Class 10 Science · Chapter 2

Acids, Bases and Salts

352 Class 10 Science questions on Acids, Bases and Salts from the NCERT textbook, NCERT Exemplar and CBSE board papers 2018–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 18 of 18

1. Q341 CBSE 2019
   A solution ‘X’ ​gives orange colour ​when a drop of it falls on pH paper, while another solution ‘Y’ gives bluish colour when a drop of it falls ​on ​pH paper. ​What is the nature of both the solutions ​? Determine the pH of solutions ‘X’ ​and ‘Y’.
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   X is acidic (pH less than 7) and Y is basic (pH more than 7).

   X is acidic: an orange colour on pH paper means $\text{pH}<7$ (an orange shade lies at roughly pH 3 to 5). Y is basic (alkaline): a bluish colour means $\text{pH}>7$ (a bluish shade lies at roughly pH 8 to 10).

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2. Q342 CBSE 2019
   How is ​the presence ​of ​an ​acid tested with a strip ​of red ​litmus paper ​?
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   Turn the red strip blue with a base, then dip it in the test solution: it turns red if an acid is present.

   An acid does not change the colour of red litmus - the strip simply stays red, as it would in a neutral solution too - because an acid turns blue litmus red. So the red strip is first converted into blue litmus: dip it in a basic solution, such as dilute sodium hydroxide, and it turns blue. Then dip this blue strip in the solution to be tested: if it turns red, an acid is present; if it stays blue, the solution is not acidic.

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3. Q343 CBSE 2018
   Write ​the name of the white powder generally added to besan to make soft and ​crisp pakoras. List two ingredients of this powder and write function of each. ​Also give ​the ​equation ​for the chemical reaction that takes place ​when this powder is heated.
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   Baking powder: baking soda releases carbon dioxide on heating, making pakoras soft and crisp; tartaric acid removes the bitter taste.

   Name: baking powder. Ingredients and their functions: 1. Sodium hydrogencarbonate (baking soda), ${\text{NaHCO}}_3$: on heating it gives off carbon dioxide, whose bubbles make the pakoras soft and crisp. 2. Tartaric acid, a mild edible acid: it reacts with the sodium hydrogencarbonate, so no bitter-tasting sodium carbonate is left; it removes the bitter taste. Equation (baking powder on heating): ${\text{NaHCO}}_3+{\text{H}}^{+}\stackrel{\text{heat}}{\to }{\text{CO}}_2+{\text{H}}_2\text{O}+$ sodium salt of the acid, the ${\text{H}}^{+}$ coming from the tartaric acid. Sodium hydrogencarbonate on its own decomposes on heating: $2{\text{NaHCO}}_3\stackrel{\text{heat}}{\to }{\text{Na}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$.

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4. Q344 CBSE 2018
   Write the ​chemical formula of bleaching powder. ​How is it ​prepared ? Write balanced chemical equation for ​the reaction ​involved. ​Write ​one important use of bleaching powder.
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   Bleaching powder is ${\text{CaOCl}}_2$, made by passing chlorine over dry slaked lime; it makes drinking water free from germs.

   Chemical formula of bleaching powder: ${\text{CaOCl}}_2$ (calcium oxychloride). Preparation: it is produced by the action of chlorine gas on dry slaked lime, that is, by passing chlorine over dry calcium hydroxide. Balanced equation: ${\text{Ca(OH)}}_2+{\text{Cl}}_2\to {\text{CaOCl}}_2+{\text{H}}_2\text{O}$ One important use: to make drinking water free from germs. (Any one other use also earns the mark: bleaching cotton and linen in the textile industry, bleaching wood pulp in paper factories, bleaching washed clothes in laundry, or as an oxidising agent in chemical industries.)

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5. Q345 CBSE 2018
   What ​is ​water ​of crystallization ? Name ​and give formula of two salts which contain water ​of ​crystallization.
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   Water of crystallisation is the fixed number of water molecules in one formula unit of a salt, e.g. Epsom salt and Glauber's salt.

   Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt; these molecules are part of the salt's crystals. Two salts (any two): 1. Magnesium sulphate heptahydrate (Epsom salt), ${\text{MgSO}}_4\cdot 7{\text{H}}_2\text{O}$ 2. Sodium sulphate decahydrate (Glauber's salt), ${\text{Na}}_2{\text{SO}}_4\cdot 10{\text{H}}_2\text{O}$ Other correct examples earn the marks equally, such as copper sulphate pentahydrate, ${\text{CuSO}}_4\cdot 5{\text{H}}_2\text{O}$, washing soda, ${\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_2\text{O}$, and gypsum, ${\text{CaSO}}_4\cdot 2{\text{H}}_2\text{O}$.

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6. Q346 CBSE 2018
   A sulphate salt of group ​two element of the periodic table is white, soft ​substance, which can be moulded into different shapes ​by making it dough. When ​this compound is left in open for sometime, it ​becomes a ​solid object and ​cannot be used for moulding purposes. Identify this sulphate salt and explain why does it show such a behaviour. Also give the equation for the ​reaction involved.
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   The salt is Plaster of Paris, ${\text{CaSO}}_4\cdot \frac{1}{2}{\text{H}}_2\text{O}$, which takes up water and sets as gypsum.

   The salt is Plaster of Paris, calcium sulphate hemihydrate, ${\text{CaSO}}_4\cdot \frac{1}{2}{\text{H}}_2\text{O}$ (calcium is a group 2 element). Reason: Plaster of Paris has only half a molecule of water of crystallisation per formula unit. Mixed with water it forms a dough that can be moulded, but on standing, or when left in the open where it takes up moisture from the air, it combines with water and changes into gypsum, a hard solid mass. So it sets and can no longer be used for moulding. Equation: ${\text{CaSO}}_4\cdot \frac{1}{2}{\text{H}}_2\text{O}+1\frac{1}{2}{\text{H}}_2\text{O}\to {\text{CaSO}}_4\cdot 2{\text{H}}_2\text{O}$ (gypsum)

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7. Q347 CBSE 2018
   2 mL of sodium hydroxide solution is ​added ​to ​a few pieces of granulated zinc metal taken in a test ​tube. ​When the contents are warmed, a gas evolves which is ​bubbled through a soap solution before testing. Write the equation ​of the chemical reaction involved and the test to detect ​the gas. Name the gas which will be evolved when the same metal reacts with dilute solution of a strong acid.
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   Zinc and warm sodium hydroxide give sodium zincate and hydrogen, which burns with a pop; dilute strong acids also give hydrogen.

   Equation: $\text{Zn}(\text{s})+2\text{NaOH}(\text{aq})\stackrel{\text{warm}}{\to }{\text{Na}}_2{\text{ZnO}}_2(\text{aq})+{\text{H}}_2(\text{g})$ (sodium zincate and hydrogen) Test: when a burning splinter or candle is brought near the gas (a gas-filled soap bubble), the gas burns with a pop sound; so the gas is hydrogen. With a dilute solution of a strong acid, the same metal also gives hydrogen gas, ${\text{H}}_2$: $\text{Zn}(\text{s})+{\text{H}}_2{\text{SO}}_4(\text{aq})\to {\text{ZnSO}}_4(\text{aq})+{\text{H}}_2(\text{g})$

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8. Q348 CBSE 2018
   1 g of solid ​sodium ​chloride is taken in a clean and dry test ​tube and 2 mL of conc. sulphuric acid is added to it. If the gas evolved is tested first with dry and then with wet ​blue litmus paper, in which case ​will the litmus paper change colour ​? ​Give reason for your ​answer. What inference can be drawn about the nature of the evolved gas ? Support your answer with chemical equation for the reaction.
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   Only wet blue litmus turns red; the gas is hydrogen chloride, which shows acidic behaviour (forms ${\text{H}}^{+}$ ions) only in water.

   The litmus paper changes colour only in the second case: the wet blue litmus paper turns red, while the dry blue litmus paper shows no change. Reason: the gas evolved is hydrogen chloride. Dry $\text{HCl}$ gas does not give ${\text{H}}^{+}$ ions; in the water on the wet paper it ionises and gives ${\text{H}}^{+}$ ions (as ${\text{H}}_3{\text{O}}^{+}$), which turn blue litmus red. Inference: the evolved gas ($\text{HCl}$) is acidic in nature, but it shows its acidic behaviour, forming ${\text{H}}^{+}$ ions, only in the presence of water. Equations: $\text{NaCl}+{\text{H}}_2{\text{SO}}_4\to {\text{NaHSO}}_4+\text{HCl}$ (on stronger heating, $2\text{NaCl}+{\text{H}}_2{\text{SO}}_4\to {\text{Na}}_2{\text{SO}}_4+2\text{HCl}$); $\text{HCl}+{\text{H}}_2\text{O}\to {\text{H}}_3{\text{O}}^{+}+{\text{Cl}}^{-}$.

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9. Q349 CBSE 2018
   The pH of a salt used to make tasty and ​crispy pakoras is 14. Identify the salt ​and write a chemical ​equation for ​its ​formation. List its ​two ​uses.
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   The salt is baking soda, ${\text{NaHCO}}_3$, used in baking powder and as an antacid.

   Salt: baking soda, sodium hydrogencarbonate (sodium bicarbonate), ${\text{NaHCO}}_3$. Formation: $\text{NaCl}+{\text{H}}_2\text{O}+{\text{CO}}_2+{\text{NH}}_3\to {\text{NH}}_4\text{Cl}+{\text{NaHCO}}_3$ Uses (any two): (i) For making baking powder; (ii) As an ingredient of antacids, to neutralise excess acid in the stomach; (iii) In soda-acid fire extinguishers.

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10. Q350 CBSE 2018
    1. (a)​For the preparation of cakes, baking powder is used. If at home ​your mother uses baking soda instead of baking powder, ​how will ​it affect the taste of the cake ​and why ​?
    2. (b)How is baking soda be converted into baking powder ​?
    3. (c)​What makes the cake soft and spongy ?
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    Baking soda alone makes the cake bitter; tartaric acid with it makes baking powder; the carbon dioxide released makes the cake soft and spongy.

    (a) The cake will taste bitter. On heating, baking soda (sodium hydrogencarbonate) decomposes and forms sodium carbonate, a base, which stays in the cake and gives it a bitter taste: $2{\text{NaHCO}}_3\stackrel{\text{heat}}{\to }{\text{Na}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$. (b) By mixing baking soda with a mild edible acid such as tartaric acid; the acid reacts with the sodium hydrogencarbonate, so no bitter sodium carbonate is formed. (c) Carbon dioxide, released when baking powder is heated or mixed with water (${\text{NaHCO}}_3+{\text{H}}^{+}\to {\text{CO}}_2+{\text{H}}_2\text{O}+\text{sodium salt of the acid}$), forms bubbles that make the cake rise, so it becomes soft and spongy.

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11. Q351 CBSE 2018
    A ​student dips one ​pH paper in solution A and another pH paper in solution B and ​observes that the ​pH paper turns blue in ​solution A ​and orange in solution B respectively. Identify solution A and B. Find the value ​of ​pH of solutions A and B.
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    Solution A is basic (pH more than 7); solution B is acidic (pH less than 7).

    Solution A: basic (alkaline). The pH paper turns blue, a colour on the basic side of the chart, so the pH of A is more than $7$ (blue matches about pH $9$ to $10$). Solution B: acidic. The pH paper turns orange, a colour on the acidic side of the chart, so the pH of B is less than $7$ (orange matches about pH $3$ to $4$).

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12. Q352 CBSE 2018
    State observations ​of the reactions of ​dil. $\text{HCl}$ with the ​following ​:

    1. (i)Blue litmus paper
    2. (ii)​Red ​litmus paper
    3. (iii)Zinc-granules
    4. (iv)Solid ${\text{NaHCO}}_3$
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    Blue litmus turns red, red litmus is unchanged, zinc gives hydrogen gas and sodium hydrogencarbonate gives carbon dioxide.

    (i) Blue litmus paper turns red. (ii) Red litmus paper shows no change (it stays red). (iii) Bubbles of a colourless gas, hydrogen, are given off from the zinc granules; the gas burns with a pop sound. $\text{Zn}+2\text{HCl}⟶{\text{ZnCl}}_2+{\text{H}}_2$ (iv) Brisk effervescence: carbon dioxide gas is given off, which turns lime water milky. ${\text{NaHCO}}_3+\text{HCl}⟶\text{NaCl}+{\text{H}}_2\text{O}+{\text{CO}}_2$

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