NCERT Exemplar Solutions Class 10 Science Acids, Bases and Salts — Short and Long Answer Questions Q31–48

NCERT Exemplar Problems, Class 10 Science — Acids, Bases and Salts, Short Answer Questions 31 to 42 and Long Answer Questions 43 to 48. All 18 questions solved.

Where the multiple-choice band could be answered by recognising a fact, this band asks you to produce one — a formula, a balanced equation, or the identity of a substance named only by what it does. Six of the eighteen questions are of the form “identify X”, and they are worth approaching in a fixed way: collect every clue, decide which element is involved, and only then name the compound. Guessing the compound first and checking it against the clues afterwards is how a wrong answer survives to the end of the working.

Three families of chemicals carry almost the whole band, and they are the ones to have at your fingertips:

• The sodium family from common salt — baking soda $\mathrm{NaHCO_3}$, washing soda $\mathrm{Na_2CO_3 \cdot 10H_2O}$, sodium hydroxide $\mathrm{NaOH}$, all reachable from brine.
• The calcium family from limestone — quicklime $\mathrm{CaO}$, slaked lime $\mathrm{Ca(OH)_2}$, bleaching powder, gypsum $\mathrm{CaSO_4 \cdot 2H_2O}$ and plaster of Paris $\mathrm{CaSO_4 \cdot \frac{1}{2}H_2O}$.
• The everyday organic acids — lactic in curd, acetic in vinegar, citric in lemon, oxalic in tomato, methanoic in an ant sting.

Key insight. Nearly every “identify the compound” question here is solved by one clue that fixes the element, after which the rest of the clues only choose between two or three compounds of that element. In Question 47 it is “Group 2”; in Question 45 it is that Y and Z are both calcium salts; in Question 32 it is the $\mathrm{Ca}$ in every option. Find that clue first.

Short Answer Questions

Question 31

Match the acids given in Column (A) with their correct source given in Column (B)

Column (A)	Column (B)
(a) Lactic acid	(i) Tomato
(b) Acetic acid	(ii) Lemon
(c) Citric acid	(iii) Vinegar
(d) Oxalic acid	(iv) Curd

Solution. Two of these are settled by the name alone, and once they are placed the other two have nowhere else to go.

Acetic acid → vinegar. Vinegar is a dilute solution of acetic acid, about 5 to 8 per cent; the two words are almost synonyms in the laboratory.

Citric acid → lemon. Citrus fruit is where the name comes from.

That leaves lactic and oxalic to be matched with curd and tomato. Lactic acid → curd: milk is soured by bacteria converting its lactose into lactic acid, and lact- is the Latin root for milk in both words. So oxalic acid → tomato by elimination, and it is correct on its own account — tomatoes contain oxalic acid, as do spinach and rhubarb.

Question 32

Match the important chemicals given in Column (A) with the chemical formulae given in Column (B)

Column (A)	Column (B)
(a) Plaster of Paris	(i) $\mathrm{Ca(OH)_2}$
(b) Gypsum	(ii) $\mathrm{CaSO_4 \cdot \frac{1}{2}H_2O}$
(c) Bleaching Powder	(iii) $\mathrm{CaSO_4 \cdot 2H_2O}$
(d) Slaked Lime	(iv) $\mathrm{Ca(ClO)_2}$

Solution. Every formula in Column B is a calcium compound, so calcium decides nothing. Sort them by the other part instead, and they fall into three groups: two sulphates, one hydroxide, one hypochlorite.

The hydroxide and the hypochlorite are the easy pair. Slaked lime is $\mathrm{Ca(OH)_2}$ — it is what quicklime becomes when slaked with water. Bleaching powder is $\mathrm{Ca(ClO)_2}$ — it is the only formula containing chlorine, and bleaching is done by chlorine.

The two sulphates differ only in their water of crystallisation, and that is the whole question. Gypsum is $\mathrm{CaSO_4 \cdot 2H_2O}$, the fully hydrated mineral. Plaster of Paris is $\mathrm{CaSO_4 \cdot \frac{1}{2}H_2O}$, made by heating gypsum to about $373\ \mathrm{K}$ and driving off three-quarters of that water.

The half-molecule in the formula is not a misprint: two formula units of calcium sulphate share one molecule of water between them, which is what “hemihydrate” means. Question 47 is about what happens when it takes the rest of the water back.

Question 33

What will be the action of the following substances on litmus paper?

Dry HCl gas, Moistened $\mathrm{NH_3}$ gas, Lemon juice, Carbonated soft drink, Curd, Soap solution.

Solution. Litmus responds to ions in solution, not to a substance’s name, and the first two entries are in the list precisely to test that.

Dry HCl gas — no change. This is the one that surprises people. Hydrogen chloride is an acid only when it is in water; without moisture there is nothing to ionise it, so there are no free hydrogen ions and dry litmus paper stays exactly as it was. This is why the gas is dried through a calcium chloride guard tube in Question 3.

Moistened $\mathrm{NH_3}$ gas — turns red litmus blue. Moist ammonia dissolves in the water present to give ammonium hydroxide, which is basic. The word moistened is doing the same work here that the word dry did above, in the opposite direction.

The remaining four are ordinary solutions and behave as their taste suggests:

Lemon juice contains citric acid — blue to red. Carbonated soft drink contains dissolved carbon dioxide as carbonic acid — blue to red. Curd contains lactic acid — blue to red. Soap solution is basic, which is why it feels slippery — red to blue.

Question 34

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.

Solution. The acid injected by an ant — and by a bee, and by a stinging nettle — is methanoic acid, which is better known by its older name formic acid. The old name is the more memorable of the two, because it comes from formica, the Latin for ant.

Its formula is $\mathrm{HCOOH}$. It is the simplest of the carboxylic acids, with a single carbon atom, and it is a weak organic acid in the same family as the acetic acid of vinegar.

The pain and swelling are caused by the acid, so relief comes from neutralising it with a mild base. Rubbing the sting with moist baking soda, $\mathrm{NaHCO_3}$, is the standard remedy; calamine lotion, which contains zinc carbonate, works the same way.

Note that the base has to be a mild one, for the reason set out in Question 9 of the multiple-choice band: a strong base on already irritated skin would cause an alkali burn of its own.

Question 35

What happens when nitric acid is added to egg shell?

Solution. Egg shell is almost entirely calcium carbonate, $\mathrm{CaCO_3}$ — the same compound as marble and limestone. Any acid on a carbonate gives a salt, water and carbon dioxide, so the shell fizzes vigorously and dissolves:

$$\mathrm{CaCO_3}(s) + 2\mathrm{HNO_3}(aq) \rightarrow \mathrm{Ca(NO_3)_2}(aq) + \mathrm{H_2O}(l) + \mathrm{CO_2}(g)$$

The gas can be confirmed as carbon dioxide by passing it through lime water, which turns milky.

The reaction happens in two stages that are worth seeing. The acid first displaces carbonic acid from the carbonate; carbonic acid is unstable and immediately falls apart into water and carbon dioxide. That is why every carbonate–acid reaction produces a gas, and why the gas is always carbon dioxide regardless of which acid was used.

An egg left in vinegar overnight loses its shell entirely for the same reason, which makes this a demonstration you can set up at home.

Question 36

A student prepared solutions of (i) an acid and (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?

Solution. Litmus is only one indicator among many, and the laboratory has several ways of asking the same question.

Use a different indicator. Phenolphthalein is the obvious substitute: it is colourless in an acid and turns bright pink in a base. Methyl orange works too, red in acid and yellow in base. Natural indicators are just as good — turmeric turns red in a base and stays yellow in an acid, and an extract of red cabbage or china rose petals changes colour both ways.

Use an olfactory indicator. If nothing coloured is available, a few drops of vanilla essence or onion extract will do it: the smell survives in the acid and is destroyed in the base. This is the method of Question 17.

Use a reaction rather than an indicator. Adding a pinch of sodium hydrogencarbonate to each beaker is decisive — the acid effervesces briskly with carbon dioxide and the base does nothing at all. A small piece of zinc or magnesium behaves similarly, giving hydrogen with the acid.

What she must not do is taste them or touch them. The sour-and-bitter test is how the properties are described in the textbook, not a laboratory procedure.

Question 37

How would you distinguish between baking powder and washing soda by heating?

Solution. Both are white powders and both are sodium compounds, so appearance is no help. Heating separates them completely, because one gives off a gas that can be tested and the other does not.

Baking powder contains sodium hydrogencarbonate, and sodium hydrogencarbonate decomposes on heating to give carbon dioxide:

$$2\mathrm{NaHCO_3} \xrightarrow{\text{Heat}} \mathrm{Na_2CO_3} + \mathrm{H_2O} + \mathrm{CO_2}$$

Pass the gas through lime water and it turns milky, which identifies carbon dioxide.

Washing soda is $\mathrm{Na_2CO_3 \cdot 10H_2O}$, and it is already a carbonate — there is no hydrogencarbonate left to decompose. Heating only drives off its water of crystallisation, leaving anhydrous sodium carbonate:

$$\mathrm{Na_2CO_3 \cdot 10H_2O} \xrightarrow{\text{Heat}} \mathrm{Na_2CO_3} + 10\mathrm{H_2O}$$

No carbon dioxide is produced, so the lime water stays clear. What you see instead is droplets of water condensing on the cooler part of the tube and the crystals crumbling to a white powder.

So the test is: heat the sample and pass any gas through lime water. Milky means baking powder; clear means washing soda.

One note on wording. The printed key answers this question by saying “the chemical formula of baking powder is sodium hydrogencarbonate ($\mathrm{NaHCO_3}$)”. Strictly, $\mathrm{NaHCO_3}$ alone is baking soda; baking powder is baking soda together with tartaric acid, as the same key confirms at Question 12 and explains at Question 44. Since baking powder contains sodium hydrogencarbonate, the test works for either, and the answer is unaffected — but the two names are not interchangeable and a board answer should keep them apart.

Question 38

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.

Solution. Work backwards from the clue that is least ambiguous. Gas C turns lime water milky, and only one common gas does that, so $\mathrm{C} = \mathrm{CO_2}$.

A salt that gives carbon dioxide on heating — not on treatment with acid — must be a hydrogencarbonate, and the one used in bakery products is baking soda, sodium hydrogencarbonate. So $\mathrm{A} = \mathrm{NaHCO_3}$.

Heating it gives:

$$2\mathrm{NaHCO_3} \xrightarrow{\text{Heat}} \mathrm{Na_2CO_3} + \mathrm{H_2O} + \mathrm{CO_2}$$

so $\mathrm{B} = \mathrm{Na_2CO_3}$, sodium carbonate — and the third clue confirms it, because sodium carbonate is exactly what is used to remove the hardness of water. It works by precipitating the dissolved calcium and magnesium ions as insoluble carbonates, so the soap is no longer wasted on them.

Lime water turning milky is itself a reaction worth writing out, since it explains the word “milky”:

$$\mathrm{Ca(OH)_2} + \mathrm{CO_2} \rightarrow \mathrm{CaCO_3}\downarrow + \mathrm{H_2O}$$

The milkiness is a fine suspension of insoluble calcium carbonate.

The printed key names A as “baking powder ($\mathrm{NaHCO_3}$)”. The formula is right and the name is not: $\mathrm{NaHCO_3}$ is baking soda. The distinction matters here more than usual, because baking powder is the mixture used in cakes while baking soda is the single salt, and Question 44 turns entirely on the difference.

Question 39

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.

Solution. The industrial process for sodium hydroxide is the chlor-alkali process, the electrolysis of concentrated brine:

$$2\mathrm{NaCl}(aq) + 2\mathrm{H_2O}(l) \rightarrow 2\mathrm{NaOH}(aq) + \mathrm{Cl_2}(g) + \mathrm{H_2}(g)$$

It produces three useful substances at once — sodium hydroxide in the solution, hydrogen at the cathode and chlorine at the anode. Two of them are by-products, so which is X?

The second clue decides it: X reacts with lime water to give a bleaching agent. Bleaching is done by chlorine, and hydrogen forms nothing of the kind, so $\mathrm{X} = \mathrm{Cl_2}$.

Chlorine passed over slaked lime gives bleaching powder, which is $\mathrm{Y}$:

$$\mathrm{Ca(OH)_2} + \mathrm{Cl_2} \rightarrow \mathrm{CaOCl_2} + \mathrm{H_2O}$$

That is the form the textbook uses. The printed key writes the same reaction with the products separated into calcium hypochlorite and calcium chloride:

$$2\mathrm{Ca(OH)_2} + 2\mathrm{Cl_2} \rightarrow \mathrm{Ca(ClO)_2} + \mathrm{CaCl_2} + 2\mathrm{H_2O}$$

Both are balanced and both are accepted; commercial bleaching powder is in fact a mixture, and $\mathrm{CaOCl_2}$ is a convenient single formula for it. Either equation earns the mark, but the textbook’s version is the safer one to write in a board answer.

One practical point the question glosses over: bleaching powder is made by passing chlorine over dry slaked lime, not over lime water. Question 45 states this correctly.

Question 40

Fill in the missing data in the following table

Name of the salt	Formula	Base	Acid
(i) Ammonium chloride	$\mathrm{NH_4Cl}$	$\mathrm{NH_4OH}$	—
(ii) Copper sulphate	—	—	$\mathrm{H_2SO_4}$
(iii) Sodium chloride	$\mathrm{NaCl}$	$\mathrm{NaOH}$	—
(iv) Magnesium nitrate	$\mathrm{Mg(NO_3)_2}$	—	$\mathrm{HNO_3}$
(v) Potassium sulphate	$\mathrm{K_2SO_4}$	—	—
(vi) Calcium nitrate	$\mathrm{Ca(NO_3)_2}$	$\mathrm{Ca(OH)_2}$	—

Solution. Every salt splits into a metal (or ammonium) part, which came from the base, and an acid radical, which came from the acid. Once that is seen, each row fills itself.

The metal part plus hydroxide gives the base; the acid radical plus hydrogen gives the acid. So:

• Chloride ⟶ hydrochloric acid, $\mathrm{HCl}$ (rows i and iii)
• Sulphate ⟶ sulphuric acid, $\mathrm{H_2SO_4}$ (rows ii and v)
• Nitrate ⟶ nitric acid, $\mathrm{HNO_3}$ (rows iv and vi)

And on the base side, remember the valency. Copper(II) and magnesium are divalent, so their hydroxides carry two $\mathrm{-OH}$ groups; potassium is monovalent, so potassium hydroxide is $\mathrm{KOH}$.

Row (ii) needs its formula as well. Copper sulphate is $\mathrm{Cu^{2+}}$ with $\mathrm{SO_4^{2-}}$, and the charges cancel one-for-one, so the formula is simply $\mathrm{CuSO_4}$.

Question 41

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.

Solution. The distinction is about how completely the acid ionises in water, not about how concentrated it is — and confusing those two is the commonest error in this chapter.

A strong acid ionises completely in aqueous solution. Every molecule gives up its hydrogen ion, so the solution contains a high concentration of hydronium ions and a low pH:

$$\mathrm{HCl} + \mathrm{H_2O} \rightarrow \mathrm{H_3O^+} + \mathrm{Cl^-}$$

A weak acid ionises only partly. Most of it stays in solution as undissociated molecules, in equilibrium with a small proportion of ions, so a solution of the same molar concentration has far fewer hydronium ions and a higher pH:

$$\mathrm{CH_3COOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{CH_3COO^-}$$

The sorting then follows the mineral/organic split of Question 22 almost exactly. The three mineral acids are strong and the three organic acids are weak.

Strong and concentrated are different words. A dilute solution of hydrochloric acid is still a strong acid; a concentrated solution of acetic acid is still a weak one. Strength is a property of the substance and concentration is a property of the solution you happen to have made.

Question 42

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.

Solution. The clue that fixes the gas is what it is used for. Hydrogenation of oil is the process that converts liquid vegetable oils into solid fats such as vanaspati by adding hydrogen across their double bonds, in the presence of a nickel catalyst. So the gas is hydrogen.

That is consistent with the chemistry: zinc lies above hydrogen in the reactivity series, so it displaces hydrogen from a dilute acid.

$$\mathrm{Zn}(s) + 2\mathrm{HCl}(aq) \rightarrow \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(g)$$

or, with the other common strong acid,

$$\mathrm{Zn}(s) + \mathrm{H_2SO_4}(aq) \rightarrow \mathrm{ZnSO_4}(aq) + \mathrm{H_2}(g)$$

The word strong in the question is not decoration. Dilute nitric acid, although strong, is an oxidising acid and gives no hydrogen at all, as Question 42 of the previous unit sets out — so hydrochloric or sulphuric acid is the right choice here.

The test. Bring a burning splint to the mouth of the test tube. The hydrogen ignites with a squeaky pop sound, and that pop is the identification. Note the difference from oxygen, which is tested with a glowing splint that relights: hydrogen burns, oxygen supports burning.

Long Answer Questions

Question 43

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?

(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.

Solution. The apparatus makes hydrogen by the reaction $\mathrm{Zn} + \mathrm{H_2SO_4} \rightarrow \mathrm{ZnSO_4} + \mathrm{H_2}$, and each change alters one variable. Keep separate in your mind what changes the rate and what changes the amount — they are different questions and the exemplar tests both.

(a) Zinc dust instead of granules. The same mass of zinc is present, so the same number of moles reacts and the same total amount of hydrogen is produced. What changes is the surface area: a powder exposes far more zinc to the acid than a few granules do, so the reaction goes much faster and the gas comes off in a rush.

(b) Dilute hydrochloric acid instead of dilute sulphuric acid. Both are strong acids and both give hydrogen with zinc:

$$\mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl_2} + \mathrm{H_2}$$

Notice the stoichiometry: one mole of zinc gives one mole of hydrogen with either acid — $\mathrm{H_2SO_4}$ supplies two hydrogen atoms per molecule and $\mathrm{HCl}$ one, so twice as many molecules of the hydrochloric acid are used. Since the zinc is what limits the reaction, almost the same amount of gas is evolved, at a broadly similar rate. The salt formed is zinc chloride instead of zinc sulphate.

(c) Copper turnings instead of zinc. Copper lies below hydrogen in the reactivity series, so it cannot displace hydrogen from an acid. No hydrogen is evolved and no reaction occurs at all — no bubbles, no bubbles in the soap solution, nothing to burn.

(d) Sodium hydroxide instead of dilute sulphuric acid, with heating. This is the one that surprises people, because it replaces the acid with a base. Zinc is amphoteric — it reacts with alkalis as well as acids — so hydrogen is still evolved:

$$\mathrm{Zn}(s) + 2\mathrm{NaOH}(aq) \xrightarrow{\text{Heat}} \mathrm{Na_2ZnO_2}(aq) + \mathrm{H_2}(g)$$

The salt formed is sodium zincate. Heating is specified because the reaction with alkali is slow at room temperature.

Question 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?

Solution. The three parts are one argument, so it is worth setting out the chemistry once and then reading the answers off it.

Baking soda is sodium hydrogencarbonate. In the oven it decomposes:

$$2\mathrm{NaHCO_3} \xrightarrow{\text{Heat}} \mathrm{Na_2CO_3} + \mathrm{H_2O} + \mathrm{CO_2}$$

The carbon dioxide is the point of using it — the bubbles are what make the cake rise and give it a light, open texture. But the reaction leaves sodium carbonate behind in the cake, and sodium carbonate is bitter.

(a) So a cake made with baking soda alone rises properly but tastes bitter, because of the sodium carbonate left in it after the carbon dioxide has escaped.

(b) Baking soda is converted into baking powder by mixing in an appropriate amount of a mild edible acid — tartaric acid. Baking powder is nothing more than that mixture.

(c) The tartaric acid’s role is to neutralise the sodium carbonate as it forms, so that nothing bitter is left in the finished cake. It does not interfere with the rising; the carbon dioxide is still produced, and in fact the acid helps release it.

The design of baking powder is rather elegant, and worth noticing: the acid and the base are mixed dry and do nothing to each other in the packet, because the reaction needs water. They only meet properly once the batter is made, which is why baking powder has to be kept dry and why a cake should go into the oven soon after mixing.

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.

Solution. Start with G, because it is the only substance named by a process rather than a property. In the electrolysis of brine, chlorine is the anode product — the chloride ions lose electrons there — so $\mathrm{G} = \mathrm{Cl_2}$.

Chlorine passed over dry $\mathrm{Y}$ gives $\mathrm{Z}$, which disinfects drinking water. The compound used to disinfect drinking water is bleaching powder, and bleaching powder is made from chlorine and slaked lime. So $\mathrm{Y} = \mathrm{Ca(OH)_2}$ and $\mathrm{Z}$ is bleaching powder:

$$\mathrm{Ca(OH)_2} + \mathrm{Cl_2} \rightarrow \mathrm{CaOCl_2} + \mathrm{H_2O}$$

Now the first sentence falls into place, and the argument is nice. Since $\mathrm{Y}$ is a calcium compound, and the gas from $\mathrm{X}$ turns $\mathrm{Y}$ back into the carbonate $\mathrm{X}$, that carbonate must be a calcium carbonate too. So $\mathrm{X} = \mathrm{CaCO_3}$, and the two reactions are the familiar pair:

$$\mathrm{CaCO_3} + 2\mathrm{HCl} \rightarrow \mathrm{CaCl_2} + \mathrm{H_2O} + \mathrm{CO_2}$$

$$\mathrm{Ca(OH)_2} + \mathrm{CO_2} \rightarrow \mathrm{CaCO_3}\downarrow + \mathrm{H_2O}$$

The second of those is the lime-water test, and the question has used it in an unusual direction: normally it identifies the gas, but here the fact that it returns the original carbonate is what identifies the metal.

Note the word dry in the question. Bleaching powder is made by passing chlorine over dry slaked lime; with lime water the products would be different.

Question 46

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.

Solution. Two clues, and they agree.

The base is sold as pellets, absorbs moisture from the air and turns sticky — that is the everyday description of sodium hydroxide, which is hygroscopic to the point of deliquescence and has to be weighed quickly. And sodium hydroxide is indeed a product of the chlor-alkali process. So $\mathrm{B} = \mathrm{NaOH}$.

(Strictly, sodium hydroxide is the main product of the chlor-alkali process and hydrogen and chlorine are the by-products, as Question 39 has it. The question’s wording is loose, but the identification is not in doubt.)

An acidic oxide is a non-metal oxide such as $\mathrm{CO_2}$ or $\mathrm{SO_2}$. A base reacting with an acidic oxide gives a salt and water, which is a neutralisation reaction — the same reaction as base-plus-acid, with the oxide playing the acid’s part:

$$2\mathrm{NaOH} + \mathrm{CO_2} \rightarrow \mathrm{Na_2CO_3} + \mathrm{H_2O}$$

Sulphur dioxide behaves the same way, $2\mathrm{NaOH} + \mathrm{SO_2} \rightarrow \mathrm{Na_2SO_3} + \mathrm{H_2O}$.

Watch the coefficient. Sodium is monovalent and the carbonate ion carries two negative charges, so two formula units of sodium hydroxide are needed — writing $\mathrm{NaOH} + \mathrm{CO_2} \rightarrow \mathrm{Na_2CO_3} + \mathrm{H_2O}$ is the common slip and it is not balanced in sodium or hydrogen.

Question 47

A sulphate salt of Group 2 element of the Periodic Table is a white, soft substance, which can be moulded 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 moulding purposes. Identify the sulphate salt and why does it show such a behaviour? Give the reaction involved.

Solution. “Group 2” fixes the metal as an alkaline earth, and “moulded into different shapes by making its dough” fixes the compound: this is plaster of Paris, calcium sulphate hemihydrate, $\mathrm{CaSO_4 \cdot \frac{1}{2}H_2O}$.

The half-molecule of water in the formula is real and is the reason the substance behaves as it does. Two formula units of calcium sulphate share one molecule of water between them, which leaves the solid soft and workable — and leaves it thirsty, because the fully hydrated form takes two molecules of water per formula unit.

Left in the open, it takes that water from the moisture in the air and sets to gypsum, a hard solid mass that can no longer be moulded:

$$\mathrm{CaSO_4 \cdot \tfrac{1}{2}H_2O} + 1\tfrac{1}{2}\mathrm{H_2O} \rightarrow \mathrm{CaSO_4 \cdot 2H_2O}$$

$$\text{Plaster of Paris (soft)} \qquad\qquad \text{Gypsum (hard)}$$

This is exactly the reaction that makes the substance useful — it is why a plaster cast sets hard around a broken limb, and why plaster of Paris is used for making statues and mouldings — and it is also why it must be stored in a moisture-proof container. The same reaction that sets the cast will ruin the sack if the air gets to it.

Note that the process is reversible in principle: heating gypsum to about $373\ \mathrm{K}$ drives three-quarters of the water back off and returns plaster of Paris. Heating it higher gives anhydrous calcium sulphate, which will no longer set, so the temperature matters.

Question 48

Identify the compound X on the basis of the reactions given below. Also, write the name and chemical formulae of A, B and C.

Solution. Take the three reactions in the order that gives the most information, which is not the order printed.

Reactions 2 and 3 are the decisive ones. Compound X reacts with hydrochloric acid to give a salt and water, and with acetic acid to give a salt and water. Producing a salt and water with an acid is the definition of a base. So X is a base.

Reaction 1 narrows it further. X reacts with zinc to give hydrogen. Only a few bases do that — it is the amphoteric behaviour of zinc, met in Question 43(d) — and it tells us X is a strong alkali. The one the syllabus uses throughout is sodium hydroxide, so $\mathrm{X} = \mathrm{NaOH}$.

Now write the three reactions out and read off A, B and C.

$$\mathrm{Zn} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2ZnO_2} + \mathrm{H_2}$$

so A is sodium zincate, $\mathrm{Na_2ZnO_2}$.

$$\mathrm{NaOH} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}$$

so B is sodium chloride, $\mathrm{NaCl}$ — ordinary common salt.

$$\mathrm{NaOH} + \mathrm{CH_3COOH} \rightarrow \mathrm{CH_3COONa} + \mathrm{H_2O}$$

so C is sodium acetate, $\mathrm{CH_3COONa}$.

The pattern worth extracting is that all three products keep the sodium from X and take on whatever the other reactant brought — the zincate, the chloride, the acetate. Once X is identified, the three products write themselves.

Common mistakes

• Using the names baking soda and baking powder interchangeably. Baking soda is $\mathrm{NaHCO_3}$; baking powder is that plus tartaric acid. Questions 37, 38 and 44 all turn on the difference, and the printed key itself slips on it twice.
• Saying dry HCl gas turns litmus red, in Question 33. No water means no ionisation, no hydrogen ions and no colour change. The words dry and moistened in that list are the question, not scene-setting.
• Confusing strong with concentrated, in Question 41. Strength is complete versus partial ionisation and is a property of the acid; concentration is how much acid is in the solution. Dilute hydrochloric acid is still a strong acid.
• Forgetting the coefficient 2 in Question 46. $2\mathrm{NaOH} + \mathrm{CO_2} \rightarrow \mathrm{Na_2CO_3} + \mathrm{H_2O}$ — the carbonate ion is doubly charged, so two sodiums are needed.
• Expecting no reaction with sodium hydroxide in Question 43(d). Zinc is amphoteric and gives hydrogen with alkalis as well as acids. Assuming only acids liberate hydrogen from a metal is the single most common error in that question.
• Confusing rate with amount in Question 43(a) and (b). Zinc dust changes how fast the hydrogen comes off, not how much; changing the acid changes neither, because the zinc is what runs out.
• Getting the half-water wrong in Question 32 and Question 47. Plaster of Paris is $\mathrm{CaSO_4 \cdot \frac{1}{2}H_2O}$ and gypsum is $\mathrm{CaSO_4 \cdot 2H_2O}$. Swapping them reverses the whole answer to Question 47, since the soft one is the hemihydrate.
• Naming a compound before collecting every clue in the “identify X” questions. In Question 45 the identity of X follows only from noticing that Y and Z are both calcium compounds; in Question 47 it follows from “Group 2”. Guessing first and checking afterwards is how a plausible wrong answer survives.

Practise next

• Exemplar Multiple Choice Questions — thirty questions on the same chemicals in recognition form, and the fastest way to fix the formulae that Questions 32 and 40 above depend on.
• NCERT textbook Chapter 2 — Acids, Bases and Salts for the end-of-chapter exercise, where the family of salts and the chlor-alkali process are set out in full.
• Textbook in-text questions for the short section checks on pH, plaster of Paris and bleaching powder — the background to Questions 39, 45 and 47.
• Textbook activities for the hydrogen preparation of Question 43 and the indicator tests of Question 36, both of which are performed there rather than described.