CBSE Class 10 · Chemistry
CBSE Class 10 Chemistry Acids, Bases and Salts Previous Year Questions
Help your child master CBSE Class 10 Chemistry Acids, Bases and Salts Previous Year Questions with this curated collection sourced from real board papers spanning 2015–2025. Every question comes with a detailed step-by-step solution, building your child’s confidence with the pH scale, neutralization reactions, and the chemistry of salts — concepts that carry consistent marks in the board exam.
CBSE Class 10 Chemistry Acids, Bases and Salts — Questions with Solutions
The following table shows the pH values of four solutions A, B, C and D on a pH scale:
The solutions A, B, C and D respectively are of a?
The solutions A, B, C and D respectively are of a?
Solution
Answer: Option (D) is correct.
The pH scale measures acidity or basicity:
• pH < 7: Acidic — Strong acids: pH 1–3; Weak acids: pH 4–6
• pH = 7: Neutral
• pH > 7: Basic — Weak bases: pH 8–10; Strong bases: pH 11–14
From the table: A (weak acid), B (neutral), C (strong base), D (strong acid) — matching Option (D).
The pH scale measures acidity or basicity:
• pH < 7: Acidic — Strong acids: pH 1–3; Weak acids: pH 4–6
• pH = 7: Neutral
• pH > 7: Basic — Weak bases: pH 8–10; Strong bases: pH 11–14
From the table: A (weak acid), B (neutral), C (strong base), D (strong acid) — matching Option (D).
Consider the following reactions:
(i) Dilute hydrochloric acid reacts with sodium hydroxide.
(ii) Magnesium oxide reacts with dilute hydrochloric acid.
(iii) Carbon dioxide reacts with sodium hydroxide.
It is found that in each case?
(i) Dilute hydrochloric acid reacts with sodium hydroxide.
(ii) Magnesium oxide reacts with dilute hydrochloric acid.
(iii) Carbon dioxide reacts with sodium hydroxide.
It is found that in each case?
Solution
Answer: Option (A) is correct.
1. Dilute HCl + NaOH → NaCl + H₂O — Neutralization → salt + water
2. MgO + 2HCl → MgCl₂ + H₂O — Basic oxide + acid → salt + water
3. CO₂ + 2NaOH → Na₂CO₃ + H₂O — Acidic oxide + base → salt + water
In each reaction, salt and water are formed — the hallmark of neutralization reactions.
1. Dilute HCl + NaOH → NaCl + H₂O — Neutralization → salt + water
2. MgO + 2HCl → MgCl₂ + H₂O — Basic oxide + acid → salt + water
3. CO₂ + 2NaOH → Na₂CO₃ + H₂O — Acidic oxide + base → salt + water
In each reaction, salt and water are formed — the hallmark of neutralization reactions.
In one formula unit of salt ‘X’, seven molecules of water of crystallisation are present. The salt ‘X’ is?
Solution
Answer: Option (C) is correct.
FeSO₄ (Iron(II) sulphate) occurs as a heptahydrate: FeSO₄·7H₂O — 7 molecules of water of crystallisation.
• CuSO₄ → 5 water molecules • Na₂CO₃ → 10 water molecules • CaSO₄ → 2 water molecules
FeSO₄ (Iron(II) sulphate) occurs as a heptahydrate: FeSO₄·7H₂O — 7 molecules of water of crystallisation.
• CuSO₄ → 5 water molecules • Na₂CO₃ → 10 water molecules • CaSO₄ → 2 water molecules
The warning sign shown in the given figure must invariably be displayed/pasted on the containers which contain hydroxide of?
Solution
Answer: Option (C) is correct.
The warning sign indicates a strong, dangerous reaction. Sodium hydroxide (NaOH) reacts exothermically with water, producing hydrogen gas and large amounts of heat, causing severe burns. Aluminium, calcium, and magnesium hydroxides are far less reactive and dangerous by comparison.
The warning sign indicates a strong, dangerous reaction. Sodium hydroxide (NaOH) reacts exothermically with water, producing hydrogen gas and large amounts of heat, causing severe burns. Aluminium, calcium, and magnesium hydroxides are far less reactive and dangerous by comparison.
The body of human beings works within the pH range of?
Solution
Answer: Option (C) is correct.
The pH of human blood typically ranges from 7.35 to 7.45, making it slightly alkaline. This pH range is tightly regulated by the body to maintain homeostasis, falling within the broader working range of 7.0 to 7.8.
The pH of human blood typically ranges from 7.35 to 7.45, making it slightly alkaline. This pH range is tightly regulated by the body to maintain homeostasis, falling within the broader working range of 7.0 to 7.8.
Which of the given options represents a family of salts?
Solution
Answer: Option (B) is correct.
A family of salts shares a common anion or cation. K₂SO₄, Na₂SO₄, and CaSO₄ all contain the common SO₄²⁻ anion. In the other options, at least one salt has a different anion, breaking the family.
A family of salts shares a common anion or cation. K₂SO₄, Na₂SO₄, and CaSO₄ all contain the common SO₄²⁻ anion. In the other options, at least one salt has a different anion, breaking the family.
In order to prepare dry hydrogen chloride gas in a humid atmosphere, the gas produced is passed through a guard tube (drying tube) which contains?
Solution
Answer: Option (A) is correct.
Calcium chloride is a hygroscopic drying agent — it absorbs moisture from the humid atmosphere, drying the hydrogen chloride gas passing through it.
Calcium chloride is a hygroscopic drying agent — it absorbs moisture from the humid atmosphere, drying the hydrogen chloride gas passing through it.
Assertion (A): Concentrated nitric acid is diluted by adding water slowly to acid with constant stirring.
Reason (R): Concentrated nitric acid is easily soluble in water.
Reason (R): Concentrated nitric acid is easily soluble in water.
Solution
Answer: Option (B) is correct.
Both A and R are true. However, the real reason for adding water slowly is the exothermic nature of dilution — rapid addition causes dangerous splattering. Solubility alone does not explain this, so R is not the correct explanation of A.
Both A and R are true. However, the real reason for adding water slowly is the exothermic nature of dilution — rapid addition causes dangerous splattering. Solubility alone does not explain this, so R is not the correct explanation of A.
A common feature observed in the crystals of washing soda, copper sulphate, gypsum, and ferrous sulphate is that all:
Solution
Answer: Option (C) is correct.
All four salts contain water of crystallization — fixed water molecules chemically incorporated into their crystal lattice: Washing Soda (10), CuSO₄ (5), Gypsum (2), FeSO₄ (7). They are not all the same colour or nature, making only Option (C) correct.
All four salts contain water of crystallization — fixed water molecules chemically incorporated into their crystal lattice: Washing Soda (10), CuSO₄ (5), Gypsum (2), FeSO₄ (7). They are not all the same colour or nature, making only Option (C) correct.
The chlorine produced during the electrolysis of brine solution is used in the manufacture of:
Solution
Answer: Option (B) is correct.
During electrolysis of brine, chlorine gas is produced at the anode. Chlorine is a strong oxidizing agent widely used in the manufacture of disinfectants such as bleach and in water purification products.
During electrolysis of brine, chlorine gas is produced at the anode. Chlorine is a strong oxidizing agent widely used in the manufacture of disinfectants such as bleach and in water purification products.
When a mixture of baking soda and tartaric acid is heated (or mixed in water), a product ‘X’ is formed, which is responsible for making breads and cakes soft and spongy. The product ‘X’ is?
Solution
Answer: Option (A) is correct.
Baking soda (NaHCO₃) reacts with tartaric acid to produce carbon dioxide (CO₂) gas, which leavens the batter and makes breads and cakes soft and spongy. $$\text{NaHCO}_3 + \text{C}_4\text{H}_6\text{O}_6 \rightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{Na}_2\text{C}_4\text{H}_4\text{O}_6$$
Baking soda (NaHCO₃) reacts with tartaric acid to produce carbon dioxide (CO₂) gas, which leavens the batter and makes breads and cakes soft and spongy. $$\text{NaHCO}_3 + \text{C}_4\text{H}_6\text{O}_6 \rightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{Na}_2\text{C}_4\text{H}_4\text{O}_6$$
You have three aqueous solutions A, B, and C:
A — Potassium nitrate B — Ammonium chloride C — Sodium carbonate
The ascending order of the pH of these solutions is:
A — Potassium nitrate B — Ammonium chloride C — Sodium carbonate
The ascending order of the pH of these solutions is:
Solution
Answer: Option (D) is correct.
• B (NH₄Cl) — salt of weak base + strong acid → Acidic, pH < 7
• A (KNO₃) — salt of strong acid + strong base → Neutral, pH = 7
• C (Na₂CO₃) — salt of strong base + weak acid → Basic, pH > 7
Ascending order: B < A < C
• B (NH₄Cl) — salt of weak base + strong acid → Acidic, pH < 7
• A (KNO₃) — salt of strong acid + strong base → Neutral, pH = 7
• C (Na₂CO₃) — salt of strong base + weak acid → Basic, pH > 7
Ascending order: B < A < C
Juice of tamarind turns blue litmus to red. It is because of the presence of a chemical compound called?
Solution
Answer: Option (D) is correct.
Tamarind juice contains tartaric acid, a weak organic acid that turns blue litmus paper red, confirming its acidic nature.
Tamarind juice contains tartaric acid, a weak organic acid that turns blue litmus paper red, confirming its acidic nature.
The water of crystallization is present in:
(i) Bleaching Powder (ii) Plaster of Paris (iii) Washing Soda
(i) Bleaching Powder (ii) Plaster of Paris (iii) Washing Soda
Solution
Answer: Option (A) is correct.
• Bleaching Powder Ca(OCl)₂ — does not contain water of crystallization
• Plaster of Paris CaSO₄·½H₂O — contains ½ molecule
• Washing Soda Na₂CO₃·10H₂O — contains 10 molecules
Water of crystallization is present in (ii) and (iii) only.
• Bleaching Powder Ca(OCl)₂ — does not contain water of crystallization
• Plaster of Paris CaSO₄·½H₂O — contains ½ molecule
• Washing Soda Na₂CO₃·10H₂O — contains 10 molecules
Water of crystallization is present in (ii) and (iii) only.
The formula of washing soda is:
Solution
Answer: Option (D) is correct.
Washing soda is sodium carbonate decahydrate: $Na_2CO_3 \cdot 10H_2O$, containing 10 molecules of water of crystallization per formula unit.
Washing soda is sodium carbonate decahydrate: $Na_2CO_3 \cdot 10H_2O$, containing 10 molecules of water of crystallization per formula unit.
Two substances ‘A’ and ‘B’ are burnt in air separately. For ‘A’, ashes are collected and dissolved in water to get solution ‘X’, while for ‘B’, fumes produced are passed through water to get solution ‘Y’. Both are tested with pH paper.
(a) If ‘X’ gives light blue colour and ‘Y’ gives orange colour to the pH paper, write the nature and range of pH of ‘X’ and ‘Y’.
(b) Which one of the two — A and B — is a metal? Justify your answer.
(a) If ‘X’ gives light blue colour and ‘Y’ gives orange colour to the pH paper, write the nature and range of pH of ‘X’ and ‘Y’.
(b) Which one of the two — A and B — is a metal? Justify your answer.
Answer
(a) Nature and pH Range:
Solution ‘X’ (light blue): Basic (alkaline) | pH range: 8–10
Solution ‘Y’ (orange): Acidic | pH range: 4–6
(b) ‘A’ is the metal.
When ‘A’ burns in air, it produces ashes that dissolve in water to form a basic solution — characteristic of metals (metal oxides are basic oxides). ‘B’ is a non-metal — non-metal oxides form acidic solutions with water.
Solution ‘X’ (light blue): Basic (alkaline) | pH range: 8–10
Solution ‘Y’ (orange): Acidic | pH range: 4–6
(b) ‘A’ is the metal.
When ‘A’ burns in air, it produces ashes that dissolve in water to form a basic solution — characteristic of metals (metal oxides are basic oxides). ‘B’ is a non-metal — non-metal oxides form acidic solutions with water.
What is observed when hydrated ferrous sulfate crystals are heated in a dry boiling tube? Give balanced chemical equation(s) of the reaction(s) that occur(s).
Answer
Observations:
• Colour changes from green → white (loss of water of crystallization)
• Further heating: turns reddish-brown (formation of Fe₂O₃) with SO₂ and SO₃ gases released
Balanced Equations:
1. Dehydration: $$\text{FeSO}_4\cdot7\text{H}_2\text{O} \xrightarrow{\text{heat}} \text{FeSO}_4 + 7\text{H}_2\text{O}$$ 2. Decomposition: $$2\text{FeSO}_4 \xrightarrow{\text{heat}} \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3$$
• Colour changes from green → white (loss of water of crystallization)
• Further heating: turns reddish-brown (formation of Fe₂O₃) with SO₂ and SO₃ gases released
Balanced Equations:
1. Dehydration: $$\text{FeSO}_4\cdot7\text{H}_2\text{O} \xrightarrow{\text{heat}} \text{FeSO}_4 + 7\text{H}_2\text{O}$$ 2. Decomposition: $$2\text{FeSO}_4 \xrightarrow{\text{heat}} \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3$$
(a) Write the formula of the ions which (i) acids, and (ii) bases generate in water solutions.
(b) Dry HCl gas does not change the colour of dry litmus paper. Why?
(b) Dry HCl gas does not change the colour of dry litmus paper. Why?
Answer
(a)
(i) Acids release hydrogen ions ($\text{H}^+$) or hydronium ions ($\text{H}_3\text{O}^+$) in water.
(ii) Bases release hydroxide ions ($\text{OH}^-$) in water.
(b)
Dry HCl gas has no water to facilitate ionization. Without forming $\text{H}^+$ ions, the litmus paper shows no colour change. Only when HCl dissolves in water does it dissociate and turn moist litmus red.
(i) Acids release hydrogen ions ($\text{H}^+$) or hydronium ions ($\text{H}_3\text{O}^+$) in water.
(ii) Bases release hydroxide ions ($\text{OH}^-$) in water.
(b)
Dry HCl gas has no water to facilitate ionization. Without forming $\text{H}^+$ ions, the litmus paper shows no colour change. Only when HCl dissolves in water does it dissociate and turn moist litmus red.
(a) State in brief the method of preparation of (i) Sodium hydroxide, and (ii) Sodium hydrogen carbonate from common salt. Write balanced chemical equations.
OR
(b) Design an experimental set-up to demonstrate that “Alcohol and glucose contain hydrogen but are not categorised as acids”. Give reason to justify this fact.
OR
(b) Design an experimental set-up to demonstrate that “Alcohol and glucose contain hydrogen but are not categorised as acids”. Give reason to justify this fact.
Answer
(a)(i) Sodium Hydroxide — Chlor-Alkali Process: $$2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \xrightarrow{\text{electricity}} \text{Cl}_2(g) + \text{H}_2(g) + 2\text{NaOH}(aq)$$
(a)(ii) Sodium Hydrogen Carbonate — Solvay Process: $$\text{NaCl} + \text{NH}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{NaHCO}_3 + \text{NH}_4\text{Cl}$$
(b) Conductivity Experiment:
Set up a circuit with a 6V battery, bulb, and electrodes dipped in different solutions.
• HCl/NaOH solutions → bulb glows (ions present, conducts electricity)
• Glucose/Ethanol solutions → bulb does not glow (no ions formed)
Conclusion: Glucose and alcohol do not produce H⁺ ions in water, so despite containing hydrogen, they are not classified as acids.
(a)(ii) Sodium Hydrogen Carbonate — Solvay Process: $$\text{NaCl} + \text{NH}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{NaHCO}_3 + \text{NH}_4\text{Cl}$$
(b) Conductivity Experiment:
Set up a circuit with a 6V battery, bulb, and electrodes dipped in different solutions.
• HCl/NaOH solutions → bulb glows (ions present, conducts electricity)
• Glucose/Ethanol solutions → bulb does not glow (no ions formed)
Conclusion: Glucose and alcohol do not produce H⁺ ions in water, so despite containing hydrogen, they are not classified as acids.
Giving reason, state the advantage of using baking powder over baking soda for the preparation of bread or cakes.
Answer
Baking powder is preferred over baking soda because:
• Baking soda (NaHCO₃) requires an additional acidic ingredient to produce CO₂ and may leave a bitter taste if not fully neutralized.
• Baking powder contains both an acid and NaHCO₃ — it releases CO₂ at a controlled rate with just moisture and heat, requiring no additional acidic ingredient.
This produces a more uniform texture and even rise in breads and cakes.
• Baking soda (NaHCO₃) requires an additional acidic ingredient to produce CO₂ and may leave a bitter taste if not fully neutralized.
• Baking powder contains both an acid and NaHCO₃ — it releases CO₂ at a controlled rate with just moisture and heat, requiring no additional acidic ingredient.
This produces a more uniform texture and even rise in breads and cakes.
A chemical compound ‘X’ is used to bleach washed clothes in laundry as well as to make drinking water free from germs. Identify ‘X’, how is it represented, and write the method of its preparation along with the chemical equation.
Answer
‘X’ is Bleaching Powder, represented as CaOCl₂ (Calcium oxychloride).
Preparation: Reacting chlorine gas with slaked lime: $$Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$$
Uses: Bleaching clothes and textile fabrics; disinfecting drinking water; oxidizing agent in chemical industries.
Preparation: Reacting chlorine gas with slaked lime: $$Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$$
Uses: Bleaching clothes and textile fabrics; disinfecting drinking water; oxidizing agent in chemical industries.
Write chemical formula of washing soda. How is it obtained from baking soda? List two uses of washing soda.
Answer
Chemical Formula: $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ (Sodium carbonate decahydrate)
Preparation from Baking Soda:
Step 1 — Heating baking soda: $$2\text{NaHCO}_3 \xrightarrow{\text{heat}} \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$$ Step 2 — Recrystallization: $$\text{Na}_2\text{CO}_3 + 10\text{H}_2\text{O} \rightarrow \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$$
Two Uses:
1. Used in the glass, soap, and paper industries.
2. Used as a cleaning agent for domestic purposes.
Preparation from Baking Soda:
Step 1 — Heating baking soda: $$2\text{NaHCO}_3 \xrightarrow{\text{heat}} \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$$ Step 2 — Recrystallization: $$\text{Na}_2\text{CO}_3 + 10\text{H}_2\text{O} \rightarrow \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$$
Two Uses:
1. Used in the glass, soap, and paper industries.
2. Used as a cleaning agent for domestic purposes.
(a) State the chemical property on which the following uses of baking soda are based:
(i) As an anti-acid (ii) As a constituent in making baking powder (iii) In soda-acid fire-extinguishers
OR
(b) Write chemical equations when an acid reacts with: (i) Metal (ii) Base (iii) Carbonate
(i) As an anti-acid (ii) As a constituent in making baking powder (iii) In soda-acid fire-extinguishers
OR
(b) Write chemical equations when an acid reacts with: (i) Metal (ii) Base (iii) Carbonate
Answer
(a)(i) As an anti-acid — Mild basic nature: $$\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\uparrow$$
(a)(ii) In baking powder — Decomposes on heating to release CO₂: $$2\text{NaHCO}_3 \xrightarrow{\text{heat}} \text{Na}_2\text{CO}_3 + \text{CO}_2\uparrow + \text{H}_2\text{O}$$
(a)(iii) In fire extinguishers — Reacts with acid to produce CO₂ that smothers fire: $$\text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{H}_2\text{O} + \text{CO}_2\uparrow$$
OR
(b)(i) Metal: $\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2\uparrow$
(b)(ii) Base: $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$
(b)(iii) Carbonate: $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\uparrow$
(a)(ii) In baking powder — Decomposes on heating to release CO₂: $$2\text{NaHCO}_3 \xrightarrow{\text{heat}} \text{Na}_2\text{CO}_3 + \text{CO}_2\uparrow + \text{H}_2\text{O}$$
(a)(iii) In fire extinguishers — Reacts with acid to produce CO₂ that smothers fire: $$\text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{H}_2\text{O} + \text{CO}_2\uparrow$$
OR
(b)(i) Metal: $\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2\uparrow$
(b)(ii) Base: $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$
(b)(iii) Carbonate: $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\uparrow$
Write chemical equations to show what happens when an acid reacts with:
(i) Metal (ii) Base (iii) Carbonate
(i) Metal (ii) Base (iii) Carbonate
Answer
(i) Acid + Metal → Salt + Hydrogen gas $$\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$$
(ii) Acid + Base → Salt + Water (Neutralization) $$\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$$
(iii) Acid + Carbonate → Salt + Water + Carbon dioxide $$\text{Na}_2\text{CO}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$$
(ii) Acid + Base → Salt + Water (Neutralization) $$\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$$
(iii) Acid + Carbonate → Salt + Water + Carbon dioxide $$\text{Na}_2\text{CO}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$$
In an experiment a student dipped pH papers in four different solutions A, B, C, and D and reported his observations:
(i) In which solution is the concentration of (1) Hydrogen/Hydronium ions; (2) Hydroxyl ions maximum?
(ii) Give one example of each of the two solutions identified in (i).
(iii) What would be the pH of the resultant mixture when these two solutions are mixed in equal proportions? Justify.
(i) In which solution is the concentration of (1) Hydrogen/Hydronium ions; (2) Hydroxyl ions maximum?
(ii) Give one example of each of the two solutions identified in (i).
(iii) What would be the pH of the resultant mixture when these two solutions are mixed in equal proportions? Justify.
Answer
(i)
• Hydrogen/Hydronium ions are maximum in Solution D (red pH paper — most acidic).
• Hydroxyl ions are maximum in Solution B (blue pH paper — most basic).
(ii) Examples:
• Solution B (strongly basic): Sodium hydroxide (NaOH)
• Solution D (strongly acidic): Hydrochloric acid (HCl)
(iii) pH of resultant mixture:
Equal amounts of strong acid (HCl) and strong base (NaOH) undergo complete neutralization: $$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$ The resulting solution is neutral with pH ≈ 7.
• Hydrogen/Hydronium ions are maximum in Solution D (red pH paper — most acidic).
• Hydroxyl ions are maximum in Solution B (blue pH paper — most basic).
(ii) Examples:
• Solution B (strongly basic): Sodium hydroxide (NaOH)
• Solution D (strongly acidic): Hydrochloric acid (HCl)
(iii) pH of resultant mixture:
Equal amounts of strong acid (HCl) and strong base (NaOH) undergo complete neutralization: $$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$ The resulting solution is neutral with pH ≈ 7.
(i) Name the gas liberated when an acid reacts with a metal. How is this gas tested?
(ii) Write the chemical equation for the reaction of zinc metal with hydrochloric acid and sodium hydroxide.
(ii) Write the chemical equation for the reaction of zinc metal with hydrochloric acid and sodium hydroxide.
Answer
(i) The gas liberated is Hydrogen gas (H₂).
Test: A burning matchstick brought near the mouth of the test tube produces a characteristic ‘pop’ sound, confirming hydrogen gas.
(ii) Zinc with HCl: $$\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2\uparrow$$ Zinc chloride and hydrogen gas are formed.
Zinc with NaOH: $$\text{Zn} + 2\text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2\uparrow$$ Sodium zincate and hydrogen gas are formed.
Test: A burning matchstick brought near the mouth of the test tube produces a characteristic ‘pop’ sound, confirming hydrogen gas.
(ii) Zinc with HCl: $$\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2\uparrow$$ Zinc chloride and hydrogen gas are formed.
Zinc with NaOH: $$\text{Zn} + 2\text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2\uparrow$$ Sodium zincate and hydrogen gas are formed.
Acid-base indicators show different colours at different pH values. Answer the following:
(a) Solution P is a strong acid while solution Q is a strong base. Where would you place P and Q on the pH scale?
(b) A solution has pH 7. Name a compound to (i) increase its pH, and (ii) decrease its pH.
(c)(i) When pH decreases from 4 to 2, what effect on hydronium ion concentration? State colour change on pH paper.
(c)(ii) A person has pain due to indigestion. What could be the pH of stomach fluid? Name the medicine used for remedy. Give chemical name of “milk of magnesia”.
(a) Solution P is a strong acid while solution Q is a strong base. Where would you place P and Q on the pH scale?
(b) A solution has pH 7. Name a compound to (i) increase its pH, and (ii) decrease its pH.
(c)(i) When pH decreases from 4 to 2, what effect on hydronium ion concentration? State colour change on pH paper.
(c)(ii) A person has pain due to indigestion. What could be the pH of stomach fluid? Name the medicine used for remedy. Give chemical name of “milk of magnesia”.
Answer
(a)
• Solution P (strong acid): pH 1 to 3
• Solution Q (strong base): pH 12 to 14
(b)
(i) To increase pH: Add a base — e.g., NaOH or Ca(OH)₂
(ii) To decrease pH: Add an acid — e.g., HCl or H₂SO₄
(c)(i)
Each pH unit decrease = 10× increase in H₃O⁺. pH 4 → 2 = 100 times increase in hydronium ion concentration.
Colour change: Orange (pH 4) → Red (pH 2)
(c)(ii)
• pH of stomach fluid: likely below 3 (excess HCl)
• Medicine: Antacid
• Chemical name of milk of magnesia: Magnesium hydroxide — Mg(OH)₂
• Solution P (strong acid): pH 1 to 3
• Solution Q (strong base): pH 12 to 14
(b)
(i) To increase pH: Add a base — e.g., NaOH or Ca(OH)₂
(ii) To decrease pH: Add an acid — e.g., HCl or H₂SO₄
(c)(i)
Each pH unit decrease = 10× increase in H₃O⁺. pH 4 → 2 = 100 times increase in hydronium ion concentration.
Colour change: Orange (pH 4) → Red (pH 2)
(c)(ii)
• pH of stomach fluid: likely below 3 (excess HCl)
• Medicine: Antacid
• Chemical name of milk of magnesia: Magnesium hydroxide — Mg(OH)₂
Common salt is an important raw material for chemicals of daily use.
(a) Write balanced chemical equations for products formed during electrolysis of brine.
(b) List two uses of any one product obtained during electrolysis of brine.
(c)(i) A mild non-corrosive basic salt ‘A’ (used in faster cooking) is heated to produce compound ‘B’. Identify A and B.
OR
(c)(ii) Define water of crystallisation. Give two examples.
(a) Write balanced chemical equations for products formed during electrolysis of brine.
(b) List two uses of any one product obtained during electrolysis of brine.
(c)(i) A mild non-corrosive basic salt ‘A’ (used in faster cooking) is heated to produce compound ‘B’. Identify A and B.
OR
(c)(ii) Define water of crystallisation. Give two examples.
Answer
(a) Electrolysis of Brine: $$2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \xrightarrow{\text{Electricity}} 2\text{NaOH}(aq) + \text{Cl}_2(g) + \text{H}_2(g)$$
(b) Two uses of Chlorine (Cl₂):
1. Disinfection of drinking water and swimming pools.
2. Manufacture of PVC plastic, bleaching agents, and disinfectants.
(c)(i)
• ‘A’ is Sodium hydrogen carbonate (NaHCO₃) — baking soda.
• ‘B’ is Sodium carbonate (Na₂CO₃). $$2\text{NaHCO}_3 \xrightarrow{\text{heat}} \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$$
OR
(c)(ii) Water of Crystallisation:
The fixed number of water molecules chemically bound within the crystal structure of a salt, essential for maintaining the crystalline form.
Examples: Washing soda $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ and Blue vitriol $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$
(b) Two uses of Chlorine (Cl₂):
1. Disinfection of drinking water and swimming pools.
2. Manufacture of PVC plastic, bleaching agents, and disinfectants.
(c)(i)
• ‘A’ is Sodium hydrogen carbonate (NaHCO₃) — baking soda.
• ‘B’ is Sodium carbonate (Na₂CO₃). $$2\text{NaHCO}_3 \xrightarrow{\text{heat}} \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$$
OR
(c)(ii) Water of Crystallisation:
The fixed number of water molecules chemically bound within the crystal structure of a salt, essential for maintaining the crystalline form.
Examples: Washing soda $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ and Blue vitriol $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$
Common salt is an important chemical compound used as a raw material for caustic soda, washing soda, baking soda etc.
(i) Name the acid and the base from which common salt can be obtained.
(ii) State the nature (acidic/basic/neutral) of sodium chloride with reason.
(iii)(A) What happens when electric current is passed through brine? Name the products and where each is obtained in the electrolytic cell.
OR
(iii)(B) How is washing soda obtained from sodium chloride? Give chemical equations.
(i) Name the acid and the base from which common salt can be obtained.
(ii) State the nature (acidic/basic/neutral) of sodium chloride with reason.
(iii)(A) What happens when electric current is passed through brine? Name the products and where each is obtained in the electrolytic cell.
OR
(iii)(B) How is washing soda obtained from sodium chloride? Give chemical equations.
Answer
(i) Acid: Hydrochloric acid (HCl) | Base: Sodium hydroxide (NaOH) $$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$
(ii) Nature: Neutral
Reason: NaCl is formed from a strong acid (HCl) and a strong base (NaOH). Salts of strong acid + strong base are neutral and do not affect pH.
(iii)(A) Electrolysis of Brine: $$2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \xrightarrow{\text{Electricity}} 2\text{NaOH}(aq) + \text{Cl}_2(g) + \text{H}_2(g)$$ • Anode (+): Chlorine gas (Cl₂)
• Cathode (−): Hydrogen gas (H₂)
• Solution (middle): Sodium hydroxide (NaOH)
OR
(iii)(B) Washing Soda from NaCl:
Step 1: $2\text{NaHCO}_3 \xrightarrow{\text{heat}} \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$
Step 2: $\text{Na}_2\text{CO}_3 + 10\text{H}_2\text{O} \rightarrow \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ (Washing Soda)
(ii) Nature: Neutral
Reason: NaCl is formed from a strong acid (HCl) and a strong base (NaOH). Salts of strong acid + strong base are neutral and do not affect pH.
(iii)(A) Electrolysis of Brine: $$2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \xrightarrow{\text{Electricity}} 2\text{NaOH}(aq) + \text{Cl}_2(g) + \text{H}_2(g)$$ • Anode (+): Chlorine gas (Cl₂)
• Cathode (−): Hydrogen gas (H₂)
• Solution (middle): Sodium hydroxide (NaOH)
OR
(iii)(B) Washing Soda from NaCl:
Step 1: $2\text{NaHCO}_3 \xrightarrow{\text{heat}} \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$
Step 2: $\text{Na}_2\text{CO}_3 + 10\text{H}_2\text{O} \rightarrow \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ (Washing Soda)
(i) The pH of tomato juice is 4.6. How is this juice likely to taste? Give reason.
(ii) How do we differentiate between a strong acid and a weak base in terms of ion-formation in aqueous solutions?
(iii) How can acid rain make the survival of aquatic animals difficult?
(ii) How do we differentiate between a strong acid and a weak base in terms of ion-formation in aqueous solutions?
(iii) How can acid rain make the survival of aquatic animals difficult?
Answer
(i) pH 4.6 indicates an acidic solution. Acidic substances taste sour. Tomato juice tastes sour due to the presence of natural acids like citric acid.
(ii)
• A strong acid ionizes completely: $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$
• A weak base ionizes only partially: $\text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$
Strong acids produce more ions; weak bases produce fewer ions.
(iii) Acid rain contains diluted H₂SO₄ and HNO₃. When it enters rivers and lakes, it lowers their pH. Most aquatic organisms survive only in a narrow pH range (6.5–8.5). Acidic water disrupts respiratory systems, damages gills, and can cause death — making the environment unsuitable for aquatic life.
(ii)
• A strong acid ionizes completely: $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$
• A weak base ionizes only partially: $\text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$
Strong acids produce more ions; weak bases produce fewer ions.
(iii) Acid rain contains diluted H₂SO₄ and HNO₃. When it enters rivers and lakes, it lowers their pH. Most aquatic organisms survive only in a narrow pH range (6.5–8.5). Acidic water disrupts respiratory systems, damages gills, and can cause death — making the environment unsuitable for aquatic life.
A compound which is prepared from gypsum has the property of hardening when water is mixed in right quantity with it:
(i) Write its common name and chemical name.
(ii) Give chemical equation for its preparation.
(iii) List its two uses.
(i) Write its common name and chemical name.
(ii) Give chemical equation for its preparation.
(iii) List its two uses.
Answer
(i)
• Common Name: Plaster of Paris (POP)
• Chemical Name: Calcium sulphate hemihydrate
• Formula: $\text{CaSO}_4 \cdot \dfrac{1}{2}\text{H}_2\text{O}$
(ii) Preparation (heating gypsum at 373 K): $$2(\text{CaSO}_4 \cdot 2\text{H}_2\text{O}) \xrightarrow{373\text{K}} 2\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + 3\text{H}_2\text{O}$$
(iii) Two Uses:
1. Making moulds for statues, toys, and decorative materials.
2. Used by doctors to set fractured bones in the correct position.
• Common Name: Plaster of Paris (POP)
• Chemical Name: Calcium sulphate hemihydrate
• Formula: $\text{CaSO}_4 \cdot \dfrac{1}{2}\text{H}_2\text{O}$
(ii) Preparation (heating gypsum at 373 K): $$2(\text{CaSO}_4 \cdot 2\text{H}_2\text{O}) \xrightarrow{373\text{K}} 2\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + 3\text{H}_2\text{O}$$
(iii) Two Uses:
1. Making moulds for statues, toys, and decorative materials.
2. Used by doctors to set fractured bones in the correct position.
(a)(i) A compound ‘X’ prepared from gypsum hardens when mixed with proper quantity of water. Identify ‘X’ and write its chemical formula.
(ii) State the difference in chemical composition between baking soda and baking powder.
OR
(b) Write balanced chemical equations for: (i) Heating blue copper sulphate crystals, and (ii) Heating sodium hydrogen carbonate during cooking.
(ii) State the difference in chemical composition between baking soda and baking powder.
OR
(b) Write balanced chemical equations for: (i) Heating blue copper sulphate crystals, and (ii) Heating sodium hydrogen carbonate during cooking.
Answer
(a)(i) ‘X’ is Plaster of Paris (POP): $\text{CaSO}_4 \cdot \dfrac{1}{2}\text{H}_2\text{O}$
$$\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + \frac{3}{2}\text{H}_2\text{O}$$
(a)(ii) Difference:
• Baking soda = NaHCO₃ only (single compound)
• Baking powder = NaHCO₃ + mild acid (e.g., tartaric acid) + starch (a mixture)
OR
(b)(i) $$\text{CuSO}_4 \cdot 5\text{H}_2\text{O} \xrightarrow{\text{Heat}} \text{CuSO}_4 + 5\text{H}_2\text{O}$$ Blue → white colour change.
(b)(ii) $$2\text{NaHCO}_3 \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$$
(a)(ii) Difference:
• Baking soda = NaHCO₃ only (single compound)
• Baking powder = NaHCO₃ + mild acid (e.g., tartaric acid) + starch (a mixture)
OR
(b)(i) $$\text{CuSO}_4 \cdot 5\text{H}_2\text{O} \xrightarrow{\text{Heat}} \text{CuSO}_4 + 5\text{H}_2\text{O}$$ Blue → white colour change.
(b)(ii) $$2\text{NaHCO}_3 \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$$
Salts play a very important role in our daily life.
(a) Identify the acid and base from which Sodium chloride is formed.
(b) Find the cation and the anion present in Calcium sulphate.
(c) “Sodium chloride and washing soda both belong to the same family of salts.” Justify this statement.
OR
(c) Define the term pH scale. Name the salt obtained by the reaction of KOH and H₂SO₄ and give the pH value of its aqueous solution.
(a) Identify the acid and base from which Sodium chloride is formed.
(b) Find the cation and the anion present in Calcium sulphate.
(c) “Sodium chloride and washing soda both belong to the same family of salts.” Justify this statement.
OR
(c) Define the term pH scale. Name the salt obtained by the reaction of KOH and H₂SO₄ and give the pH value of its aqueous solution.
Answer
(a) Acid: $HCl$ | Base: $NaOH$
$$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$
(b) Cation: $\text{Ca}^{2+}$ | Anion: $\text{SO}_4^{2-}$
(c)(i) Justification:
NaCl contains Na⁺ as cation. Na₂CO₃ (washing soda) also contains Na⁺ as cation. Both share the same sodium (Na⁺) cation, placing them in the sodium salt family.
OR
(c)(ii) pH Scale:
The pH scale measures hydrogen ion concentration in a solution, ranging from 0–14 (0–6 acidic, 7 neutral, 8–14 basic).
Salt formed: $2\text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$ → Potassium sulphate (K₂SO₄)
pH of aqueous solution: 7 (Neutral)
(b) Cation: $\text{Ca}^{2+}$ | Anion: $\text{SO}_4^{2-}$
(c)(i) Justification:
NaCl contains Na⁺ as cation. Na₂CO₃ (washing soda) also contains Na⁺ as cation. Both share the same sodium (Na⁺) cation, placing them in the sodium salt family.
OR
(c)(ii) pH Scale:
The pH scale measures hydrogen ion concentration in a solution, ranging from 0–14 (0–6 acidic, 7 neutral, 8–14 basic).
Salt formed: $2\text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$ → Potassium sulphate (K₂SO₄)
pH of aqueous solution: 7 (Neutral)
(a) What is meant by pH of a solution? The pH of rain water from cities ‘A’ and ‘B’ was 6.1 and 5.3 respectively. Which city’s water is more acidic? What would happen to aquatic life in a pond where rain water from city ‘B’ flows?
OR
(b)(i) Identify the acid and base from which these salts are obtained: (I) Sodium chloride (II) Ammonium sulphate
(ii) Write the nature — acidic / basic / neutral — of each salt with reason.
OR
(b)(i) Identify the acid and base from which these salts are obtained: (I) Sodium chloride (II) Ammonium sulphate
(ii) Write the nature — acidic / basic / neutral — of each salt with reason.
Answer
(a)
pH measures the acidity or basicity of a solution — lower pH means more acidic, higher pH means more basic.
City B (pH 5.3) is more acidic — lower pH means higher concentration of H⁺ ions.
Effect on aquatic life: Rain water from city B (pH 5.3) lowers the pond’s pH. Most aquatic organisms survive only in pH 6.5–8.5. Acidic water damages gills, disrupts respiration, and causes death — the aquatic life will be severely affected.
OR
(b)(i)
(b)(ii)
pH measures the acidity or basicity of a solution — lower pH means more acidic, higher pH means more basic.
City B (pH 5.3) is more acidic — lower pH means higher concentration of H⁺ ions.
Effect on aquatic life: Rain water from city B (pH 5.3) lowers the pond’s pH. Most aquatic organisms survive only in pH 6.5–8.5. Acidic water damages gills, disrupts respiration, and causes death — the aquatic life will be severely affected.
OR
(b)(i)
(b)(ii)
(i) An aqueous solution turns blue litmus red. Which solution when added in excess would reverse the change?
(1) Lemon juice (2) Magnesium hydroxide (3) Vinegar (4) Calcium sulphate
(ii) Which compound/compounds turn phenolphthalein pink? (1) $CH_3COOH$ (2) $Ca(OH)_2$ (3) HCl (4) NaOH
(iii) Name a gas whose aqueous solution is basic. Write the formula/name of this solution.
(iv) A basic substance is used to treat a honey-bee sting. Why?
(v) Name the acid present in (1) Tomato (2) Tamarind
(1) Lemon juice (2) Magnesium hydroxide (3) Vinegar (4) Calcium sulphate
(ii) Which compound/compounds turn phenolphthalein pink? (1) $CH_3COOH$ (2) $Ca(OH)_2$ (3) HCl (4) NaOH
(iii) Name a gas whose aqueous solution is basic. Write the formula/name of this solution.
(iv) A basic substance is used to treat a honey-bee sting. Why?
(v) Name the acid present in (1) Tomato (2) Tamarind
Answer
(i) Magnesium hydroxide (Option 2) — a base that neutralizes the acid and turns red litmus back to blue.
(ii) Ca(OH)₂ and NaOH — both are bases; phenolphthalein turns pink in basic solutions only.
(iii) Ammonia (NH₃) — its aqueous solution is Ammonium hydroxide (NH₄OH): $$NH_3 + H_2O \rightarrow NH_4^+ + OH^-$$
(iv) A bee sting injects formic acid. A basic substance like baking soda (NaHCO₃) neutralizes this acid, reducing pain and irritation.
(v)
• Tomato: Oxalic acid
• Tamarind: Tartaric acid
(ii) Ca(OH)₂ and NaOH — both are bases; phenolphthalein turns pink in basic solutions only.
(iii) Ammonia (NH₃) — its aqueous solution is Ammonium hydroxide (NH₄OH): $$NH_3 + H_2O \rightarrow NH_4^+ + OH^-$$
(iv) A bee sting injects formic acid. A basic substance like baking soda (NaHCO₃) neutralizes this acid, reducing pain and irritation.
(v)
• Tomato: Oxalic acid
• Tamarind: Tartaric acid
(i) Define water of crystallisation.
(ii) Write the chemical name and formula of a compound having water of crystallisation and appearing blue.
(iii) Write the chemical formula of bleaching powder. Write a balanced chemical equation for its preparation. List its three uses.
(ii) Write the chemical name and formula of a compound having water of crystallisation and appearing blue.
(iii) Write the chemical formula of bleaching powder. Write a balanced chemical equation for its preparation. List its three uses.
Answer
(i) Water of crystallisation is the fixed number of water molecules present in a crystalline substance, chemically combined in a definite proportion with the salt, essential for maintaining the crystalline structure.
(ii)
• Chemical Name: Copper(II) sulphate pentahydrate
• Chemical Formula: CuSO₄·5H₂O (appears blue due to 5 water molecules of crystallisation)
(iii)
• Formula of Bleaching Powder: CaOCl₂ $$Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$$
Three Uses:
1. Disinfecting drinking water to make it free from germs.
2. Bleaching agent in textile and paper industries.
3. Oxidizing agent in chemical industries.
(ii)
• Chemical Name: Copper(II) sulphate pentahydrate
• Chemical Formula: CuSO₄·5H₂O (appears blue due to 5 water molecules of crystallisation)
(iii)
• Formula of Bleaching Powder: CaOCl₂ $$Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$$
Three Uses:
1. Disinfecting drinking water to make it free from germs.
2. Bleaching agent in textile and paper industries.
3. Oxidizing agent in chemical industries.
(i) Give the relation between hydrogen ion concentration of an aqueous solution and its pH.
(ii) Concentrated acids should not be diluted by adding water to acid. Why?
(iii) Why do the same molar concentrations of HCl and acetic acid not produce the same amounts of hydrogen ions?
(ii) Concentrated acids should not be diluted by adding water to acid. Why?
(iii) Why do the same molar concentrations of HCl and acetic acid not produce the same amounts of hydrogen ions?
Answer
(i) Relation between H⁺ concentration and pH: $$\text{pH} = -\log [\text{H}^+]$$
As H⁺ concentration increases, pH decreases (acidic). As H⁺ concentration decreases, pH increases (basic).
(ii) Why not add water to concentrated acid?
Diluting a concentrated acid is highly exothermic. Adding water rapidly causes the acid to splatter dangerously, causing burns. Acid must always be added slowly to water, which absorbs the heat safely.
(iii) HCl vs Acetic acid at same molar concentration:
• HCl is a strong acid — ionizes completely: $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$
• CH₃COOH is a weak acid — ionizes partially: $\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$
HCl releases more H⁺ ions despite equal molar concentration, making it a stronger acid.
(ii) Why not add water to concentrated acid?
Diluting a concentrated acid is highly exothermic. Adding water rapidly causes the acid to splatter dangerously, causing burns. Acid must always be added slowly to water, which absorbs the heat safely.
(iii) HCl vs Acetic acid at same molar concentration:
• HCl is a strong acid — ionizes completely: $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$
• CH₃COOH is a weak acid — ionizes partially: $\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$
HCl releases more H⁺ ions despite equal molar concentration, making it a stronger acid.
Give reason:
(i) Crystals of washing soda change to white powder on exposure to air.
(ii) Plaster of Paris should be stored in moisture-proof containers.
(iii) Baking soda can be used on bee sting area to get relief.
(iv) Distilled water does not conduct electricity.
(v) Farmers treat the soil of their fields with quick lime.
(i) Crystals of washing soda change to white powder on exposure to air.
(ii) Plaster of Paris should be stored in moisture-proof containers.
(iii) Baking soda can be used on bee sting area to get relief.
(iv) Distilled water does not conduct electricity.
(v) Farmers treat the soil of their fields with quick lime.
Answer
(i) Washing soda (Na₂CO₃·10H₂O) loses its water of crystallization on exposure to air (efflorescence), turning into white anhydrous sodium carbonate (Na₂CO₃).
(ii) Plaster of Paris (CaSO₄·½H₂O) absorbs moisture from air and hydrates to gypsum (CaSO₄·2H₂O), causing it to harden prematurely and lose effectiveness.
(iii) Baking soda (NaHCO₃) is a mild base. A bee sting injects formic acid; baking soda neutralizes this acid, reducing pain and irritation.
(iv) Distilled water is pure — contains no dissolved ions. Electrical conduction requires free ions; without them, distilled water cannot conduct electricity.
(v) Quick lime (CaO) is a strong base used to raise the pH of acidic soils, neutralizing the acidic components and improving soil quality for better plant growth.
(ii) Plaster of Paris (CaSO₄·½H₂O) absorbs moisture from air and hydrates to gypsum (CaSO₄·2H₂O), causing it to harden prematurely and lose effectiveness.
(iii) Baking soda (NaHCO₃) is a mild base. A bee sting injects formic acid; baking soda neutralizes this acid, reducing pain and irritation.
(iv) Distilled water is pure — contains no dissolved ions. Electrical conduction requires free ions; without them, distilled water cannot conduct electricity.
(v) Quick lime (CaO) is a strong base used to raise the pH of acidic soils, neutralizing the acidic components and improving soil quality for better plant growth.
What is the chlor-alkali process? Write the chemical equation. Name the electrode at which (i) chlorine gas, and (ii) hydrogen gas is given off.
Answer
The chlor-alkali process is the industrial electrolysis of brine (aqueous NaCl) to produce chlorine gas, hydrogen gas, and sodium hydroxide.
$$2NaCl(aq) + 2H_2O(l) \xrightarrow{\text{electric current}} Cl_2(g) + H_2(g) + 2NaOH(aq)$$
(i) Chlorine gas is given off at the Anode (positive electrode).
(ii) Hydrogen gas is given off at the Cathode (negative electrode).
(i) Chlorine gas is given off at the Anode (positive electrode).
(ii) Hydrogen gas is given off at the Cathode (negative electrode).
Assertion (A): Sodium sulphate is a neutral salt.
Reason (R): Sodium sulphate is a salt formed by a strong acid and a strong base.
Reason (R): Sodium sulphate is a salt formed by a strong acid and a strong base.
Solution
Answer: Option (A) is correct.
Na₂SO₄ is neutral (A is true) because it is formed from a strong acid (H₂SO₄) and a strong base (NaOH) — which is exactly what R states, and this is the correct reason why the salt is neutral. Both A and R are true and R correctly explains A.
Na₂SO₄ is neutral (A is true) because it is formed from a strong acid (H₂SO₄) and a strong base (NaOH) — which is exactly what R states, and this is the correct reason why the salt is neutral. Both A and R are true and R correctly explains A.
Juice of tamarind turns blue litmus to red. It is because of the presence of an acid called:
Solution
Answer: Option (D) is correct.
Tamarind juice contains tartaric acid — a weak organic acid responsible for turning blue litmus paper red. Oxalic acid is found in spinach; lactic acid in dairy; methanoic acid in ants — none are the acid of tamarind.
Tamarind juice contains tartaric acid — a weak organic acid responsible for turning blue litmus paper red. Oxalic acid is found in spinach; lactic acid in dairy; methanoic acid in ants — none are the acid of tamarind.
Which one of the following oxides reacts with HCl as well as NaOH to form corresponding salt and water?
Solution
Answer: Option (B) is correct.
$Al_2O_3$ is an amphoteric oxide — it reacts with both acids and bases:
With HCl: $Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$
With NaOH: $Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2NaAlO_2 + 3H_2O$
CaO, Fe₃O₄, K₂O are all basic oxides — they react only with acids, not with bases.
$Al_2O_3$ is an amphoteric oxide — it reacts with both acids and bases:
With HCl: $Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$
With NaOH: $Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2NaAlO_2 + 3H_2O$
CaO, Fe₃O₄, K₂O are all basic oxides — they react only with acids, not with bases.
Name and state in brief the process to obtain sodium hydroxide from brine. In this process, along with NaOH, two gases A and B are produced at the two electrodes. Name A and B specifying their respective electrodes. One of these gases reacts with dry calcium hydroxide to produce compound C, used as an oxidising agent and to make drinking water free from germs. Name C and write the chemical equation for its formation.
Answer
Process: Chlor-Alkali Process
Brine (concentrated aqueous NaCl) is electrolysed to produce NaOH, Cl₂, and H₂.
Gases at electrodes:
• Gas A — Chlorine (Cl₂) — at the Anode (+)
• Gas B — Hydrogen (H₂) — at the Cathode (−)
Compound C — Bleaching Powder Ca(OCl)₂:
Chlorine gas (A) reacts with dry calcium hydroxide: $$Cl_2 + Ca(OH)_2 \rightarrow Ca(OCl)_2 + H_2O$$ Bleaching powder is used as an oxidizing agent and for disinfecting drinking water.
Brine (concentrated aqueous NaCl) is electrolysed to produce NaOH, Cl₂, and H₂.
Gases at electrodes:
• Gas A — Chlorine (Cl₂) — at the Anode (+)
• Gas B — Hydrogen (H₂) — at the Cathode (−)
Compound C — Bleaching Powder Ca(OCl)₂:
Chlorine gas (A) reacts with dry calcium hydroxide: $$Cl_2 + Ca(OH)_2 \rightarrow Ca(OCl)_2 + H_2O$$ Bleaching powder is used as an oxidizing agent and for disinfecting drinking water.
When dry crystals of ferrous sulphate were heated in a dry boiling tube, some tiny water droplets were observed.
(i) State the source of these water droplets.
(ii) Write the colour change observed during heating.
(iii) Write the balanced chemical equation of the reaction during heating.
(iv) How many water molecules are attached per formula unit of ferrous sulphate crystal?
(v) Write the molecular formula of crystalline form of (I) copper sulphate, and (II) sodium carbonate.
(i) State the source of these water droplets.
(ii) Write the colour change observed during heating.
(iii) Write the balanced chemical equation of the reaction during heating.
(iv) How many water molecules are attached per formula unit of ferrous sulphate crystal?
(v) Write the molecular formula of crystalline form of (I) copper sulphate, and (II) sodium carbonate.
Answer
(i) The water droplets come from the water of crystallization in FeSO₄·7H₂O, released as steam and condensed inside the tube.
(ii) Colour change:
Green (FeSO₄·7H₂O) → White (anhydrous FeSO₄) → Reddish-brown (Fe₂O₃)
(iii) Balanced equation: $$\text{FeSO}_4\cdot7\text{H}_2\text{O} \xrightarrow{\text{heat}} \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3 + 7\text{H}_2\text{O}$$
(iv) 7 water molecules per formula unit (FeSO₄·7H₂O).
(v)
(I) Copper sulphate: CuSO₄·5H₂O
(II) Sodium carbonate: Na₂CO₃·10H₂O
(ii) Colour change:
Green (FeSO₄·7H₂O) → White (anhydrous FeSO₄) → Reddish-brown (Fe₂O₃)
(iii) Balanced equation: $$\text{FeSO}_4\cdot7\text{H}_2\text{O} \xrightarrow{\text{heat}} \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3 + 7\text{H}_2\text{O}$$
(iv) 7 water molecules per formula unit (FeSO₄·7H₂O).
(v)
(I) Copper sulphate: CuSO₄·5H₂O
(II) Sodium carbonate: Na₂CO₃·10H₂O
(a) Write the name and chemical formula of a sodium compound added for faster cooking. How is it produced from NaCl? Give chemical equation.
(b) The compound mentioned in (a) is also an ingredient of antacids. Why?
(b) The compound mentioned in (a) is also an ingredient of antacids. Why?
Answer
(a) The compound is Sodium bicarbonate (Baking soda) — NaHCO₃
Produced from NaCl: $$NaCl + NH_3 + CO_2 + H_2O \rightarrow NaHCO_3 + NH_4Cl$$ $$Na_2CO_3 + CO_2 + H_2O \rightarrow 2NaHCO_3$$
(b) NaHCO₃ is a mild alkaline substance. It neutralizes excess HCl in the stomach: $$NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$$ This relieves heartburn and acid indigestion by reducing acidity in the stomach.
Produced from NaCl: $$NaCl + NH_3 + CO_2 + H_2O \rightarrow NaHCO_3 + NH_4Cl$$ $$Na_2CO_3 + CO_2 + H_2O \rightarrow 2NaHCO_3$$
(b) NaHCO₃ is a mild alkaline substance. It neutralizes excess HCl in the stomach: $$NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$$ This relieves heartburn and acid indigestion by reducing acidity in the stomach.
(i) On adding dilute HCl to sodium carbonate, an effervescence is observed. Explain and give chemical equation. How is the gas tested in a school laboratory?
(ii) What happens when excess carbon dioxide passes through lime water? Write chemical equations to explain.
(ii) What happens when excess carbon dioxide passes through lime water? Write chemical equations to explain.
Answer
(i) The effervescence is due to release of carbon dioxide gas:
$$\text{Na}_2\text{CO}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$$ Test for CO₂: Pass the gas through lime water — it turns milky due to formation of CaCO₃ precipitate:
$$\text{Ca(OH)}_2(aq) + \text{CO}_2(g) \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}(l)$$
(ii) Excess CO₂ through lime water:
Step 1 — Lime water turns milky (CaCO₃ forms):
$\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}$
Step 2 — On excess CO₂, the milkiness disappears (CaCO₃ dissolves): $$\text{CaCO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{Ca(HCO}_3)_2(aq)$$ Final observation: Milky solution turns clear (soluble calcium bicarbonate forms).
(ii) Excess CO₂ through lime water:
Step 1 — Lime water turns milky (CaCO₃ forms):
$\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}$
Step 2 — On excess CO₂, the milkiness disappears (CaCO₃ dissolves): $$\text{CaCO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{Ca(HCO}_3)_2(aq)$$ Final observation: Milky solution turns clear (soluble calcium bicarbonate forms).
(i) A small amount of copper oxide is taken in a beaker and dilute HCl is added. Name the compound formed and state the colour of the solution. Write the balanced chemical equation.
(ii) A compound $X$ on heating at 373 K loses water molecules and becomes $Y$. Substance $Y$ is used by doctors for supporting fractured bones.
I. Identify $X$ and $Y$. II. How can $X$ be obtained from $Y$?
(ii) A compound $X$ on heating at 373 K loses water molecules and becomes $Y$. Substance $Y$ is used by doctors for supporting fractured bones.
I. Identify $X$ and $Y$. II. How can $X$ be obtained from $Y$?
Answer
(i)
• Compound formed: Copper(II) chloride (CuCl₂)
• Colour of solution: Blue $$\text{CuO}(s) + 2\text{HCl}(aq) \rightarrow \text{CuCl}_2(aq) + \text{H}_2\text{O}(l)$$
(ii) I. Identification:
• $X$ is Gypsum (CaSO₄·2H₂O)
• $Y$ is Plaster of Paris (CaSO₄·½H₂O) $$\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + \frac{3}{2}\text{H}_2\text{O}$$
II. Obtaining X (Gypsum) from Y (Plaster of Paris):
Add water to Plaster of Paris — it rehydrates to form Gypsum (setting of plaster): $$\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}(s) + \frac{3}{2}\text{H}_2\text{O}(l) \rightarrow \text{CaSO}_4 \cdot 2\text{H}_2\text{O}(s)$$
• Compound formed: Copper(II) chloride (CuCl₂)
• Colour of solution: Blue $$\text{CuO}(s) + 2\text{HCl}(aq) \rightarrow \text{CuCl}_2(aq) + \text{H}_2\text{O}(l)$$
(ii) I. Identification:
• $X$ is Gypsum (CaSO₄·2H₂O)
• $Y$ is Plaster of Paris (CaSO₄·½H₂O) $$\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + \frac{3}{2}\text{H}_2\text{O}$$
II. Obtaining X (Gypsum) from Y (Plaster of Paris):
Add water to Plaster of Paris — it rehydrates to form Gypsum (setting of plaster): $$\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}(s) + \frac{3}{2}\text{H}_2\text{O}(l) \rightarrow \text{CaSO}_4 \cdot 2\text{H}_2\text{O}(s)$$
Explore More CBSE Class 10 Chemistry Chapters
Frequently Asked Questions
What does the Acids, Bases and Salts chapter cover in CBSE Class 10 Chemistry? ⌄
The Acids, Bases and Salts chapter covers properties of acids and bases, the pH scale and its significance, neutralization reactions, water of crystallization, and key industrial chemicals such as baking soda, washing soda, bleaching powder, Plaster of Paris, and the chlor-alkali process. All these topics regularly appear across all question types in the CBSE board exam.
How many marks does Acids, Bases and Salts carry in the CBSE Class 10 board exam? ⌄
Acids, Bases and Salts is one of the highest-weightage chapters in CBSE Class 10 Science, typically contributing 7 to 10 marks in the board paper. Questions appear as 1-mark MCQs, 2-mark short answers, 3-mark application-based questions, and 5-mark case study or long-answer questions — making thorough preparation across all question types essential.
What are the most important topics in Acids, Bases and Salts for board exams? ⌄
The most frequently tested topics are the pH scale and its applications, neutralization reactions with balanced equations, properties and uses of baking soda, washing soda, bleaching powder, and Plaster of Paris, the chlor-alkali process, water of crystallization with examples, and Assertion-Reason questions. Your child should be able to write all balanced chemical equations for major reactions in this chapter.
What common mistakes do students make in Acids, Bases and Salts questions? ⌄
A very common error is confusing the number of water molecules in different salts — for instance, mixing up CuSO₄·5H₂O with FeSO₄·7H₂O. Students also frequently make errors in balancing chemical equations for neutralization reactions and the chlor-alkali process, and confuse the properties of baking soda and baking powder. Practicing previous year questions with step-by-step solutions helps avoid these errors under exam pressure.
How does Angle Belearn help students score well in Acids, Bases and Salts? ⌄
Angle Belearn’s CBSE specialists carefully curate chapter-wise question banks drawn from real board papers, each paired with clear step-by-step solutions that show exactly how to earn full marks — including how to write balanced equations and justify answers. Regular practice with these verified previous year questions builds both the conceptual understanding and the exam confidence your child needs on board exam day.
