Question 3

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

Answer

Pb(NO3)2 + 2NaCl ⟶ PbCl2 + 2NaNO3

2008

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.

2009

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)]

2010

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

2011

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

2012

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

2013

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

2014

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 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+ + SO42-

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+ + SO42-
   H2SO4 + 2H2O ⇌ 2H3O+ + SO42-

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

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 —

NaCl+H2SO4[conc.]→<200°CNaHSO4+HClNaCl+H2​SO4​[conc.]<200°C​NaHSO4​+HCl

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

NaNO3+H2SO4[conc.]→<200°CNaHSO4+HNO3NaNO3​+H2​SO4​[conc.]<200°C​NaHSO4​+HNO3​

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:

NaCl+H2SO4[conc.]→<200°CNaHSO4+HClNaCl+H2​SO4​[conc.]<200°C​NaHSO4​+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.

Select correct acid/base/salt

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 .

Unit Test Paper 3A — Acids, Bases & Salts

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.

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.