Class 10 Science · Chapter 2

Acids, Bases and Salts

208 Class 10 Science questions on Acids, Bases and Salts from the NCERT textbook, NCERT Exemplar and CBSE board papers 2024–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 4 of 11

1. Q61 Exemplar · Multiple Choice Questions
   Which of the following is(are) true when $\text{HCl}(g)$ is passed through water?

   1. (i)It does not ionise in the solution as it is a covalent ​compound.
   2. (ii)It ionises in the solution
   3. (iii)It gives ​both hydrogen ​and hydroxyl ion in the solution
   4. (iv)It forms hydronium ion in the ​solution due to the ​combination of hydrogen ion ​with water molecule

   1. (a)(i) only
   2. (b)(iii) only
   3. (c)(ii) and (iv)
   4. (d)(iii) and (iv)
   Show answer

   (c) (ii) and (iv)

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2. Q62 Exemplar · Multiple Choice Questions
   Which ​of the ​following statements ​is ​true ​for ​acids?

   1. (a)Bitter and change red litmus to blue
   2. (b)Sour and change red litmus to blue
   3. (c)Sour and change blue litmus to red
   4. (d)Bitter and change blue litmus to red
   Show answer

   (c) Sour and change blue litmus to red

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3. Q63 Exemplar · Multiple Choice Questions
   Which ​of ​the ​following ​are present in ​a dilute aqueous ​solution ​of ​hydrochloric acid?

   1. (a)${\text{H}}_3{\text{O}}^{+}+{\text{Cl}}^{-}$
   2. (b)${\text{H}}_3{\text{O}}^{+}+{\text{OH}}^{-}$
   3. (c)${\text{Cl}}^{-}+{\text{OH}}^{-}$
   4. (d)unionised $\text{HCl}$
   Show answer

   (a) ${\text{H}}_3{\text{O}}^{+}+{\text{Cl}}^{-}$

   Step-by-step solution →
4. Q64 Exemplar · Multiple Choice Questions
   Identify the correct representation ​of reaction ​occurring ​during chloralkali ​process

   1. (a)$2\text{NaCl}(l)+2{\text{H}}_2\text{O}(l)\to 2\text{NaOH}(l)+{\text{Cl}}_2(g)+{\text{H}}_2(g)$
   2. (b)$2\text{NaCl}(aq)+2{\text{H}}_2\text{O}(aq)\to 2\text{NaOH}(aq)+{\text{Cl}}_2(g)+{\text{H}}_2(g)$
   3. (c)$2\text{NaCl}(aq)+2{\text{H}}_2\text{O}(l)\to 2\text{NaOH}(aq)+{\text{Cl}}_2(aq)+{\text{H}}_2(aq)$
   4. (d)$2\text{NaCl}(aq)+2{\text{H}}_2\text{O}(l)\to 2\text{NaOH}(aq)+{\text{Cl}}_2(g)+{\text{H}}_2(g)$
   Show answer

   (d) $2\text{NaCl}(aq)+2{\text{H}}_2\text{O}(l)\to 2\text{NaOH}(aq)+{\text{Cl}}_2(g)+{\text{H}}_2(g)$

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5. Q65 Exemplar · Short Answer Questions
   Match the acids ​given ​in ​Column ​(A) with their correct source ​given ​in ​Column (B)
   Show answer

   (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i). That is, lactic acid comes from curd, acetic acid from vinegar, citric acid from lemon and oxalic acid from tomato.

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6. Q66 Exemplar · Short Answer Questions
   Match the important chemicals given in ​Column (A) ​with ​the ​chemical formulae given ​in Column ​(B)
   Show answer

   (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i). That is, plaster of Paris is ${\text{CaSO}}_4\cdot \frac{1}{2}{\text{H}}_2\text{O}$, gypsum is ${\text{CaSO}}_4\cdot 2{\text{H}}_2\text{O}$, bleaching powder is ${\text{Ca(ClO)}}_2$ and slaked lime is ${\text{Ca(OH)}}_2$.

   Step-by-step solution →
7. Q67 Exemplar · Short Answer Questions
   What will be the action of the ​following substances on ​litmus ​paper? ​Dry $\text{HCl}$ gas, ​Moistened ${\text{NH}}_3$ gas, Lemon juice, Carbonated soft ​drink, ​Curd, Soap solution.
   Show answer

   Dry $\text{HCl}$ has no effect; moist ${\text{NH}}_3$ and soap turn red litmus blue; lemon juice, soft drink and curd turn blue litmus red.

   Dry $\text{HCl}$ gas: no action on dry litmus paper, because it cannot ionise without water. Moistened ${\text{NH}}_3$ gas: basic, turns red litmus blue. Lemon juice: acidic, turns blue litmus red. Carbonated soft drink: acidic, turns blue litmus red. Curd: acidic, turns blue litmus red. Soap solution: basic, turns red litmus blue.

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8. Q68 Exemplar · Short Answer Questions
   Name the ​acid present in ant sting and give its chemical ​formula. Also ​give the ​common ​method to get relief from the discomfort caused by ​the ant sting.
   Show answer

   Ant stings contain methanoic (formic) acid, $\text{HCOOH}$; rubbing with baking soda or calamine solution neutralises it.

   The acid present in an ant sting is methanoic acid (formic acid), formula $\text{HCOOH}$. Relief is obtained by rubbing the affected area with a mild base — baking soda, that is sodium hydrogencarbonate (${\text{NaHCO}}_3$), or moist calamine solution — which neutralises the acid: $\text{HCOOH}+{\text{NaHCO}}_3\to \text{HCOONa}+{\text{H}}_2\text{O}+{\text{CO}}_2$.

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9. Q69 Exemplar · Short Answer Questions
   What ​happens ​when ​nitric ​acid is added ​to egg shell?
   Show answer

   Brisk effervescence of carbon dioxide occurs, forming calcium nitrate and water, and the calcium carbonate shell slowly dissolves.

   Egg shell is made of calcium carbonate. Nitric acid reacts with it to give calcium nitrate, water and carbon dioxide, so brisk effervescence is seen and the shell slowly dissolves: ${\text{CaCO}}_3(s)+2{\text{HNO}}_3(aq)\to {\text{Ca(NO}}_3{\text{)}}_2(aq)+{\text{H}}_2\text{O}(l)+{\text{CO}}_2(g)$. The gas evolved turns lime water milky, which confirms it is carbon dioxide.

   Step-by-step solution →
10. Q70 Exemplar · Short Answer Questions
    A student prepared solutions of

    1. (i)an acid and
    2. (ii)a base in two separate beakers. ​She ​forgot to label the solutions and litmus paper is ​not ​available in the laboratory. Since both the solutions ​are colourless, how will she distinguish between the two?
    Show answer

    Use another indicator (turmeric, china rose, phenolphthalein) or add sodium hydrogencarbonate: only the acid gives effervescence of carbon dioxide.

    She can use any other indicator, or a chemical test. (i) Indicators: turmeric paper stays yellow in the acid and turns red in the base; china rose extract turns dark pink in the acid and green in the base; phenolphthalein stays colourless in the acid and turns pink in the base. (ii) Chemical test: add a pinch of sodium hydrogencarbonate to a little of each solution — the acid gives brisk effervescence of carbon dioxide, ${\text{NaHCO}}_3+\text{HCl}\to \text{NaCl}+{\text{H}}_2\text{O}+{\text{CO}}_2$, while the base gives no gas. The gas turns lime water milky, confirming the beaker it came from holds the acid.

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11. Q71 Exemplar · Short Answer Questions
    How ​would ​you distinguish between ​baking powder and ​washing ​soda by heating?
    Show answer

    Heat both and pass the gas through lime water: baking powder gives carbon dioxide that turns it milky; washing soda does not.

    Heat a little of each solid in a separate dry test tube and pass the gas evolved through lime water. Baking powder contains sodium hydrogencarbonate, which decomposes on heating to give carbon dioxide: $2{\text{NaHCO}}_3\stackrel{\Delta }{\to }{\text{Na}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$, so the lime water turns milky. Washing soda, ${\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_2\text{O}$, only loses its water of crystallisation on heating: ${\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_2\text{O}\stackrel{\Delta }{\to }{\text{Na}}_2{\text{CO}}_3+10{\text{H}}_2\text{O}$, and gives no carbon dioxide, so the lime water stays clear. The solid that turns lime water milky is baking powder.

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12. Q72 Exemplar · Short Answer Questions
    Salt A 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.
    Show answer

    A is baking soda (${\text{NaHCO}}_3$), B is sodium carbonate (${\text{Na}}_2{\text{CO}}_3$), and C is carbon dioxide (${\text{CO}}_2$).

    A is baking soda, sodium hydrogencarbonate (${\text{NaHCO}}_3$). B is sodium carbonate (${\text{Na}}_2{\text{CO}}_3$), sold as washing soda, which removes the hardness of water. C is carbon dioxide (${\text{CO}}_2$). The reaction is $2{\text{NaHCO}}_3\stackrel{\Delta }{\to }{\text{Na}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$.

    Step-by-step solution →
13. Q73 Exemplar · Short Answer Questions
    In one of the industrial processes used ​for manufacture of sodium hydroxide, a gas X is formed as by product. The gas ​X reacts ​with lime water to give a ​compound Y which is used as a bleaching ​agent in chemical industry. Identify ​X ​and Y giving the chemical ​equation of the reactions involved.
    Show answer

    X is chlorine (${\text{Cl}}_2$) from the chlor-alkali process; Y is bleaching powder (${\text{CaOCl}}_2$), formed when chlorine reacts with slaked lime.

    The process is the chlor-alkali process, the electrolysis of brine: $2\text{NaCl}(aq)+2{\text{H}}_2\text{O}(l)\to 2\text{NaOH}(aq)+{\text{Cl}}_2(g)+{\text{H}}_2(g)$. X is chlorine, ${\text{Cl}}_2$, liberated at the anode. Passed over slaked lime it gives Y, bleaching powder (calcium oxychloride), ${\text{CaOCl}}_2$: ${\text{Ca(OH)}}_2+{\text{Cl}}_2\to {\text{CaOCl}}_2+{\text{H}}_2\text{O}$.

    Step-by-step solution →
14. Q74 Exemplar · Short Answer Questions
    Fill ​in the ​missing ​data in the following ​table

    Salt obtained from — the printed dash marks each blank to be filled

    Name of the salt	Formula	Salt obtained from: Base	Salt obtained from: Acid
    (i) Ammonium chloride	${\text{NH}}_4\text{Cl}$	${\text{NH}}_4\text{OH}$	—
    (ii) Copper sulphate	—	—	${\text{H}}_2{\text{SO}}_4$
    (iii) Sodium chloride	$\text{NaCl}$	$\text{NaOH}$	—
    (iv) Magnesium nitrate	${\text{Mg(NO}}_3{\text{)}}_2$	—	${\text{HNO}}_3$
    (v) Potassium sulphate	${\text{K}}_2{\text{SO}}_4$	—	—
    (vi) Calcium nitrate	${\text{Ca(NO}}_3{\text{)}}_2$	${\text{Ca(OH)}}_2$	—

    Show answer

    Row by row the missing entries are $\text{HCl}$; ${\text{CuSO}}_4$ and ${\text{Cu(OH)}}_2$; $\text{HCl}$; ${\text{Mg(OH)}}_2$; $\text{KOH}$ and ${\text{H}}_2{\text{SO}}_4$; ${\text{HNO}}_3$.

    (i) Acid: $\text{HCl}$. (ii) Formula: ${\text{CuSO}}_4$; Base: ${\text{Cu(OH)}}_2$. (iii) Acid: $\text{HCl}$. (iv) Base: ${\text{Mg(OH)}}_2$. (v) Base: $\text{KOH}$; Acid: ${\text{H}}_2{\text{SO}}_4$. (vi) Acid: ${\text{HNO}}_3$.

    Step-by-step solution →
15. Q75 Exemplar · Short Answer Questions
    What are strong ​and weak acids? ​In the ​following list of ​acids, ​separate strong acids from weak acids. Hydrochloric acid, citric acid, ​acetic acid, nitric ​acid, ​formic acid, sulphuric acid.
    Show answer

    Strong acids ionise completely (hydrochloric, nitric, sulphuric); weak acids ionise partially (citric, acetic, formic).

    A strong acid is one that ionises completely in aqueous solution and therefore gives a high concentration of hydrogen ions; a weak acid ionises only partially and gives a low concentration of hydrogen ions. Strong acids in the list: hydrochloric acid, nitric acid and sulphuric acid. Weak acids in the list: citric acid, acetic acid and formic acid.

    Step-by-step solution →
16. Q76 Exemplar · Short Answer Questions
    When zinc metal ​is treated with a dilute solution of a ​strong acid, a ​gas is evolved, which is utilised in the hydrogenation of ​oil. Name the gas evolved. Write ​the chemical equation of the ​reaction involved and also ​write a test to detect ​the gas formed.
    Show answer

    The gas is hydrogen: $\text{Zn}+{\text{H}}_2{\text{SO}}_4\to {\text{ZnSO}}_4+{\text{H}}_2$; it burns with a pop sound near a flame.

    The gas is hydrogen, ${\text{H}}_2$. The reaction is $\text{Zn}(s)+{\text{H}}_2{\text{SO}}_4(aq)\to {\text{ZnSO}}_4(aq)+{\text{H}}_2(g)$ with dilute sulphuric acid, or $\text{Zn}(s)+2\text{HCl}(aq)\to {\text{ZnCl}}_2(aq)+{\text{H}}_2(g)$ with dilute hydrochloric acid. Test: bring a burning candle or splinter near the mouth of the tube — the gas burns with a pop sound, which confirms hydrogen.

    Step-by-step solution →
17. Q77 Exemplar · Long Answer Questions
    In the following schematic diagram for the preparation of hydrogen gas as shown in Figure 2.3, what would happen if following changes are made?

    1. (a)In place of zinc ​granules, ​same amount ​of zinc dust ​is taken in ​the test tube
    2. (b)Instead of dilute sulphuric acid, dilute hydrochloric acid is taken
    3. (c)In place of zinc, copper turnings are ​taken
    4. (d)Sodium hydroxide is taken in place of dilute sulphuric acid and the tube is ​heated.

    Show answer

    Zinc dust reacts faster; dilute HCl changes nothing; copper gives no reaction; heated NaOH still gives hydrogen, as zinc is amphoteric.

    (a) The reaction becomes much faster, because zinc dust exposes a far larger surface area than granules of the same mass; hydrogen is evolved more rapidly, but the total volume finally collected is unchanged. (b) No change in the observation — hydrogen is still evolved and still gives a pop sound; only the salt differs, zinc chloride instead of zinc sulphate: $\text{Zn}+2\text{HCl}\to {\text{ZnCl}}_2+{\text{H}}_2$. (c) No reaction occurs. Copper lies below hydrogen in the reactivity series and cannot displace hydrogen from a dilute acid, so no gas is evolved. (d) Hydrogen is still evolved, because zinc is amphoteric and reacts with alkalis as well as with acids: $\text{Zn}+2\text{NaOH}\stackrel{\Delta }{\to }{\text{Na}}_2{\text{ZnO}}_2+{\text{H}}_2$.

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18. Q78 Exemplar · Long Answer Questions
    For making cake, ​baking powder is taken. If at home your mother uses baking soda instead of baking powder ​in cake,

    1. (a)how will ​it affect the ​taste ​of the cake and why?
    2. (b)how can baking soda be converted into baking ​powder?
    3. (c)what is the role of tartaric acid added to baking ​soda?
    Show answer

    Baking soda leaves bitter sodium carbonate; adding tartaric acid makes baking powder, and the acid neutralises that carbonate.

    (a) The cake will taste bitter. On heating, baking soda decomposes to sodium carbonate, water and carbon dioxide, $2{\text{NaHCO}}_3\stackrel{\Delta }{\to }{\text{Na}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$; the carbon dioxide escapes and raises the cake, but the sodium carbonate left behind is strongly alkaline and bitter. (b) Baking soda is converted into baking powder by mixing a suitable amount of a mild edible acid, tartaric acid, with it. (c) The tartaric acid neutralises the sodium carbonate formed on heating, so no bitter residue remains and the cake does not taste bitter.

    Step-by-step solution →
19. Q79 Exemplar · Long Answer Questions
    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. Identity X, Y, G and Z.
    Show answer

    X is ${\text{CaCO}}_3$, Y is ${\text{Ca(OH)}}_2$, G is ${\text{Cl}}_2$ and Z is bleaching powder, ${\text{CaOCl}}_2$.

    X is calcium carbonate, ${\text{CaCO}}_3$. Y is calcium hydroxide, ${\text{Ca(OH)}}_2$ (slaked lime, used here as lime water). G is chlorine, ${\text{Cl}}_2$, liberated at the anode. Z is bleaching powder, calcium oxychloride, ${\text{CaOCl}}_2$. The reactions are ${\text{CaCO}}_3+2\text{HCl}\to {\text{CaCl}}_2+{\text{H}}_2\text{O}+{\text{CO}}_2$; ${\text{CO}}_2+{\text{Ca(OH)}}_2\to {\text{CaCO}}_3\downarrow +{\text{H}}_2\text{O}$; $2\text{NaCl}(aq)+2{\text{H}}_2\text{O}(l)\to 2\text{NaOH}(aq)+{\text{Cl}}_2(g)+{\text{H}}_2(g)$; and ${\text{Ca(OH)}}_2+{\text{Cl}}_2\to {\text{CaOCl}}_2+{\text{H}}_2\text{O}$.

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20. Q80 Exemplar · Long Answer Questions
    A ​dry pellet ​of a common base B, when kept ​in open absorbs moisture and turns sticky. The compound is also a ​by–product of ​chloralkali process. Identify B. What type ​of ​reaction occurs when B is treated with an acidic oxide? Write a balanced chemical ​equation for one such solution.
    Show answer

    B is sodium hydroxide, $\text{NaOH}$; its reaction with an acidic oxide such as ${\text{CO}}_2$ is neutralisation, giving a salt and water.

    B is sodium hydroxide, $\text{NaOH}$ — its dry pellets are deliquescent, absorbing moisture from the air and turning sticky, and it is obtained from the chlor-alkali process. With an acidic oxide it undergoes a neutralisation reaction, giving a salt and water. For example: $2\text{NaOH}(aq)+{\text{CO}}_2(g)\to {\text{Na}}_2{\text{CO}}_3(aq)+{\text{H}}_2\text{O}(l)$.

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