Selina ICSE Class 10 Chemistry Solutions Chapter 3 Acids Bases and Salts

Intext – Question- 1

Question 1: What do you understand by the terms acid and base?
Solution 1:
(a) Acids are defined as substances that contain one or more hydrogen atoms and form hydronium ions (H₃O⁺), the only positively charged ions, when dissolved in water.
(b) Hydroniumion
(c) H₃O⁺

Question 2: Explain the formation of hydronium ion. Write the ionization of sulphuric acid showing the formation of hydronium ion.
Solution 2:
Hydronium ions are created when H⁺ (from an acid) reacts with water. When hydronium ion comes into contact with water, it formsH₃O⁺ (Hydronium ion).
H⁺ + H₂O → H₃O⁺
The production of the hydronium ion during the ionisation of sulphuric acid:

Question 3: Water is never added to acid in order to dilute it why?
Solution 3:
When water is added to a concentrated acid, the heat produced causes the mixture to splash out, resulting in severe burns. As a result, water is never added to acid to dilute it.

Question 4: Define the term ‘basicity’ of an acid. Give the basicity of: nitric acid, sulphuric acid and phosphoric acid.
Solution 4:
The number of hydronium ions (H₃O₊) that can be created by ionizing one molecule of an acid in aqueous solution is known as the basicity of the acid.
Basicity is defined as the number of replaceable or ionizable hydrogen atoms contained in acid molecule.
The basicity of the substances listed below is:
Nitric acid: Basicity = 1
Sulphuric acid: Basicity = 2
Phosphoric acid: Basicity = 3

Question 5: Give two examples of each of the following:
(a) oxy-acid
(b) hydracids
(c) monobasic acid
(d) dibasicacid
(e) tribasicacid
Solution 5:
(a) Examples of Oxyacids: – HNO₃, H₂SO₄
(b) Examples of Hydracid:- HCl, HBr
(c) Examples of Monobasic acid:- HCl, HBr
(d) Examples of Dibasic acid: – H₂SO₄, H₂CO₃
(e) Examples of Tribasic acid:- H₃PO₄, H₃PO₃

Question 6: Name the:
(a) acidic anhydride of the following acids:
(i) sulphurous acid
(ii) nitric acid
(iii) phosphoric acid
(iv) carbonic acid
(b) acids present in vinegar, grapes and lemon
(c) (i) ion that turns blue litmus red,
(ii) ion that turns red litmus blue.
Solution 6:
(a) The anhydride of the acids listed below is:
(i) Sulphurous acid: SO₂
(ii) Nitric acid : N₂O₅
(iii) Phosphoric acid : P2O₅
(iv) Carbonic acid : CO₂
(b) Acids found in the following foods:
Vinegar: Acetic acid
Grapes: Tartaric acid and malice acid Lemon: Citric acid
(c) (i) H+ ion turns blue litmus red.
(ii) OH- ion turns red litmus blue.

Question 7: What do you understand by the statement ‘acetic acid is a monobasic acid?
Solution 7:
Acetic acid is a monobasic acid that produces one hydronium ion per molecule when ionized in water.

Question 8: (a) Give a balanced equation for reaction of nitrogen dioxide with water.
(b) How many types of salts does dibasic acid produce when it reacts with caustic soda solution? Give equation(s).
Solution 8:
(a) 2NO₂ + H₂O ⟶ HNO₂ + HNO₃
(b) When dibasic acid combines with caustic soda, two types of salts are formed. One is acidic salt, whereas the other is ordinary salt.
Acid salts:
H₂SO₄ + NaOH ⟶ NaHSO₄ + H₂O
Normal salts:
H₂SO₄ + 2NaOH ⟶ Na₂SO₄ + 2H₂O

Question 9: Carbonic acid gives an acid salt but hydrochloric acid does not. Explain.
Solution 9:
The extent to which an acid ionizes or dissociates in water determines its strength.
The degree of ionization and concentration of hydronium ions [H₃O⁺] produced by an acid in aqueous solution determines its strength.

Question 10: What do you understand by the strength of an acid? On which factor does the strength of an acid depend?
Solution 10:
(a) Carbonic acid is a dibasic acid containing two hydrogen ions that can be replaced, resulting in one acid salt or one normal salt.
Because hydrogen chloride is a monobasic acid with only one replaceable hydrogen ion, it only generates one normal salt.
(b) An acid’s strength is defined as the amount of hydronium ions it creates in its aqueous solution. In comparison to concentrated acetic acid, dilute HCl creates a large concentration of hydronium ion. As a result, dilute HCl is a more powerful acid than highly concentrated acetic acid.
(c) H₃PO₃ is not a tribasic acid because hydrogen atoms coupled to oxygen atoms are replaceable in phosphorus oxy-acids. Directly bound hydrogen atoms to phosphorus atoms cannot be replaced.

(c) Because the salt formed is insoluble in the solution, the reaction fails. As a result, we don’t anticipate lead carbonate reacting with hydrochloric acid.
(d) NO₂ is known as double acid anhydride because it produces two acids when it combines with water: nitrous acid and nitric acid.
2NO₂ + H₂O → HNO₂ + HNO₃

Question 11: Dilute HCl acid is stronger than highly concentrated acetic acid. Explain.
Solution 11:
Acid rain is a result of a number of human activities that emit sulphur and nitrogen oxides into the atmosphere. Sulphur is produced when fossil fuels, such as coal, oil, gasoline, and diesel, are burned.
Pollutants in the air include carbon dioxide and nitrogen oxide. Polluted air also contains a high concentration of oxidizing chemicals, which create oxygen as a result of high temperatures. This oxygen reacts with sulphur and nitrogen oxides, as well as rainwater, to generate acids.
2SO₂ + O₂ + 2H₂O → 2H₂SO₄
4NO₂ + O₂ + 2H₂O → 4HNO₃

Question 12: How is an acid prepared from a
(a) Non-metal
(b) Salt?
Give an equation for each.
Solution 12:
Acids are made from non-metals that have been oxidized. Sulphur or phosphorus, for example, is oxidized by concentrated nitric acid to produce sulphuric acid or phosphoric acid.

Displacement reaction is used to make acids from salt. H₂SO₄ and sodium chloride, for example, are used to make nitric acid.

Question 13: Name an acid used:
(a) to flavor and preserve food,
(b) in a drink,
(c) to remove inkspots,
(d) as aneyewash
Solution 13:

Question 14: Give equations to show how the following are made from their corresponding anhydrides.
(a) sulphurous acid
(b) phosphoricacid,
(c) carbonicacid
(d) sulphuricacid
Solution 14:
(a) Citricacid
(b) Carbonicacid
(c) Oxalicacid
(d) Boricacid

Intext – Question – 2

Question 1: What do you understand by an alkali? Give two examples of:
(a) Strong alkalis
(b) Weak alkalis
Solution 1:
An alkali is a basic hydroxide that produces hydroxyl ions (OH-) as the only negatively charged ions when dissolved in water.
(a) Strong alkalis: Sodium hydroxide , Potassium hydroxide
(b) Weak alkalis: Calcium hydroxide , Ammonium hydroxide

Question 2: What is the difference between:
(a) an alkali and a base,
(b) an alkali and a metal hydroxide?
Solution 2:
(a) A base and an alkali:
1. Alkalis are water soluble, whereas bases are either water soluble or not.
2. All bases are alkalis, but not all alkalis are bases.
(b) A metal hydroxide and an alkali:
1. Alkalis are water soluble, whereas metal hydroxides may or may not be water soluble.

Question 3: Define in terms of ionization:
(a) an acid
(b) an alkali
Solution 3:
(a) An acid: Acids are chemicals that create hydronium ions when dissolved in water.
(b) An alkali: Alkalis are chemicals that create hydroxyl ions when dissolved in water.

Question 4: Name the ions furnished by:
(a) bases in solution,
(b) a weak alkali
(c) an acid
Solution 4:
(a) In solution, bases produce the hydroxide ion.
(b) Hydroxide ions are produced by weak alkali.
(c) A hydronium ion is produced by an acid.

Question 5: Give one example in each case:
(a) A basic oxide which is soluble in water,
(b) A hydroxide which is highly soluble in water,
(c) A basic oxide which is insoluble in water,
(d) a hydroxide which is insoluble in water,
(e) A weak mineral acid,
(f) a base which is not an alkali
(g) An oxide which is a base,
(h) A hydrogen containing compound which is not an acid,
(i) A base which does not contain a metal ion.
Solution 5:
(a) Example of basic oxide which is soluble in water Barium oxide.
(b) Example of hydroxide which is highly soluble in water Sodium hydroxide.
(c) Example of basic oxide which is insoluble in water Manganese oxide.
(d) Example of hydroxide which is insoluble in water Copper hydroxide.
(e) Example of weak mineral acid Carbonic acid.
(f) Example of base which is not an alkali Ferrichydr oxide.
(g) Example of oxide which is a base Copper oxide.
(h) Example of hydrogen containing compound which is not an acid Ammonia.
(i) Example of base which does not contain a metal ion Ammonium hydroxide.

Question 6: You have been provided with three test tubes. One of them contains distilled water and the other two have an acidic solution and a basic solution respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?
Solution 6:
The red litmus paper is unaffected by the test tube containing distilled water. The red litmus paper is unaffected by the acidic solution in the test tube. The test tube holding the basic solution, on the other hand, turns red litmus paper blue.

Question 7: HCl, HNO₃, C₂H₅OH, C₆H₁₂O₆ all contain H atoms but only HCI and HNO₃ show acidic character. Why?
Solution 7:
HCl, HNO₃, C₂H₅OH, C₆H₁₂O₆ all contain H atoms but only HCI and HNO₃ show acidic character this is due to the fact that HCl and HNO₃ ionize in aqueous solution, whereas ethanol and glucose do not.

Question 8: Dry HCI gas does not change the colour of dry litmus paper. Why?
Solution 8:
Dry HCI gas does not change the colour of dry litmus paper because HCl only ionizes in aqueous solution, this is the case.

Question 9: Is PbO₂ a base or not? Comment.
Solution 9:
Lead oxide is not a base because it produces chlorine, salt, and water when it combines with acid. As a result, it is not included in the class bases.

Question 10: (a) What effect does the concentration of [H₃O⁺] ion have on solution?
(b) Do basic solutions also have H+ (aq)? Why are they basic?
Solution 10:
(a) As the concentration of [H₃O⁺] in solution rises, the pH falls. As a result, the solution’s acidity increases.
(b) Yes, H + (aq) ions can be found in basic solutions. The concentration of H + (aq) ions in basic solutions is smaller than that of OH (aq) ions.

Question 11: How would you obtain:
(a) a base from other base
(b) an alkali from a base,
(c) salt from another salt?
Solution 11:
(a) Double decomposition can be used to create a base from another base. The appropriate metallic hydroxide is precipitated from an aqueous solution of salts with base.

Question 12: Write balanced equations to satisfy each statement:
(a) Acid + Active metal ⟶ Salt + hydrogen
(b) Acid + Base ⟶ Salt + water
(c) Acid + carbonate Or bicarbonate ⟶ Salt + water + carbon dioxide
(d) Acid + sulphite Or bisulphate ⟶ Salt + water + sulphur dioxide
(e) Acid + sulphide ⟶ Salt + Hydrogen sulphide
Solution 12:
(a) Mg + 2HCl → MgCl₂+ H₂
(b) HCl + NaOH → NaCl + H₂O
(c) CaCO₃+ 2HCl → CaCl₂ + H₂O + CO₂
(d) CaSO₃ + 2HCl → CaCl₂ + H₂O + SO₂
(e) ZnS +2HCl → ZnCl₂ + H₂S

Question 13: The skin has and needs natural oils. Why is it advisable to wear gloves while working with strong allkalis?
Solution 13:
Alkalis react with oil to create soap, as we all know. Because our skin is oily, a reaction occurs when we come into contact with powerful alkalis, resulting in a soapy solution. As a result, gloves must be worn.

Question 14: Why are alkalis soapy to touch? What do you understand by PH value?
Solution 14:
Alkalis have a soapy feel to them because they react with our skin’s oils to generate soap.
The negative logarithm to the base 10 of hydrogen ion concentration given in moles per litre is the pH of a solution.

Question 15: Complete the table:

Indicator	Neutral	Acidic	Alkaline
Litmus Phenolphthalein	Purple Colourless	————— — ————— —	—————– – —————– –

Solution 15:

Indicator	Neutral	Acidic	Alkaline
Litmus Phenolphthalein	Purple Colourless	Blue to red Colourless	Red to blue Pink

Question 16: Two solutions X and Y have pH values of 4 and 10 respectively. Which one of these two will give a pink colour with phenolphthalein indicator?
Solution 16:
When two solutions X and Y have pH values of 4 and 10 respectively, and with phenolphthalein indicator solution Y, produces pink colour and solution X will be colourless.

Question 17: You are supplied with five solutions: A, B, C, D and E with Ph values as follows: A= 1.8, B = 7, C= 8.5, D = 13, and E=5
Classify these solutions as neutral, slightly or strongly acidic and slightly or strongly alkaline. Which solution would be most likely to liberate hydrogen with:
(a) Magnesium powder,
(b) powdered zinc metal. Give a word equation for each reaction.
Solution 17:
Solution A will be strongly acidic.
Solution B will be neutral.
Solution C will be weakly alkaline.
Solution D will be strongly alkaline.
Solution E will be weakly acidic.

Question 18: (a) What are the acidic range and the alkaline range in the pH scale?
(b) State one advantage of using ‘pH paper’ for measuring the pH value of an unknown solution.
Solution 18:
(a) The pH scale ranges from 0 to 14.
If pH is 7, solution is neutral.
If pH is less than 7, solution is acidic.
If pH is more than 7, solution is basic.
(b) One advantage of utilizing pH paper to determine the pH of an unknown solution is that we can determine whether the solution is acidic, basic or neutral without wasting the solution.

Question 19: Distinguish between:
(a) a common acid base indicator and a universal indicator,
(b) acidity of bases and basicity of acids,
(c) acid and alkali (other than indicators).
Solution 19:
(a) A universal acid-base indicator:
Litmus, for example, is an acid-base indicator that merely tells us if a chemical is acid or basic. The universal indication indicates whether a substance is acidic or basic. With solutions with various pH values, a universal indicator produces various colours.
(b) Bases’ acidity and acids’ basicity: Bases’ acidity In an aqueous solution, the number of hydroxyl ions that can be created per molecule of base.
Acid’s baseness: The amount of hydronium ions created by ionizing one molecule of an acid in aqueous solution is referred to as the basicity of the acid.
(c) Acids and alkalis: An acid is a chemical that, when dissolved in water, produces H+ ions. When dissolved in water, an alkali produces OH-ions.

Question 20: How does tooth enamel get damaged? What should be done to prevent it?
Solution 20:
Bacteria present in our mouth breakdown substances like chocolate and sweets. Tooth decay begins when the pH falls below 5.5. Enamel on our teeth is the toughest substance in our bodies, and it corrodes. Saliva produced by salivary glands is somewhat alkaline, which helps to raise the pH to a degree, but tooth paste containing a basic ingredient is used to neutralize excess acid in the mouth.

Intext – Question – 3

Question 1: Define an acidic salt, a normal salt and a mixed salt. Give two examples in each case of: (a) a normal salt, (b) an acid salt, (c) a mixed salt.
Solution 1:
Acid salts are generated when the ionizable hydrogen atoms of a polybasic acid are partially replaced by a metal or an ammonium ion.
Normal salts are salts created when the ionizable hydrogen atoms of an acid are completely replaced by a metallic or ammonium ion.
Mixed Salts that include more than one basic or acid radical are known as mixed salts.
Examples:
(a) A Normal salt : Na₂SO₄ , NaCl
(b) An acid salt : NaHSO₄ , Na₂HPO₄
(c) A mixed salt : NaKCO₃ , CaOCl₂

Question 2: Answer the following questions related to salts and their preparations:
(a) What is a ‘salt’?
(b) What kind of salt prepared by direct combination. Write an equation for the reaction that takes place in preparing the salt you have named.
(c) Name a salt prepared by direct combination. Write an equation for the equation for the reaction that takes place in preparing the salt you have named.
(d) Name the procedure used to prepare a sodium salt such as sodium sulphate.
Solution 2:
(a) (a) Salt is a chemical generated when the ionizable hydrogen atoms of an acid are partially or completely replaced by a metallic ion or an ammonium ion.
(b) Precipitation can be used to make an insoluble salt.
(c) Iron (III) chloride is a salt made through direct combination.
(d) Reaction:
2Fe + 3Cl₂ → 2FeCl₃The name of the procedure used to prepare a sodium salt such as sodium sulphate is Neutralization of acid with base

Question 3: How are the following salts prepared:
(a) Calcium sulphate from calcium carbonate,
(b) Lead (II) oxide from lead,
(c) Lead carbonate from lead nitrate,
(d) Sodium nitrate from sodium hydroxide,
(e) Magnesium carbonate from magnesium chloride,
(f) Copper (II) sulphate from copper (II) oxide?
Solution 3:
(a) Calcium sulphate from calcium carbonate: Calcium carbonates are decomposed by acids, resulting in calcium sulphate.
(b) Lead (II) oxide from lead: Lead oxide can be made by combining lead and oxygen.
(c) Lead carbonate from lead nitrate: Lead carbonate is made by precipitating lead nitrate with sodium carbonate (double decomposition).
(d) Sodium nitrate from sodium hydroxide: Sodium nitrate is made by neutralizing sodium hydroxide with nitric acid.
(e) Magnesium carbonate from Magnesium chloride: Magnesium carbonate can be made from Magnesium chloride by decomposing it twice with Sodium carbonate.
(f) Copper (II) sulphate from copper oxide: Copper sulphate is made by reacting copper oxide with sulphuric acid.

Question 4: (a) How is lead sulphate prepared in the laboratory?
(b) Why lead sulphate cannot be prepared by the action of dilute H₂SO₄ on leadoxide?
Solution 4:
(a) Insoluble lead oxide is converted to soluble lead nitrate with dilute nitric acid, and the resultant solution is treated with sulphuric acid to get white ppt. of lead sulphate.
PbO + 2HNO₃ ⟶ Pb(NO₃)₂ + H₂O
Pb(NO₃)₂ + H₂SO₄ ⟶ PbSO₄ + 2HNO₃
(a) Because lead sulphate is insoluble, it interacts only on the surface when mixed with sulfuric acid. The lead is covered with insoluble lead sulphate, which prevents the inner lead from reacting. As a result, dilute sulfuric acid cannot be used to make lead sulphate.

Question 5: Describe giving all practical details, how would you prepare:
(a) Copper sulphate crystals from a mixture of charcoal and black copper oxide,
(b) zinc sulphate crystals from zinc dust (powdered zinc and zinc oxide),
(c) lead sulphate from metallic lead,
(d) sodium hydrogen carbonate crystals.
Solution 5:
(a) Crystals of copper sulphate made from a mix of charcoal and black copper oxide:
Because of the carbon in the charcoal, the black copper oxide is reduced to reddish-brown copper. If the lid is removed before the crucible has cooled, the hot copper will be re-oxidized by air.
In a beaker, boil dilute sulfuric acid over wire gauze. Add little amounts of cupric oxide at a time, stirring constantly until no more dissolves and the surplus compound sinks to the bottom.
Filter it while it’s still hot, then collect the filtrate in a China dish. Allow the filtrate to cool and collect the copper sulphate pentahydrate crystals after evaporating it to the point of crystallization.
Reaction: CuO + H₂SO₄⟶ CuSO₄ + H₂O CuSO₄ + 5H₂O ⟶ CuSO₄. 5H₂O
(b) Zinc sulphate crystals from Zinc dust:
In a beaker, boil dilute sulfuric acid over wire gauze. With continual swirling, add several granular zinc bits. Add until the Zinc has settled to the bottom of the beaker. Effervescence is caused by the release of hydrogen gas. When effervescence ceases, it means that all of the acid has been consumed. The surplus zinc is removed through filtration. To make crystals, collect the solution in a china dish and evaporate it. Filter them, wash them in water, and dry them between paper folds. The white needle crystals are Zinc sulphate hydrate.
Reaction:
Zn + H₂SO₄ ⟶ ZnSO₄ + H₂
ZnSO₄ + 7 H₂O ⟶ ZnSO₄. 7H₂O
(c) Lead sulphate from metallic lead:
Oxidation converts metallic lead to lead oxide. The insoluble lead oxide is then converted into soluble lead nitrate, which is then used to make lead sulphate. The lead nitrate solution is then processed with sulphuric acid to produce white lead sulphate ppt.
Reaction:
PbO + 2HNO₃ ⟶ Pb(NO₃)₂ + H₂O
Pb(NO₃)₂ + H₂SO₄⟶ PbSO₄ + 2HNO₃
(d) Sodium hydrogen carbonate crystals:
In a flask, dissolve 5 grammes of anhydrous sodium carbonate in 25 mL distilled water. Keep the flask in a freezing combination to cool the solution. Fill the solution with carbon dioxide gas. After some time, sodium bicarbonate crystals will appear. Filter the crystals and dry them in filter paper folds.
Reaction:
Na₂CO₃ + CO₂ + H₂O ⟶ 2NaHCO₃

Question 6: The following is a list of methods for the preparation of salts.
A – direct combination of two elements.
B – reaction of a dilute acid with a metal.
C – reaction of a dilute acid with an insoluble base.
D – titration of a dilute acid with a solution of soluble base.
E – reaction of two solutions of salts to form a precipitate.
Choose from the above list A to E, the best method of preparing the following salts by giving a suitable equation in each case:
1. Anhydrous ferric chloride,
2. Lead chloride,
3. Sodium sulphate.
4. Copper sulphate.
Solution 6:
1. Anhydrous ferric chloride: A (Direct combination of two elements)
2Fe + 3Cl₂ ⟶ 2FeCl₃
2. Lead chloride: E (Reaction of two solutions of salts to form aprecipitate)
Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ + 2HNO₃
3. Sodium sulphate: D (Titration of dilute acid with a solution of soluble base)
2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O
4. Copper sulphate: C (reaction of dilute acid with an insoluble base) Cu(OH)₂ + H₂SO₄ ⟶CuSO₄ + 2H₂O

Question 7: Name:
(a) a chloride which is insoluble in cold water but dissolves in hot water,
(b) a chloride which is in soluble,
(c) two sulphates which are in soluble,
(d) a basic salt,
(e) an acidic salt,
(f) a mixed salt,
(g) a complex salt,
(h) a double salt,
(i) two salts whose solubility increases with temperature,
(j) a salt whose solubility decreases with temperature.
Solution 7:
(a) Lead chloride is insoluble in cold water but dissolves in hot water.
(b) Silver chloride is in soluble.
(c) Barium sulphate and lead sulphate are two sulphates which are in soluble.
(d) Basic lead chloride is a basic salt.
(e) Sodium hydrogen sulphate is an acidic salt.
(f) Sodium potassium carbonate is a mixed salt.
(g) Sodium argentocyanide is a complex salt.
(h) Potash alum is a double salt.
(i) Potassium bromide and potassium chloride are two salts whose solubility increases with temperature.
(k) Calcium sulphate is a salt whose solubility decreases with temperature.

Question 8: Explain ‘salt hydrolysis’ name two salts which are:
(a) acidic
(b) basic
(c) neutral, when dissolved in water.
Solution 8:
Salt hydrolysis is the process by which a salt generated by a weak acid and a strong base, or a strong acid and a weak base, combines with water to produce an acidic or alkaline solution.
(a) Acidic : Iron chloride, Copper sulphate
(b) Basic: Sodium carbonate, potassium acetate
(c) Neutral: Sodium chloride, sodium sulphate

Question 9: What would you observe when:
(a) blue litmus is introduced into a solution of ferric chloride,
(b) red litmus paper is introduced into a solution of sodium sulphate,
(c) red litmus paper is introduced in sodium carbonate solution?
Solution 9:
(a) Blue litmus will turn red, indicating that the fluid of ferric chloride is acidic in nature.
(b) There will be no change, if red litmus paper is introduced into a solution of sodium sulphate.
(c) When red litmus is introduced in sodium carbonate solution it will turn blue. It means the solution is basic in nature.

Question 10: Write the balanced equations for the preparation of the following salts in the laboratory:
(a) A soluble sulphate by the action of an acid on an in soluble base,
(b) An insoluble salt by the action of an acid on another salt,
(c) An insoluble base by the action of a soluble base on a soluble salt,
(d) A soluble sulphate by the action of an acid on a metal.
Solution 10:
(a) MgCO₃ + H₂SO₄ ⟶ MgSO₄ + H₂O +CO₂
(b) Pb(NO₃)₂ + H₂SO₄ ⟶ PbSO₄ + 2HNO₃
(c) Pb(NO₃)₂ + Na₂CO₃ ⟶ PbCO₃ + 2NaNO₃
(d) Zn + H₂SO₄ ⟶ ZnSO₄ + H₂

Question 11: Give the preparation of the salt shown in the left column by matching with the methods given in the right column. Write a balanced equation for each preparation.
Salt Method of Preparation
Zinc sulphate Precipitation
Ferrous sulphide Oxidation
Barium Sulphate Displacement
Ferric Sulphate Neutralisation
Sodium sulphate Synthesis
Solution 11:
Zinc Sulphate – Displacement
Zn(OH)₂ + H₂SO₄ ⟶ ZnSO₄ + 2H₂O
Ferrous sulphide – synthesis
Fe + S ⟶ FeS
Barium sulphate – Precipitation
BaCI₂ + H₂SO₄ ⟶ BaSO₄ + 2HCI
Ferric sulphate – Oxidation
Fe + H₂SO₄ ⟶ FeSO₄ + H₂
Sodium sulphate – Neutralisation
2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

Question 12: You are provided with the following chemicals:
NaOH, Na₂CO₃, H₂O, Zn(OH)₂, CO₂, HCI, Fe, H₂SO₄, CI₂, Zn.
Using the suitable chemicals from the given list only, state briefly how you would prepare:
(a) iron (III) chloride,
(b) sodium sulphate,
(c) sodium zincate
(d) iron (II) sulphate,
(e) sodium chloride?
Solution 12:
(a) Iron (III) Chloride: Iron chloride is created by combining elements in a direct manner.
2Fe + 3Cl₂ ⟶ 2FeCI₃
(b) Sodium sulphate: Caustic soda is neutralised with weak sulphuric acid to produce sodium sulphate.
2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O
(c) Sodium zincate: Metals react with alkalis to form sodium zincate.
Zn + 2NaOH ⟶Na₂ZnO₂+ H₂
(d) Iron (II) sulphate: Iron sulphate is made when a dilute acid reacts with an active metal.
Fe + H₂SO₄ ⟶ FeSO₄ + H₂
(e) Sodium chloride: The neutralising reaction of a strong acid with a strong base produces sodium chloride.
NaOH + HCI ⟶ NaCI + H₂O

Question 13: Define the term neutralization:
(a) Give a reaction, mentioning clearly acid and base used in the reaction.
(b) If one mole of a strong acid reacts with one mole of a strong base, the heat produced is always the same. Why?
Solution 13:
The process of neutralisation is when the H+ ions of an acid react entirely with the [OH]^– ions of a base to produce just salt and water.
(a) NaOH + HCl ⟶ NaCI + H₂O
(b) Neutralization is merely the reaction of H⁺ ions produced by a strong acid and OH^– ions produced by a strong base. The quantity of H⁺ and OH^– ions produced by one mole of a strong acid or strong base is always the same for all strong acids and strong bases. As a result, the heat required to neutralise a strong acid with a strong base is always the same.

Question 14: Explain why:
(a) It is necessary to find out the ration of reactants required in the preparation of sodium sulphate.
(b) Fused calcium chloride is used in the preparation of FeCI₃?
Solution 14:
(a) Because sodium hydroxide and sulphuric acid are both soluble, filtering will not remove an excess of either. As a result, the ratio of solutions of the two reactants must be determined on a tiny scale.
(a) Because iron chloride is highly deliquescent, fused calcium chloride is used to keep it dry.

Question 15: Give the pH value of pure water. Does it change if common salt it added to it?
Solution 15:
At 25^oC, the pH of pure water is 7. When common salt is introduced, the pH does not alter.

Question 16: Classify the following solutions as acids, bases or salts ammonium hydroxide, barium chloride, sodium chloride, sodium hydroxide, H₂SO₄ and HNO₃
Solution 16:
Acids: H₂SO₄ and HNO₃
Bases: Ammonium hydroxide and sodium hydroxide. Salts: Barium chloride and sodium chloride.

Intext – Question – 4

Question 1: What do you understand by water of crystallization?
Give four substances which contain water of crystallization and write their common names.
Solution 1:
While crystallizing out of their solutions, some salts combine with a specific quality of water known as crystallization water.
Four substances which contain water of crystallization: Na₂CO₃ . 10H₂O Washing soda
MgSO₄ . 7H₂O Epsom salt
K₂SO₄ . Al₂(SO₄)₃ . 24H₂O Potash alum Na₂SO₄ . 10H₂O Glauber’s salt

Question 2: (a) Define efflorescence. Give examples.
(b) Define deliquescence. Give examples.
Solution 2:
(a) Efflorescence is the property of some compounds to lose all or part of their crystallisation water when exposed to dry air, even for a short period of time.
Here are several examples: Epsom salt, washing soda, Glauber’s salt
(b) When exposed to the atmosphere at room temperature, certain water-soluble compounds absorb moisture from the air, becoming wet, and eventually dissolving in the absorbed water, forming a saturated solution. Consider the following scenario: Caustic potash and Caustic soda.

Question 3: Distinguish between drying and dehydrating agent.
Solution 3:

Drying agent	Dehydrating agent
They remove moisture from other substances.They are used to dry gases like chlorine, Sulphur dioxide. They are used in desiccators to keep substances dry.They represent physical Change.	They remove chemically combined elements of water in the ratio of 2:1 from a compound.They prepare substances like carbon monoxide, sugar charcoaletc.They represent chemical Change.

Question 4: Explain clearly how conc, H2SO4 is used as dehydrating as well as drying agent.
Solution 4:
Conc. H2SO4 may extract water molecules from blue vitriol and remove moisture from gases. As a result, conc. H2SO4 is employed as a dehydrating and drying agent.

Question 5: M is an element in the form of a powder. M burns in oxygen and the product obtained is soluble in water. The solution is tested with litmus. Write down only the word which will correctly complete each of the following sentences.
(i) If M is a metal, then the litmus will turn .
(ii) If M is a non-metal, then the litmus will turn .
(iii) If M is are active metal, then will be evolved when M reacts with dilute sulphuric acid.
(iv) If M is a metal, it will form oxide, which will form solution with water.
(v) If M is a non-metal, it will not conduct electricity in the form of .
Solution 5:
(i) If M is a metal, then the litmus will turn Blue.
(ii) If M is a non-metal, then the litmus will turn Red.
(iii) If M is are active metal, then hydrogen will be evolved when M reacts with dilute sulphuric acid.
(iv) If M is a metal, it will form basic oxide, which will form alkaline solution with water.
(v) If M is a non-metal, it will not conduct electricity in the form of Graphite.

Question 6: Give reasons for the following:
(a) Sodium hydrogen sulphate is not an acid but it dissolves in water to give hydrogen ions, according to the equation
NaHSO₄ ⇆ H⁺ + Na⁺ + SO₄^2-
(b) Anhydrous calcium chloride is used in a desiccator.
Solution 6:
(a) Although sodium hydrogen sulphate is not an acid, it does undergo partial replacement of the ionizable hydrogen atom and acts as an acidic salt, releasing H+ ions.
(b) Calcium chloride is employed as a drying agent in desiccators because it absorbs moisture and keeps the compound dry.

Question 7: State whether a sample of each of the following would increase or decrease in a mass if exposed to air.
(a) Solid NaOH
(b) Solid CaCI₂
(c) Solid Na₂CO₃ 10H₂O
(d) Conc, sulphuric acid
(e) Iron (III) Chloride
Solution 7:
(a) Increases
(b) Increase
(c) Decrease
(d) Increases
(e) Increases

Question 8: (a) Why does common salt get wet during the rainy season?
(b) How can this impurity be removed?
(c) Name a substance which changes the blue colour of copper sulphate crystals to white.
(d) Name two crystalline substances which do not contain water of crystallization.
Solution 8:
(a) Common salt get wet during the rainy season because impurities such as magnesium chloride, which are deliquescent compounds, are found in common salt. Table salt becomes moist when exposed to air, especially during the rainy season, despite the fact that sodium chloride is not deliquescent.
(b) By passing a current of dry hydrogen chloride gas through a saturated solution of the affected salt, this impurity can be eliminated. A precipitate of pure sodium chloride is formed, which can be recovered by filtering and washing with water and then alcohol.
(c) H₂SO₄ can change the colour of copper sulphate from blue to white.
(d) Common salt, nitre, and Sugar are two crystalline substances that do not contain water of crystallisation.

Miscellaneous Questions Based On Icse Examinations

Question 1: For each of the salt: A, B, C and D, suggest a suitable method of its preparation.
(a) A is a sodium salt.
(b) B is an insoluble salt.
(c) C is a soluble salt of copper.
(d) D is a soluble salt of zinc.
Solution 1:
(a) A is a sodium salt, which is made by neutralising a base with acids.
(b) B is an insoluble salt: An insoluble salt is made by decomposing another insoluble salt twice. The insoluble salt must first be transformed to a soluble salt before being utilised to make the desired salt.
(c) C is a copper soluble salt: The soluble salt of copper is made by acid breakdown of carbonates.
(d) D is Zinc’s soluble salt: Zinc’s soluble salt can be made by decomposing chlorides in conc. H2SO4.

Question 2: (a) A solution has a Ph of 7. Explain how you would: (i) increase its Ph; (ii) decrease its pH.
(b) If a solution changes the colour of litmus from red to blue, what can you say about its pH?
(c) What can you say about the Ph of a solution that liberates carbon dioxide from sodium carbonate?
Solution 2:
(a) (i)The addition of base raises the pH.
(ii) The addition of acid lowers the pH.
(b) The pH shows the presence of base if the solution changes the colour of litmus from red to blue.
(c) The pH of the solution releasing carbon dioxide from sodium carbonate is less than 7.

Question 3: Answer the questions below, relating your answers only to salts in the following list: sodium chloride, anhydrous calcium chloride, copper sulphate – water.
(a) What name is given to the water in the compound copper sulphate – 5 -water?
(b) If copper sulphate 5 – water is heated, anhydrous copper sulphate is formed. What is its colour?
(c) By what means, other than heating, could you dehydrate copper sulphate – 5 – water and obtain anhydrous copper sulphate?
(d) Which one of the salts in the given list is deliquescent?
Solution 3:
(a) Water of crystallisation is the term given to the water in the combination copper sulphate-5-water.
(b) The colour of anhydrous copper sulphate is white.
(c) By adding dehydrating compounds such conc. sulphuric acid, which eliminate the crystallisation water.
(d) calcium chloride, anhydrous

Question 4: Solution P has a PH of 13, solution Q had a pH of 6 and solution R has a pH of 2. Which solution:
(a) will liberate ammonia from ammonium sulphate on heating?
(b) is a strong acid?
(c) contains molecules as well as ions?
Solution 4:
(a) Solution P will liberate ammonia from ammonium sulphate on heating.
(b) Solution R is a strong acid.
(c) Solution Q contains molecules as well as ions.

Question 5: (a) Outline the steps that would be necessary to convert insoluble lead (II) oxide into soluble lead chloride.
write the balanced equations for the reactions, to convert insoluble lead (II) oxide into soluble lead choride.
(b) A solution of iron (III) chloride has a PH less than 7. Is the solution acidic or alkaline?
Solution 5:
(a) To make soluble Lead nitrate, lead oxide is treated with dilute nitric acid. To make insoluble Lead chloride, this Lead nitrate is mixed with soluble Metallic chloride or dilute hydrochloric acid.
PbO + 2HNO₃ (dil) ⟶ Pb(NO₃)₂ +H₂O
Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃
(c) Solution of iron (III) chloride has a PH less than 7 the solution is acidic in nature.

Question 6: Choosing only substances from the list given in the box below, write equations which you would use in the laboratory to obtain:
(a) Sodium sulphate,
(b) Copper sulphate
(c) Iron (II) sulphate
(d) Zinc carbonate

Dilute Sulphuric acid	Copper	Copper Carbonate
	Iron	Sodium Carbonate
	Sodium
	Zinc

Solution 6:
(a) Sodium sulphate:
Na₂CO₃ + H₂SO₄ (dil) ⟶ Na₂SO₄ + H₂O + CO₂
(b) Copper sulphate:
CuCO₃ + H₂SO₄ (dil) ⟶ CuSO₄ + H₂O + CO₂
(c) Iron (II) sulphate:
Fe + H₂SO₄ (dil) ⟶FeSO₄ + H₂
(d) Zinc Carbonate:
Zn + H₂SO₄(dil) ⟶ZnSO₄ + H₂
ZnSO₄ + Na₂CO₃ ⟶ ZnCO₃ + Na₂SO₄

Question 7: From the formula listed below, choose one, in each case corresponding to the salt having the given description:-
AgCl, CuCO₃, CuSO₄, 5H₂O, KNO₃, NaCI, NaHSO₄, Pb(NO₃)₂, ZnCO₃, ZnSO₄, 7H₂O
(a) an acid salt
(b) an insoluble chloride
(c) on treating with concentrated sulphuric acid, this salt changes from blue to white.
(d) On heating, this salt changes from green to black
(e) this salt gives nitrogen dioxide on heating.
Solution 7:
(a) NaHSO₄
(b) AgCl
(c) CuSO₄ . 5H₂O
(d) CuCO₃
(e) Pb(NO₃)₂

Question 8: Ca(H₂PO₄)₂ is an example of a compound called ____(acid salt/ basic salt/ normal salt)
Solution 8:
Ca(H₂PO₄)₂ is an example of a compound called Acid salt.

Question 9: Write the balanced equation for the reaction of: A names acid and a named alkali.
Solution 9:
2NaOH + H₂SO 4 ⟶ Na₂SO₄ + 2H₂O

Question 10: State the terms defined by the following sentences:
(a) A soluble base
(b) The insoluble solid formed when two solutions are mixed together.
(c) An acidic solution in which there is only partial ionization of the solute molecules.
Solution 10:
(a) Alkali is a soluble base.
(b) Precipitate is insoluble solid formed when two solutions are mixed together.
(c) Weak acid is an acidic solution in which there is only partial ionization of the solute molecules.

Question 11: Differentiate between the chemical nature of an aqueous solution of HCI and an aqueous solution of NH₃.
Solution 11:

Aqueous solution of HCl	Aqueous solution of NH₃
1. It is acidic in nature. 2. It turns blue litmus to red. 3. It gives Hydronium ions in the solution.	1. It is basic in nature. 2. It turns red litmus to blue. 3. It gives hydroxyl ions in the solution.

Question 12: Write the balanced equations for the preparation of the following compounds (as the major product) starting from iron and using only one other substance:
(a) Iron (II) chloride
(b) Iron (III)chloride
(c) Iron (II) sulphate
(d) Iron (II) Sulphide.
Solution 12:
(a) Fe + 2HCl (dil) ⟶ FeCl₂ + H₂
(bi) 2Fe (heated) + 3Cl₂ (dry) ⟶ 2FeCl₃
(c) Fe + H₂SO₄ (dil) ⟶ FeSO₄ + H₂

Question : Which of the following methods, A, B, C, D or E is generally used for preparing the chlorides listed below from (i)to(v),Answer by writing down the chloride and the letter pertaining to the corresponding method each letter is to be used only once.
A. Action of acid on metal
B. Action of an acid on an oxide or carbonate
C. C Direct combination
D. Neutralization of an alkali by an acid E Precipitation (double decomposition)
(i) Copper (II) chloride
(ii) Iron (II) chloride
(iii) Iron (III)chloride
(iv) Lead (II)chloride
(v) Sodium chloride
Solution :
(i) Copper (II)chloride   –   (B) Action of an acid on an oxide or carbonate
(ii) Iron(II)chloride        –   (A) Action of acid on metal
(iii) Iron(III)chloride    –    (C) Direct combination
(iv) Lead(II)chloride      –    (E) Precipitation (double decomposition)
(v) Sodium chloride      –   (D) Neutralization of an alkali by anacid

Question : The preparation of lead sulphate from lead carbonate is a two – step process. (lead sulphate cannot be prepared by adding dilute sulphuric acid to lead carbonate.)
(a) What is the first step that is required to prepare lead sulphate from lead carbonate?
(b) Write the equation for the reaction that will take place when this first step is carried out.
(c) Why is the direct addition of dilute sulphuric acid to lead carbonate an impractical method of preparing lead sulphate?
Solution :
(a) The first step is to treat insoluble lead carbonate with dilute nitric acid to convert it to soluble lead nitrate.
(b) PbCO₃ (s) + 2HNO₃(dil) ⟶ Pb(NO₃)₂ (aq) + H₂O (l) + CO₂↑
(c) When dilute sulfuric acid is introduced directly to lead carbonate, the resultant lead sulphate is deposited on solid lead carbonate, effectively separating the lead carbonate from the sulphuric acid.

Question 2: Fill in the blanks with suitable words:
An acid is a compound which when dissolved in water forms hydronium ions as the only
…………………. Ions. A base is a compound which is soluble in water contains …………..
ions. A base reacts with an acid to form a and water only. This type of reaction is
known as ……………………..
Solution 2:
An acid is a compound which when dissolved in water forms hydronium ions as the only positive Ions. A base is a compound which is soluble in water contains charged ions. A base reacts with an acid to form a negatively charged salt and water only. This type of reaction is known as Neutralization reaction.

Question 1: From the list given below, select the word (s) required to correctly complete blanks (a) to (e) in the following passage:
Ammonia, ammonium, carbonate, carbon-dioxide, hydrogen, hydronium, hydroxide, precipitate, salt, water.
A solution X turns blue litmus red, so it must contain
(a) ions; another solution Y turns red litmus blue and therefore, must contain.
(b) ions, When solutions X and Y are mixed together, the products will be
And
(c) if a piece of magnesium were put into solution X.
(d) Gas would be evolved.
(Note: words chosen from the list are to be used only once. Write the answers as (1) (a), (b), (c) and so on. Do not copy the passage).
Solution 1:
(a) hydronium
(b) hydroxide
(c) salt
(d) water
(e) Hydrogen

Question 2: Match the following:
Column A Column B
(a) Acid salt A Sodium potassium carbonate
(b) Mixed salt B Alum
(c) Complex salt C Sodium carbonate
(d) Double salt D Sodium zincate
(e) Normal salt E Sodium hydrogen carbonate
Solution 2:
(a) – (E)
(b) – (A)
(c) – (D)
(d) – (B)
(e) – (C)
Aqueous solution of HCl aqueous solution of NH3
1. It is acidic in nature.
2. It turns blue litmus to red.
3. It gives Hydronium ions in the solution.
4. It is basic in nature.
5. It turns red litmus to blue.
6. It gives hydroxyl ions in the solution.

Question 3: Write balanced equations for the following reactions:
(a) Lead sulphate from lead nitrate solution and dilute sulphuric acid,
(b) Copper sulphate from copper and concentrated sulphuric acid.
(c) Lead chloride from lead nitrate solution and sodium chloride solution,
(d) Ammonium sulphate from ammonia and dilute sulphuric acid,
(e) Sodium chloride from sodium carbonate solution and dilute hydrochloric acid
Solution 3:
(a) Pb(NO₃)₂ + H₂SO₄ ⟶ PbSO₄ + 2HNO₃
(b) Cu + H₂SO₄ ⟶ CuSO₄ + H₂
(C) Pb(NO₃)₂ + 2NaCI ⟶ PbCI₂ + 2NaNO₃
(d) 2NH₃ + H₂SO₄ ⟶ (NH₄)₂SO₂
(e) Na₂CO₃ + 2HCI ⟶ 2NaCI + H₂O + CO₂

Question 1: What are the terms define by the following?
(i) A salt containing a metal ion surrounded by other ions or molecules,
(ii) A base which is soluble in water.
Solution 1:
(i) Complex salts
(ii) Alkali

Question 2: Making use only of substances chosen from those given below:
(i) Dilute sulphuric acid sodium carbonate
(ii) Zinc sodium sulphite
(iii) Lead Calcium carbonate
Give the equations for the reactions by which you could obtain:
(i) Hydrogen
(ii) Sulphur-dioxide
(iii) Carbon-dioxide,
(iv) zinc carbonate (two steps required).
Solution 2:
(i) Nn + H₂SO₄ ⟶ ZnSO₄ + H₂
(ii) Na₂SO₃ ⟶ Na₂O + SO₂
(iii) Na₂CO₃+ H₂SO₄ ⟶ Na₂SO₄ + H₂O + CO₂
(iv) Zn + H₂SO₄ ⟶ ZnSO₄ + H₂ZnS
(v) Na₂CO₃ ⟶ Na₂SO₄ + ZnCO₃