ +0.242 V, 1 M ≈ +0.280 V, 0.1 M ≈ +0.334 V vs SHE at 25 °C.
• Silver–silver chloride electrode: Ag wire coated with AgCl dipped in KCl solution: AgCl + e^– ⇌ Ag + Cl^–; E° = +0.222 V (about +0.197 V in saturated KCl). Compact, mercury-free; used as internal electrode of the glass electrode.

Ionic product of water and pH scale

H₂O ⇌ H⁺ + OH^–; K_w = [H⁺][OH^–] = 1.0 × 10^–14 at 25 °C (increases with temperature). pH = −log[H⁺]; pOH = −log[OH^–]; pH + pOH = 14. Neutral pH 7, acidic < 7, basic > 7. Example: 0.001 M HCl → pH 3; 0.01 M NaOH → pOH 2, pH 12.

Buffer solutions, types and buffering action

A buffer resists change in pH on adding small amounts of acid or base or on dilution.

• Acidic buffer: weak acid + its salt with strong base (CH₃COOH + CH₃COONa), pH < 7.
• Basic buffer: weak base + its salt with strong acid (NH₄OH + NH₄Cl), pH > 7.

Buffering action (acetate buffer): added H⁺ is removed by acetate ions: CH₃COO^– + H⁺ → CH₃COOH; added OH^– is removed by the acid: CH₃COOH + OH^– → CH₃COO^– + H₂O. So [H⁺] hardly changes. Buffer capacity is maximum when [salt] = [acid] (pH = pK_a).

Henderson–Hasselbalch equation and its application

For an acidic buffer: pH = pK_a + log([salt]/[acid]); for a basic buffer: pOH = pK_b + log([salt]/[base]).

Derivation in brief: K_a = [H⁺][A^–]/[HA] → [H⁺] = K_a[HA]/[A^–]; taking −log and taking [A^–] ≈ [salt], [HA] ≈ [acid] gives the equation.

Example: buffer of 0.1 M CH₃COOH and 0.2 M CH₃COONa, pK_a = 4.76: pH = 4.76 + log 2 = 4.76 + 0.30 = 5.06.

Uses: preparing buffers of desired pH, finding pK_a (pH at half-neutralisation = pK_a), biochemistry (blood pH 7.4 kept by H₂CO₃/HCO₃^– buffer).

Theory of potentiometric titrations and titration curves

In a potentiometric titration the emf of a cell (indicator electrode + reference electrode) is measured after each addition of titrant. Near the equivalence point the concentration of the ion changes enormously, so the emf (or pH) changes sharply. The end point is found from: (1) the E (pH) vs V curve — steepest point; (2) the first derivative ΔE/ΔV vs V — a peak; (3) the second derivative Δ²E/ΔV² vs V — crosses zero.

• Monoprotic strong acid vs strong base (HCl vs NaOH): pH starts at about 1, rises slowly, jumps sharply from about 4 to 10 at the equivalence point (pH 7), then levels off.
• Weak acid vs strong base (CH₃COOH vs NaOH): starts near pH 3, buffer region with pH = pK_a at half-neutralisation, smaller jump, equivalence point above 7 (about 8.7).
• Weak base vs strong acid: equivalence point below 7.
• Polyprotic acid (H₃PO₄ vs NaOH): separate steps for each proton when successive K_a values differ widely — first end point near pH 4.7 and second near pH 9.7; the third proton is too weak to give a clear break.
• Mixture of acids (HCl + CH₃COOH vs NaOH): the strong acid is neutralised first (first break), then the weak acid (second break); volumes between breaks give each acid.

Advantages: no indicator needed; works in coloured/turbid solutions; accurate; applicable to acid–base, redox, precipitation and complexometric titrations; can be automated.

Likely exam questions

1. (Long, 8 marks) Explain the construction and working of the glass electrode and the calomel electrode. How is pH measured potentiometrically?
2. (Long, 8 marks) Discuss the theory of potentiometric titrations. Draw and explain the titration curves of a monoprotic and a polyprotic acid with NaOH.
3. (Short, 4 marks) Derive the Henderson–Hasselbalch equation for an acidic buffer.
4. (Short, 4 marks) Calculate the pH of a buffer containing 0.1 M acetic acid and 0.1 M sodium acetate (pKa = 4.76).
5. (Short, 4 marks) What is buffering action? Explain with an example.

Syllabus topics (RTMNU)

• Paper chromatography: Principle and Technique
• Types: Ascending, descending, ascending-descending and two dimensional
• Applications
• Numerical based on Rf values
• Column Chromatography: Principle, technique and applications
• Ion-Exchange: Principle, Types of ion exchange resins, ion exchange equilibria
• Ion exchange capacity and its experimental determination
• Factors affecting ion exchange equilibrium
• Application in separation of binary mixtures
• Solvent Extraction: Principle and technique
• Types of extraction techniques
• Factors influencing extraction efficiency
• Applications in chemistry
• Numerical based on extraction efficiency

Paper chromatography — principle and technique

Chromatography separates components of a mixture by their different distribution between a stationary phase and a mobile phase.

Principle of paper chromatography: it is mainly partition chromatography. Water held in the cellulose fibres of the filter paper (Whatman No. 1) is the stationary phase; an organic solvent mixture moving through the paper by capillary action is the mobile phase. Components more soluble in the mobile phase move faster; those more soluble in water move slower.

Technique:

1. Draw a pencil base line about 2 cm from the bottom of the paper strip; apply a small spot of the sample with a capillary and dry it.
2. Hang the paper in a closed jar saturated with solvent vapour, with the lower edge dipping in the solvent (spot above solvent level).
3. Let the solvent rise till near the top; mark the solvent front and dry the paper.
4. Locate colourless spots with a spraying (locating) reagent — ninhydrin for amino acids (purple), H₂S/dithizone/rubeanic acid for metal ions — or under UV light.
5. Calculate R_f values.

Types of paper chromatography

• Ascending: solvent kept at the bottom, rises upward by capillarity. Simple, common.
• Descending: solvent in a trough at the top; flows down aided by gravity — faster, useful for substances with small R_f.
• Ascending–descending: paper folded over a rod; solvent first rises, then descends — longer path, better separation.
• Radial (circular): solvent spreads from the centre of a circular paper; components form concentric rings.
• Two-dimensional: square paper; run with solvent 1, dry, turn 90° and run with solvent 2 — separates complex mixtures (e.g. amino acids of a protein hydrolysate).

Applications: separation and identification of amino acids, sugars, metal ions (Cu²⁺, Ni²⁺, Co²⁺), plant pigments, dyes, drugs; checking purity; forensic and food analysis.

Numericals based on R_f values

Example 1. Solvent front moves 12.0 cm; spot moves 4.2 cm. R_f = 4.2/12.0 = 0.35.

Example 2. R_f of glycine = 0.26 and solvent front moves 15 cm. Distance moved by glycine = 0.26 × 15 = 3.9 cm.

Column chromatography — principle, technique and applications

Principle: adsorption chromatography — components are adsorbed to different extents on a solid adsorbent (stationary phase, e.g. silica gel, alumina) and are washed down (eluted) by a liquid (mobile phase) at different rates; weakly adsorbed components come out first.

Technique: a glass column with a stopcock is plugged with glass wool and packed uniformly with a slurry of adsorbent; the sample is loaded at the top as a narrow band; eluents of increasing polarity (e.g. hexane → chloroform → methanol) are passed; separated bands are collected as fractions and identified (TLC, colour, spectra).

Applications: separation of plant pigments (chlorophylls and carotenes), purification of reaction products and natural products (alkaloids), separation of isomers.

Ion exchange — principle and types of resins

Principle: reversible exchange of ions between a solution and an insoluble polymer (resin) carrying fixed charged groups. R–SO₃^–H⁺ + Na⁺ ⇌ R–SO₃^–Na⁺ + H⁺.

Most resins are sulphonated or aminated polystyrene cross-linked with divinylbenzene (DVB).

Ion exchange equilibria, capacity and its determination

For R–H + M⁺ ⇌ R–M + H⁺, the selectivity (equilibrium) coefficient K = [R–M][H⁺]/[R–H][M⁺]. A larger K means the ion is held more strongly. Generally affinity increases with charge (Th⁴⁺ > Al³⁺ > Ca²⁺ > Na⁺) and, for the same charge, with smaller hydrated size (Cs⁺ > K⁺ > Na⁺ > Li⁺).

Ion exchange capacity: number of milliequivalents of ions exchanged per gram of dry resin (meq g^–1).

Experimental determination (cation exchanger, H⁺ form): pack a known weight (say 1 g) of dry resin in a column; pass excess NaCl solution; the H⁺ released (as HCl) is collected and titrated with standard NaOH. Capacity (meq g^–1) = (N_NaOH × V_NaOH in mL)/weight of resin in g. Example: 1.0 g resin; eluate needs 25.0 mL of 0.18 N NaOH → capacity = 0.18 × 25.0/1.0 = 4.5 meq g^–1.

Factors affecting ion exchange equilibrium; separation of binary mixtures

• Charge and hydrated size of ions; concentration of the solution; pH; degree of cross-linking (more DVB → more selective but slower); temperature; presence of complexing agents; particle size and flow rate.

Separation of a binary mixture (e.g. Zn²⁺ and Cd²⁺, or Cl^– and Br^–): the mixture is loaded on the column and eluted with a suitable eluent (e.g. HCl of changing concentration, which forms chloro-complexes to different extents); the less strongly held ion elutes first and each fraction is estimated by titration. Other uses: water softening and demineralisation, recovery of metals, separation of lanthanides and amino acids.

Solvent extraction — principle, techniques and factors

Principle (Nernst distribution law): at constant temperature a solute distributes itself between two immiscible solvents in a constant ratio: K_D = C_org/C_aq. The distribution ratio D = total concentration in organic phase / total concentration in aqueous phase (all forms).

Percentage extraction: %E = 100 D / (D + V_aq/V_org).

Types of extraction techniques: (1) batch extraction in a separating funnel (for large D); (2) continuous extraction (Soxhlet for solids, continuous liquid–liquid extractors for small D); (3) counter-current extraction (Craig) for substances with close D values. Extractions are also classed as chelate extraction (e.g. Ni–dimethylglyoxime in chloroform, Cu–dithizone) and ion-association extraction (e.g. Fe(III) as HFeCl₄ into ether).

Factors influencing efficiency: value of D (choice of solvent), pH of aqueous phase, concentration of chelating agent, masking agents, salting-out agents, phase volume ratio, number of extractions, temperature, shaking time.

Applications: separation and pre-concentration of metal ions before analysis (AAS, spectrophotometry), extraction of natural products (caffeine from tea, oils), purification in organic synthesis, nuclear chemistry (uranium with TBP).

Numericals based on extraction efficiency

Fraction remaining in aqueous phase after n extractions: q_n = [V_aq/(D V_org + V_aq)]ⁿ.

Example. D = 10; 100 mL aqueous solution. (a) One extraction with 50 mL solvent: q = 100/(10 × 50 + 100) = 100/600 = 0.167 → 83.3% extracted. (b) Two extractions with 25 mL each: q₂ = [100/(10 × 25 + 100)]² = (0.2857)² = 0.0816 → 91.8% extracted.

Likely exam questions

1. (Long, 8 marks) Explain the principle and technique of paper chromatography. Describe its types and applications.
2. (Long, 8 marks) What is solvent extraction? Explain its principle, types and factors affecting extraction efficiency. Show that multiple extractions are more efficient than a single extraction.
3. (Short, 4 marks) In a paper chromatogram the solvent front moved 10 cm and a spot moved 6.5 cm. Calculate the Rf value.
4. (Short, 4 marks) How is the ion exchange capacity of a cation exchange resin determined?
5. (Short, 4 marks) Write the principle of column chromatography.