# NCERT Exemplar Problems and their Solutions

Reviewed By:
Krishna Kant Majee
M.Sc., B.Ed.

### Chapter 2-Acid Bases and Salts-Long Answers

Q.43. In the following schematic diagram for the preparation of hydrogen gas as shown in the figure, what would happen if the following changes are made?

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

(a) Since zinc dust has larger surface area than zinc granules, so reaction will be faster and hydrogen gas will evolve with greater speed.
(b) Both hydrochloric acid and sulphuric acid are strong acids so almost same amount of gas is evolved.
(c) Copper is a less reactive metal and does not react with dilute acids so there will be no reaction between copper turnings and dilute HCl.
(d) Reaction of zinc with sodium hydroxide forms sodium zincate $(Na_{2}ZnO_{2})$with evolution of hydrogen gas.
$Zn + 2NaOH \rightarrow Na_{2}ZnO_{2} + H_{2}\uparrow$

Q.44. For making cake, baking powder is taken. If at home your mother uses baking soda instead of baking powder in cake.
(a) How will it affect the taste of the cake and why?
(b) How can baking soda be converted into baking powder?
(c) What is the role of tartaric acid added to baking soda?

(a) Baking soda is sodium hydrogen carbonate which decomposes to sodium carbonate, water and carbon dioxide on heating. Baking powder is a mixture of sodium hydrogen carbonate with tartaric acid which readily reacts with sodium carbonate and neutralizes it. Therefore use of baking soda will give a bitter taste to cake due to presence of sodium carbonate.
(b) Baking powder is formed by addition of tartaric acid to baking soda.
(c) Presence of tartaric acid in baking powder neutralizes the effect of sodium carbonate formed during decomposition of baking soda.

Question. 45. 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.

Here X is calcium carbonate which is a metal carbonate which reacts with dilute HCl to form calcium carbonate $(CaCO_{3})$, water and carbon dioxide gas.
$CaCO_{3} + dil. 2HCl \rightarrow CaCl_{2} + H_{2}O + CO_{2}\uparrow$
When carbon dioxide gas is passed through lime water or calcium hydroxide (Ca(OH)2), it gives the calcium carbonate back. So Y is calcium hydroxide.
$Ca(OH)_{2} + CO_{2}\rightarrow CaCO_{3}\downarrow + H_{2}O$
Brine is saturated solution of sodium chloride (NaCl). Electrolysis of brine forms hydrogen gas at cathode and chlorine gas at anode.
$\ce{2NaCl + 2H2O->[{Electrolysis}] 2NaOH + Cl2(g)+ H2(g)}$
So G is chlorine gas which is passed through dry calcium hydroxide $(Ca(OH)_{2})$, produces bleaching powder, CaOCl2 (Z).
$Ca(OH)_{2} + Cl_{2}\rightarrow CaOCl_{2} +H_{2}O$
Hence, X= Calcium carbonate; Y= Lime water; G= Chlorine gas and Z =Calcium oxychloride or bleaching powder

Q.46. A dry pellet of a common base B, when kept in open absorbs moisture and turns sticky. The compound is also a product of chlor-alkali 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.

Here common base B is sodium hydroxide which is hygroscopic and absorbs moisture from the atmosphere. It is also formed as a product of chlor-alkali process. Chlor-alkali process is electrolysis of brine (saturated solution of sodium hydroxide) that forms aqueous sodium hydroxide with hydrogen gas and chlorine gas.
Reaction of sodium hydroxide with acidic oxide like carbon dioxide forms salt (sodium carbonate) and water.
$2NaOH + CO_{2}\rightarrow Na_{2}CO_{3} + H_{2}O$

Q.47. A sulphate salt of group-2 element of the periodic table is a white, soft substance, which can be molded into different shapes by making its dough. When this compound is left in open for some time, it becomes a solid mass and cannot be used for molding purposes. Identify the sulphate salt and why does it show such a behaviour? Give the reaction involved.

The sulphate salt which can be molded into different shapes by making its dough is calcium sulphate hemi-hydrate $(CaSO_{4}.1/2H_{2}O)$. It is commonly known as plaster of Paris. Here, two molecules of calcium sulphate share one molecule of water.
It readily reacts with atmospheric moisture and forms hard solid mass. This solid mass is known as gypsum $(CaSO_{4}.2H_{2}O)$. Conversion of plaster of Paris to gypsum can be represented as given below:
$CaSO_{4}.1/2H_{2}O + ½ H_{2}O \rightarrow CaSO_{4}.2H_{2}O$
$2NaOH + Zn \rightarrow Na_{2}ZnO_{2}+ H_{2}$
$NaOH + H\rightarrow Na_{2}ZnO_{2} + H_{2}$
$NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O$