CBSE Class 10 Science HOTs Acids Bases And Salts Set 06

This Class 10 Science page gives 21 higher order thinking questions (HOTS) from Chapter 2, Acids, Bases and Salts, each with a step-by-step answer, a simple-words summary and a teacher's note. The questions test reasoning on dibasic acids, the strength of acids, the pH of salt solutions, amphoteric behaviour and the identification of unknown compounds from clues.

Reasoning Based HOTS Questions

Question 1. Sulphuric acid is a dibasic acid. Write the reaction equations to justify this statement and name the reaction products in the two cases.
Answer:
Sulphuric acid has two replaceable hydrogen atoms, so it neutralises a base in two steps.
With one mole of NaOH: NaOH + H₂SO₄ → NaHSO₄ + H₂O. Products: sodium hydrogensulphate and water.
With two moles of NaOH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. Products: sodium sulphate and water.
In simple words: A dibasic acid gives two types of salts.

Teacher's Note:
1. Basicity is the number of replaceable H atoms.
2. Show both equations.

Question 2. Why does a solution of a strong acid conduct electricity better than a solution of a weak acid having the same concentration?
Answer:
A strong acid ionises almost completely, so its solution contains many ions.
A weak acid ionises only slightly, so its solution contains few ions. More ions carry more current, so the strong acid conducts better.
In simple words: More ions mean better conduction.

Teacher's Note:
1. Compare HCl with CH₃COOH.
2. Conduction depends on the number of free ions.

Question 3. How is chloride of lime chemically different from calcium chloride? Why does chloride of lime gradually lose its chlorine when kept exposed to air?
Answer:
Chloride of lime (bleaching powder) is CaOCl₂, calcium oxychloride, whereas calcium chloride is CaCl₂.
Chloride of lime reacts with carbon dioxide of the air and slowly gives out chlorine: CaOCl₂ + CO₂ → CaCO₃ + Cl₂↑.
In simple words: Bleaching powder keeps losing chlorine in air.

Teacher's Note:
1. Store it in an airtight container.
2. Write both formulae clearly.

Question 4. Suggest an explanation for why people once washed their hands with ashes.
Answer:
Ashes contain potassium carbonate, which in water forms a basic solution (potassium hydroxide, a strong base, and carbonic acid, a weak acid). Being basic, it feels soapy and dissolves grease and oil, and it also neutralises acids on the skin, so it helped to clean the hands.
In simple words: Ash is basic and works like soap.

Teacher's Note:
1. Basic solutions are soapy and remove grease.
2. Soap is also a basic substance.

Question 5. What happens to the corrosive properties of an acid and a base after they neutralize each other? Defend your answer.
Answer:
The corrosive properties of both are almost lost.
Neutralisation forms a salt and water. The harmful H⁺ ions of the acid and OH⁻ ions of the base combine to form water. Salts are generally much less corrosive than the acid and base from which they are formed.
In simple words: Neutralisation removes H⁺ and OH⁻, so corrosion is reduced.

Teacher's Note:
1. Corrosive action comes from the ions.
2. Salt and water are usually harmless.

Question 6. Explain why an aqueous solution of ammonium chloride is acidic in nature.
Answer:
Ammonium chloride is the salt of a strong acid (HCl) and a weak base (NH₄OH).
In water, HCl ionises completely and gives more H⁺ ions, while NH₄OH ionises only slightly and gives fewer OH⁻ ions. So H⁺ ions are in excess and the solution is acidic (pH less than 7).
In simple words: The strong acid part decides the nature of the solution.

Teacher's Note:
1. Strong acid + weak base gives an acidic salt solution.
2. Compare with sodium carbonate.

Question 7. Is it possible for a salt to behave as a base? Explain.
Answer:
Yes. A salt of a strong base and a weak acid behaves as a base in water. For example, sodium acetate, CH₃COONa, is the salt of the weak acid CH₃COOH and the strong base NaOH.
In water it gives CH₃COOH (weakly ionised) and NaOH (completely ionised), so OH⁻ ions exceed H⁺ ions and the solution is basic (pH more than 7). Sodium carbonate behaves in the same way.
In simple words: Salts of a strong base and weak acid give basic solutions.

Teacher's Note:
1. The solution of such a salt turns red litmus blue.
2. Give a named salt.

Question 8. Which of the two soil, A (pH = 8) or B (pH = 4) should be treated with powdered chalk to adjust its pH and why?
Answer:
Soil B (pH = 4) should be treated with powdered chalk.
Soil B is acidic, and most plants grow best near neutral pH. Powdered chalk is calcium carbonate, which is basic and reacts with the excess acid in the soil and raises the pH. Soil A is basic, so chalk would not help it.
In simple words: Basic chalk is used for acidic soil.

Teacher's Note:
1. Choose the soil with pH less than 7.
2. Explain with neutralisation.

Question 9. Why is calcium sulphate hemihydrate called Plaster of Paris? You know only half a water molecule is shown to be attached as water of crystallisation in the formula of calcium sulphate hemihydrate. How can you get half a water molecule?
Answer:
Calcium sulphate hemihydrate is called Plaster of Paris because it was first made by heating gypsum found at Montmartre near Paris (in France), and it is used for plastering.
A half water molecule cannot exist. The formula CaSO₄·½H₂O means that two formula units of CaSO₄ share one molecule of water, that is, (CaSO₄)2·H₂O.
In simple words: Half water molecule means one water molecule for every two CaSO₄ units.

Teacher's Note:
1. Hemi means half.
2. Write the formula as (CaSO₄)₂·H₂O.

Question 10. A salt X when dissolved in distilled water gives a clear solution which turns red litmus blue. Explain this phenomenon.
Answer:
A solution that turns red litmus blue is basic, so X is the salt of a weak acid and a strong base.
For example, sodium carbonate in water gives sodium hydroxide (a strong base, fully ionised, giving many OH⁻ ions) and carbonic acid (a weak acid, giving few H⁺ ions). The OH⁻ ions are in excess, so the solution is basic: Na₂CO₃ + 2H₂O → 2NaOH + H₂CO₃.
In simple words: Salt of a strong base and a weak acid is basic.

Teacher's Note:
1. Name the acid and base forming the salt.
2. Basic salt turns red litmus blue.

Application Based HOTS Questions

Question 11. Consider the following chemical reactions: (I) HCl + Metal → Salt + Hydrogen gas (II) NaOH + Metal → Salt + Hydrogen gas (A) Do all the metals displace hydrogen from the hydrochloric acid and form hydrogen gas? Justify your answer by giving examples.
(B) Is the reaction (II) between NaOH and metal possible for all the metals? Illustrate your answer with the help of an example.
Answer:
(A) No. Only metals more reactive than hydrogen (such as Mg, Zn, Fe) displace hydrogen from dilute HCl: Zn + 2HCl → ZnCl₂ + H₂. Metals such as copper and silver, which are below hydrogen in the reactivity series, do not react.
(B) No. Only a few metals such as zinc and aluminium react with NaOH: 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂. Metals such as iron and copper do not react with NaOH.
In simple words: Only some metals react with acids, and still fewer with bases.

Teacher's Note:
1. Use the reactivity series for part (A).
2. Zn and Al are the metals that react with NaOH.

Question 12. (A) List the different forms of calcium carbonate. Which gas is liberated on heating these forms?
(B) What happens on passing carbon dioxide through the lime water?
(C) Metal carbonates and metal bicarbonates react with acids and produce carbon dioxide gas. It was found that the amount of carbon dioxide gas formed in both the reactions was the same. Is this finding correct? Justify your answer.
Answer:
(A) Limestone, chalk and marble are forms of calcium carbonate. On heating they give carbon dioxide: CaCO₃ → CaO + CO₂.
(B) Lime water turns milky: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O. On passing excess CO₂, the milkiness disappears due to calcium hydrogencarbonate.
(C) Yes, when equal moles of the carbonate and the hydrogencarbonate are taken with excess acid, one mole of each gives one mole of CO₂: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂; NaHCO₃ + HCl → NaCl + H₂O + CO₂. (The carbonate uses twice as much acid.)
In simple words: Equal moles of carbonate and hydrogencarbonate give equal CO₂ with excess acid.

Teacher's Note:
1. Compare mole ratios.
2. The statement is true only for equal moles of salt.

Question 13. (A) Demonstrate an activity showing that copper oxide is basic in nature.
(B) Astitva carried a reaction between metallic oxide and dil. HCl in a beaker with a glass rod.
(i) What is the colour of the solution after the chemical reaction?
(ii) What has happened to copper oxide?
(iii) Write a balanced chemical equation for the above chemical reaction.
Answer:
(A) Take a little black copper oxide powder in a beaker and add dilute hydrochloric acid slowly while stirring with a glass rod. The black powder dissolves and the solution turns blue-green, showing the oxide reacted with an acid like a base. Metal oxide + acid → salt + water.
(B)(i) The solution is blue-green.
(ii) Copper oxide dissolved (reacted with the acid) and formed copper(II) chloride and water.
(iii) CuO(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l)
In simple words: Copper oxide reacts with an acid to give blue-green copper chloride.

Teacher's Note:
1. Metal oxides are basic.
2. Write the general word equation first and then the example.

Question 14. Equal lengths of magnesium ribbons are taken in test tube A and B. Hydrochloric acid (HCl) is added to test tube A while acetic acid (CH₃COOH) is added to test tube B. In which case fizzing occurs more vigorously and why?
Answer:
Fizzing occurs more vigorously in test tube A.
Hydrochloric acid is a strong acid; it ionises completely and gives a higher concentration of H⁺ ions. Acetic acid is a weak acid and gives few H⁺ ions, so it reacts more slowly. Fizzing is due to hydrogen gas: Mg + 2HCl → MgCl₂ + H₂↑.
In simple words: A strong acid reacts faster with a metal.

Teacher's Note:
1. Fizzing = hydrogen gas bubbles.
2. Compare ionisation of HCl and CH₃COOH.

Question 15. A milkman adds very small amount of baking soda to fresh milk.
(a) Why does he shift the pH of milk from 6 to slightly alkaline?
(b) Why does this milk take a long time to set as curd?
Answer:
(a) Baking soda makes the milk slightly alkaline, so it neutralises the lactic acid that forms on standing and prevents the milk from turning sour (and curdling) quickly.
(b) Setting of curd needs an acidic medium. The alkaline milk must first have its base neutralised by the lactic acid formed, so it takes a long time to become acidic enough to set as curd.
In simple words: A little baking soda delays souring and also delays setting.

Teacher's Note:
1. Think about neutralisation of lactic acid.
2. Both parts rely on the same idea.

Question 16. Five solutions A, B, C, D and E when tested with universal indicator show pH as 4, 2, 12, 7 and 9 respectively. Which solution is (a) Neutral (b) Strongly alkaline (c) Strongly acidic (d) Weakly alkaline (e) Weakly acidic. Arrange the pH in increasing order of H⁺ ion concentration.
Answer:
(a) Neutral: D (pH 7).
(b) Strongly alkaline: C (pH 12).
(c) Strongly acidic: B (pH 2).
(d) Weakly alkaline: E (pH 9).
(e) Weakly acidic: A (pH 4).
The higher the pH, the lower the H⁺ concentration. So the increasing order of H⁺ concentration is C < E < D < A < B, that is pH 12, 9, 7, 4, 2.
In simple words: Lower pH means more H⁺ ions.

Teacher's Note:
1. Compare each value with 7.
2. Arrange pH from highest to lowest for increasing H⁺.

Question 17. Aditi finds that a mixture of an acid and a base does not change the colour of either red or blue litmus paper. Compare the amounts of H⁺ and OH⁻ ions in the solution. pH is measured on a scale of 0 to 14, with lower values indicating high hydrogen ion concentration (more acidic) and higher values indicating low hydrogen ion concentration (less acidic). A pH of 7 is considered as neutral. Every whole unit in pH represents a ten-fold increase or decrease in hydrogen ion concentration. What would be the hydrogen ion concentration of a solution of pH 4 compared to a solution of pH 8?
Answer:
Since neither red nor blue litmus changes colour, the mixture is neutral, so the amount of H⁺ ions is equal to the amount of OH⁻ ions.
Each unit of pH is a tenfold change. From pH 8 to pH 4 is a change of 4 units, so the hydrogen ion concentration is 10 × 10 × 10 × 10 = 10⁴ = 10,000 times higher in the solution of pH 4 than in the solution of pH 8.
In simple words: A neutral mixture has equal H⁺ and OH⁻; pH 4 has 10,000 times more H⁺ than pH 8.

Teacher's Note:
1. Count the pH units between the two solutions.
2. [H⁺] = 10^−pH.

Identification and Analysis Questions

Question 18. Salt commonly used in bakery products on heating gets converted into another salt B which itself is used for removal of hardness of water and a gas C is evolved. The gas C when passed through lime water, turns it milky. Identify A, B and C.
Answer:
A is sodium hydrogencarbonate, NaHCO₃ (baking soda), used in bakery products.
On heating: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂↑. So B is sodium carbonate, Na₂CO₃, and C is carbon dioxide, CO₂.
CO₂ turns lime water milky due to the formation of calcium carbonate: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O.
In simple words: A = NaHCO₃, B = Na₂CO₃, C = CO₂.

Teacher's Note:
1. Hardness removal uses washing soda, made from B.
2. Baking soda is NaHCO₃ and washing soda is Na₂CO₃·10H₂O.

Question 19. A metal carbonate X on reacting with an acid gives a gas which when passed through a solution Y gives the carbonate back. On the other hand, a gas G that is obtained at anode during electrolysis of brine is passed on dry Y, it gives a compound Z, used for disinfecting drinking water. Identify X, Y, G and Z.
Answer:
X is calcium carbonate, CaCO₃. With acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂↑.
The gas CO₂ passed through lime water Y, Ca(OH)₂, gives CaCO₃ back: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O.
Gas G obtained at the anode during electrolysis of brine is chlorine, Cl₂. Passed over dry slaked lime Y it gives Z, bleaching powder: Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O.
Thus X = CaCO₃, Y = Ca(OH)₂, G = Cl₂, Z = CaOCl₂.
In simple words: X is limestone, Y is slaked lime, G is chlorine and Z is bleaching powder.

Teacher's Note:
1. Work backwards from bleaching powder.
2. Write each step as an equation.

Question 20. Mohit was given a substance X and asked to identify it. He conducted various tests on the substance and recorded the results as: (P) It turns universal indicator blue. (Q) It can be prepared from brine solution when electricity is passed through it. (R) The amphoteric oxides of some metals react with it to form salt and water.
(A) Identify the substance X and write its two uses.
(B) How can substance X be manufactured in industries? What is the name given to such a chemical reaction and why?
(C) What are amphoteric oxides? Illustrate by chemical reaction the formation of salt and water when substance X reacts with an amphoteric oxide.
Answer:
(A) X is sodium hydroxide, NaOH. Uses: making soap and detergents; making paper; (also artificial fibres and dyes).
(B) It is manufactured by passing electricity through brine: 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g). It is called the chlor-alkali process because chlorine and an alkali are formed.
(C) Amphoteric oxides are metal oxides that react with both acids and bases to form salt and water, for example Al₂O₃ and ZnO. Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O (sodium aluminate and water).
In simple words: NaOH is made by electrolysis of brine and reacts with amphoteric oxides such as Al₂O₃.

Teacher's Note:
1. Amphoteric means acts as both acid and base.
2. Name the process and the products.

Question 21. Identify the compound X on the basis of the reactions given below. Also, write the name and chemical formulae of A, B and C. X + Zn → A + H₂(g); X + HCl → B + H₂O; X + CH₃COOH → C + H₂O
Answer:
X is sodium hydroxide, NaOH. It gives hydrogen with zinc, and it reacts with acids to give salt and water.
A is sodium zincate, Na₂ZnO₂: 2NaOH + Zn → Na₂ZnO₂ + H₂.
B is sodium chloride, NaCl: NaOH + HCl → NaCl + H₂O.
C is sodium acetate, CH₃COONa: NaOH + CH₃COOH → CH₃COONa + H₂O.
In simple words: NaOH reacts with zinc and with acids.

Teacher's Note:
1. Use the reaction with zinc to decide.
2. Write the name and formula of each.