Acids, Bases and Salts

Question 1

Mention the colour changes observed when the following indicators are added to acids:

(i) Alkaline phenolphthalein solution.

(ii) Methyl orange solution

(iii) Neutral litmus solution

Answer

(i) Pink solution becomes colourless.

(ii) Orange solution changes to pink colour.

(iii) Purple solution changes to red colour.

Question 2

Which of the following hydroxides is not an alkali — [Choose from the choices A, B, C and D]

    (A) ammonium hydroxide

    (B) calcium hydroxide

    (C) copper hydroxide

    (D) sodium hydroxide

Answer

Copper hydroxide [Cu(OH)₂]
Reason — Copper hydroxide [Cu(OH)₂] is an example of insoluble base and it is not an alkali.

Question 1

Complete the blanks from the list given:

Ammonia, Ammonium, Carbonate, Carbon dioxide, Hydrogen, Hydronium, Hydroxide, Precipitate, Salt, Water.

A solution X turns blue litmus red, so it must contain (i) ............... ions; another solution Y turns red litmus blue and therefore, must contain (ii) ............... ions. When solutions X and Y are mixed together the products will be a (iii) ............... and (iv) ............... . If a piece of magnesium were put into solution X, (v) ............... gas would be evolved.

Answer

A solution X turns blue litmus red, so it must contain (i) hydronium ions; another solution Y turns red litmus blue and therefore, must contain (ii) hydroxide ions. When solutions X and Y are mixed together the products will be a (iii) salt and (iv) water. If a piece of magnesium were put into solution X, (v) hydrogen gas would be evolved.

Question 2

Match the following:

Answer

Question 3

Write balanced equation for formation of PbCl₂ from Pb(NO₃)₂ soln. and NaCl soln.

Answer

Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃

Question 1

What is the term defined : i) A base which is soluble in water.

Answer

Alkali is a base which is soluble in water.

Question 1

The acid which contains four hydrogen atoms —

1. Formic acid
2. Sulphuric acid
3. Nitric acid
4. Acetic acid

Answer

Acetic acid contains four hydrogen atoms.

Question 2

A black coloured solid which on reaction with dilute sulphuric acid forms a blue coloured solution is:

1. Carbon
2. Manganese [IV] oxide
3. Lead [II] oxide
4. Copper [II] oxide

Answer

Copper [II] oxide
Reason — Copper [II] oxide is black in colour and the following reaction takes place when it is treated with dilute sulphuric acid —

CuO + H₂SO₄ ⟶ CuSO₄ + H₂O

CuSO₄ is a blue coloured soln.

Question 3

Solution A is a strong acid
Solution B is a weak acid
Solution C is a strong alkali

(i) Which solution contains solute molecules in addition to water molecules?

(ii) Which solution will give a gelatinous white precipitate with zinc sulphate solution? The precipitate disappears when an excess of the solution is added.

(iii) Which solution could be glacial acetic acid solution?

(iv) Give example of a soln. of a weak alkali.

Answer

(i) Solution B — weak acid
Reason — Weak Acid is an acid which dissociates only partially in aqueous solution thereby producing a low concentration of hydrogen [H⁺] ions [or H₃O⁺ ions]. For example — CH₃COOH ⇌ CH₃COO^- + H⁺ [contains molecules and ions]

(ii) Solution C — strong alkali
Reason — Alkalis react with certain salt solutions to precipitate insoluble hydroxide. Hence,
ZnSO₄ + 2NaOH ⟶ Na₂SO₄ + Zn(OH)₂ [gelatinous white precipitate]

(iii) Solution B — weak acid
Reason — Anhydrous acetic acid on cooling forms crystals of glacial acetic acid and acetic acid is a weak acid.

(iv) Ammonium hydroxide (NH₄OH)

Question 4

Write the equation[s] for the reaction[s] to prepare lead sulphate from lead carbonate.

Answer

PbCO₃ + 2HNO₃ ⟶ Pb(NO₃)₂ + H₂O + CO₂

Pb(NO₃)₂ + Na₂SO₄ ⟶ PbSO₄ + 2NaNO₃

Question 5

Define the following terms — Neutralization

Answer

Neutralization — It is the process due to which [H⁺] ions of an acid react completely or combine with [OH^-] ions of a base to give salt and water only.

Acid + Base ⟶ Salt + Water

HCl + NaOH ⟶ NaCl + H₂O

H⁺Cl^- + Na⁺OH^- ⟶ Na⁺Cl^- + H₂O

[H⁺ (aq) + OH^- (aq) ⇌ H₂O (l)]

Question 1

A: Nitroso Iron [II] sulphate

B: Iron [III] chloride

C: Chromium sulphate

D: Lead [II] chloride

E: Sodium chloride.

Select from A, B, C, D and E —

(i) A compound soluble in hot water but insoluble in cold water.

(ii) A compound which in the aqueous solution state, is neutral in nature.

Answer

(i) D: Lead [II] chloride

(ii) E: Sodium chloride

Question 2

Select the correct answer from A, B, C and D –

(i) A weak organic acid is:

1. Formic acid
2. Sulphuric acid
3. Nitric acid
4. Hydrochloric acid

(ii) A complex salt is :

1. Zinc sulphate
2. Sodium hydrogen sulphate
3. Iron (II) ammonium sulphate
4. Tetrammine copper (II) sulphate

Answer

(i) Formic acid

(ii) Tetrammine copper (II) sulphate

Question 3

Give an equation for the conversions

(i) ZnSO₄ to ZnCO₃

(ii) ZnCO₃ to Zn(NO₃)₂

Answer

(i) ZnSO₄ + (NH₄)₂CO₃ ⟶ (NH₄)₂SO₄ + ZnCO₃

(ii) ZnCO₃ + 2HNO₃ ⟶ Zn(NO₃)₂ + H₂O + CO₂

Question 4

1. NaOH soln.
2. Weak acid
3. Dil. H₂SO₄

Select the one which contains solute ions and molecules.

Answer

Weak acid
Reason — An acid which dissociates only partially in aqueous solution thereby producing a low concentration of hydrogen [H⁺] ions [or H₃O⁺ ions] is a weak acid.

Example — CH₃COOH ⇌ CH₃COO^- + H⁺ [contains molecules and ions]

Question 5

Give balanced equation/s for the preparation of the following salts:

1. Copper [II] sulphate from CuO.
2. Iron [III] chloride from Fe.
3. K₂SO₄ from KOH soln.
4. Lead [II] chloride from PbCO₃ [give two equations].

Answer

1. CuO + H₂SO₄ ⟶ CuSO₄ + H₂O

2. 2Fe + 3Cl₂ ⟶ 2FeCl₃

3. 2KOH + H₂SO₄ [dil.] ⟶ K₂SO₄ + 2H₂O

4. PbCO₃ + 2HNO₃ ⟶ Pb(NO₃)₂ + H₂O + CO₂
   Pb(NO₃)₂ + 2NaCl ⟶ 2NaNO₃ + PbCl₂

Question 1

Write the balanced chemical equation : Lead nitrate solution is added to sodium chloride solution

Answer

Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃

Question 2

Name the method used from the list:

A : Simple displacement
B : Neutralization
C : Decomposition by acid
D : Double decomposition
E : Direct synthesis

For preparation of the following salts –

(i) Sodium nitrate

(ii) Iron (III) chloride

(iii) Lead chloride

(iv) Zinc sulphate

(v) Sodium hydrogen sulphate.

Answer

(i) Sodium nitrate — B : Neutralization

(ii) Iron (III) chloride — E : Direct synthesis

(iii) Lead chloride — D : Double decomposition

(iv) Zinc sulphate — A : Simple displacement

(v) Sodium hydrogen sulphate — C : Decomposition by acid

Question 1

Match the following i.e.,

(1) Acid salt

(2) Double salt — with the correct choice from — A and B

A : Ferrous ammonium sulphate

B : Sodium hydrogen sulphate

Answer

1. Acid salt — B : Sodium hydrogen sulphate
2. Double salt — A : Ferrous ammonium sulphate

Question 1

Select the word/s given below which are required to correctly complete the blanks — [ammonia, ammonium, carbonate, carbon dioxide, hydrogen, hydronium, hydroxide, precipitate, salt water] :

(i) A solution M turns blue litmus red, so it must contain (i) ............... ions ; another solution O turns red litmus blue and hence, must contain, (ii) ............... ions.

(ii) When solution M and O are mixed together, the prod­ucts will be (iii) ............... and (iv) ............... .

(iii) If a piece of magnesium was put into a solution M,(v) ............... gas would be evolved.

Answer

(i) A solultion M turns blue litmus red, so it must contain (i) hydronium ions ; another solution O turns red litmus blue and hence, must contain, (ii) hydroxide ions.

(ii) When solution M and O are mixed together, the prod­ucts will be (iii) salt and (iv) water.

(iii) If a piece of magnesium was put into a solution M, (v) hydrogen gas would be evolved.

Question 2

Give a suitable chemical term for:

(i) A salt formed by incomplete neutralisation of an acid by a base.

(ii) A definite number of water molecules bound to some salts.

Answer

(i) Acid salt

(ii) Water of crystallization

Question 3

Choosing the substances from the list given:

dil. Sulphuric acid, Copper, Iron, Sodium, Copper (II) carbonate, Sodium carbonate, Sodium chloride, Zinc nitrate

Write balanced equations for the reactions which would be used in the laboratory to obtain the following salts:

1. Sodium sulphate
2. Zinc carbonate
3. Copper (II) sulphate
4. Iron (II) sulphate.

Answer

1. Sodium sulphate
   Na₂CO₃ + H₂SO₄ (dil.) ⟶ Na₂SO₄ + H₂O + CO₂

2. Zinc carbonate
   Zn(NO₃)₂ + Na₂CO₃⟶ 2NaNO₃ + ZnCO₃

3. Copper (II) sulphate
   CuCO₃ + H₂SO₄ (dil.) ⟶ CuSO₄ + H₂O + CO₂

4. Iron (II) sulphate
   Fe + H₂SO₄ (dil.) ⟶ FeSO₄ + H₂

Question 4

Identify: An acid which is present in vinegar.

Answer

Acetic acid

Question 1

Fill in the blank from the choices given:

The basicity of acetic acid is ............... [3, 1, 4].

Answer

The basicity of acetic acid is 1.

Question 2

Draw the structure of the stable positive ion formed when an acid dissolves in water.

Answer

Hydronium ion is the stable positive ion formed when an acid dissolves in water. Its structure is shown below:

Question 3

State the inference drawn from the observation:

Salt S is prepared by reacting dilute sulphuric acid with copper oxide. Identify S.

Answer

Salt S is Copper sulphate CuSO₄

CuO + H₂SO₄ (dil.) ⟶ CuSO₄ + H₂O

Question 4

Give balanced chemical equations for the preparation of the following salts:

1. Lead sulphate — from lead carbonate.
2. Sodium sulphate — using dilute sulphuric acid.
3. Copper chloride — using copper carbonate.

Answer

1. Lead sulphate from lead carbonate.
   PbCO₃ + 2HNO₃ ⟶ Pb(NO₃)₂ + H₂O + CO₂
   Pb(NO₃)₂ + Na₂SO₄ ⟶ 2NaNO₃ + PbSO₄

2. Sodium sulphate using dilute sulphuric acid.
   Na₂CO₃ + H₂SO₄(dil.) ⟶ Na₂SO₄ + H₂O+ CO₂

3. Copper chloride using copper carbonate.
   CuCO₃ + 2HCl (dil) ⟶ CuCl₂ + H₂O + CO₂

Question 1

Give a balanced chemical equation for the following conversion.

Fe ⟶ FeCl₃

Answer

2Fe + 3Cl₂ ⟶ 2FeCl₃

Question 2

From the list of salts — AgCl, MgCl₂, NaHSO₄, PbCO₃
Choose the salt that most appropriately fits the description given below :
An insoluble chloride.

Answer

AgCl

Question 3

From — SO₂, SiO₂, Al₂O₃, MgO, CO, Na₂O — Select an oxide which dissolves in water forming an acid.

Answer

SO₂

The following reaction takes place.

SO₂ + H₂O ⟶ H₂SO₃

Question 1

Fill in the blank:
Higher the pH value of a solution, the more ............... [acidic/alkaline] it is.

Answer

Higher the pH value of a solution, the more alkaline it is.

Question 2

Match the following salts given below:

(i) Pb(NO₃)₂ from PbO

(ii) MgCl₂ from Mg

(iii) FeCl₃ from Fe

(iv) NaNO₃ from NaOH

(v) ZnCO₃ from ZnSO₄

With their correct method of preparation from: A, B, C, D and E.

(A) Simple displacement

(B) Titration

(C) Neutralization

(D) Precipitation

(E) Combination

Answer

(i) Pb(NO₃)₂ from PbO — (C) Neutralization

(ii) MgCl₂ from Mg — (A) Simple displacement

(iii) FeCl₃ from Fe — (E) Combination

(iv) NaNO₃ from NaOH — (B) Titration

(v) ZnCO₃ from ZnSO₄ — (D) Precipitation

Question 1

Fill in the blanks from the choices given in brackets —

When a metallic oxide is dissolved in water, the solution formed has a high concentration of ............... ions. [H⁺, H₃O⁺, OH^-]

Answer

When a metallic oxide is dissolved in water, the solution formed has a high concentration of OH^- ions.

Question 2

Choose the correct answer from the options —

(i) To increase the pH value of a neutral solution, we should add :

1. An acid
2. An acid salt
3. An alkali
4. A salt

(ii) Anhydrous iron [III] chloride is prepared by:

1. Direct combination
2. Simple displacement
3. Decomposition
4. Neutralization

Answer

(i) An alkali
Reason — The pH of alkali solutions is more than 7 therefore in order to increase the pH value of a neutral solution, an alkali should be added.

(ii) Direct combination
Reason — Anhydrous iron [III] chloride is prepared by direct combination of iron and chloride as follows —
2Fe + 3Cl₂ ⟶ 2FeCl₃

Question 3

Write a balanced chemical equation for the preparation of each of the following salts:

(i) Copper carbonate

(ii) Ammonium sulphate crystals.

Answer

(i) CuSO₄ + Na₂CO₃ ⟶ Na₂SO₄ + CuCO₃

(ii) 2NH₄OH + H₂SO₄ ⟶ (NH₄)₂SO₄ + 2H₂O

Question 1

Give one word or a phrase for the statement :
The property by which certain hydrated salts, when left exposed to the atmosphere, lose their water of crystallization and crumble into powder.

Answer

Efflorescence

Question 2

State one relevant observation for the following :
Anhydrous calcium chloride is exposed to air for some time.

Answer

It absorbs moisture from the atmosphere, dissolve in the same and change into a solution. This property is known as deliquescence

Question 3

Fill in the blank with the correct choice given in the bracket —
The salt prepared by the method of direct combination is ............... [iron [II] chloride / iron [III] chloride]

Answer

The salt prepared by the method of direct combination is iron [III] chloride

Question 4

Three solutions P, Q, and R have pH value of 3.5, 5.2 and 12.2 respectively. State which one of these is a:

(i) weak acid

(ii) strong alkali

Answer

(i) Q
Reason — On a pH scale, acids have pH less than 7 whereas weak acids have pH towards 7. Hence, Q will be a weak acid with pH 5.2 .

(ii) R
Reason — On a pH scale, alkali have pH more than 7 and alkalinity increases as the pH value moves away from 7. Hence, R will be a strong alkali with pH 12.2 .

Question 5

Write a balanced equation for the preparation of each of the following salts :

(i) Copper [II] sulphate from copper carbonate

(ii) Zinc carbonate from zinc sulphate

Answer

(i) CuCO₃ + H₂SO₄ ⟶ CuSO₄ + H₂O + CO₂

(ii) ZnSO₄ + (NH₄)₂CO₃ ⟶ (NH₄)₂SO₄ + ZnCO₃

Question 1

Give the appropriate term defined by the statement given :
The substance that releases hydronium ion as the only positive ion when dissolved in water.

Answer

Acid
Reason — An acid is a compound which when dissolved in water yields hydronium ions [H₃O⁺] as the only positively charged ion.

HCl (aq) ⇌ H⁺ + Cl^-

H⁺ + H₂O ⇌ H₃O⁺ [hydronium ion]

HCl + H₂O ⇌ H₃O⁺ + Cl^-

Question 2

The pH values of three solutions A, B, C are given .

Solution A : pH value 12

Solution B : pH value 2

Solution C : pH value 7

Answer the following questions :

(i) Which solution will have no effect on litmus solution.

(ii) Which solution will liberate CO₂ when reacted with sodium carbonate.

(iii) Which solution will turn red litmus solution blue.

Answer

(i) Solution C : pH value 7

(ii) Solution B : pH value 2

(iii) Solution A : pH value 12

Question 3

Choose the method of preparation of the following salts, from the methods given in the list:

List —
A: Neutralization
B: Precipitation
C: Direct combination
D: Substitution

(i) Lead chloride
(ii) Iron [II] Sulphate
(iii) Sodium nitrate
(iv) Iron [III] chloride

Answer

(i) Lead chloride — B: Precipitation

(ii) Iron [II] Sulphate — D: Substitution

(iii) Sodium nitrate — A: Neutralization

(iv) Iron [III] chloride — C: Direct combination

Question 1

Fill in the blanks from the choices given : A salt prepared by displacement reaction is ............... [Ferric chloride, ferrous chloride, silver chloride]

Answer

Ferrous chloride

Question 2

Complete the following by selecting the correct options from the choices :
pH of acetic acid is greater than dilute sulphuric acid. So, acetic acid contains ............... concentration of H⁺ ions. [greater, same, low]

Answer

pH of acetic acid is greater than dilute sulphuric acid. So, acetic acid contains low concentration of H⁺ ions.

Question 3

Differentiate between the following pairs based on the information given in the brackets :
Acid and alkali [formation of type of ions]

Answer

Question 4

Write balanced chemical equations, for the preparation of given salts (i) to (iii) by using the methods A to C respectively.

A: Neutralization
B: Precipitation
C: Titration

(i) Copper sulphate
(ii) Zinc carbonate
(iii) Ammonium sulphate

Answer

1. Preparation of copper sulphate by neutralization
   CuO + H₂SO₄ ⟶ CuSO₄ + H₂O

2. Preparation of zinc carbonate by precipitation
   Zn(NO₃)₂ + Na₂CO₃ ⟶ 2NaNO₃ + ZnCO₃

3. Preparation of ammonium sulphate by titration
   2NH₄OH + H₂SO₄ ⟶ (NH₄)₂SO₄ + 2H₂O

Question 1

Define the following as per ionic theory with examples and ionic equations wherever relevant :

(i) acid (ii) base (iii) alkali (iv) neutralization

Answer

(i) Acid — An acid is a compound which when dissolved in water yields hydronium ions [H₃O⁺] as the only positively charged ion.

HCl (aq) ⇌ H⁺ + Cl^-

H⁺ + H₂O ⇌ H₃O⁺ [hydronium ion]

HCl + H₂O ⇌ H₃O⁺ + Cl^-

(ii) Base — A base is a compound which reacts with hydronium ions of an acid to give salt and water only.

CuO + 2HCl ⟶ CuCl₂ + H₂O

Cu(OH)₂ + H₂SO₄ ⟶ CuSO₄ + 2H₂O

• Bases are oxides or hydroxides of a metal [including ammonium hydroxide]

• Examples of insoluble bases [i.e., not alkalis] — ZnO, PbO, CuO, Fe(OH)₂, Pb(OH)₂, Cu(OH)₂

(iii) Alkali — An alkali is a compound which when dissolved in water yields hydroxyl ions [OH^-] as the only negatively charged ions.

NaOH [aq.] ⇌ Na⁺ + OH^- [Hydroxyl or hydroxide ion]

Alkali is a base, soluble in water. [All alkalis are bases, but all bases are not alkalis.]

Examples of soluble bases [i.e., alkalis] — KOH, NaOH [strong alkalis] , Ca(OH)₂, NH₄OH (weak alkalis).

(iv) Neutralization — It is the process due to which [H⁺] ions of an acid react completely or combine with [OH^-] ions of a base to give salt and water only.

Acid + Base ⟶ Salt + Water

HCl + NaOH ⟶ NaCl + H₂O

H⁺Cl^- + Na⁺OH^- ⟶ Na⁺Cl^- + H₂O

[H⁺ (aq) + OH^- (aq) ⇌ H₂O (l)]

Question 2

Differentiate between:

(i) Organic and inorganic acids.

(ii) Hydracids and oxyacids with examples.

Answer

(i) Difference between organic and inorganic acids are as follows :

(ii) Difference between hydracids and oxyacids are as follows :

Question 3

State on what basis does the strength of an acid and an alkali depend on.

Answer

Strength of acids depends on the concentration of hydronium ion [H₃O⁺] present in an aqueous solution of an acid.

Strength of alkali depends on the concentration of the hydroxyl ions [OH^-] present in an aqueous solution of the alkali.

Question 4

Differentiate between (i) strong and weak acid (ii) strong and weak alkali with suitable examples and ionic equations.

Answer

(i) Differences between strong and weak acid are as follows :

(ii) Differences between strong alkali and and weak alkali are as follows :

Question 5

Name the ions formed when — HCl; HNO₃; H₂SO₄; CH₃COOH; NaOH and NH₄OH ionise in aq. soln.

Answer

(i) When HCl is dissolved in water, it is ionised into hydrogen ion [or H₃O⁺ ion] and chloride ion.

HCl ⟶ H⁺ + Cl^-

The H⁺ cannot exist independently, therefore, it combines with water molecule to form hydronium ion (H₃O⁺)

H⁺ + H₂O ⟶ H₃O⁺

(ii) When HNO₃ is dissolved in water, it is ionised into hydronium ion and nitrate ion.

HNO₃ ⟶ H⁺ + NO₃^-

H⁺ + H₂O ⟶ H₃O⁺

(iii) When H₂SO₄ is dissolved in water, it is ionised into hydronium ion and sulphate ion.

H₂SO₄ ⟶ 2H⁺ + SO₄^2-

H⁺ + H₂O ⟶ H₃O⁺

(iii) When CH₃COOH is dissolved in water, it is ionised into hydronium ion and acetate ion.

CH₃COOH ⟶ CH₃COO^- + H⁺

H⁺ + H₂O ⟶ H₃O⁺

(iii) When NaOH is dissolved in water, it is ionised into sodium ion and hydroxyl ion.

NaOH ⟶ Na⁺ + OH^-

(iv) When NH₄OH is dissolved in water, it is ionised into ammonium ion and hydroxyl ion.

NH₄OH ⟶ NH₄⁺ + OH^-

Question 6

State giving reasons which is a stronger acid — dil. HCl or conc. H₂CO₃.

Answer

Dilute HCl is a stronger acid than concentrated H₂CO₃
Reason — HCl dissociates almost completely in aqueous solution and produces a high concentration of H⁺ ions and Cl^- ions, hence is a strong acid. Whereas, H₂CO₃ is a weak acid because it dissociates partially yielding H⁺ ions and bicarbonate HCO₃^- ion and hence, contains ions as well as molecules. Therefore, dil. HCl is a stronger acid than conc. H₂CO₃.

Question 7

State why the basicity of acetic acid is one and acidity of calcium hydroxide is two.

Answer

Basicity of acid — is the number of hydrogen ions [H⁺] which can be produced per molecule of the acid in aq. soln. Acetic acid [CH₃COOH] ionises in aq. soln. and gives one hydrogen ion per molecule of the acid, hence acetic acid is monobasic i.e., it's basicity is one.

Acidity of base — is the number of hydroxyl ions [OH^-] which can be produced per molecule of the base in aq. soln. Calcium hydroxide Ca(OH)₂ ionises in aq. soln. and gives two hydroxyl ions per molecule of the base, hence calcium hydroxide is a diacidic base i.e., it's acidity is two.

Question 8

Give three reasons with equations wherever required, why Sulphuric acid is a dibasic acid.

Answer

Sulphuric acid (H₂SO₄) is a dibasic acid as :

1. It ionises in aq. soln. to produce two hydrogen ions per molecule of the acid.
2. It dissociate in two steps in aq. soln. as shown below:
   H₂SO₄ + H₂O ⇌ H₃O⁺ + HSO₄^-
   HSO₄^- + H₂O ⇌ H₃O⁺ + SO₄^2-
   H₂SO₄ + 2H₂O ⇌ 2H₃O⁺ + SO₄^2-
3. It contains two replaceable hydrogen ions per molecule of the acid so forms two types of salt [acid and normal salt] :
   NaOH + H₂SO₄ ⟶ NaHSO₄ (Acid Salt) + H₂O
   2NaOH + H₂SO₄ ⟶ Na₂SO₄ (Normal Salt) + 2H₂O

Question 9

State how acids are defined as per Arrhenius's and Lowry – Bronsted's theory.

Answer

Arrhenius Theory — Acids are substances which dissociate in aqueous solution to give H⁺ ions.
Strong acids dissociate almost completely, while weak acids dissociate partially.

Lowry – Bronsted's theory — Acids are proton donors and bases are proton acceptors [proton = H⁺].
HCl [aq.] ⟶ H⁺ + Cl^- [acid - proton donors]
NH₃ + H⁺ ⟶ NH₄⁺ [bases - proton acceptors]

Question 10

Oxygen atom in water has two 'lone pair of electrons'. Explain the meaning of the term in italics. With the help of an electron dot diagram show the formation of hydronium ion and ammonium ion from a water molecule and an ammonia molecule respectively.

Answer

Oxygen atom in water has two 'lone pair of electrons' implies that two pairs of electrons on oxygen are not shared with any other atom as shown below:

Formation of hydronium ion

Formation of ammonium ion

Question 11

State how you would obtain:

1. Sulphuric acid from an acidic oxide
2. KOH from a basic oxide.

Answer

1. Acidic oxides dissolve in water to give an acid.
   SO₃ + H₂O ⟶ H₂SO₄
2. Basic oxides [soluble] dissolve in water to give a base i.e., alkali.
   K₂O + H₂O ⟶ 2KOH

Question 12

State two chemical properties each with equations of a solution containing

(i) H⁺ ions

(ii) OH^- ions

Answer

(i) Chemical properties of a solution containing H⁺ (acids) are as follows —

1. Neutralization — acids neutralizes base to give salt and water only.
   Example : CuO + H₂SO₄ ⟶ CuSO₄ + H₂O
2. Reaction with active metals — Acids react with active metals [e.g., Mg, Al, Zn, Fe] to liberate hydrogen.
   Example : Zn + 2HCl ⟶ ZnCl₂ + H₂

(ii) Chemical properties of a solution containing OH^- (bases/alkalis) are as follows —

1. Neutralization — Alkalis neutralize acids to form salt and water.
   Example :
   Ca(OH)₂ + 2HCl ⟶ CaCl₂ + 2H₂O
2. Reaction with metallic salt — Alkalis react with certain matallic salt solutions to precipitate insoluble hydroxide.
   Example :
   FeCl₃ + 3NaOH ⟶ 3NaCl + Fe(OH)₃ ↓ red brown ppt.

Question 13

Give equations for the decomposition of a metallic (i) chloride (ii) nitrate with conc. H₂SO₄.

Answer

(i) Decomposition of a metallic chloride —

$\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}] \xrightarrow{\lt 200 \degree\text{C}}\text{NaHSO}_4 + \text{HCl}$

(ii) Decomposition of a metallic nitrate with conc. H₂SO₄ —

$\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}] \xrightarrow{\lt 200 \degree\text{C}}\text{NaHSO}_4 + \text{HNO}_3$

Question 14

State in the above reactions a reason for the formation of the respective acids from conc. H₂SO₄.

Answer

As H₂SO₄ is a less volatile acid and displaces the more volatile acid on heating with the salt, hence the respective acids are formed.

AB [salt I] + HX [acid I] ⟶ AX [salt II] + HB [acid II]

Question 15

Convert (i) NaHCO₃ (ii) Na₂CO₃ to unstable carbonic acid by action with dil. H₂SO₄.
State the reason why ammonia is evolved when an ammonium salt and alkali are heated.

Answer

2NaHCO₃ + H₂SO₄ [dil.] ⟶ Na₂SO₄ + 2H₂O + 2CO₂ ⟶ H₂CO₃ (unstable)

Na₂CO₃ + H₂SO₄ [dil.] ⟶ Na₂SO₄ + H₂O + CO₂ ⟶ H₂CO₃ (unstable)

Ammonia is evolved when an ammonium salt and alkali are heated because a less volatile base (e.g.,NaOH) displace the more volatile base, NH₄OH and we get the products NH₃ and H₂O.

NH₄Cl + NaOH ⟶ NaCl + H₂O + NH₃

Question 16

Define pH value. What would you say about the pH of a solution in which

(i) H⁺ aq. ions = OH^- ions

(ii) evolves CO₂ when heated with Na₂CO₃

(iii) OH^-ions > H⁺ aq. ions.

Answer

pH is defined as the negative logarithm [to the base 10] of the hydrogen ion concentration expressed in moles/litre. Thus, pH = -log₁₀H⁺. It represents the strength of acids and alkalis, expressed in terms of hydrogen ion concentration [H⁺ aq.]

(i) When, H⁺ ions = OH^- ions, the solution is neutral with pH = 7.

(ii) When the solution evolves CO₂ when heated with Na₂CO₃, it is acidic in nature with pH less than 7.

(iii) when, OH^-ions > H⁺ aq. ions, the solution is basic in nature with pH more than 7.

Question 17

State whether litmus is a common acid-base indicator or a universal indicator.

Answer

Litmus is a common acid-base indicator and not a universal indicator. It only indicates whether a solution is acidic or alkaline. It cannot be utilized for determining the strength of the acidic or alkaline solution.

Question 18

State the colour change in a neutral litmus in presence of (i) acidic (ii) alkaline medium.

Answer

(i) A neutral litmus is purple in colour. In presence of an acidic medium, the colour of neutral litmus changes from purple to red.

(ii) In an alkaline medium, colour of neutral litmus changes from purple to blue.

Question 19

State the colour change in a universal indicator e.g. pH paper on

(i) slightly acidic soil

(ii) slightly alkaline soil

(iii) dairy milk

(iv) human blood tested for medical diagnosis.

Answer

(i) The pH paper changes to yellow colour on a slightly acidic soil.

(ii) The pH paper changes to blue colour on a slightly alkaline soil.

(iii) The pH paper changes to green colour in a dairy milk.

(iv) The pH paper changes to green colour in human blood tested for medical diagnosis.

Question 20

Define (i) salt (ii) normal salt (iii) acid salt – with relevant examples and equations.

Answer

(i) Salt — A salt is a compound formed by partial or complete replacement of the replaceable hydrogen ions of an acid by a metallic ion or ammonium ion [basic radical].

NaOH + H₂SO₄ ⟶ NaHSO₄ + H₂O [Partial replacement]

2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O [Complete replacement]

(ii) Normal salt — The salt formed by complete replacement of the replaceable hydrogen ion of an acid molecule by a basic radical [metallic or ammonium ion].
For example,

2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

2NaOH + H₂SO₃ ⟶ Na₂SO₃ + 2H₂O

[Both H ions in sulphuric and sulphurous acid are replaced by metallic radical — sodium. ]

(3) Acid salt — The salt formed by partial replacement of the replaceable hydrogen ion of an acid molecule by a basic radical [metallic or ammonium ion]. For example,

NaOH + H₂SO₄ ⟶ NaHSO₄ + H₂O

NaOH + H₂SO₃ ⟶ NaHSO₃ + H₂O

[Only one H ion in sulphuric and sulphurous acid is replaced by metallic radical — sodium. ]

Question 21

State (i) the formation (ii) the components of – a basic salt.

State which of following salts is an – acid, normal or basic salt.

(i) bleaching powder

(ii) potassium mercuric iodide

(iii) sodium sulphite

(iv) sodium hydrogen sulphite

(v) sodium silver cyanide

(vi) basic lead nitrate

(vii) potassium zincate

(viii) alum

(ix) calcium bicarbonate

(x) basic copper chloride

(xi) trisodium phosphate.

Answer

(i) Formation of a basic salt — A basic salt is formed by partial replacement of hydroxyl radicals of a diacidic or triacidic base with an acid radical.

(ii) Components of a basic salt — A basic salt contains a cation [metallic], a hydroxyl ion [of a base] and an anion [of an acid].

For example — Basic copper nitrate Cu[OH]NO₃ , basic copper chloride Cu[OH]Cl

Acid, normal or basic salt —

(i) bleaching powder — Normal salt (Mixed salt)

(ii) potassium mercuric iodide — Normal salt (Complex salt)

(iii) sodium sulphite — Normal salt

(iv) sodium hydrogen sulphite — Acid salt

(v) sodium silver cyanide — Normal salt (Complex salt)

(vi) basic lead nitrate — Basic salt

(vii) potassium zincate — Normal salt

(viii) alum — Normal salt (Double salt)

(ix) calcium bicarbonate — Acid salt

(x) basic copper chloride — Basic salt

(xi) trisodium phosphate — Normal salt

Question 22

Name three (i) sulphates (ii) chlorides insoluble in water and two (i) oxides (ii) carbonates soluble in water.

Answer

Insoluble in Water

(i) Three sulphates insoluble in water are:

1. Lead sulphate (PbSO₄)
2. Calcium sulphate (CaSO₄)
3. Barium sulphate (BaSO₄).

(ii) Three chloride insoluble in water are:

1. Silver chloride (AgCl)
2. Lead chloride (PbCl₂)
3. Mercury chloride (Hg₂Cl₂).

Soluble in Water

(i) Two oxides soluble in water are:

1. Sodium oxide (Na₂O)
2. Potassium oxide (K₂O)

(ii) Two carbonates soluble in water are:

1. Sodium carbonate (Na₂CO₃)
2. Ammonium carbonate [(Na₄)₂CO₃]

Question 23

State the method only, generally used for the preparation of the following salts

(i) Zn(NO₃)₂

(ii) NH₄Cl

(iii) ZnSO₄

(iv) ZnS

(v) CaCO₃

(vi) FeCl₃

(vii) PbCl₂

(viii) Pb(NO₃)₂

Answer

Method of preparation are as follows :

(i) Zn(NO₃)₂ — By action of dilute acids on carbonate.

(ii) NH₄Cl — By neutralisation of an alkali (titration)

(iii) ZnSO₄ — By displacement of active metal and acid

(iv) ZnS — Direct combination (Synthesis)

(v) CaCO₃ — By precipitation (double decomposition)

(vi) FeCl₃ — Direct combination (Synthesis)

(vii) PbCl₂ — Precipitation (double decomposition)

(viii) Pb(NO₃)₂ — Action of dil. acid on carbonates and bicarbonates

Question 24

Give balanced equations for the preparation of the following salts –

(a) (i) CuSO₄

(ii) NaHSO₄

(iii) Na₂SO₄

(iv) FeSO₄

(v) BaSO₄

(vi) PbSO₄ — using dil. H₂SO₄

(b) (i) NaHSO₄

(ii) CuSO₄ — using conc. H₂SO₄

Answer

(a) Using dil. H₂SO₄ :

1. Preparation of CuSO₄:
   CuO + H₂SO₄ ⟶ CuSO₄ + H₂O
2. Preparation of NaHSO₄:
   NaOH + H₂SO₄ ⟶ NaHSO₄ + H₂O
3. Preparation of Na₂SO₄:
   Na₂CO₃ + H₂SO₄ ⟶ Na₂SO₄ + H₂O + CO₂
4. Preparation of FeSO₄:
   Fe + H₂SO₄ ⟶ FeSO₄ + H₂
5. Preparation of BaSO₄:
   BaCl₂ + H₂SO₄ ⟶ BaSO₄ + 2HCl
6. Preparation of PbSO₄:
   Pb(NO₃)₂ + H₂SO₄ ⟶ PbSO₄ + 2HNO₃

(b) Using conc. H₂SO₄ :

1. Preparation of NaHSO₄:

$\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}] \xrightarrow{\lt 200 \degree\text{C}}\text{NaHSO}_4 + \text{HCl}$

2. Preparation of CuSO₄:
   CuO + H₂SO₄ ⟶ CuSO₄ + H₂O

Question 25

Starting from insoluble ZnO how would you obtain insoluble ZnCO₃ by precipitation.

Answer

ZnO + 2HCl ⟶ ZnCl₂ + H₂O

ZnCl₂ + Na₂CO₃ ⟶ ZnCO₃ + 2NaCl

Dissolve zinc oxide in dil. HCl. Add to it a saturated solution of Na₂CO₃. The precipitate formed by the interchange of radicals is filtered. It is dried to obtain the zinc carbonate.

Question 26

Give balanced equations for the action of a dilute acid on

(i) zinc carbonate,

(ii) potassium bicarbonate for the preparation of the respective salt.

Answer

(i) ZnCO₃ + 2HNO₃ (dil) ⟶ Zn(NO₃)₂ + H₂O + CO₂

(ii) 2KHCO₃ + H₂SO₄ (dil) ⟶ K₂SO₄ + 2H₂O + 2CO₂

Question 27

Give balanced equations for the decomposition of

1. calcium bicarbonate by dil. HCl
2. calcium carbonate by dil. HNO₃
3. sodium sulphite by dil. H₂SO₄
4. zinc sulphide by dil. H₂SO₄.

Answer

1. Decomposition of calcium bicarbonate by dil. HCl
   Ca(HCO₃)₂ + 2HCl ⟶ CaCl₂ + 2H₂O+ 2CO₂
2. Decomposition of calcium carbonate by dil. HNO₃
   CaCO₃ + 2HNO₃ ⟶ Ca(NO₃)₂ + H₂O+ CO₂
3. Decomposition of sodium sulphite by dil. H₂SO₄
   Na₂SO₃ + H₂SO₄ (dil.) ⟶ Na₂SO₄ + H₂O + SO₂
4. Decomposition of zinc sulphide by dil. H₂SO₄
   ZnS + H₂SO₄ ⟶ ZnSO₄ + H₂S

Question 28

State what will be the effect of each of the following solution on blue litmus —

(i) K₂CO₃ soln.

(ii) KCl soln.

(iii) NH₄NO₃ soln.

Answer

(i) As K₂CO₃ is a salt of a strong base (KOH) and weak acid (H₂CO₃), it hydrolyses in water to give alkaline solutions. They have pH more than 7, hence it will have no effect on blue litmus.

(ii) KCl is a salt of a strong acid (HCl) and a strong base (KOH). Hence it's aqueous solution will be neutral in nature and will have no effect on blue litmus.

(iii) NH₄NO₃ is a salt of a strong acid (HNO₃) and weak base (NH₄OH), it hydrolyses in water to give an acidic solution. They have pH less than 7, hence will turn blue litmus red.

Question 1

Answer

Question 2

Answer

Question 3

Answer

Question 1

An example of an acid derived from a mineral is ............... [citric acid / nitric acid / acetic acid]

Answer

An example of an acid derived from a mineral is nitric acid.

Question 2

An example of a base which is not an alkali is ............... [caustic soda / zinc hydroxide / liquor ammonia / caustic potash]

Answer

An example of a base which is not an alkali is zinc hydroxide.

Question 3

An example of a strong acid is dilute ............... [acetic acid / sulphuric acid / tartaric acid / carbonic acid]

Answer

An example of a strong acid is dilute sulphuric acid.

Question 4

An example of a weak alkali is ............... [potassium hydroxide / calcium hydroxide / sodium hydroxide] solution.

Answer

An example of a weak alkali is calcium hydroxide solution.

Question 5

An acid having basicity 1 is ............... [carbonic acid / acetic acid / sulphurous acid]

Answer

An acid having basicity 1 is acetic acid.

Question 6

An acid obtained by dissolving sulphur trioxide in water is ............... [sulphurous acid / sulphuric acid / oleum]

Answer

An acid obtained by dissolving sulphur trioxide in water is sulphuric acid.

Question 7

A volatile acid obtained when nitre reacts with non­-volatile concentrated sulphuric acid on heating is ............... [hydrochloric acid / sulphuric acid / nitric acid]

Answer

A volatile acid obtained when nitre reacts with non­-volatile concentrated sulphuric acid on heating is nitric acid.

Question 8

A base obtained when lead nitrate undergoes thermal decomposition is ............... [trilead tetroxide / lead (IV) oxide/ lead (II) oxide].

Answer

A base obtained when lead nitrate undergoes thermal decomposition is lead (II) oxide.

Question 9

An acid obtained when concentrated nitric acid is heated with sulphur is ............... [sulphurous acid / sulphuric acid / nitrous acid]

Answer

An acid obtained when concentrated nitric acid is heated with sulphur is sulphuric acid.

Question 10

The more volatile acid obtained when the less volatile acid reacts with sodium bicarbonate is ............... [sulphuric acid / carbonic acid / nitric acid]

Answer

The more volatile acid obtained when the less volatile acid reacts with sodium bicarbonate is carbonic acid.

Question 11

The insoluble base obtained when sodium hydroxide reacts with iron (III) chloride is ............... [iron (II) hydroxide / iron (III) hydroxide / iron (II) oxide]

Answer

The insoluble base obtained when sodium hydroxide reacts with iron (III) chloride is Iron (III) hydroxide.

Question 12

A solution whose pH is above 7 is ............... [vinegar/milk/liquor ammonia].

Answer

A solution whose pH is above 7 is liquor ammonia.

Question 13

The salt formed when sulphuric acid reacts with excess caustic soda solution is ............... [sodium bisulphite / sodium sulphate / sodium sulphite / sodium bisulphate].

Answer

The salt formed when sulphuric acid reacts with excess caustic soda solution is sodium sulphate.

Question 14

An example of an acid salt is ............... [CH₃COONa/NaNO₃/Na₂HPO₄/NaKCO₃]

Answer

An example of an acid salt is Na₂HPO₄.

Question 15

An example of a soluble salt is ............... (AgCl/PbSO₄/CaSO₄/CaCl₂)

Answer

An example of a soluble salt is CaCl₂.

Question 16

An example of an insoluble salt is ............... (Na₂CO₃/K₂CO₃/MgCO₃/(NH₄)₂CO₃)

Answer

An example of an insoluble salt is MgCO₃.

Question 17

A salt prepared by neutralization in which titration is involved is ............... [MgCl₂/CaCl₂/NH₄Cl/CuCl₂]

Answer

A salt prepared by neutralization in which titration is involved is NH₄Cl.

Question 18

An insoluble salt prepared by direct combination or synthesis is ............... [FeCl₃/FeSO₄/FeS/Fe(NO₃)₂]

Answer

An insoluble salt prepared by direct combination or synthesis is FeS.

Question 19

A salt prepared by precipitation i.e. by double decomposition of two salt solutions is ............... [Na₂SO₄/PbSO₄/ZnSO₄/CuSO₄]

Answer

A salt prepared by precipitation i.e. by double decomposition of two salt solutions is PbSO₄.

Question 20

A salt prepared by simple displacement i.e. action of dilute acid on a metal is ............... [PbCl₂/CuCl₂/AlCl₃/HgCl]

Answer

A salt prepared by simple displacement i.e. action of dilute acid on a metal is AlCl₃.

Question 21

Decomposition of calcium hydrogen carbonate with ............... [dil. HNO₃ / dil. HCl / dil. H₂SO₄] results in formation of calcium chloride.

Answer

Decomposition of calcium hydrogen carbonate with dil. HCl results in formation of calcium chloride.

Question 22

Action of dilute acid on a metallic sulphide results in evolution of ............... [SO₂/H₂S/CO₂] gas.

Answer

Action of dilute acid on a metallic sulphide results in evolution of H₂S gas.

Question 23

A salt which on hydrolysis produces a neutral solution is ............... (sodium chloride / ammonium chloride / sodium carbonate)

Answer

A salt which on hydrolysis produces a neutral solution is sodium chloride .

Question 1

Name the following:

1. A basic solution which does not contain a metallic element.
2. A normal salt of sodium formed from acetic acid.
3. A base which reacts with an acid to give a salt which on hydrolysis gives a slightly acidic solution.
4. An ion which combines with a polar covalent molecule to form an ammonium ion.
5. A soluble salt formed by direct combination between a light metal & a greenish yellow gas.

Answer

1. Ammonium Hydroxide (NH₄OH) is a basic solution which does not contain a metallic element.
2. Sodium acetate (CH₃COONa) is a normal salt of sodium formed from acetic acid.
3. Ammonium hydroxide (NH₄OH) is a base which reacts with an acid to give a salt which on hydrolysis gives a slightly acidic solution.
4. Hydrogen ion (H⁺) is an ion which combines with a polar covalent molecule to form an ammonium ion.
5. Aluminium chloride (AlCl₃) is a soluble salt formed by direct combination between a light metal & a greenish yellow gas where, light metal is Al and greenish yellow gas is Cl.

Question 2

Identify which of the following terms matches with the appropriate descriptions 1 to 5.

A: Hydracid

B: Monobasic acid

C: Less volatile acid

D: Weak acid

E: Tribasic acid

F: Dibasic acid

G: More volatile acid

1. An acid having basicity 1 and having only one replaceable hydrogen ion per molecule of the acid.
2. An acid which dissociates to give a low concentration of H⁺ ions.
3. An acid containing hydrogen and a non-metallic element other than oxygen.
4. The type of acid which generally displaces another acid when the acid is heated with a salt.
5. The type of acid which reacts with a base to give an acid salt and a normal salt.

Answer

1. B: Monobasic acid
2. D: Weak acid
3. A: Hydracid
4. C: Less volatile acid
5. F: Dibasic acid

Question 3

State which of the following methods is generally used for preparing the salts 1 to 5 given below:

A: Neutralisation — insoluble base and dil. acid

B: Neutralisation — alkali and dil. acid

C: Simple displacement — active metal and dil. acid

D: Direct combination

E: Precipitation [double decomposition]

1. PbCO₃
2. Zn(NO₃)₂
3. NaCl
4. Cu(NO₃)₂
5. FeS

Answer

1. E: Precipitation (Double decomposition)
2. C: Simple displacement — active metal and dil. acid
3. B: Neutralisation — alkali and dil. acid
4. A: Neutralisation — insoluble base and dil. acid
5. D: Direct combination

Question 4

Give balanced equations for the preparation of the following salts:

1. Calcium oxide ⟶ Calcium chloride ⟶ Calcium carbonate
2. Zinc sulphide ⟵ Zn ⟶ Zinc sulphate
3. Iron [II] chloride ⟵ Fe ⟶ Iron [III] chloride
4. Lead [II] oxide ⟶ Lead nitrate ⟶ Lead sulphate
5. Copper [II] oxide ⟶ Copper [II] sulphate ⟵ Copper [II] hydroxide

Answer

1. Calcium oxide ⟶ Calcium chloride ⟶ Calcium carbonate
   1. CaO + 2HCl ⟶ CaCl₂ + H₂O
   2. CaCl₂ + Na₂CO₃ ⟶ 2NaCl + CaCO₃
2. Zinc sulphide ⟵ Zn ⟶ Zinc sulphate
   1. Zn + S ⟶ ZnS
   2. Zn + H₂SO₄ ⟶ ZnSO₄ + H₂
3. Iron [II] chloride ⟵ Fe ⟶ Iron [III] chloride
   1. Fe + 2HCl ⟶ FeCl₂ + H₂
   2. 2Fe + 3Cl₂ ⟶ 2FeCl₃
4. Lead [II] oxide ⟶ Lead nitrate ⟶ Lead sulphate
   1. PbO + 2HNO₃ ⟶ Pb(NO₃)₂ + H₂O
   2. Pb(NO₃)₂ + Na₂SO₄ ⟶ 2NaNO₃ + PbSO₄
5. Copper [II] oxide ⟶ Copper [II] sulphate ⟵ Copper [II] hydroxide
   1. CuO + H₂SO₄ ⟶ CuSO₄ + H₂O
   2. Cu(OH)₂ + H₂SO₄ ⟶ CuSO₄ + 2H₂O

Question 5

The diagram represents the preparation of sodium sulphate salt from dil. H₂SO₄ acid and sodium hydroxide.

1. Name the apparatus 'A'.
2. Name the substance 'X' placed in 'A' and the substance 'Y' placed in B.
3. State the reason for conducting the titration using the apparatus 'A' and 'B'.
4. State which solution is transferred to the evaporating dish and evaporated to point of crystallization for obtaining the salt.
5. State why titration is not conducted for the preparation of copper [II] sulphate crystals by neutralization.

Answer

1. Burette
2. Dil. sulphuric acid (H₂SO₄) is substance 'X' placed in 'A' and sodium hydroxide (NaOH) is the substance 'Y' placed in B.
3. Titration is conducted to determine the completion of the neutralization reaction, i.e. to determine the amount of sulphuric acid required to neutralize a known amount of sodium hydroxide.
4. Aqueous sodium sulphate (Na₂SO₄.10H₂O) is transferred to the evaporating dish and evaporated to point of crystallisation for obtaining the salt.
5. As copper (II) oxide is not soluble in water hence titration is not conducted for the preparation of copper (II) sulphate crystals by neutralization.

Question 6.1

Give reasons for the following:

Concentrated sulphuric acid is a weaker acid compared to dilute sulphuric acid.

Answer

Strength of an acid is determined by the amount of hydronium ions it produces in it's aqueous solution. As dilute sulphuric acid dissociates almost completely thereby producing a high concentration of hydronium ions hence is a stronger acidic than concentrated sulphuric acid.

Question 6.2

Give reasons for the following:

An aqueous solution of the salt ammonium chloride is acidic in nature while an aqueous solution of sodium chloride is neutral.

Answer

As ammonium chloride is a salt of a weak alkali and strong acid, hence it's aqueous solution is acidic in nature. On the other hand, sodium chloride is a product of strong alkali and strong acid hence, it's aqueous solution is neutral in nature.

Question 6.3

Give reasons for the following:

In the preparation of an insoluble salt from another insoluble salt by precipitation [double decomposition], dilute nitric acid and not dilute sulphuric acid is generally used.

Answer

In the preparation of an insoluble salt from another insoluble salt by precipitation, if dil. sulphuric acid is directly used then it forms an insoluble precipitate and slows the reaction.

For example — Direct addition of dil. sulphuric acid to PbCO₃ is an impractical method of preparing lead sulphate since PbSO₄ is insoluble and forms a coating on PbCO₃, thereby the reaction slowly comes to a stop.

Question 6.4

Give reasons for the following:

Acetic acid does not form an acid salt but forms a normal salt.

Answer

Acetic acid (CH₃COOH) is a monobasic acid and contains only one replaceable hydrogen atom, hence it can only form normal salt and not acid salts.

Question 6.5

Give reasons for the following:

Sulphurous acid forms two types of salts on reaction with an alkali.

Answer

Sulphurous acid (H₂SO₃) is a dibasic acid, i.e., it contains two replaceable hydrogen ions per molecule of the acid. Hence, it can form normal salt as well as acid salt on reaction with an alkali.