Cations are separated and identified in a fixed, systematic sequence of groups, each precipitated selectively by a specific group reagent under carefully controlled conditions — this ensures that ions from one group don't interfere with the identification of ions in a later group.
| Group | Group Reagent | Cations Precipitated |
|---|---|---|
| I | Dilute HCl | Pb²⁺, Ag⁺, Hg₂²⁺ |
| II | H₂S in dilute HCl (acidic medium) | Cu²⁺, Cd²⁺, Pb²⁺, Bi³⁺, As³⁺, Sb³⁺, Sn²⁺ |
| III | NH₄OH + NH₄Cl (basic medium) | Fe³⁺, Al³⁺, Cr³⁺ |
| IV | H₂S in ammoniacal medium | Zn²⁺, Mn²⁺, Ni²⁺, Co²⁺ |
| V | (NH₄)₂CO₃ in ammoniacal medium | Ca²⁺, Sr²⁺, Ba²⁺ |
| VI | No group reagent (confirmed by flame test/other tests) | Mg²⁺, Na⁺, K⁺, NH₄⁺ |
Group II cations must be precipitated in an acidic medium (using dilute HCl to suppress H₂S ionisation, keeping [S²⁻] low) so that only the least soluble sulphides precipitate. Group IV cations need an ammoniacal (basic) medium instead, where [S²⁻] is much higher, allowing the more soluble sulphides of this group to precipitate as well — this pH control is the entire basis for separating groups II and IV, which would otherwise precipitate together.
Explain why H₂S is passed in acidic medium for group II precipitation but in ammoniacal (basic) medium for group IV precipitation.
H₂S is a weak diprotic acid, and its ionisation to give S²⁻ ions is suppressed in the presence of excess H⁺ (common ion effect) — this keeps the sulphide ion concentration very low in acidic medium, just enough to precipitate only the extremely insoluble sulphides of group II cations (like CuS, PbS), while leaving the more soluble group IV sulphides (like ZnS, NiS) still in solution. Switching to an ammoniacal (basic) medium removes H⁺ ions, allowing H₂S to ionise much more fully and raise [S²⁻] considerably — this higher sulphide concentration is now sufficient to exceed the solubility product of the comparatively more soluble group IV metal sulphides, precipitating them as well.
Anions like $\text{CO}_3^{2-}$, $\text{S}^{2-}$, $\text{SO}_3^{2-}$, and $\text{NO}_2^-$ react with dilute HCl or dilute H₂SO₄ to evolve a characteristic gas, often visible/detectable at room temperature or with mild warming — e.g., carbonate gives $\text{CO}_2$ (turns lime water milky), sulphide gives $\text{H}_2\text{S}$ (rotten-egg smell, blackens lead acetate paper).
Anions like $\text{Cl}^-$, $\text{Br}^-$, $\text{I}^-$, and $\text{NO}_3^-$ require concentrated H₂SO₄ (a stronger dehydrating/oxidising acid) to show a reaction, since they don't react noticeably with the dilute acid alone. Chloride gives colourless, choking HCl fumes; bromide gives reddish-brown Br₂ fumes; and nitrate (with a copper turning) gives brown fumes of NO₂.
Sulphate ($\text{SO}_4^{2-}$) does not evolve a gas with either dilute or concentrated acid, and is instead confirmed with $\text{BaCl}_2$ solution, giving a white precipitate of $\text{BaSO}_4$ that is insoluble even in dilute HCl (distinguishing it from other Ba-precipitating anions like carbonate or sulphite, whose barium salts dissolve in dilute acid).
A salt evolves a gas with dilute HCl that turns lime water milky, and the milkiness disappears on passing excess gas. Identify the anion and gas involved.
The anion is carbonate ($\text{CO}_3^{2-}$), releasing CO₂ gas: $\text{CO}_3^{2-} + 2\text{H}^+ \rightarrow \text{H}_2\text{O} + \text{CO}_2$. The initial milkiness is insoluble $\text{CaCO}_3$ forming with lime water; on passing excess CO₂, this converts to soluble calcium bicarbonate, $\text{Ca(HCO}_3)_2$, clearing the milkiness — a classic confirmatory observation for CO₂.
Mohr's salt is a double salt with the formula $\text{FeSO}_4 \cdot (\text{NH}_4)_2\text{SO}_4 \cdot 6\text{H}_2\text{O}$ — ferrous ammonium sulphate hexahydrate. It's valued in the laboratory because, unlike plain ferrous sulphate, it resists aerial oxidation of Fe²⁺ to Fe³⁺ much better, making it a reliable primary standard for volumetric analysis involving Fe²⁺.
Equimolar quantities of ferrous sulphate and ammonium sulphate are dissolved together in a minimum amount of water acidified with a little dilute sulphuric acid (to prevent hydrolysis and premature oxidation of Fe²⁺), and the solution is then allowed to crystallise slowly on cooling — pale green crystals of Mohr's salt separate out, since a double salt of two compounds is often less soluble in combination than the individual constituents, once the solution is sufficiently concentrated and cooled.
Why is Mohr's salt preferred over plain ferrous sulphate as a primary standard in volumetric analysis?
Plain ferrous sulphate readily oxidises in air (Fe²⁺ → Fe³⁺), and also tends to undergo partial hydrolysis, both of which change its effective composition/purity over time, making its Fe²⁺ content unreliable for precise titrations. Mohr's salt, being a double salt, is significantly more resistant to atmospheric oxidation (the ammonium sulphate component provides a more stable crystal lattice environment) and remains stable and well-defined in composition over longer storage — making it a dependable primary standard whose exact Fe²⁺ content can be trusted for calculations.
Potash alum is a double salt, potassium aluminium sulphate: $\text{K}_2\text{SO}_4 \cdot \text{Al}_2(\text{SO}_4)_3 \cdot 24\text{H}_2\text{O}$, forming characteristic large, colourless octahedral crystals — widely used historically as a water purifier/coagulant and in traditional dyeing/tanning processes.
Equimolar solutions of potassium sulphate and aluminium sulphate (or aluminium metal/its oxide reacted first with dilute sulphuric acid to form aluminium sulphate in situ) are mixed and acidified slightly with dilute sulphuric acid, then concentrated by evaporation and allowed to cool slowly — well-formed, large octahedral crystals of potash alum separate on cooling, since this double salt has different (typically lower, under these controlled conditions) solubility than a simple mixture of its individual constituent salts at that temperature.
Explain why the solution is cooled slowly (rather than rapidly) during the crystallisation step of potash alum preparation.
Slow cooling allows crystal growth to proceed gradually and in an orderly manner, giving large, well-formed, higher-purity octahedral crystals of potash alum. Rapid cooling would instead cause many small crystals to form simultaneously and haphazardly (rapid, uncontrolled nucleation), often trapping impurities within the smaller crystal structures and generally giving a lower-quality (and sometimes lower-yield) product.
Aniline is heated with excess acetic anhydride (or glacial acetic acid) to give acetanilide ($\text{C}_6\text{H}_5\text{NHCOCH}_3$), via acylation of the amine group. This reaction serves the practical purpose of "protecting" and moderating the powerfully activating -NH₂ group of aniline before further ring substitution reactions are attempted.
Acetanilide is subjected to controlled nitration (using a nitrating mixture of conc. HNO₃ and conc. H₂SO₄, under cold conditions) to give predominantly p-nitroacetanilide, since the acetanilide's $-\text{NHCOCH}_3$ group is an ortho/para-directing (activating) group, and the para position is generally favoured over ortho due to less steric hindrance from the bulky acetamido group.
Why is aniline first converted to acetanilide before nitration, rather than nitrating aniline directly?
Direct nitration of aniline is problematic for two reasons: (1) aniline's free -NH₂ group is so strongly activating that nitration tends to be poorly controlled, giving a complex mixture of multiply-substituted products, and (2) under the strongly acidic conditions of nitration, aniline's -NH₂ group can itself be oxidised, leading to unwanted side reactions and tarry byproducts. Converting aniline to acetanilide first moderates the ring-activating strength of the nitrogen substituent (still o,p-directing, but less powerfully activating) and also protects the nitrogen from oxidation during nitration, giving a much cleaner, more controllable mono-nitration reaction.
Aspirin (acetylsalicylic acid) is prepared by the acetylation of salicylic acid (2-hydroxybenzoic acid) using acetic anhydride, typically with a trace of concentrated sulphuric acid or phosphoric acid as a catalyst — the phenolic -OH group of salicylic acid is acetylated to an ester group ($-\text{OCOCH}_3$), while the carboxylic acid group remains unchanged.
Why is acetic anhydride generally preferred over acetyl chloride for the acetylation step in aspirin synthesis at a small/lab scale?
Acetic anhydride reacts somewhat more gently and controllably with the phenolic -OH group compared to the more vigorously reactive acetyl chloride, and importantly, acetic anhydride's by-product is simply acetic acid (a comparatively mild, easy-to-handle by-product), whereas acetyl chloride's reaction releases HCl gas, which is corrosive and more hazardous to handle safely, especially in a typical school/college laboratory setting.
Methyl orange, a well-known acid-base indicator dye, is prepared via a diazonium coupling reaction: sulphanilic acid is first diazotised (converted to its diazonium salt), which is then coupled with N,N-dimethylaniline (an electron-rich, activated aromatic amine) under mildly acidic/neutral conditions — this forms the characteristic azo (-N=N-) linkage responsible for methyl orange's intense colour.
The $-\text{N(CH}_3)_2$ group is a very strong electron-donor (activating group), making the ring highly nucleophilic and reactive towards the electrophilic diazonium ion — coupling occurs predominantly at the position para to the dimethylamino group, since that position is both directed by the group and typically the most accessible (para, not ortho, is the major site here mainly because it's less hindered and the electronic activation is strongest there through the conjugated system).
Explain why the colour of methyl orange changes with pH, making it useful as an acid-base indicator.
Methyl orange's structure includes a basic nitrogen (part of the $-\text{N(CH}_3)_2$ group) that can be protonated in acidic solution. This protonation significantly alters the molecule's electron delocalisation across the conjugated azo system, changing the wavelength of light it absorbs — shifting its observed colour from yellow (in basic/neutral solution, unprotonated form) to red (in acidic solution, protonated form), which is exactly why it functions as a useful visual indicator of pH change around its specific transition range.
A solution of unknown concentration is reacted with a solution of precisely known concentration (the standard/titrant), added gradually until the reaction reaches its stoichiometric completion (the equivalence point) — an indicator is used to visually signal when this point has been reached (the end point), which should be as close as possible to the true equivalence point.
An indicator's own colour-change (transition) pH range must fall within the sharp vertical portion of the titration's pH curve near the equivalence point. Phenolphthalein (transition range ~8.3-10, colourless to pink) suits titrations where the equivalence point is basic (e.g., weak acid vs strong base). Methyl orange (transition range ~3.1-4.4, red to yellow) suits titrations where the equivalence point is acidic (e.g., strong acid vs weak base). For a strong acid-strong base titration, either indicator works well, since the pH change near equivalence is extremely sharp and steep, covering a wide pH range in a tiny volume of titrant.
Explain why phenolphthalein, not methyl orange, is the appropriate indicator for titrating acetic acid (a weak acid) against sodium hydroxide (a strong base).
At the equivalence point of a weak acid-strong base titration, the solution contains sodium acetate — the salt of a weak acid and a strong base, which hydrolyses to make the solution mildly basic (pH greater than 7, roughly 8-9). Phenolphthalein's colour-change range (8.3-10) aligns well with this basic equivalence point pH, giving a sharp, accurate end point. Methyl orange's transition range (3.1-4.4) is far too acidic and would change colour well before the true equivalence point is reached, giving an inaccurate, premature end point.
In a redox titration, the titrant and analyte undergo an oxidation-reduction reaction rather than a simple acid-base neutralisation. A classic and very common example is the titration of a standard oxalic acid solution against potassium permanganate (in acidic medium), which is self-indicating — the deep purple colour of $\text{MnO}_4^-$ disappears as it's reduced to nearly colourless $\text{Mn}^{2+}$, so the end point is signalled by the first permanent faint pink colour persisting in the solution (indicating all oxalate has been consumed and excess unreacted $\text{MnO}_4^-$ remains).
This particular titration is initially slow at room temperature, so the oxalic acid solution is typically warmed (to around 60-70°C) to speed up the reaction rate, which then becomes autocatalytic — the $\text{Mn}^{2+}$ product itself catalyses further reaction, causing the rate to noticeably speed up partway through the titration. Dilute sulphuric acid is used to maintain acidic conditions specifically, since HCl could be partially oxidised by $\text{MnO}_4^-$ (introducing an unwanted side reaction and error), and HNO₃ is itself an oxidising acid that could interfere.
Why is dilute sulphuric acid specifically preferred over dilute hydrochloric acid for acidifying the oxalic acid-KMnO₄ titration?
Potassium permanganate is a strong oxidising agent and can oxidise chloride ions (from HCl) to chlorine gas as an unwanted competing side reaction, which would consume extra permanganate not accounted for in the oxalic acid calculation, introducing significant titration error. Dilute sulphuric acid provides the acidic medium needed for the main reaction without itself being oxidised or interfering, making it the appropriate, non-reactive acid to use here.
Iodimetry is a direct titration where a standard iodine solution itself acts as the oxidising titrant, reacting directly with a reducing agent being estimated. Iodometry is an indirect method: an oxidising agent (being estimated) first liberates iodine from excess potassium iodide, and this liberated iodine is then titrated separately against a standard reducing agent (typically sodium thiosulphate).
Freshly prepared starch solution is used as the indicator in iodine-related titrations — it forms an intensely blue-black complex with iodine, which is a highly sensitive way to detect the presence of even a small amount of remaining iodine. Starch is always added near the end of the titration (once most of the iodine has already reacted), since adding it too early can cause the iodine-starch complex to bind so tightly that it's slow to release iodine back, making the end point sluggish and less sharp.
Sodium thiosulphate reduces iodine to iodide, itself being oxidised to sodium tetrathionate — the endpoint is reached when the blue-black starch-iodine colour disappears completely (turning colourless), since all free iodine has now been consumed.
Explain why starch indicator is added only near the end point of an iodometric titration, rather than right at the beginning.
Starch forms a very stable, tightly-bound blue-black complex with iodine. If added early, while a large amount of iodine is still present, this complex can become difficult to fully "release" its bound iodine back for reaction with the thiosulphate titrant as the titration nears completion — this can make the final colour change sluggish, less sharp, and prone to error (giving a false, delayed end point). Adding starch only when the iodine concentration is already low (near the true end point) avoids this problem, giving a sharp, immediate colour change exactly when the last traces of iodine are consumed.