Soap is basic in nature. When soap solution (a base) is applied to the curry stain, turmeric (a natural indicator present in curry) changes from yellow to reddish-brown, indicating a basic medium.

Water is neutral (or very slightly acidic due to dissolved CO₂). When the cloth is rinsed with plenty of water, the base is washed away and the medium returns to neutral/acidic, so turmeric changes back to yellow.

Turmeric is a natural acid-base indicator — it is yellow in acidic or neutral conditions and reddish-brown in basic conditions. Since this colour change is reversible, the indicator returns to its original colour once the base is removed by rinsing.

Source: Chapter 2, Introduction (Acids, Bases and Salts)

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• Key point 1: Soap is basic → turmeric turns reddish-brown (base indicator response).
• Key point 2: Water is neutral → rinsing removes the base, turmeric reverts to yellow.
• Key point 3: Examiners expect you to name turmeric as a natural indicator and state that its colour change is reversible — this directly answers "why the colour reverses."
• Avoid confusing turmeric with litmus; turmeric is yellow in neutral/acidic and reddish-brown in basic medium (opposite pattern from litmus in one direction).
• Do not write about micelles or soap chemistry — that is not what this question tests.

Key limitation: Both litmus and turmeric can only tell whether a solution is acidic or basic (qualitative result); they cannot indicate the strength (degree) of acidity or basicity.

Effect on practical usefulness: In a laboratory, knowing merely that a solution is acidic is often insufficient. A universal indicator, by contrast, shows a range of colours corresponding to pH values (0–14), allowing the chemist to determine whether an acid is weak or strong and to compare the strengths of different solutions accurately.

Source: Chapter 2, Section 2.3 — How Strong Are Acid or Base Solutions?

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• Examiners expect you to name the limitation clearly: natural indicators give only a yes/no (acid/base) answer, not a pH value or strength.
• The contrast with the universal indicator (which shows different colours at different H⁺ concentrations on the pH scale 0–14) is what earns the second mark.
• Activity 4.7 (Chapter 4) directly illustrates this: litmus cannot distinguish dilute acetic acid from dilute HCl, but the universal indicator can, because it shows their different pH values.
• Keep the answer focused — one limitation + one practical consequence = 2 marks.

The gas evolved is hydrogen (H₂).

$$\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{H}_2(g)$$

Confirmation: When a burning candle/matchstick is brought near the mouth of the test tube collecting the gas, it burns with a pop sound, confirming it is hydrogen gas.

Source: Chapter 1, Section 1.1.2 (Balanced Chemical Equations)

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• 1 mark: Correctly naming the gas as hydrogen and writing the equation.
• 1 mark: Describing the test — "pop sound" with a burning splint/candle is the standard textbook test for hydrogen. Examiners expect both the method and the observation (pop sound) for full credit. Simply writing "it burns" is insufficient.

Curd and sour substances contain acids (such as lactic acid). These acids react with brass (copper-zinc alloy) or copper vessels to form harmful salts, which can cause food poisoning. Storing such foods in these vessels makes the food toxic and unsafe for consumption.

Source: Chapter 3, Section 3.5 (Corrosion)

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• The key point examiners expect: acids in sour food react with the metal/alloy to produce poisonous/harmful compounds, making food unsafe.
• Mention that curd/sour substances contain acid — just saying "acidic" is enough; naming lactic acid is a bonus.
• Do NOT just say "it causes corrosion" — you must link it to harmful salts or food becoming poisonous, as that is the reason for the practical concern.
• Two marks: one for the reason (acid reacts with metal) and one for the consequence (harmful/toxic compounds formed).

Reaction 1 – Zinc with dilute sulphuric acid:

$$\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\uparrow$$

The other product formed is zinc sulphate (ZnSO₄), a salt.

Reaction 2 – Zinc with sodium hydroxide solution:

$$\text{Zn} + 2\text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2\uparrow$$

The other product formed is sodium zincate (Na₂ZnO₂).

What this reveals about zinc oxide:

In Reaction 1, zinc behaves like a typical metal (reacting with acid to give a salt and hydrogen). In Reaction 2, zinc also reacts with a base (NaOH) to produce hydrogen, which means it shows acidic character as well.

This tells us that zinc oxide (ZnO) is an amphoteric oxide — it reacts with both acids and bases to form salt and water. Zinc oxide is therefore an exception to the general rule that metallic oxides are basic. Most metal oxides are basic, but some (like ZnO and Al₂O₃) show both acidic and basic properties and are known as amphoteric oxides.

Source: Metals and Non-metals, Section 3.2.1 and 3.2.3; Chemical Equations, Section 1.1.2

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• The examiner expects both balanced equations written correctly — these fetch 1 mark each.
• Naming the other product in each reaction is explicitly asked — 1 mark each.
• The final point on amphoteric oxides ties everything together for the 5th mark. Use the word "amphoteric" — it is the textbook term.
• The equation for zinc + NaOH is less commonly drilled than zinc + acid, so practise writing sodium zincate (Na₂ZnO₂) correctly.
• Do not confuse the question: it asks what zinc's dual behaviour reveals about metallic oxides in general — the answer is that not all metallic oxides are purely basic; some are amphoteric.

When CO₂ is passed through lime water, it reacts with calcium hydroxide Ca(OH)₂ to form a white precipitate of calcium carbonate (CaCO₃):

$$\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}$$

When excess CO₂ is passed, the insoluble CaCO₃ reacts with CO₂ and water to form calcium bicarbonate Ca(HCO₃)₂, which is soluble in water, causing the precipitate to dissolve:

$$\text{CaCO}_3(s) + \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{Ca(HCO}_3)_2(aq)$$

Source: Chapter 1, Section 1.2.1 (Combination Reaction)

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• Examiners expect both equations — formation of CaCO₃ and its conversion to soluble Ca(HCO₃)₂.
• Key term: calcium bicarbonate (soluble) — this is why the precipitate disappears.
• The second equation is the key chemical change being asked about; don't omit it.
• State symbols (s) and (aq) add value but are not strictly penalised if missing in short answers.

Chemical equation:

$$\text{CuO (s) + 2HCl (aq)} \rightarrow \text{CuCl}_2\text{(aq) + H}_2\text{O (l)}$$

The black copper oxide dissolves to form copper chloride (which gives the blue-green colour) and water. Copper chloride is a salt, and water is formed — these are the products of a reaction between a base and an acid.

Since copper oxide reacts with hydrochloric acid to form salt and water, it behaves as a base. Therefore, metallic oxides should be classified as basic oxides (also called basic in nature). Some metallic oxides, such as aluminium oxide and zinc oxide, react with both acids and bases and are called amphoteric oxides.

Source: Chapter 3, Section 3.2.1

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• Examiners award 1 mark for the correct balanced equation, 1 mark for identifying the products as salt + water (indicating a base–acid reaction), and 1 mark for the correct classification: metallic oxides are basic oxides (with the amphoteric example as a bonus/completing point).
• Write the equation with state symbols for full credit.
• The key phrase the examiner looks for is "metal oxides are basic in nature" — this must appear explicitly.
• Mentioning amphoteric oxides (Al₂O₃, ZnO) shows a more complete answer and is mentioned directly in the textbook in this context.

Observation: When dilute HCl solution is tested, the bulb glows, indicating it conducts electricity. When glucose solution is tested, the bulb does not glow, showing glucose solution does not conduct electricity.

Reason: HCl ionises in water to produce H⁺(aq) and Cl⁻ ions. These ions carry electric current through the solution. Glucose, despite containing hydrogen, does not ionise in water and produces no ions.

Conclusion: The presence of hydrogen alone does not make a substance acidic. A substance is acidic only if it produces H⁺(aq) ions (hydronium ions, H₃O⁺) in aqueous solution. This is what makes HCl an acid while glucose is not.

Source: Chapter 2, Section 2.2 (Activity 2.8)

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• Examiners expect: the specific observation (bulb glows / does not glow), the reason (ions / no ions), and the conclusion about what makes a substance acidic.
• Key phrase to use: "ionisation" and "H⁺(aq) ions in solution."
• Don't just say "HCl is an acid" — explain why in terms of ion production.
• This question tests Activity 2.8 directly; the three-part structure (observation → reason → conclusion) is the ideal format for 3 marks.

Acidic properties of HCl are due to the production of H⁺(aq) or hydronium ions (H₃O⁺) in solution. In dry HCl gas, water is absent, so ionisation cannot occur:

$$\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-$$

Since H⁺ ions are not produced without water, dry HCl shows no acidic properties.

Source: Chapter 2, Section 2.2.1

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• The key concept is that acids produce H⁺(aq) ions only in the presence of water — this is what the examiner is looking for.
• State clearly that H⁺ cannot separate from HCl in the absence of water.
• Writing the equation earns a mark; don't skip it.
• The Activity 2.9 observation (dry litmus doesn't change colour, wet litmus does) is the experimental basis — mention it if the question asks for an activity, but here the conceptual explanation + equation is sufficient for 2 marks.

(B) NaOH dissolves in water to produce Na⁺ and OH⁻ ions, which carry the electric current.

Source: Acids, Bases and Salts (Chapter 2), Section 2.2.1

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The textbook explicitly shows: NaOH(s) →(H₂O) Na⁺(aq) + OH⁻(aq). These free ions act as charge carriers, enabling conduction. Option A is wrong because intact molecules cannot carry charge. Option C is fabricated. Option D is incorrect — density has no role in ionic conduction. Examiners look for the key idea: ionic dissociation produces free ions → conduction.

(a) Ionic equation for neutralisation:

$$\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)}$$

(b) Why the resulting solution is basic:

HCl provides H⁺(aq) ions and NaOH provides OH⁻(aq) ions. During neutralisation, H⁺ and OH⁻ combine to form water. Since NaOH is added in excess, all the H⁺ ions are used up, but surplus OH⁻ ions remain in solution. These extra OH⁻ ions make the solution basic (pH > 7).

(c) Universal indicator colours:

• Original HCl solution: The colour will be red, because HCl is a strong acid producing a high concentration of H⁺(aq) ions, giving a very low pH (around 1–2).
• Final solution after excess NaOH: The colour will be violet/purple, because excess OH⁻ ions remain, making the solution strongly alkaline with a high pH (above 10).

Source: Chapter 2, Section 2.2 and 2.2.1

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• Examiners expect the net ionic equation in (a) — not the full molecular equation. Na⁺ and Cl⁻ are spectator ions and must be omitted.
• In (b), the key phrase is "excess OH⁻ ions remain" — link the ionic concept directly to basicity. Simply saying "more NaOH was added" without mentioning ions will lose marks.
• In (c), remember the universal indicator colour scale: red/orange = strongly/weakly acidic; green = neutral; blue/violet = alkaline. HCl (strong acid) → red; strongly alkaline solution → violet. Always give a reason linked to ions or pH.

Comparison of H⁺(aq) ion concentrations:
On the pH scale, higher H⁺ ion concentration means lower pH. Since pH 3 < pH 11, the solution with pH 3 has a much higher concentration of H⁺(aq) ions than the solution with pH 11. In fact, the difference of 8 pH units means the H⁺ concentration of the pH 3 solution is 10⁸ times greater.

Nature of solutions:

• pH 3 → less than 7 → acidic solution (higher H⁺ ion concentration)
• pH 11 → greater than 7 → basic (alkaline) solution (lower H⁺ ion concentration, higher OH⁻ ion concentration)

A neutral solution has pH = 7.

Source: Chapter 2, Section 2.3 — How Strong Are Acid or Base Solutions?

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• The key relationship to state: higher H⁺ concentration = lower pH value (directly from the passage: "Higher the hydronium ion concentration, lower is the pH value").
• Examiners expect you to explicitly compare the two concentrations AND identify each solution's nature (acidic/basic).
• For 3 marks: 1 mark for comparison, 1 mark for identifying pH 3 as acidic, 1 mark for identifying pH 11 as basic. Don't skip any part.
• Mentioning the neutral pH (7) as a reference point shows understanding of the scale and is good practice.

When equal concentrations of HCl and acetic acid are tested, HCl shows a lower pH, meaning it has a higher concentration of H⁺(aq) ions in solution.

This tells us that HCl is a strong acid and acetic acid (ethanoic acid) is a weak acid.

The reason is that HCl is a mineral acid and completely ionises in water, producing a large number of H⁺ ions. Acetic acid, being a carboxylic acid, is only partially ionised, producing fewer H⁺ ions at the same concentration. Since lower pH means higher H⁺ ion concentration, HCl registers a lower pH value.

Source: Chapter 2, Section 2.3; Chapter 4, Section 4.4.2

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• Examiners expect three clear points: (1) HCl has lower pH → more H⁺ ions, (2) HCl = strong acid / acetic acid = weak acid, (3) reason = complete vs. partial ionisation.
• The textbook explicitly states: "unlike mineral acids like HCl, which are completely ionised, carboxylic acids are weak acids" — use this reasoning directly.
• Avoid vague language like "HCl is stronger" without linking it to ionisation and H⁺ ion concentration.

Acid rain increases the concentration of H⁺(aq) ions in the river water, lowering its pH to below 7, making it acidic. Living organisms carry out their metabolic activities within an optimal pH range. A sustained drop in pH disrupts these processes, making survival impossible for most aquatic life.

Source: Chapter 2, Acids, Bases and Salts

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• The key chemical change to mention is the increase in H⁺(aq) ions and the resulting fall in pH below 7.
• Link the threat directly to the textbook point: "Living beings carry out their metabolic activities within an optimal pH range."
• Don't just say "water becomes acidic" — always connect it to pH scale and biological consequence for full marks.
• For 2 marks: 1 mark for the chemical change (pH drops/acidic), 1 mark for the biological threat (optimal pH disrupted → aquatic life harmed).

The stomach produces hydrochloric acid (HCl) for digestion. During excess acid production, the pH of the stomach falls too low, causing pain and irritation.

Milk of magnesia is magnesium hydroxide — Mg(OH)₂ — a mild base. When taken, it neutralises the excess hydrochloric acid in the stomach:

$$\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}$$

This neutralisation reaction raises the stomach's pH back to a comfortable level, relieving pain. Since Mg(OH)₂ is a mild base, it does not make the stomach too alkaline, making it a safe antacid.

Source: Acids, Bases and Salts (Chapter 2), Section 2.3.1 — pH in our digestive system

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• 3 marks = 3 key points: (1) stomach produces excess HCl → low pH, (2) Mg(OH)₂ is a mild base/antacid, (3) neutralisation reaction raises pH and relieves pain.
• Always write the balanced equation — examiners expect it and it can fetch a dedicated mark.
• The word "mild base" is important; it shows you understand why Mg(OH)₂ is safe to use (won't over-alkalise the stomach).
• Directly grounded in the textbook passage: "Magnesium hydroxide (Milk of magnesia), a mild base, is often used for this purpose. These antacids neutralise the excess acid."

Sequence of events leading to tooth enamel corrosion:

1. When we eat sugary food, bacteria present in the mouth act on the sugars (simple carbohydrates) and produce acids.
2. These acids attack and demineralise the tooth enamel, causing gradual softening — this is called dental caries (tooth decay).
3. Bacterial cells, along with food particles, stick to the teeth forming dental plaque, which covers the tooth surface.
4. The plaque prevents saliva from reaching the tooth surface. Normally, saliva would neutralise the acid, but plaque blocks this protective action.
5. If untreated, the microorganisms penetrate deeper, invading the pulp, causing inflammation and infection.

Prevention:
Brush teeth immediately after eating to remove plaque before the bacteria can produce acids, allowing saliva to neutralise any remaining acid and protect the enamel.

Source: Chapter 5, Section 5.2.4 — "More to Know: Dental Caries"

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• The examiner expects the full chain: sugar → bacteria → acid → demineralisation → plaque formation → saliva blocked → pulp infection.
• All key terms (dental caries, demineralise, plaque, saliva) should appear — they are directly from the "More to Know" box and fetch marks.
• One clear prevention point (brushing after eating) is sufficient; don't list multiple vague ones.
• The word "corrosion" in the question is a hook to the chemistry chapter, but the actual answer is entirely from Chapter 5's dental caries box — stay grounded there.

The nature of a salt solution (acidic, basic, or neutral) is determined by the relative strengths of the acid and base used to form it.

• Salts of a strong acid and strong base are neutral (pH = 7). Example: NaCl (HCl + NaOH).
• Salts of a strong acid and weak base are acidic (pH < 7). Example: copper sulphate (H₂SO₄ + Cu(OH)₂).
• Salts of a strong base and weak acid are basic (pH > 7). Example: sodium acetate (NaOH + CH₃COOH).

Potassium nitrate (strong acid + strong base) is neutral; sodium acetate (weak acid + strong base) is basic.

Source: Chapter 2, Section 2.4.2

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• The key principle examiners expect: strength of parent acid and base determines the pH of the salt solution.
• Mention all three cases (neutral, acidic, basic) with the rule and at least one example each — this covers all 3 marks.
• Using examples from the question itself (copper sulphate, sodium acetate) shows application and earns full marks.
• Do not confuse "strong/weak" acid-base with "concentrated/dilute" — these are different concepts.

Chlor-alkali process involves electrolysis of brine (NaCl solution):

• At cathode: Hydrogen gas (H₂) is released. Used as fuel and in margarine production.
• At anode: Chlorine gas (Cl₂) is released. Used in water treatment (disinfection) and PVC manufacturing.
• In solution: Sodium hydroxide (NaOH) is formed. Used in degreasing, soap making, and paper industry.

Why called 'chlor-alkali': The process produces chlorine (chlor) and sodium hydroxide — an alkali (alkali) — hence the name chlor-alkali process.

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• Examiners expect you to name all three products (Cl₂, H₂, NaOH), locate them correctly (anode/cathode/solution), give at least one use each, and explain the name derivation.
• "Chlor" = chlorine; "alkali" = NaOH. That two-part etymology directly explains the name — don't skip it.
• This topic is from Chapter 2 (Acids, Bases and Salts), specifically the section on salts and their uses. It is a frequent 3-mark question.

No, the reaction does not work with wet slaked lime or with calcium oxide.

• Wet slaked lime contains moisture, and chlorine reacts with water to form hypochlorous acid and hydrochloric acid instead of forming bleaching powder. The product obtained is not bleaching powder.
• Calcium oxide (quicklime) reacts vigorously with water and heat; it does not react with chlorine in the same way to produce bleaching powder.

Only dry slaked lime [Ca(OH)₂] reacts with chlorine gas to give bleaching powder (calcium oxychloride):

$$\text{Ca(OH)}_2 + \text{Cl}_2 \rightarrow \text{CaOCl}_2 + \text{H}_2\text{O}$$

Source: Chapter 2, Acids, Bases and Salts

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• Examiners expect: (1) clear rejection of both alternatives with reasons, (2) the correct balanced equation.
• The key point is that dry Ca(OH)₂ is required — moisture causes side reactions; CaO reacts with water first before it can react with Cl₂ meaningfully to form bleaching powder.
• Write the chemical name of bleaching powder (calcium oxychloride) or its formula (CaOCl₂) for full marks.
• The balanced equation shown is the standard one expected in board exams.

Baking soda (NaHCO₃) decomposes on heating to release CO₂ gas:

$$2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2\uparrow$$

This CO₂ makes the cake rise, giving it a light, fluffy texture.

Washing soda (Na₂CO₃·10H₂O) is sodium carbonate, a strongly alkaline salt. It does not decompose on baking to release CO₂. Without CO₂ gas being produced, no air bubbles are formed in the batter, so the cake does not rise and remains flat and dense.

Source: Chapter 2, Uses of Salts / Baking Soda and Washing Soda

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• Examiners want you to name both compounds correctly (NaHCO₃ and Na₂CO₃), state the decomposition equation for baking soda, and clearly link CO₂ production to the cake rising.
• The key contrast: baking soda → releases CO₂ on heating; washing soda → no such decomposition, so no leavening action.
• Writing the chemical equation earns a dedicated mark; don't skip it.

(B) Moisture reacts with Plaster of Paris to convert it back to gypsum, making it hard and unusable.

Plaster of Paris is calcium sulphate hemihydrate (CaSO₄·½H₂O). When it absorbs moisture, it rehydrates to form gypsum (CaSO₄·2H₂O), setting hard — making it useless for further application. Options A, C, and D describe incorrect reactions. Examiners expect students to recall the reversible relationship between gypsum and Plaster of Paris.

Water of crystallisation is the fixed number of water molecules chemically bonded within a crystal. This experiment shows that water of crystallisation is essential for the blue colour of copper sulphate crystals, and its removal/addition causes a reversible change.

Forward change (heating):
$$\text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \xrightarrow{\Delta} \text{CuSO}_4(s) + 5\text{H}_2\text{O}(g)$$
$$\text{(Blue)} \hspace{4.5cm} \text{(White)}$$

Reverse change (adding water):
$$\text{CuSO}_4(s) + 5\text{H}_2\text{O}(l) \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s)$$
$$\text{(White)} \hspace{3.5cm} \text{(Blue)}$$

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• The key concept tested is water of crystallisation — examiners expect you to define it and link it to the colour change.
• Write both balanced equations; each is worth a mark in a 2-mark question.
• Note the state symbols and the $\cdot 5\text{H}_2\text{O}$ notation — these show the examiner you understand the chemical formula of hydrated copper sulphate (blue vitriol).
• The reaction is reversible, which is also a good point to mention briefly.

NaHCO₃ in Fire Extinguishers:
When NaHCO₃ is heated (or reacts with acid in the extinguisher), it decomposes to release CO₂ gas. CO₂ being heavier than air cuts off the oxygen supply to the fire, extinguishing it.

$$2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2\uparrow$$

In soda-acid extinguishers, NaHCO₃ reacts with H₂SO₄:

$$2\text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} + 2\text{CO}_2\uparrow$$

NaHCO₃ as Antacid:
When excess acid (HCl) is produced in the stomach, NaHCO₃ neutralises it, relieving indigestion. It is a mild, non-corrosive base, safe for consumption.

$$\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\uparrow$$

Thus, the same compound acts as a CO₂ source in fire-fighting and as a mild base (antacid) in medicine.

Source: Acids, Bases and Salts, Chapter 2; uses of baking soda

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• 5 marks → ~5 key points: name the two uses, give the mechanism for each, write correct balanced equations (2 equations minimum), and conclude.
• Examiners look for both equations and the reason (CO₂ smothers fire; base neutralises stomach acid).
• Note: in soda-acid extinguishers the acid triggers the reaction; thermal decomposition is the other route — mentioning either is fine, but both scores higher.
• The word "mild/non-corrosive base" is important for the antacid explanation — it distinguishes NaHCO₃ from stronger, harmful bases.
• "What you have learnt" (Ch. 2) confirms: acid + hydrogencarbonate → salt + CO₂ + water. Use that reaction pattern for the antacid equation.

Steps to obtain washing soda from NaCl:

1. Electrolyse brine (NaCl solution) → NaOH (chlor-alkali process).
2. React NaCl + H₂O + CO₂ + NH₃ → NaHCO₃ (baking soda).
3. Heat NaHCO₃: $2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$
4. Recrystallise Na₂CO₃: $\text{Na}_2\text{CO}_3 + 10\text{H}_2\text{O} \rightarrow \text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}$ (washing soda).

Recrystallisation is essential to incorporate the water of crystallisation and obtain the pure crystalline form of washing soda.

Water of crystallisation is the fixed number (10) of water molecules present per formula unit. It gives washing soda its crystalline structure; the compound is not wet, but the water is chemically bound within the crystal lattice.

Source: Chapter 2, Section 2.4.3 & 2.4.4

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• Examiners expect the sequence to be clear: NaCl → baking soda → Na₂CO₃ (by heating) → washing soda (by recrystallisation). Missing any step loses marks.
• The equation for recrystallisation and for heating NaHCO₃ are commonly asked — write them.
• For water of crystallisation, the key point is that it is chemically fixed, not free water, so the salt is not "wet" — this directly addresses the textbook question "Does 10H₂O make Na₂CO₃ wet?"
• Avoid padding; three tight points covering steps, recrystallisation purpose, and water of crystallisation role are sufficient for 3 marks.

(a) The balanced chemical equation is:

$$\text{H}_2\text{SO}_4 \text{(dil.)} + 2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} + 2\text{CO}_2\uparrow$$

(b) The gas produced is carbon dioxide (CO₂).

Two properties that make it effective:

1. CO₂ does not support combustion — it neither burns nor allows other substances to burn, so it cuts off the fuel's oxygen supply.
2. CO₂ is heavier than air — it forms a blanket over the burning material, cutting off the supply of oxygen and smothering the fire.

Source: Chapter 2, Acids Bases and Salts; Chapter 1, Chemical Reactions and Equations

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• The equation is the most commonly asked part — ensure it is balanced (coefficient 2 before NaHCO₃, H₂O, and CO₂).
• For part (b), examiners expect two distinct properties — non-supporter of combustion AND heavier than air (blanket effect). Mentioning only one will cost a mark.
• Avoid vague answers like "it puts out fire" — name the specific property (non-supportive of combustion, density higher than air).

Zinc reacts with dilute H₂SO₄ as well as NaOH solution to produce hydrogen gas, which shows that zinc is amphoteric in nature — it behaves as a metal (reacting with acid) and also as a non-metal (reacting with base).

With dilute H₂SO₄:
$$\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\uparrow$$
The salt formed is zinc sulphate, containing the simple $\text{SO}_4^{2-}$ anion from the acid.

With NaOH:
$$\text{Zn} + 2\text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2\uparrow$$
The salt formed is sodium zincate, which contains the oxyanion $\text{ZnO}_2^{2-}$ — a negative ion composed of the metal and oxygen.

This difference arises because when a base reacts with a metal, the salt formed has a negative ion composed of the metal and oxygen (an oxyanion), whereas an acid simply donates its own anion ($\text{SO}_4^{2-}$) to the salt.

Source: Chapter 2 (Acids, Bases and Salts), What you have learnt; Chapter 3 (Metals and Non-metals), What you have learnt

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• Key concept tested: Amphoteric nature of zinc — it reacts with both acids and bases. The textbook explicitly states that zinc oxide is an amphoteric oxide, implying zinc itself is amphoteric.
• Oxyanion point: The "What you have learnt" summary in Chapter 2 directly states: "When a base reacts with a metal, along with the evolution of hydrogen gas a salt is formed which has a negative ion composed of the metal and oxygen." Quote or paraphrase this for full marks.
• Do not confuse ZnSO₄ (sulphate anion comes from the acid, not from zinc) with Na₂ZnO₂ (zincate anion is built from zinc + oxygen — hence the oxyanion).
• Write both equations; one equation alone will lose a mark.

Acidic behaviour is due to the formation of H⁺(aq) ions in solution. Dry HCl gas has no water, so it cannot ionise — the separation of H⁺ from HCl cannot occur in the absence of water. Therefore, dry HCl produces no H⁺ ions and does not affect dry litmus paper.

When HCl dissolves in water, the reaction occurs:

$$\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-$$

H⁺ ions (as hydronium ions, H₃O⁺) are produced, which turn blue litmus red.

Since HCl shows acidic properties only in the presence of water, it is incorrect to call HCl itself an acid. It is the aqueous solution of HCl that is the acid.

Source: Chapter 2, Section 2.2.1

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• The key concept examiners look for: ionisation requires water; without it, no H⁺(aq) forms.
• Always include the equation HCl + H₂O → H₃O⁺ + Cl⁻ for full marks.
• Conclude clearly that acidic behaviour belongs to the solution, not the pure compound — this directly answers the "why incorrect" part.
• The term hydronium ion (H₃O⁺) should be used; writing only H⁺ is acceptable but mentioning both is better.

(i) A low pH (around 4) means the soil is highly acidic. Most crops grow best in a neutral to slightly acidic pH (near 7). Highly acidic soil reduces the availability of essential nutrients and is harmful to beneficial soil organisms, causing poor crop growth.

(ii) Slaked lime is calcium hydroxide, Ca(OH)₂, which is a base. When added to acidic soil, it undergoes a neutralisation reaction with the acids present:

$$\text{Ca(OH)}_2 + 2\text{H}^+ \rightarrow \text{Ca}^{2+} + 2\text{H}_2\text{O}$$

The base (slaked lime) neutralises the excess acid in the soil, raising the pH.

(iii) Yes, a salt has been formed. Neutralisation is the reaction between an acid and a base to produce a salt and water. When Ca(OH)₂ reacts with the acids in the soil, the calcium ions (Ca²⁺) combine with the negative ions of the acid to form a salt (e.g., calcium sulphate or calcium chloride, depending on the acid present). A pH of 7 confirms that the reaction has reached neutrality, which is characteristic of a neutral salt formed from a strong acid and a strong base.

Source: Chapter 2, Acids, Bases and Salts — Introduction; Section 2.4.2 pH of Salts

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• (i) Living beings carry out metabolic activities within an optimal pH range — this is the key idea. Low pH disrupts nutrient availability and soil biology.
• (ii) Examiners expect you to name the reaction as neutralisation, write the ionic equation (or word equation), and identify Ca(OH)₂ as the base. Even a word equation with a balanced formula earns marks.
• (iii) This is a justify question — stating "yes" alone is insufficient. You must link it to the definition: acid + base → salt + water. The textbook states: "Salts of a strong acid and strong base are neutral with pH = 7," which supports the justification. Always connect the pH observation back to the chemistry.

(A) Both A and R are true, and R is the correct explanation of A.

When excess CO₂ is passed through milky lime water, CaCO₃ reacts with H₂O and CO₂ to form Ca(HCO₃)₂ (calcium hydrogencarbonate), which is soluble, making the solution clear again. R correctly explains A.

The key chemistry here: CO₂ + H₂O + CaCO₃ → Ca(HCO₃)₂. Both statements are factually correct, and R directly explains the mechanism behind A — the dissolving of the white precipitate is caused by formation of soluble calcium hydrogencarbonate. This is a standard "excess CO₂ with lime water" observation tested in CBSE boards. Option A is correct.

(i) Steps from NaCl to each product:

Baking soda (NaHCO₃): NaCl is treated with water, CO₂, and NH₃ (Solvay-type reaction):
$$\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 + \text{NH}_3 \rightarrow \text{NH}_4\text{Cl} + \text{NaHCO}_3$$

Washing soda (Na₂CO₃·10H₂O): Baking soda is heated to give Na₂CO₃, which is then recrystallised with water:
$$2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$$
$$\text{Na}_2\text{CO}_3 + 10\text{H}_2\text{O} \rightarrow \text{Na}_2\text{CO}_3.10\text{H}_2\text{O}$$

(ii) Uses linked to chemical nature:

• Baking soda is a mild non-corrosive base (NaHCO₃). Being alkaline, it neutralises excess acid in the stomach, making it effective as an antacid.
• Washing soda is a basic salt. It removes permanent hardness of water by reacting with dissolved calcium/magnesium salts.

(iii) Chemical relationship:

Baking soda (NaHCO₃) and washing soda (Na₂CO₃·10H₂O) are chemically related — washing soda is produced from baking soda itself. Heating NaHCO₃ gives Na₂CO₃, which on recrystallisation becomes washing soda. Both are sodium salts of carbonic acid; baking soda has one Na⁺ (acid salt) while washing soda has two Na⁺ (normal salt).

Source: Chapter 2, Section 2.4.3

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• Examiners expect the equations for both products — write them clearly for full marks.
• For part (ii), always link the use to the chemical property (basic nature → neutralises acid; basic salt → softens water). Don't just list uses.
• For part (iii), the key point is that washing soda is made from baking soda, and both are sodium carbonate salts differing in the number of Na⁺ ions and water of crystallisation. This shows their direct chemical relationship.
• Keep equations balanced; the heating arrow (Δ) over the arrow is expected.

Student A is correct.

HCl is a strong acid — it ionises completely in water, producing a high concentration of H⁺ ions. CH₃COOH (acetic/ethanoic acid) is a weak acid — it ionises only partially, producing fewer H⁺ ions.

Since pH depends on H⁺ ion concentration (higher H⁺ → lower pH), equal moles of HCl give a lower pH than equal moles of CH₃COOH at the same concentration.

The concept is degree of ionisation (strong vs. weak acids). Equal moles do not mean equal H⁺ ions — it depends on how completely the acid ionises.

Source: Chapter 2, Section 2.3; Chapter 4, Section 4.4.2

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• Examiners want three things: (1) identify the correct student, (2) define strong vs. weak acid using ionisation, (3) link it to pH/H⁺ concentration. Each point is roughly 1 mark.
• Key phrase to use: "completely ionised" for HCl vs. "partially ionised / weak acid" for CH₃COOH — the textbook uses exactly this language in Section 4.4.2.
• Do not say Student B is "completely wrong" without explaining why equal moles ≠ equal H⁺; that explanation is the core of the answer.

(i) Bacteria in the mouth act on sugars (from food) and produce acids. These acids lower the pH of the mouth below 5.5, causing corrosion of tooth enamel.

(ii) Toothpaste is basic (alkaline) in nature. It neutralises the acid produced in the mouth, raising the pH above 5.5 and preventing damage to the enamel.

(iii) Rinsing with water immediately dilutes and washes away the acids before they can act on the enamel. This provides quick, temporary relief even before brushing can remove the bacterial plaque.

Source: Chapter 2 (Acids, Bases and Salts); Chapter 5, Section 5.2.4 (Dental caries — More to Know!)

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• (i) The key point is bacteria + sugars → acid. The source passage (Dental caries box) states bacteria acting on sugars produce acids.
• (ii) The examiner wants the word basic/alkaline and the concept of neutralisation.
• (iii) This part tests application: water dilutes acid quickly. Plaque prevents saliva from neutralising acid (per the passage), so rinsing physically removes it.
• Do not write more than what is shown above — three focused points is exactly what a 3-mark answer needs (1 mark each).

Both changes involve water of crystallisation.

• Copper sulphate (CuSO₄·5H₂O) is blue. On strong heating, it loses water of crystallisation and becomes anhydrous CuSO₄, which is white — water is removed.
• Plaster of Paris (CaSO₄·½H₂O) gains water when mixed with water, forming gypsum (CaSO₄·2H₂O), which sets hard — water is absorbed.

Thus, both processes involve loss or gain of water of crystallisation; they proceed in opposite directions.

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The examiner expects students to name "water of crystallisation" as the common principle and clearly state the direction for each case — loss of water on heating (CuSO₄) and gain of water on setting (Plaster of Paris). Mixing them up is a common error. Note: the source passages confirm the concept of water of crystallisation in hydrated salts (e.g., FeSO₄·7H₂O losing water on heating), which underpins both examples here.