NCERT Exemplar Solutions Class 10 Science Chapter 2

Strategic angle (rule out, then confirm). Two physical things always happen in a neutralisation: a salt appears and heat comes out. Check each option against those two facts and the answer falls out.

Concept used. Neutralisation Acid + Base → Salt + Water is exothermic because forming the strong O-H bonds in water releases more energy than is spent breaking up the acid and base. The standard enthalpy of neutralisation for a strong acid with a strong base is close to -57.1 kJ/mol of water.

1. Salt test. Every acid–base reaction leaves a salt. In HCl + NaOH → NaCl + H₂O the salt is NaCl. So (iv) is definitely correct, and any option missing (iv) is wrong.
2. Heat test. The energy released (≈ 57 kJ per mole of water) heats the mixture. So (i) is correct and (ii), (iii) are wrong.
3. Match. The only option that pairs (i) with (iv) is (d). Options (a), (b) and (c) each drop one of the two true statements.

Why this matters. The same warm-up is the basis of antacid relief: the base in the tablet neutralises stomach acid and you feel the burn ease as the reaction completes.

Option (d): (i) temperature increases and (iv) salt formation, both occur.

Quick reading (one acid among bases). The starting solution is basic. Three options are bases and one is an acid. Only the acid can flip the colour back. Spot the odd one out.

Concept used. Litmus changes colour with H+ and OH- balance. A base raises OH- (red litmus turns blue); an acid raises H+ (blue litmus turns red). Reversing a base-driven change needs an acid that supplies enough H+ to overshoot neutral.

1. Classify each option. Baking powder contains NaHCO₃ (mildly basic); lime is Ca(OH)₂ (basic); ammonium hydroxide NH₄OH is a weak base; hydrochloric acid HCl is a strong acid.
2. Eliminate the bases. Adding any base keeps OH- high, so the litmus stays blue. Options (a), (b) and (c) all fail.
3. Confirm the acid. Excess HCl neutralises the base and then drives the pH below 7, turning the blue colour red.

Why this matters. This is how a lab worker safely “cancels” a spilt alkali: a measured acid wash restores neutral conditions, and the indicator colour confirms when to stop.

Option (d): hydrochloric acid is the only acid, so it alone reverses the change.

Quick reading (name the job of the chemical). “Guard tube” + “calcium chloride” is textbook shorthand for “dry the gas.” CaCl₂ is a classic desiccant, so its only job here is moisture removal.

Concept used. A desiccant such as anhydrous CaCl₂ has a strong affinity for water and grabs water vapour from a passing gas. It must be chemically inert towards the gas it dries, otherwise it would consume the product.

1. Identify the problem. Humid air contributes water vapour to the freshly made HCl gas; the experiment needs dry HCl.
2. Identify the fix. CaCl₂ absorbs that water vapour as it flows through the guard tube, drying the gas.
3. Reject the traps. It does not absorb HCl (it is inert to it), does not moisten the gas (opposite of its nature), and has no mechanism to remove Cl- ions from a gas. Hence only (c) fits.

Why this matters. The same drying trick keeps reagents pure in real labs: many gases are passed through CaCl₂ or concentrated H₂SO₄ to remove water before use.

Option (c): calcium chloride dries the gas by absorbing its moisture.

Quick reading (look for the dot-water tail). Three of these salts are famous hydrates whose formulas end in “· n H₂O”. The one whose formula has no such tail is the answer.

Concept used. Water of crystallisation is water chemically locked into a crystal lattice in a definite ratio. Its presence is shown by the dot notation, e.g. CuSO₄·5H₂O. Heating drives off this water and changes the salt's colour or texture.

1. Write each formula. Blue vitriol CuSO₄·5H₂O; washing soda Na₂CO₃·10H₂O; gypsum CaSO₄·2H₂O; baking soda NaHCO₃.
2. Count the dot-water. The first three carry 5, 10 and 2 water molecules respectively. Baking soda carries none.
3. Conclude. Only NaHCO₃ lacks water of crystallisation, so option (b) is correct.

Why this matters. This is why heated blue vitriol turns from blue to white (it loses its 5 water molecules) while baking soda just decomposes to Na₂CO₃ on heating instead of dehydrating.

Option (b): baking soda, whose formula NaHCO₃ carries no water of crystallisation.

Strategic angle (trace the parents). A salt remembers its parents. Sodium carbonate is basic, so its stronger parent must be a base. Confirm which acid and base combine to give it.

Concept used. Salt hydrolysis: when a salt dissolves, the ion from the weaker parent reacts with water. For Na₂CO₃, the carbonate ion (from weak H₂CO₃) grabs H+ from water and frees OH-, so the solution turns basic. Na+ (from strong NaOH) does not hydrolyse.

1. Name the parents. Na₂CO₃ = NaOH (strong base) + H₂CO₃ (weak acid).
2. Predict the pH. The weak-acid anion CO₃^2- hydrolyses: CO₃^2- + H₂O ⇌ HCO₃- + OH-. Released OH- makes the solution basic.
3. Match the option. Weak acid plus strong base gives a basic salt, exactly option (d). Options (a)–(c) would predict neutral or acidic, which contradicts the observed basicity.

Why this matters. The same logic explains why washing soda solution (Na₂CO₃·10H₂O) feels slippery and cleans grease: its basic OH- attacks oily dirt.

Option (d): salt of weak acid H₂CO₃ and strong base NaOH, hence basic.

Strategic angle (salt of which acid and base). Enamel is a calcium salt of phosphoric acid. Calcium hydroxide is a strong base and phosphoric acid is weak, so the salt should lean basic. Confirm and match.

Concept used. The acid–base nature of a salt follows its stronger parent. Ca₃(PO₄)₂ comes from strong Ca(OH)₂ and weak H₃PO₄; a strong-base/weak-acid salt is basic. Amphoteric means it reacts as both acid and base, which calcium phosphate does not do here.

1. Write the salt and its parents. Enamel = Ca₃(PO₄)₂; base = Ca(OH)₂ (strong); acid = H₃PO₄ (weak).
2. Apply the rule. Strong base + weak acid → basic salt. So the enamel is basic.
3. Reject other options. It is not acidic or neutral by the rule above, and it is not amphoteric (that label fits oxides like Al₂O₃ or ZnO, not this phosphate).

Why this matters. Knowing enamel is basic explains the dentist's advice: rinse after acidic drinks so the acid does not sit on and dissolve your basic enamel.

Option (a): basic, since Ca₃(PO₄)₂ is a strong-base/weak-acid salt.

Quick reading (which way is the colour moving). Yellow-orange to green-blue is acidic to basic. So you need to add a base. Three options are acids or neutral; only the antacid is basic.

Concept used. The pH-paper colour scale runs red (strong acid) through yellow-orange (weak acid) to green (neutral) to blue (base). A colour shift up the scale (towards blue) requires adding OH-, i.e. a base.

1. Read the starting pH. Yellowish-orange ⇒ pH around 5 (mildly acidic soil solution).
2. Decide the direction. Greenish-blue ⇒ pH around 8 (basic). To go from 5 to 8 you must add a base.
3. Pick the base. Lemon juice and vinegar are acids (push pH down); common salt is neutral (no change); an antacid is basic and raises the pH. So the antacid is the only fit.

Why this matters. This is real agriculture: acidic soil is treated with a base such as lime or calcium carbonate to raise its pH so crops grow better.

Option (d): an antacid, being basic, raises the pH and gives the greenish-blue colour.

Strategic angle (rank by H+ released). Acidic strength is just “how much H+.” Water gives almost none, acetic acid gives some, HCl gives the most. That ordering is the answer.

Concept used. Degree of ionisation fixes acid strength. Water self-ionises to a tiny extent (K_w = 10^-14), acetic acid is a weak acid (K_a ≈ 1.810^-5), and HCl is a strong acid (K_a very large). Larger ionisation means more free H+ and higher acidity.

1. Place water. Pure water is neutral; its H+ concentration is only 10^-7 M. Weakest acid here.
2. Place acetic acid. It partly ionises (CH₃COOH ⇌ CH₃COO- + H+), giving more H+ than water but far less than a strong acid.
3. Place HCl. It ionises completely (HCl → H+ + Cl-), giving the highest H+. Final increasing order: water < acetic acid < HCl.

Why this matters. This is why vinegar (acetic acid) is safe on food while HCl is corrosive: same idea of an acid, but vastly different H+ output.

Option (a): water < acetic acid < hydrochloric acid, by increasing ionisation.

Strategic angle (safety first, mild base second). First remove the acid with water, then neutralise gently. The only option that does both safely is (b); the others either skip the wash or use a dangerous strong base.

Concept used. First aid for an acid spill: dilution by copious water lowers H+ fast, then a weak base (NaHCO₃) neutralises the residue without harming skin. Strong bases are corrosive and are never applied to skin.

1. Dilute and remove. Plenty of running water washes the acid off and dilutes what stays, cutting the damage.
2. Neutralise gently. A paste of NaHCO₃ reacts with leftover acid (NaHCO₃ + HCl → NaCl + H₂O + CO₂) and is mild enough for skin.
3. Reject unsafe choices. (a) saline does not neutralise; (c) and (d) use a strong alkali NaOH, which burns skin. Only (b) is both effective and safe.

Why this matters. This is the exact protocol on every school lab safety chart: water first, mild base second, never a strong base.

Option (b): wash with plenty of water, then apply NaHCO₃ paste.

Strategic angle (identify the gas, then test each claim). The acid–bicarbonate reaction gives CO₂. Run CO₂'s known properties against the four statements: three fit, one (pungent smell) does not.

Concept used. NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂. CO₂ is a colourless, odourless, acidic gas that turns lime water milky, does not support combustion, and is absorbed by alkalis.

1. Lime-water test. CO₂ + Ca(OH)₂ → CaCO₃ + H₂O; the white CaCO₃ makes lime water milky. Statement (i) true.
2. Combustion test. CO₂ does not burn or support burning, so it puts out a flaming splinter. Statement (ii) true.
3. Alkali test. Being acidic, CO₂ dissolves in NaOH: 2NaOH + CO₂ → Na₂CO₃ + H₂O. Statement (iii) true. But CO₂ has no smell, so (iv) is false. Answer: (i), (ii), (iii).

Why this matters. These three tests together (lime water, flame, alkali absorption) are how you positively identify CO₂ in any lab, not just this reaction.

Option (b): the evolved CO₂ satisfies statements (i), (ii) and (iii).

Strategic angle (sodium products only). Common salt is a sodium compound, so its products contain sodium. Washing soda and baking soda are sodium salts; bleaching powder and slaked lime are calcium compounds. Pick the sodium pair.

Concept used. NaCl is the feedstock for the Solvay process and the chlor-alkali process. These give Na₂CO₃, NaHCO₃, NaOH and Cl₂. Calcium chemicals (Ca(OH)₂, CaOCl₂) trace back to CaCO₃, not NaCl.

1. Baking soda. Made from NaCl via NaCl + H₂O + CO₂ + NH₃ → NaHCO₃ + NH₄Cl. So (iii) holds.
2. Washing soda. Heating that NaHCO₃ gives Na₂CO₃, then recrystallisation gives Na₂CO₃·10H₂O. So (i) holds.
3. Eliminate calcium items. Slaked lime and bleaching powder need CaCO₃/CaO, not common salt. (ii) and (iv) are out, leaving (i) and (iii).

Why this matters. It shows how one cheap mineral (rock salt) feeds a whole family of household chemicals, a core idea of the industrial chemistry in this chapter.

Option (c): common salt gives washing soda and baking soda.

Strategic angle (which acid is food-safe). The added acid has to be edible and mild. Of the four, only tartaric acid is a food acid; HCl, H₂SO₄ and even CH₃COOH are not used in baking powder.

Concept used. Baking powder = NaHCO₃ + tartaric acid. The role of the acid is to neutralise the basic Na₂CO₃ produced when NaHCO₃ decomposes, preventing a bitter taste in the baked food.

1. Recall the mixture. Baking powder always pairs NaHCO₃ with a mild solid acid.
2. Pick the edible acid. Tartaric acid is a safe, weak food acid; mineral acids HCl and H₂SO₄ are corrosive and never added to food.
3. State its job. Tartaric acid neutralises the bitter Na₂CO₃ left after heating, so the cake tastes right. Hence (b).

Why this matters. This is why a cake made with plain baking soda tastes bitter, while one made with baking powder does not: the tartaric acid does the cleanup.

Option (b): tartaric acid.

Quick reading (fight acid with base). Decay is acid damage, so the cure has to be a base. Toothpaste is therefore basic. The other options cannot neutralise acid.

Concept used. Importance of pH in everyday life: the mouth becomes acidic after eating; below pH 5.5 enamel dissolves. A basic toothpaste raises the pH back above neutral, neutralising the acid and protecting the teeth.

1. Find the problem. Acid from bacteria lowers mouth pH and corrodes enamel.
2. Find the cure. A base neutralises that acid, so the toothpaste must be basic.
3. Reject the rest. Acidic or neutral paste cannot neutralise acid; a corrosive paste would damage gums. Hence basic, option (c).

Why this matters. It connects directly to Q6: enamel is basic calcium phosphate, and a basic paste defends that basic enamel from acid attack.

Option (c): basic toothpaste neutralises mouth acid and prevents decay.

Strategic angle (pH direction for each). pH down means more acidic; pH up means more basic. Read each of the four claims through that single rule and only (ii) and (iv) survive.

Concept used. pH = -log[H+]. Smaller pH ⇒ larger H+ ⇒ stronger acid; larger pH ⇒ larger OH- ⇒ stronger base. Strength tracks the right ion concentration.

1. Acid statements. Stronger acid has lower pH, so “higher pH, weaker acid” (ii) is true and “higher pH, stronger acid” (i) is false.
2. Base statements. Stronger base has higher pH, so “lower pH, weaker base” (iv) is true and “lower pH, stronger base” (iii) is false.
3. Combine. The only correct pair is (ii) and (iv), option (d).

Why this matters. Getting this direction right is the basis of reading any pH measurement: a falling pH always signals a more acidic (less basic) solution.

Option (d): (ii) higher pH weaker acid and (iv) lower pH weaker base.

Quick reading (HCl means acidic). Gastric juice contains HCl, so it is acidic, so its pH is below 7. The trap is option (d) “equal to 0,” which is far too strong.

Concept used. Acidity in the body: gastric juice carries dilute HCl, giving a pH of roughly 1.5 to 3. Any pH under 7 is acidic; pH 0 would mean 1 M H+, much stronger than stomach acid.

1. Identify the acid. Gastric juice has HCl, so it is acidic.
2. Place it on the scale. Acidic means pH below 7; the measured value is about 1.5–3.
3. Reject extremes. pH 0 (option d) is far stronger than real gastric acid; basic (b) and neutral (c) are wrong. So “less than 7” (a) is the safe, correct answer.

Why this matters. This is why antacids work: they are bases that raise the stomach pH toward neutral, relieving the discomfort of excess HCl.

Option (a): gastric juice has pH less than 7 (acidic, due to HCl).

Strategic angle (no base, no neutralisation). Only acid and water are present, so anything needing a base is impossible. That rules out neutralisation and salt formation, leaving ionisation and dilution.

Concept used. Dissolving an acid in water: the polar water molecules pull the acid apart into ions (ionisation) and, since a small amount of acid spreads through a lot of water, its concentration drops (dilution). Neutralisation/salt require an acid and a base.

1. Ionisation. Water ionises the acid: HCl + H₂O → H₃O+ + Cl-. Statement (i) holds.
2. Dilution. The same small amount of acid in more water is less concentrated. Statement (iii) holds.
3. Rule out the rest. With no base added, neither neutralisation (ii) nor salt formation (iv) can occur. So the answer is (i) and (iii).

Why this matters. The dilution step is exothermic, which is the chemical reason behind the lab rule “add acid to water, slowly and with stirring.”

Option (b): ionisation and dilution occur; no base means no neutralisation or salt.

Quick reading (smell, not sight). A visually impaired student needs a clue they can detect without seeing. Only an olfactory indicator fits, and vanilla essence is the olfactory one here.

Concept used. Olfactory indicators change their smell with acidity. Vanilla, onion and clove oil lose their odour in a basic solution; this smell change is perceivable without vision, unlike the colour change of litmus, turmeric or petunia.

1. Set the requirement. The indicator's signal must be non-visual, i.e. detected by smell.
2. Classify the options. Litmus, turmeric, petunia leaves are colour indicators (visual). Vanilla essence is an olfactory indicator (smell).
3. Choose. Only vanilla essence gives a smell change, so it alone suits a visually impaired student. Option (c).

Why this matters. It shows that indicators are not only about colour: the chapter's idea of detecting acids and bases extends to smell, making chemistry accessible to all students.

Option (c): vanilla essence, whose smell changes, is the olfactory indicator.

Strategic angle (carbonate or not). CO₂ comes only from a carbonate or hydrogencarbonate reacting with acid. Three options carry a carbonate group; lime (CaO) does not. So lime is the odd one out.

Concept used. Acid + carbonate/hydrogencarbonate → salt + water + CO₂. An oxide such as CaO reacts with acid to give only salt + water, because it has no CO₃ or HCO₃ group to supply carbon.

1. Carbonate items. Marble and limestone are CaCO₃; with HCl they give CaCl₂ + H₂O + CO₂. CO₂ released.
2. Hydrogencarbonate item. Baking soda NaHCO₃ with acid gives NaCl + H₂O + CO₂. CO₂ released.
3. Oxide item. Lime CaO with acid gives only CaCl₂ + H₂O, no carbon source, so no CO₂. Hence (d).

Why this matters. The lime-water test for CO₂ uses Ca(OH)₂, not CaO; this question checks that you separate the oxide from the carbonate, a frequent exam trap.

Option (d): lime (CaO), being an oxide, gives no CO₂.

Quick reading (find the sour one). Acidic things taste sour and have pH <7. Lime juice is the sour citrus drink; blood, lime water and antacid are all on the basic side. So lime juice is acidic.

Concept used. pH classification: pH <7 acidic, =7 neutral, >7 basic. Citrus juices carry citric acid (acidic). Blood is buffered slightly basic; Ca(OH)₂ and antacids are basic.

1. Lime juice. Contains citric acid, pH ≈ 2, so acidic. This is our candidate.
2. Blood and lime water. Blood pH ≈ 7.4 (slightly basic); lime water Ca(OH)₂ is basic.
3. Antacid. Antacids are bases by design. So the only acidic option is lime juice, (a).

Why this matters. It is a clean test of the “lime juice vs lime water” trap and of reading pH categories, both core to this chapter.

Option (a): lime juice (citric acid) is the acidic substance.

Picture-first (does it glow, and why). The circuit is complete and the electrolyte is NaOH, a strong base full of ions. Ions carry current, so the bulb glows. Only statement (ii) says exactly that.

Concept used. Electrolytic conduction: a solution conducts when it contains mobile ions. Strong electrolytes like NaOH dissociate completely, giving many Na+ and OH- ions. Being basic does not stop conduction; only the presence of ions matters.

1. Check the electrolyte. NaOH → Na+ + OH- fully, so the solution is rich in ions and conducts well.
2. Check the circuit. Battery, bulb, switch and the two nail electrodes in Fig. 2.1 form a closed loop, so current can flow and the bulb glows. Statement (ii) is correct.
3. Reject the others. (i) wrongly says only acids conduct; (iii) wrongly says the circuit is open; (iv) wrongly says it will not glow. All false, so (ii) only.

Why this matters. It corrects a frequent misconception that only acids conduct. Any strong electrolyte, acid or base or salt, conducts because the test is simply “are there free ions?”

Option (c): only statement (ii); the strong base NaOH supplies ions and the bulb glows.

Strategic angle (gold needs the special mix). Gold is too noble for any ordinary acid. Only the special HCl/HNO₃ blend, aqua regia, can take it into solution. So the answer must be (d).

Concept used. Aqua regia = 3 conc. HCl : 1 conc. HNO₃. The nitric acid acts as the oxidiser (removing electrons from gold) while chloride ions stabilise the resulting Au³⁺ as the [AuCl₄]- complex, allowing dissolution.

1. Why single acids fail. Gold sits below hydrogen and is very unreactive, so HCl, H₂SO₄ and HNO₃ alone leave it untouched.
2. Why the mixture works. In aqua regia, HNO₃ oxidises gold and HCl supplies Cl- to complex the gold ions, together overcoming gold's inertness.
3. Conclude. Only aqua regia dissolves gold, so option (d) is correct.

Why this matters. This is used in real refining and testing of gold jewellery: aqua regia separates and purifies precious metals on an industrial scale.

Option (d): aqua regia dissolves gold; no single acid can.

Strategic angle (which comes from a fruit). Mineral acids are lab/mineral acids; organic acids come from living sources. Citric acid is from citrus fruit, so it is the organic one, the odd one out.

Concept used. Classification of acids by source: mineral acids (HCl, H₂SO₄, HNO₃) are inorganic, usually strong, and made from minerals; organic acids (citric, acetic) are carbon-based and sourced from plants or animals.

1. Tag the mineral acids. HCl, H₂SO₄ and HNO₃ are classic inorganic mineral acids.
2. Tag the organic acid. Citric acid is a carbon-based acid obtained from citrus fruits, hence organic.
3. Answer the negative. The question wants the non-mineral acid, which is citric acid, option (b).

Why this matters. The mineral/organic split explains why citric acid is safe in food and drinks while mineral acids like H₂SO₄ are strictly lab chemicals.

Option (b): citric acid, an organic acid, is not a mineral acid.

Strategic angle (does it give OH-). A base must release OH- in water. Three of these are metal/ammonium hydroxides that do; ethanol's -OH is an alcohol group that does not. So ethanol is not a base.

Concept used. Arrhenius bases release OH- in water. NaOH, KOH and NH₄OH ionise to give OH-. In an alcohol like C₂H₅OH, the O-H bond is covalent to a carbon chain and does not ionise, so no free OH- appears.

1. Strong/weak hydroxides. NaOH, KOH (strong bases) and NH₄OH (weak base) all furnish OH-.
2. Alcohol case. C₂H₅OH keeps its -OH as part of the molecule; it does not dissociate into OH-.
3. Pick the non-base. Only ethanol fails the OH- test, so option (d) is correct.

Why this matters. It guards against the trap of reading every “OH” as a base, a distinction that returns in organic chemistry where alcohols behave very differently from hydroxide bases.

Option (d): C₂H₅OH does not give OH- ions, so it is not a base.

Strategic angle (test each statement). Three statements match standard reactions; one breaks the “metal oxides are basic” rule. Find the rule-breaker.

Concept used. Metal oxides are basic and give bases with water (Na₂O + H₂O → 2NaOH). Non-metal oxides are acidic and give acids (CO₂ + H₂O → H₂CO₃). Carbonates with acid give salt + water + CO₂; reactive metals with acid give salt + H₂.

1. Verify (a), (c), (d). Each is a standard, correct reaction (carbonate+acid; metal+acid; non-metal oxide+ water).
2. Examine (b). Metal oxides react with water to give a base, not “salt and acid.” This contradicts the basic nature of metal oxides.
3. Conclude. The incorrect statement is (b), so it is the answer.

Why this matters. This single rule, “metal oxide = basic, non-metal oxide = acidic,” explains acid rain (SO₂, NO₂) and the basicity of quicklime, both real-world consequences.

Option (b): the false statement; metal oxides give bases with water, not salt and acid.

Strategic angle (match the surest links first). Pin the applications you are certain of, then the rest fall into place. Antacid = baking soda and H₂/Cl₂ = brine are the anchors.

Concept used. Uses of common salts: bleaching powder CaOCl₂ decolourises fabric and paper; NaHCO₃ is a mild antacid; Na₂CO₃ is a flux in glass manufacture; electrolysis of NaCl brine yields H₂, Cl₂ and NaOH.

1. Anchor B and D. Baking soda is the antacid, so B–(iv); brine gives H₂ and Cl₂, so D–(ii).
2. Place A and C. Bleaching powder decolourises, so A–(iii); washing soda is used in glass, so C–(i).
3. Read the option. A–(iii), B–(iv), C–(i), D–(ii) matches option (c).

Why this matters. These four salts and their uses recur across the chapter's “chemicals from common salt” section, so memorising the pairings pays off in several questions.

Option (c): A–(iii), B–(iv), C–(i), D–(ii).

Strategic angle (equal strong acid and base = neutral). Equal moles of a strong acid and a strong base cancel exactly, leaving a neutral salt solution at pH 7. On the colour guide, pH 7 is green (yellowish green).

Concept used. HCl + NaOH → NaCl + H₂O. With equal volumes and concentrations, moles of H+ equal moles of OH-, so they neutralise completely. NaCl is a strong-acid/strong-base salt, hence neutral (pH 7), which reads green on a pH chart.

1. Count moles. Equal volume × equal concentration gives equal moles of HCl and NaOH.
2. Neutralise. They react 1:1 to give NaCl and water, with no acid or base left over, so pH = 7.
3. Read the colour. Fig. 2.2 shows pH 7 as yellowish green. Red (acidic) and blue (basic) are ruled out by neutrality, so option (c).

Why this matters. This is the principle of a titration endpoint: when acid exactly neutralises base, the indicator shows the neutral colour, telling you the reaction is complete.

Option (c): neutral pH 7 NaCl solution gives yellowish green.

Strategic angle (covalent yet ionises in water). Covalent HCl still ionises in water, and its H+ attaches to water as H₃O+. So the true statements describe ionisation and hydronium formation, (ii) and (iv).

Concept used. HCl + H₂O → H₃O+ + Cl-. The polar water molecules ionise the covalent HCl; the released proton does not stay free but bonds to water as the hydronium ion H₃O+. HCl produces no OH-.

1. Ionisation happens. Despite being covalent, HCl ionises in water, making (ii) true and (i) false.
2. Hydronium forms. H+ + H₂O → H₃O+, so (iv) is true.
3. No hydroxyl. HCl gives H₃O+ and Cl-, never OH-, so (iii) is false. Hence (ii) and (iv), option (c).

Why this matters. This explains why acids must be dissolved in water to act as acids: the hydronium ion, not a bare proton, is the real acidic species in solution.

Option (c): HCl ionises and forms hydronium ions, statements (ii) and (iv).

Quick reading (sour + blue-to-red). Acids have two signature properties: a sour taste and turning blue litmus red. Only one option states both correctly.

Concept used. Properties of acids: sour taste, turn blue litmus red, react with metals to give H₂. Bases are bitter, soapy, and turn red litmus blue. Matching the taste and the litmus action picks the right option.

1. Taste filter. Acids are sour, so drop the “bitter” options (a) and (d).
2. Litmus filter. Acids turn blue litmus red, so drop “red litmus to blue” (option b).
3. Result. The surviving option is (c): sour and blue litmus to red.

Why this matters. These are the defining sensory and indicator tests for acids used throughout the chapter and in the lab, so getting the pairing right is foundational.

Option (c): acids are sour and turn blue litmus red.

Strategic angle (full ionisation of a strong acid). HCl is strong, so it splits almost entirely into H₃O+ and Cl-. No OH- forms and barely any whole HCl remains.

Concept used. HCl + H₂O → H₃O+ + Cl-. A strong acid has a near-complete degree of ionisation, so the dominant species are the ions, not the molecule, and there is no source of OH-.

1. Ionise fully. HCl converts to H₃O+ and Cl- almost completely.
2. No hydroxide. HCl gives no OH-, so any option with OH- (b, c) is wrong.
3. Negligible molecules. Unionised HCl (d) is negligible for a strong acid. So the answer is (a).

Why this matters. This is why dilute HCl is a strong electrolyte and conducts well: it is almost wholly a solution of free ions, linking directly to Q20's conductivity idea.

Option (a): H₃O+ and Cl- are the species in dilute HCl.

Strategic angle (check the state symbols). All four options have the same formulas; only the state symbols differ. Get the states right and the answer is forced to (d).

Concept used. Electrolysis of aqueous NaCl (brine): 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g). Cl₂ is liberated at the anode and H₂ at the cathode, both gases, while NaOH stays dissolved.

1. Fix the reactants. Brine is NaCl(aq) (not molten (l) as in option a), and water is H₂O(l) (not (aq) as in option b).
2. Fix the products. Cl₂ and H₂ escape as gases (g), not dissolved (aq) as in option c; NaOH is (aq).
3. Select. Only (d) has all states correct, so it is the right representation.

Why this matters. The chlor-alkali process is the industrial source of NaOH, Cl₂ and H₂, three feedstocks behind soap, bleach and many other products in this chapter.

Option (d): 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g), with correct states.

Strategic angle (name tells the source). Each acid's everyday name points to where it is found. Lactic to curd, acetic to vinegar, citric to lemon, oxalic to tomato. Match by familiarity.

Concept used. Natural organic acids are classified by source: foods and biological materials each contain a characteristic acid. Knowing these pairings is part of the chapter's “acids in daily life” section.

1. Dairy acid. Lactic acid forms when milk sours, so it belongs to curd: (a)–(iv).
2. Kitchen acids. Acetic acid is vinegar's acid (b)–(iii); citric acid is the lemon's acid (c)–(ii).
3. Vegetable acid. Oxalic acid is present in tomato, so (d)–(i). Full match: a-iv, b-iii, c-ii, d-i.

Why this matters. These source pairings show that acids are not just lab chemicals; they are part of everyday food, which is the intuition this chapter builds.

(a)–(iv), (b)–(iii), (c)–(ii), (d)–(i).

Strategic angle (recall each formula). All four are calcium chemicals. Match each name to its formula by recalling the water content and the anion.

Concept used. Formulas of key calcium salts: POP CaSO₄·1/2H₂O, gypsum CaSO₄·2H₂O, bleaching powder CaOCl₂, slaked lime Ca(OH)₂. The sulphates differ only in water of crystallisation.

1. Sulphate pair. POP carries half a water molecule (a)–(ii); gypsum carries two (b)–(iii).
2. Chloride salt. Bleaching powder is CaOCl₂, so (c)–(iv).
3. Hydroxide. Slaked lime is Ca(OH)₂, so (d)–(i). Final: a-ii, b-iii, c-iv, d-i.

Why this matters. These formulas appear again in the long-answer questions (POP, gypsum, bleaching powder), so the matching here is a springboard for the rest of the chapter.

(a)–(ii), (b)–(iii), (c)–(iv), (d)–(i).

Strategic angle (acid, base, or no ions). Sort each substance: acids redden blue litmus, bases blue the red, and a dry gas does nothing because it has no water to ionise in.

Concept used. Litmus is an acid–base indicator that needs ions in a moist medium. Dry gases give no ions; acidic solutions give H+; basic solutions give OH-.

1. The dry gas. Dry HCl has no water, releases no H+, so litmus shows no change.
2. The base. Moist NH₃ forms NH₄OH; soap solution is basic; both turn red litmus blue.
3. The acids. Lemon juice, carbonated drink and curd are acidic; each turns blue litmus red.

Why this matters. The dry-gas case is a classic test of the deeper idea that acidic behaviour requires water, the same point made in Q27 about HCl ionising in water.

Dry HCl: no change; moist NH₃ and soap: red→blue; lemon, soft drink, curd: blue→red.

Strategic angle (acid in, base on). The sting is acidic (formic acid), so the cure is a mild base applied on the skin to neutralise it. Name the acid, give its formula, name the base.

Concept used. Neutralisation in daily life: an ant injects methanoic acid HCOOH; a mild base (NaHCO₃) reacts with it (HCOOH + NaHCO₃ → HCOONa + H₂O + CO₂), removing the acid and the sting.

1. Name and formula. The sting acid is methanoic (formic) acid, HCOOH.
2. Why it hurts. The acid irritates the skin, causing burning and itching.
3. The relief. Apply baking soda NaHCO₃ (a mild base); it neutralises the acid, easing the discomfort.

Why this matters. This is the everyday face of neutralisation: the same acid–base reaction that fills a textbook also soothes a real ant bite.

Methanoic (formic) acid HCOOH; apply a mild base like baking soda to neutralise it.

Strategic angle (shell is a carbonate). Egg shell is CaCO₃, so adding any acid gives salt, water and CO₂. Write the balanced reaction with HNO₃ and describe the fizzing.

Concept used. Acid + carbonate → salt + water + CO₂. With HNO₃: CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂. The CO₂ causes effervescence and turns lime water milky.

1. Identify the shell. Egg shell is calcium carbonate CaCO₃.
2. Write the reaction. CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂, giving calcium nitrate, water and carbon dioxide.
3. Observe. Bubbles of CO₂ appear; passing the gas through lime water turns it milky, confirming CO₂.

Why this matters. It is a neat demonstration that shells, marble and chalk are all CaCO₃: any acid makes them fizz, which is also why acid rain damages marble monuments.

Effervescence of CO₂: CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂.

Strategic angle (any indicator will do). Litmus is just one indicator. Phenolphthalein alone solves it: pink means base, colourless means acid. Natural indicators are a backup.

Concept used. Acid–base indicators change colour with pH. Phenolphthalein is colourless in acidic/neutral solution and pink in basic solution; turmeric is yellow in acid and red in base. Either distinguishes the two beakers.

1. Pick an indicator. Add phenolphthalein to a sample from each beaker.
2. Read the colour. The base turns it pink; the acid leaves it colourless.
3. Backup test. If no synthetic indicator is at hand, turmeric paper turns red in the base and stays yellow in the acid.

Why this matters. It teaches that the concept “indicator” is broader than litmus, which is the practical point of the chapter's section on natural and synthetic indicators.

Add phenolphthalein: pink identifies the base, colourless identifies the acid.

Strategic angle (only one gives CO₂). On heating, the hydrogencarbonate (baking soda) gives CO₂ while the carbonate (washing soda) does not. The lime-water test tells them apart at once.

Concept used. Thermal decomposition: NaHCO₃ breaks down to Na₂CO₃ + H₂O + CO₂, but Na₂CO₃·10H₂O only dehydrates to Na₂CO₃ + 10H₂O. CO₂ turns lime water milky; water does not.

1. Heat and trap the gas. Warm each solid and lead any gas into lime water.
2. Baking powder. Its NaHCO₃ gives CO₂ (2NaHCO₃ → Na₂CO₃ + H₂O + CO₂), turning lime water milky.
3. Washing soda. It only loses water; lime water stays clear. The milky result identifies the baking powder.

Why this matters. The same decomposition is what makes baking powder leaven a cake: the CO₂ released on heating puffs up the batter, linking lab test to kitchen use.

Baking powder gives CO₂ on heating (lime water milky); washing soda only loses water.

Strategic angle (follow the clue chain). Bakery salt that decomposes on heating is baking soda; its product removes water hardness (sodium carbonate); the gas that fogs lime water is CO₂. Three clues, three identities.

Concept used. On heating, 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂. Sodium carbonate softens hard water by precipitating Ca²⁺/Mg²⁺; CO₂ turns lime water milky via CaCO₃.

1. Find A. A bakery salt decomposing on heating is baking soda, NaHCO₃.
2. Find B. Heating gives Na₂CO₃, used to remove hardness of water. So B is sodium carbonate.
3. Find C. The gas that turns lime water milky is CO₂. Thus A = NaHCO₃, B = Na₂CO₃, C = CO₂.

Why this matters. It ties three chapter ideas together (baking soda's use, washing/sodium carbonate softening water, and the CO₂ test) into one clue-based identification.

A = NaHCO₃, B = Na₂CO₃, C = CO₂.

Strategic angle (by-product to bleach). NaOH comes from brine electrolysis, whose by-product is chlorine. Chlorine plus slaked lime is bleaching powder. So X is Cl₂, Y is CaOCl₂.

Concept used. Chlor-alkali process: 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g). Bleaching powder formation: Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O.

1. Name X. The by-product gas of NaOH manufacture is chlorine, Cl₂ (hydrogen is the other by-product).
2. Form Y. Cl₂ with slaked lime gives bleaching powder: Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O.
3. Identify Y. Y is calcium oxychloride CaOCl₂, the bleaching agent.

Why this matters. It shows how one industrial process feeds another: chlorine from NaOH production becomes the raw material for bleaching powder, a theme of the chapter's industrial chemistry.

X = chlorine Cl₂; Y = bleaching powder CaOCl₂; Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O.

Strategic angle (parent base and parent acid). Each salt is a base part plus an acid part. Identify the metal/ammonium (from the base) and the acid radical (from the acid), and the blanks fill themselves.

Concept used. Salt formation: Base + Acid → Salt + Water. The cation traces to a hydroxide base, the anion traces to an acid. So CuSO₄ = Cu(OH)₂ + H₂SO₄, and so on.

1. Chlorides. NH₄Cl and NaCl both take their chloride from HCl, so the acid is HCl in (i) and (iii).
2. Sulphates. CuSO₄ comes from Cu(OH)₂ + H₂SO₄ (giving formula CuSO₄); K₂SO₄ comes from KOH + H₂SO₄.
3. Nitrates. Mg(NO₃)₂ takes Mg from Mg(OH)₂; Ca(NO₃)₂ takes NO₃ from HNO₃.

Why this matters. Reading a salt back to its acid and base is the core skill for naming and predicting salts, used throughout inorganic chemistry.

(i) HCl; (ii) CuSO₄, base Cu(OH)₂; (iii) HCl; (iv) Mg(OH)₂; (v) KOH and H₂SO₄; (vi) HNO₃.

Strategic angle (mineral acids strong, food acids weak). The three mineral acids (HCl, HNO₃, H₂SO₄) ionise fully and are strong; the three carbon-based acids (citric, acetic, formic) ionise partly and are weak.

Concept used. Acid strength = degree of ionisation. Strong acids ionise nearly completely (large H+); weak acids ionise only a small fraction, shown by an equilibrium arrow, giving far less H+ at the same concentration.

1. Define both. Strong acid: full ionisation, high H+. Weak acid: partial ionisation, low H+.
2. Sort the strong. HCl, HNO₃ and H₂SO₄ ionise fully, so they are strong.
3. Sort the weak. Citric, acetic and formic acids ionise only partly, so they are weak.

Why this matters. This is why dilute H₂SO₄ is dangerous while equally dilute acetic acid (vinegar) is safe to eat: same concentration, very different H+ output.

Strong: hydrochloric, nitric, sulphuric. Weak: citric, acetic, formic.

Strategic angle (metal + acid gives H₂). Reactive metals displace hydrogen from acids. The gas, H₂, is the one used to hydrogenate oils. Write the reaction and the pop-sound test.

Concept used. Metal + dilute acid → salt + H₂. Zinc with HCl gives ZnCl₂ and H₂. Hydrogen is detected by its characteristic pop sound on burning, and industrially it adds across C=C bonds in oils (hydrogenation).

1. Name the gas. Zinc with a dilute strong acid gives hydrogen, H₂, the gas used in hydrogenation of oils.
2. Write the reaction. Zn + 2HCl → ZnCl₂ + H₂ (zinc chloride plus hydrogen).
3. Test it. A burning splinter held at the tube mouth makes the H₂ burn with a pop sound, confirming the gas.

Why this matters. The same H₂ that fizzes out of a simple metal–acid reaction is what hardens vegetable oil into vanaspati, linking a basic lab reaction to a real food industry.

Hydrogen H₂; Zn + 2HCl → ZnCl₂ + H₂; detected by the pop sound on burning.

Strategic angle (rate, equality, reactivity, amphoterism). Each change tests one idea: surface area (a), acid swap (b), reactivity order (c), and zinc's amphoteric reaction with hot alkali (d). Handle them one at a time.

Concept used. Metal + dilute acid → salt + H₂. Reaction rate rises with surface area; the amount of H₂ depends on moles of metal and acid; a metal below hydrogen (copper) gives no H₂; amphoteric zinc reacts with hot NaOH to give Na₂ZnO₂ + H₂.

1. (a) Faster with dust. Powdered zinc has far more surface area, so Zn + H₂SO₄ → ZnSO₄ + H₂ proceeds faster; the gas comes off more quickly.
2. (b) Similar with HCl. HCl is also a dilute strong acid: Zn + 2HCl → ZnCl₂ + H₂. For the same moles of zinc, nearly the same volume of H₂ is produced.
3. (c) None with copper. Copper is below hydrogen, so it cannot displace hydrogen from dilute acid; no gas forms.
4. (d) H₂ with hot NaOH. Zinc is amphoteric: Zn + 2NaOH → Na₂ZnO₂ + H₂ on heating, so hydrogen is still evolved, now with the alkali.

Why this matters. Part (d) shows zinc's special amphoteric character: it reacts with both acids and hot alkalis, a property that separates it from ordinary metals.

(a) faster; (b) about the same amount; (c) no hydrogen; (d) hydrogen via Zn + 2NaOH → Na₂ZnO₂ + H₂.

Strategic angle (the missing acid). Baking soda alone leaves bitter Na₂CO₃; baking powder adds tartaric acid to cancel it. All three parts revolve around that one idea.

Concept used. 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂ on heating; Na₂CO₃ is bitter. Tartaric acid neutralises this carbonate, and baking powder is the ready mix of NaHCO₃ + tartaric acid.

1. (a) Why bitter. With only NaHCO₃, heating yields bitter Na₂CO₃, so the cake tastes bitter.
2. (b) Make powder. Mix a measured amount of tartaric acid into the baking soda to get baking powder.
3. (c) Acid's job. During baking the tartaric acid neutralises the bitter Na₂CO₃, keeping the taste pleasant while CO₂ still makes the cake rise.

Why this matters. It explains a real kitchen mistake: swap powder for soda and the cake turns bitter, all because the neutralising acid is missing.

(a) bitter, from Na₂CO₃; (b) add tartaric acid; (c) the acid neutralises the bitter carbonate.

Strategic angle (carbonate, lime water, chlorine, bleach). Read each clue in turn: carbonate giving CO₂ is CaCO₃; the solution returning the carbonate is lime water; the anode gas from brine is chlorine; chlorine on dry lime is bleaching powder.

Concept used. Carbonate + acid → CO₂; Ca(OH)₂ + CO₂ → CaCO₃ + H₂O; brine electrolysis gives Cl₂ at the anode; Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O (bleaching powder).

1. Find X. A metal carbonate giving CO₂ with acid is CaCO₃ (CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂).
2. Find Y. The CO₂ passed through Y returns the carbonate, so Y is lime water Ca(OH)₂.
3. Find G and Z. The anode gas from brine is chlorine G = Cl₂; on dry Ca(OH)₂ it gives bleaching powder Z = CaOCl₂, used to disinfect water.

Why this matters. It weaves together the lime cycle and the chlor-alkali process to explain how everyday bleaching powder, the disinfectant in drinking water, is actually made.

X = CaCO₃, Y = Ca(OH)₂, G = Cl₂, Z = CaOCl₂.

Strategic angle (sticky base from brine). A common base that absorbs moisture and comes from the chlor-alkali process is NaOH. With an acidic oxide it neutralises to give salt and water.

Concept used. NaOH is deliquescent (absorbs water and turns sticky) and is produced alongside Cl₂ and H₂ in the chlor-alkali process. Base + acidic (non-metal) oxide → salt + water, a neutralisation: 2NaOH + CO₂ → Na₂CO₃ + H₂O.

1. Identify B. The hygroscopic base that is a chlor-alkali by-product is sodium hydroxide, NaOH.
2. Type of reaction. NaOH with an acidic oxide is a neutralisation, giving salt and water.
3. Balanced equation. With CO₂: 2NaOH + CO₂ → Na₂CO₃ + H₂O.

Why this matters. This is why NaOH pellets must be stored in airtight bottles: left open they absorb water and even CO₂ from air, slowly turning into Na₂CO₃.

B = NaOH; neutralisation with an acidic oxide: 2NaOH + CO₂ → Na₂CO₃ + H₂O.

Strategic angle (soft now, hard later). A soft, mouldable calcium sulphate is POP; left open it grabs water and sets into hard gypsum. Name it, explain the water uptake, write the setting reaction.

Concept used. POP is calcium sulphate hemihydrate CaSO₄·1/2H₂O, made by gently heating gypsum. It reacts with water to re-form gypsum CaSO₄·2H₂O, a hard mass: CaSO₄·1/2H₂O + 3/2 H₂O → CaSO₄·2H₂O.

1. Identify the salt. A white, soft, mouldable Group 2 sulphate is Plaster of Paris, CaSO₄·1/2H₂O.
2. Explain the change. In open air it absorbs moisture and sets, turning into hard gypsum, so it can no longer be moulded.
3. Write the reaction. CaSO₄·1/2H₂O + 3/2 H₂O → CaSO₄·2H₂O (POP → gypsum).

Why this matters. This setting reaction is why POP is used for casts and moulds: mix with water, shape it fast, and it hardens into gypsum, which is also why POP must be kept tightly sealed.

Plaster of Paris CaSO₄·1/2H₂O; it sets to gypsum on absorbing water: CaSO₄·1/2H₂O + 3/2 H₂O → CaSO₄·2H₂O.

Strategic angle (one base, three reactions). The H₂ clue with zinc plus the salt-and-water clue with two acids both fit a single base, NaOH. Once X is fixed, A, B and C follow from the three reactions.

Concept used. NaOH is a strong base: with amphoteric zinc it gives sodium zincate and hydrogen (Zn + 2NaOH → Na₂ZnO₂ + H₂); with acids it neutralises to a salt and water (NaOH + HCl → NaCl + H₂O; NaOH + CH₃COOH → CH₃COONa + H₂O).

1. Fix X from the zinc clue. A base that gives H₂ with zinc is NaOH; the product A is sodium zincate Na₂ZnO₂ (Zn + 2NaOH → Na₂ZnO₂ + H₂).
2. Get B from the HCl clue. NaOH + HCl → NaCl + H₂O, so B is sodium chloride NaCl.
3. Get C from the acetic-acid clue. NaOH + CH₃COOH → CH₃COONa + H₂O, so C is sodium acetate CH₃COONa.

Why this matters. It pulls together two strands of the chapter: zinc's amphoteric reaction (hydrogen with alkali) and the standard neutralisation of NaOH by a strong and a weak acid.

X = NaOH; A = Na₂ZnO₂; B = NaCl; C = CH₃COONa.