Equilibrium Questions and Answers AP Inter 1st Year Chemistry Chapter 6

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 6th Lesson Equilibrium Class 11 Textbook Exercise Questions and Answers.

Equilibrium Class 11 Questions and Answers AP Inter 1st Year Chemistry 6th Lesson

I. Multiple Choice Questions (1 Mark)

Question 1.
The pH of a solution of hydrochloric acid is 4. The molarity of the solution is:
(1) 4.0 M
(2) 0.4 M
(3) 0.0001 M
(4) 0.04 M
Answer:
(3) 0.0001 M

Question 2.
The strong conjugate base in the following is
(1) N\(\mathrm{O}_3^{-}\)
(2) \(\mathrm{Cl}^{-}\)
(3) \(\mathrm{SO}_4^{2-}\)
(4) CH₃COO^–
Answer:
(4) CH₃COO^–

Question 3.
When NH₄Cl is added to NH₄OH solution the dissociation of ammonium hydroxide is reduced. It is due to:
(1) common ion effect
(2) hydrolysis
(3) oxidations
(4) reduction
Answer:
(1) common ion effect

Question 4.
The following reaction is at equilibrium
Fe³⁺_(aq) + SCN^– \(\text { ⇋ }\) [Fe(SCN)]²⁺_(aq)
In the above reaction, colour intensity of red colour can be increased by
(1) addition of oxalic acid which reacts with Fe³⁺ ions
(2) addition of Hg²⁺ ions which reacts with SCN^– ions
(3) addition to KSCN
(4) red colour intensity cannot be changed
Answer:
(3) addition to KSCN

Question 5.
K_sp of a sparingly soluble salt AB₂ is 4 × 10^-12. The solubility of the salt is
(1) 2 × 10^-6M
(2) 4 × 10^-4 M
(3) 1 × 10^-12 M
(4) 1 × 10^-4 M
Answer:
(4) 1 × 10^-4 M

Question 6.
An aqueous solution of ammonia consists of
(1) H⁺ only
(2) OH^– only
(3) N\(\mathrm{H}_4^{+}\) only
(4) N\(\mathrm{H}_4^{+}\) and OH^–
Answer:
(4) N\(\mathrm{H}_4^{+}\) and OH^–

Question 7.
The pH of a 0.05 M of H₂SO₄ solution is
(1) 0.05
(2) 1
(3) -1
(4) 1.3010
Answer:
(2) 1

Question 8.
If the molar Solubility of Zirconium phosphate Zr₃(PO₄)₄ is ‘S’. Solubility product K_sp of Zirconium phosphate is
(1) 6912 (S)⁵
(2) 6912 (S)⁷
(3) 108 (S)⁷
(4) 27 (S)⁵
Answer:
(2) 6912 (S)⁷

Question 9.
For a reversible reaction, if the concentration of reactants is increased, the equilibrium constant of the reaction.
(1) Increases
(2) Remains constant
(3) Decreases
(4) Depends on amount of reactant
Answer:
(2) Remains constant

Question 10.
The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively. The pH of ammonium acetate solution is
(1) 6.098
(2) 7.005
(3) 8
(4) 5.0
Answer:
(2) 7.005

Question 11.
Which among the following salt solutions is basic in nature ?
(1) Ammonium chloride
(2) Ammonium sulphate
(3) Ammonium nitrate
(4) Sodium acetate
Answer:
(4) Sodium acetate

Question 12.
Observe the following equilibrium at T (Kelvin)
H_2(g) + I_2(g) ⇌ 2HI_(g)
Which one of the following does not disturb the above equilibrium at constant volume ?
(1) Addition of H_2(g)
(2) Removal of HI_(g)
(3) Addition of I_2(g)
(4) Addition of He_(g)
Answer:
(4) Addition of He_(g)

Question 13.
Solution of 0.1 N NH₄OH and 0.1 N NH₄Cl has pH 9.25. Find out pK_b of NH₄OH.
(1) 9.25
(2) 4.75
(3) 3.75
(4) 8.25
Answer:
(2) 4.75

Question 14.
For the following process the expression for the K_C ?

(1) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^2}{[\mathrm{Ni}][\mathrm{CO}]^4}\)
(2) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)
(3) K_c = \(\frac{[\mathrm{Ni}][(\mathrm{CO})]^{-}}{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^{2-}}\)
(4) K_c = \(\frac{[\mathrm{CO}]^4}{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^2}\)
Answer:
(2) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)

Question 15.
K₁ and K₂ are equilibrium constants for reactions (i) & (ii). Then the relation between K₁ and K₂.

(1) K₁ = \(K_2^2\)
(2) K₁ = \(\frac{1}{\mathrm{~K}_2}\)
(3) K₁ = \(\mathrm{k}_2^0\)
(4) K_c = \(\frac{[\mathrm{CO}]^4}{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^2}\)
Answer:
(4) K_c = \(\frac{[\mathrm{CO}]^4}{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^2}\)

II. Fill in the Blanks (1 Mark)

Question 1.
Acidity of BF₃ can be explained on the basis of ………. Theory of acids and bases.
Answer:
Lewis

Question 2.
The nature of aqueous solution of ammonium chloride (NH₄Cl) is ……….
Answer:
Acidic

Question 3.
K_p fór the following equation: ………………………… .
Answer:
Kp = P_CO2

Question 4.
The molarity of pure ‘water is ………………………… .
Answer:
55.5M

Question 5.
If K_C is in the range of ……. appreciable concentrations of both reactants and products arê present.
Answer:
10^-3 to 10³

III. One Word Answer Questions (1 Mark)

Question 1.
What is Bronsted base?
Answer:
The substance which accepts a proton from the other substance is called Bronsted base.
Example: NH₃, H₂O, O\(\mathrm{H}^{\ominus}\), CH₃CO\(\mathrm{O}^{\ominus}\) etc.

Question 2.
What is Lewis acid ?
Answer:
A substance which can accept an electron pair to form a coordinate covalent bond with donor is called lewis acid.
Example: \(\mathrm{H}^{\oplus}\), BF₃, SnCl₂ etc.

Question 3.
What is the value of K_W at 25° C ?
Answer:
At 25°C, the value of Ionic product of water, L_W is 1.0 × 10^-14 mole²/lit².

Question 4.
What is conjugate acid-base pair ?
Answer:
A pair of Bronsted acid and base that differs by one proton (\(\mathrm{H}^{\oplus}\)) is known as conjugate acid-base pair.
Example: NH₃ and N\(\mathrm{H}_4^{\oplus}\), H₂O and H₃\(O^{\oplus}\) etc.

Question 5.
Write the relation between Gibbs energy and equilibrium constant.
Answer:
The relation between Gibbs free energy and the equilibrium constant of a chemical reaction is given by the following fundamental thermodynamic equation
∆G° = – RT ln k
Where, ∆G° – Standard Gibbs free energy change
R – Universal gas constant
T – Temperature is kelvin
K – Equilibrium constant of the reaction

IV. Very Short Answer Questions (2 Marks)

Question 1.
State law of ehemical equilibrium
Answer:
The law of chemical equilibrium equilibrium states that “At a given temperature, the ratio of the product of the concentration of the products to the product of the concentrations of the reactants each raised to the power of their respective stoichiometric coefficients is constant at equilibrium”.
For a general reversible reaction,
aA + bB ⇌ cd + dD
At equilibrium, the equilibrium constant, K_c is given by
K_C = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{~A}]^{\mathrm{a}}[\mathrm{~B}]^{\mathrm{b}}}\)
Where, [A], [B], [C], [D] are the equilibrium concentrations of the respective species.
a, b, c, d are the stoichiometric coefficients.

Question 2.
What is homogeneous equilibrium ? Give two examples.
Answer:
A chemical equilibrium in which all thd reactants and products are present in the same phase is called homogeneous equilibrium.
Examples: N₂(g) 3H₂(g) ⇌ 2NH₃(g)
CH₃COOH(aq) ⇌ CH₃CO\(O^{\Theta}\) (aq) + \(\mathrm{H}^{\oplus}\)(aq)

Question 3.
What is heterogeneous equilibrium ? Give two examples.
Answer:
A chemical equilibrium in Which the reactants and / or products are present in more than one phase is called Heterogeneous equilibrium.
Examples: CaCO₃(s) ⇌ CaO(s) + CO₂(g)
CO₂(g) + C(s) ⇌ 2CO(g)

Question 4.
What is meant by dynamic equilibrium ?
Answer:
Dynamic equilibrium is a state in a reversible chemical reaction, where the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of reactants and products remain constant over time.

Question 5.
Give two chemical equilibrium reactions for which K_p > K_c.
Answer:
Relation between K_p and K_c is K_p = K_c (RT)^∆n
where R – Gas constant
T – Temperature in Kelvin
∆n – Number of moles of gaseous products – Number of moles of gaseous reactants – n_p – n_R
If ∆n > 0, then K_p > K_c.

Examples:

1) Decomposition of N₂O₄:
N₂O₄(g) ⇌ 2NO₂(g)
∆n = n_p – n_R
= 2 – 1
= 1
∴ K_p > K_c

2) Decomposition of CaCO₃
CaO₃(s) ⇌ CaO(s) + CO₂(g)
∆n = n_p – n_R = 1 – 0 = 1
∴ K_p > K_c

Question 6.
Give two chemical equilibrium reactions for which K_p < K_c.
Answer:
Relation between K_p and K_c is K_p = K_c. (RT)^∆n
∆n = Number of moles of gaseous products – Number moles of gaseous reactants.
= n_p – n_R

Examples:
1) Synthesis of Ammonia by Haber process
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
∆n = n_p – n_R = 2 – (1 + 3) = 2 – 4 = – 2
∴ K_p < K_c

2) Formation of surphur Trioxide.
2SO₃(g) + O₂(g) ⇌ 2SO₃(g)
∆n = n_p – n_R = 2 – (2 + 1) = 2 – 3 = -1
∴ K_p < K_c

Question 7.
Write the equations for the conversion of K_c to K_p for the following reaction.
CO_(g) + H₂O_(g) ⇌ CO_2(g) + H_2(g)
Answer:
Relation between k_p and K_c is K_p = K_c (RT)^∆n
Given reaction
CO(g) + H₂O_(g) ⇌ CO_2(g) + H_2(g)
∴ ∆h = n_p – n_R
= (1 + 1) – (1 + 1)
= 2 – 2
= 0
∴ K_p = K (RT)^∆n = K_c (RT)⁰ = K_c
⇒ K_p = K_c.

Question 8.
What are the factors which influence the chemical equilibrium ?
Answer:
The factors influencing the chemical equilibrium are

1. Concentration of reactants and the products
2. Temperature of reaction
3. Pressure of reaction
4. Addition of inert gas
5. Catalysts.

Question 9.
What is meant by ionic product of water ?
Answer:
The product of concentrations of \(\mathrm{H}^{\oplus}\) and O\(\mathrm{H}^{\oplus}\) ions in water at a given temperature is called Ionic product of water. It is denoted by K_w.
∴ K_w = [O\(\mathrm{H}^{\oplus}\)] [O\(\mathrm{H}^{\ominus}\)]
The value of ionic product of water at 25°C is
1.0 × 10^-14 mole² / lit².

Question 10.
Give two examples of salts whose aqueous solutions are acidic.
Answer:
The following are the examples of salts whose aqueous solutions are acidic.

1) Ammonium chloride (NH₄Cl): The N\(\mathrm{H}_4^{\oplus}\) ion from NH₄Cl is a weak acid that donates \(\mathrm{H}^{\oplus}\) ions in water making the solution acidic
N\(\mathrm{H}_4^{\oplus}\) + H₂O ⇌ N\(\mathrm{H}_3{ }^{\oplus}\) + H₃\(0^{\oplus}\)

2) Aluminium chloride (AlGl₃): The A\(l_3^{\oplus}\) ion hydrolizes in water, forming hydrated Aluminium complexes that release \(\mathrm{H}^{\oplus}\) ions, resulting in an acidic solution.
Al³⁺ + 6H₂O ⇌ [Al(H₂O)₅OH]²⁺ + H₃\(O^{\oplus}\)

V. Short Answer Questions (4 Marks)

Question 1.
Derive the relation between K_p and K_c for the equilibrium reaction.
Answer:
Gives equilibrium reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Relation between K_p and K_c is K_p = K_c (RT)^∆n
Where, K_p – Equilibrium constant in terms of partial pressures.
K_c – Equilibrium constant in terms of molar concentrations
R – Gas constant
T – Temperature in Kelvin
∆_n – Number of moles of gaseous products number of moles of gaseous reactants = n_p – n_R
For the given equation,
∆_n = n_p – n_R
= 2 – (1 + 3)
= 2 – 4
= -2
∴ K_p = K_c (RT)^∆n
⇒ K_p = K_p (RT)^-2–
This shows that K_p < K_c

Question 2.
State Le Chateliers’s Principle. Discuss the application of Le Chatelier’s principle for the industrial synthesis of Ammonia by Haber’s process.
Answer:
Statement: “If a system at equilibrium is subjected to the change of pressure, temperature (or) concentration, the system is shifted in such away as to nullify the effect of change”.
Applications of Le Chatelier’s principle for the industrial synthesis of Ammonia by Haber’s process:

Nitrogen and Hydrogen combine to form ammonia. The formation of ammonia is reversible and exothermic reaction. It is accompanied by decrease in volume.

Effect of Pressure: 1 volume of N₂ combines with 3 volumes of H₂ to form 2 volumes of NH₃. There is decrease in volume in the forward reaction (4 volumes to 2 volumes). According to Le Chatelier’s principle increase of pressure favours the reaction where there is decrease in volume. So higher the pressure, greater the yield of ammonia. In practice 200 atmospheres are used in the manufacture of ammonia by Haber’s process. Low pressures favour the reverse reaction i.e, decomposition of NH₃ already formed.

Effect of Temperature : The formation of ammonia (forward reaction) is exothermic reaction
N₂(g) + 3H₂(g) ⇌ 2 NH₃ (g), ∆H = -92kJ

Low temperatures favour the forward reaction. But at low temperatures the reaction is too slow. Therefore an optimum temperature (725 K – 775 K) is chosen in Haber’s process. To speed up the reaction, a catalyst, finely divided iron is used. To increase the activity of the catalyst molybdenum or a mixture of oxides of K and Misused as promoter. .

The reverse reaction (i.e.,) decomposition of NH₃ is an endothermic reaction. High temperatures favour the decomposition of NH₃. Therefore high temperatures are avoided in Haber’s process.

Thus the optimum conditions are, Pressure : 200 atm
Temperature : 725-775 K
Catalyst : Fe (Powdered)
Promoter : Mo or (K₂O + Al₂O₃)

Question 3.
State Le Chateliers’s Priciple. Discuss the application of Le Chatelier’s principle for the manufacture of SO₃ by contact process.
Answer:
Statement: “If a system at equilibrium is subjected to the change of pressure, temperature (or) concentration, the system is shifted in such a way as to nullify the effect of change”.

Applications of Le Chatelier’s principle for the manufacture of SO₃ by contact process:
The formation of SO₃ is reversible and exothermic reaction. It is accompanied by decrease in volume.

Effect of Pressure : Two volumes of SO₂ and one volume of O₂ combine to give 2 volumes of SO₃. According to Le Chatelier’s Principle increase of pressure favours the reaction where there is decrease in volume. The formation of SO₃ is accompanied by decrease in volume (3 volumes to 2 volumes). Higher the pressure greater is the yield of SO₃. But in contact process high pressures are not used because towers used in the manufacture are corroded by the acid at these high pressures.

Low pressures favour the decomposition of SO₃ as there is increase in the volume (2 volumes to 3 volumes). Therefore, optimum pressures are used (1.5 to 1.7 atmosphere).

Effect of temperature : The formation of SO₃ is exothermic. 189 kJ of heat is evolved. High temperatures favour the reverse reaction which is endothermic and do not favour the forward reaction which is exothermic. Low temperatures are favourable for the formation of SO₃. At low temperature, the reaction is too slow. Therefore, an optimum temperature 673 K is used. To speed up the reaction V₂O₅ is used as catalyst. Thus the optimum conditions are
Pressure : 1,5 to 17 atm
Temperature : 673 K
Catalyst : V₂O₅ (or) Platinised asbestos

Question 4.
The species H₂O, HC\(\mathrm{O}_3^{-},\) HS\(\mathrm{O}_4^{-}\) and NH₃ can act both as Bronsted acids and bases. Give the corresponding conjugate acid and base for each of them.
Answer:
Each of the given species can act as both Bronsted acid (donates a proton, H⁺) and a Bronsted base (accepts a proton, H⁺). Below are their corresponding conjugate acids and bases:

1) Water (H₂O):
As a base (accepts H⁺): H₂O + H⁺ → H₃O⁺
Conjugate acid : H₃O⁺ (hydronium ion)
As an acid (donates H⁺): H₂O → H⁺ + OH^–
Conjugate base : OH^– (hydroxide ion)

2. Bicarbonate ion (HC\(\mathrm{O}_3^{-}\)):
As a base (accepts H⁺): HC\(\mathrm{O}_3^{-}\) + H⁺ → H₂CO₃
Conjugate acid: H₂CO₃ (carbonic acid)
As an acid (donates H⁺) : HC\(O_3^{-}\) → H⁺ + C\(\mathrm{O}_3^{2-}\)
Conjugate base: \(\mathrm{CO}_3^{2-}\) (carbonate ion)

3. Bisulphate ion (HS\(O_4^{-}\)):
As a base (accepts H⁺): HS\(\mathrm{O}_4^{-}\) + H⁺ → H₂SO₄
Conjugate acid: H₂SO₄ (sulphuric acid)
As an acid (donates H⁺): HS\(\mathrm{O}_4^{-}\) → H⁺ + S\(\mathrm{O}_4^{2-}\)
Conjugate base: S\(\mathrm{O}_4^{2-}\) (sulphate ion)

4. Ammonia (NH₃):
As a base (accepts H⁺): NH₃ + H⁺ → \(\mathrm{NH}_4^{+}\)
Conjugate acid: N\(\mathrm{H}_4^{+}\) (ammonium ion)
As an acid (donates H⁺): (rare, but possible in strongly basic conditions)
NH₃ → N\(\mathrm{H}_2^{-}\) + H⁺
Conjugate base: N\(\mathrm{H}_2^{-}\) (amide ion)

Question 5.
Write notes on
(i) Common ion effect
(ii) Solubility product
Answer:
Common ion effect: The decrease in the ionisation or dissociation of a weak electrolyte, by the addition of a strong electrolyte having an ion common with the weak electrolyte, is known as common ion effect.
Example: The dissociation of NH₄OH is diminished by the addition of NH₄Cl due to the common ion, N\(\mathrm{H}_4^{+}\) ion
NH₄OH ⇌ \(\mathrm{NH}_4^{+}\) + OH^–

Importance of Common ion effect in qualitative analysis:

1. This principle is used in the systematic qualitative analysis of cations.

2. Selective precipitation of ions : In group separation, adding a common ion helps precipitate specific ions while keeping others in solution.
Example: In Group II (Acidic H₂S Group), Pb²⁺, Cu²⁺, and Cd²⁺ are precipitated as sulphides (PbS, CuS, CdS) by adding H₂S in acidic medium. The common ion (S^2-) suppresses the solubility of these sulphides, ensuring complete precipitation.

3. Confirmation Tests: The common ion effect is used to confirm the presence of ions by enhancing precipitation.
Example : Confirming Ba²⁺ After precipitation as BaCrO₄, adding K₂CrO₄(source of Cr\(\mathrm{O}_4{ }^{2-}\)) ensures all Ba²⁺ precipitates, leaving no doubt in identification.

4. The common ion effect helps control the pH and ionic concentrations to ensure only specific ions precipitate at a given time. This prevents false positives or co-precipitation of unwanted ions.

5. Separation of chemically similar ions: Ions with similar properties (e.g., Ag⁺ and Pb²⁺) can be separated using a common ion.
Example: AgCl is insoluble in dilute HCl, while PbCl₂ is slightly soluble. Adding HCl ensures Ag⁺ precipitates completely, while Pb²⁺ remains partly in solution.

6. The common ion effect principle is also used in controlling the H⁺ ion concentration in buffer solutions.

7. It is also used in the purification of common salts by passing dry HCl gas into impure salt solution.

Solubility product (K_sp): The solubility product constant is the equilibrium constant for the dissolution of a sparingly soluble ionic compound in water. It is the product of the molar concentrations of its ions at saturation point, each raised to the power of their stoichiometric coefficients. It represents the extent to which a salt dissolves in water to form its constituent ions. It Can be denoted by K_sp.
Consider dissociation of a salt AB
A_xB_y(s) ⇌ xA^y+ (aq) + y B^x-(aq)

For this reaction, the solubility product can be written as, K_sp = [A^y+]^x[B^x-]^y

Applications of K#p:

1) In predicting Precipitation:

• If the ion product > K_sp → precipitation occurs
• If ion product < K_sp → solution is unsaturated, no precipitate
• If ion product = K_sp → saturated solution (at equilibrium)

2) From known K_sp, we can determine molar solubility of salts in water.
Used in qualitative analysis to separate ions based on different K_sp values. Presence of a common ion reduces solubility, shifting equilibrium left.

Question 6.
Explain the concept of Bronsted-Lowry acids and bases. Illustrate with suitable examples.
Answer:
In order to overcome the limitations of Arrhenius theory, Bronsted and Lowry in 1923 proposed a new acid – base theory. According to this theory, acid and base can be defined as follows.
ACID: A substance which can donate a proton to other substances is called an acid.
BASE: A substance which can accept a proton from other substances is called a base.
Neutralization: Transfer of proton from acid to base is called neutralization. Consider the following reactions.

In the above equations, HCl donates a proton to water. So, HCl is an acid and water is a base. In the reverse reaction, H₃O⁺ loss a proton to act as an acid and Cl- accepts a proton to act as a base. In the second equation, NH₃ accepts a proton from H₂O. So, NH₃ is a base and water is an acid. In the reverse reaction, N\(\mathrm{H}_4^{+}\) acts as an acid and OH^– acts as a base.

In these reactions, the acid, base pair differs by one proton only. These are known as Conjugate acid, base pairs. “The acid – base pair which differs by one proton is called as conjugate acid – base pair”.

Strength of acids and bases : The strength of acid and base depends upon its ability to donate or accept proton.
Strong acid : An acid which has a greater tendency to donate proton is called strong acid.
E.g.: HCl, H₂SO₄ HNO₃ etc.
Weak acid : An acid which has less tendency to donate proton is called weak acid.
E.g.: CH₃COOH, HCOOH, H₂CO₃ etc.
Strong base : A base which has a greater tendency to accept proton is called strong base.
E.g.: CH₃COO^–, OH^–, CN^– etc.
Weak base: A base which has less tendency to accept proton is called weak base.
E.g. : Cl^–, Cl\(O_4^{-}\), HS\(\mathrm{O}_4^{-}\) etc.

Thus, the conjugate base of a weak acid is always strong and conjugate acid of a weak base is always strong.

Question 7.
Explain Lewis acids and hases with suitable’ examples. Classify the following species into lewis acids and Lewis bases.
(a) OH
(b) F^–
(c) H⁺
(d) BCl₃
Answer:
G.N. Lewis in 1923 proposed new acid — base theory on the basis of electronic theory of valency. According to this theory, acid and base can be defined as follows.
ACID : A substance that can accept an electron pair to form coordinate covalent bond iš called Acid.
E.g.: H⁺, BF₃, ZnCl₂, AlCl₃, FeCl₃ etc.

BASE: A substance that can donate an electron pair to form coordinate covalent bond is called Base.
E.g.: OH^–, CN^–, H₂O, NH₃ etc.
Neutralization: Formation of coordinate covalent bond between acid and base is called Neutralization.

E.g.:
1) Formation of NE₃ – BF₃ : Ammbnia contains one electron pair of nitrogen. So it acts as Lewis base. BF₃ acts as Lewis acid due to containing of vacant orbital in the valency orbit of Boron.

Types of Lewis acids: Lewis acids are divided into following types.

• All cations. E.g.: H+, Ag+, Cu²⁺, Fe³⁺ etc.
• Compounds in which central atom has incomplete octet (electron deficient compounds).
  E.g.: BCl₃, BF₃, AlCl₃, FeCl₃ etc.
• Compounds in which central atom has available d-orbitals and can expand their octet.
  E.g.: SiF₄, SF₄, SnCl₄, etc.
• Molecules containing double bonds between dissimilar atoms.
  E.g.: CO₂, SO₂, NO₂ etc.
• Elements with six electrons in the valence shell can act as Lewis acids.

Types of Lewis bases: Lewis bases are divided into following types.

• All anions. E.g.: Cl^–, OH^–, CN^–, O^2-, S\(\mathrm{O}_4^2\) etc.
• Molecules containing lone pair of electrons. E.g.: NH₃, H₂O, ROH, RNH₂ etc.
• Compounds containing multiple bonds. E.g.: C₂H₄, C₂H₂ etc.

Given examples
a) OH^–
b) F^–
c) H⁺
d) BCl₃

a) Hydroxyl ion (OH^–) is a Lewis base as it can donate an electron lone pair.
b) Flouride ion (F^–) acts as a Lewis base as it can donate any one of its four electron lone pairs.
c) A Proton (H⁺) is a Lewis acid as it can accept a lane pair of electrons from bases like hydroxyl ion and fluoride ion.
d) BCl₃ acts as a Lewis acid as it can accept a lone pair of electrons from species
like Ammonia or amine molecules.

Question 8.
Calculate the pH of
(a) 10^-3 M HCl
(b) 10^-3M H₂SO₄ [log 0.3010]
Answer:
a) HCl is a strong acid. It completely dissociates in water
HCl → \(\mathrm{H}^{\oplus}+\mathrm{Cl}^{\ominus}\)
Given that concentration of \(H^{\oplus}\), [] = 10^-3M.
∴ pH = – log []
= -log (10^-3) = 3 log 10
= 3 × 1
= 3

b) H₂SO₄ undergo dissociation as H₂SO₄ → 2 + S\(\mathrm{O}_4^{2 \ominus}\)
Concentration of \(\mathbf{H}^{\oplus}\), [] = 2 × 10^-3M.
∴ p^H = – log []
= – log (2 × 10^-3)
= – log2 – log(10^-3)
= – log2 – (-3 log 10)
= – log2 + 3 log 10
= – 0.3010 + 3(1)
= – 0.3010 + 3
= 2.699
= 3

Question 9.
Calculate the pH for
a. 0.001 M NaOH
b. 0.01M Ca(OH)₂
Answer:
a) p^H of 0.001M NaOH
Given that concetration of O\(H^{\ominus}\), [O\(H^{\ominus}\)] = 0.001 M p^OH = – log [O\(\mathbf{H}^{\Theta}\)]
= -log (0.001)
= -(-3)
= 3
We know that p^H + p^OH = 14
⇒ p^H = 14 – p^OH
⇒ p^H = 14 – 3
⇒ p^H = 11

b) p^H of 0.01 M Ca(OH)₂
Given that concentration of O\(\mathrm{H}^{\ominus}\), [O\(\mathrm{H}^{\ominus}\)] = 2 × 0.01
= 0.02 M
∴ p^OH = – log [-O\(\mathrm{H}^{\ominus}\)]
= – log (0.02)
= – (-1.6989)
= 1.6989.
We know that, p^H + p^OH = 14
⇒ p^H = 14 – p^OH
⇒ p^H = 14 – 1.6989
⇒ p^H = 12.3011

Question 10.
The pH of 0.1 M solution of weak mono protic acid is 4.0 calculate its [H⁺] and K_a.
Answer:
Given that, concentration of weak mono protic acid is 0.1M
P^H of the solution is 4.0
Calculation of []
Given that p^H = 4.0
We know that, p^H = -log []
⇒ [] = 10^-PH = 10^-4 M.
Calculation of K_a :
For a weak mono protic acid (HA), equilibrium is
HA ⇌ \(\mathrm{H}^{\oplus}\) + \(\mathbf{A}^{\Theta}\)
Acid dissociation constant, K_a = \(\frac{\left[\mathrm{H}^{\oplus}\right]\left[\mathrm{A}^{\ominus}\right]}{[\mathrm{HA}]}\)
[] = [] = 10^-4 M
Given that [HA] = 0.1 M
∴ K_a = \(\frac{10^{-4} \times 10^{-4}}{0.1}\) = \(\frac{10^{-8}}{10^{-1}}\) = 10^-7

Question 11.
One litre of buffer solution contains 0.1 mole of acetic acid and 1 mole of sodium acetate. Find its p^H if pK_a of CH₃COOH is 4.8.
Answer:
Given that concentration of Acetic acid = 0.1 M
Concentration of sodium acetate (salt) = 1M
pK_a of CH₃COOH = 4.8, volume = 1 lit
p^H of Buffer solution can be calculated y using the following Henderson,- Hassel- balch equation
p^H = pK_a + log \(\frac{[\text { Salt }]}{[\text { Acid }]}\)
= 4.8 + log \(\left(\frac{1}{0.1}\right)\)
= 4.8 + log(10)
= 4.8 + 1
= 5.8

VI. Long Answer Questions (8 Marks)

Question 1.
What are equilibrium process ? Explain equilibrium in Physical and Chemical processes with examples.
Answer:
Equilibrium processes are reversible processes in which the rate of the forward reaction becomes equal to the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant over time, though both reactions continue to occur.
Consider a reversible reaction taking place in a closed vessel.
A + B ⇌ C + D
At the beginning we have only the reactants A and B, and their concentrations are maximum. As the reaction proceeds the reactants A and B change into the products C and D. The concentrations of the products increase gradually.

The rate of forward reaction diminishes while the reverse reaction sets in and proceeds with increasing speed. A state is soon reached where the speeds of forward and backward reactions become equal. If the rate of forward reaction = the rate of reverse reaction, the system is said to have attained a state of equilibrium. Once equilibrium is reached, there is no further change in the composition of the system. The system appears to be stand still although it is dynamic. The products are formed by forward reaction just as fast as they change back into reactants by the reverse reaction.

The state at which the rate of forward reaction is equal to the rate of the reverse reaction in reversible reaction is known as the equilibrium state or chemical equilibrium.

Equilibrium can be classified into two types:

1. Physical Equilibrium (Involves no chemical change).
2. Chemical Equilibrium (Involves reversible chemical reactions).

1. Physical Equilibrium: This equilibrium occurs when physical processes (phase changes; dissolution) reach a balance between two phases.

Examples:
a) Solid-Liquid Equilibrium (Melting/Freezing)
Example: ice ⇌ Water at 0°C.
Forward process: Ice melts into water.
Reverse process: Water freezes into ice.
At equilibrium: Melting rate = Freezing rate.

b) Liquid-Vapour Equilibrium (Evaporation/Condensation)
Example: Water ⇌ Water vapour in a closed container.
Forward process : Water evaporates.
Reverse process: Vapour condenses.
At equilibrium: Evaporation rate = Condensation rate.

2. Chemical Equilibrium : This equilibrium occurs in reversible chemical reations when the rates of the forward and reverse reactions balance. Examples:

(a) Synthesis of Ammonia (Haber Process): N₂(g)+ 3 H₂ (g) ⇌ 2NH₃(g)
Forward: N₂ and H₂ react to form NH₃.
Reverse : NH₃ decomposes back into N₂ and H₂.
At equilibrium: Concentrations of N₂,H₂, and NH₃ remain constant and reaction rates are equal.

(b) Acid Dissociation: CH₃COOH ⇌ CH_3<COO^– + H⁺
Acetic acid partially dissociates in water, and an equilibrium is established between the undissociated acid and the ions formed.

Question 2.
What is meant by dynamic equilibrium ? Explain the suitable examples.
Answer:
Dynamic equilibrium: The forward and reverse reactions of a reversible reaction continue to take place with equal rates simultaneously at the equilibrium stage also. Hence, the equilibrium is called Dynamic equilibrium.

Explanation: In order to understand the dynamic nature of the reaction, synthesis of ammonia is carried with exactly the same starting conditions but using D₂ (Deuterium) in place of H₂. The reaction mixtures starting either with H₂ or with D₂ reach equilibrium with the same composition, except that D₂ and ND₃ are present instead of H₂ and NH₃.

After equilibrium is attained, these two mixtures (H₂, N₂, NH₃ and D₂, N₂, ND₃) are mixed together and left for a while. Later when this mixture analysed, it is found that the concentration of ammonia is just the same as before. However, when this mixture is analysed by a mass spectrometer, it is found that ammonia and all deuterium containing forms of ammonia (NH₃, NH₂D, NHD₂ and ND₃), and dihydrogen and its deuterated forms (H₂, HD, and D₂) are present. Thus, one can conclude that scrambling of H and D atoms in the molecules must result from a Continuation of the forward and reverse reactions in the mixture. If the reaction had simply stopped when they reached equilibrium, then there would have been no mixing of isotopes in this way.

Use of isotope (deuterium) in the formation of ammonia clearly indicates that chemical reaction react a state of dynamic equilibrium in which the rates of forward and reverse reactions are equal and there is no net change in composition of the equilibrium mixture. Equilibrium can be attained from both sides, whether we start reaction by taking H₂(g) and N₂(g) and get NH₃(g) or by taking NH₃(g) and decomposing it into N₂(g) and H₂(g).
N₂(g) + 3H₂(g) ⇌ 2 NH₃(g).

Question 3.
What are the important features of equilibrium constant ? Discuss any two applications of equilibrium constant.
Answer:
The ratio of product of molar concentrations of products to the product of molar concentrations of reactants each raised to the power of their coefficients in the balanced chemical equation is called equilibrium constant (K).
Consider a general reaction at equilibrium
aA + bB ⇌ cC + dD
Where, A and B are reactants, C and D are products where as a, b, c and d are coefficients.
The equilibrium constant for this reaction is given by
K = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{~A}]^{\mathrm{a}}[\mathrm{~B}]^{\mathrm{b}}}\)

Important Features of Equilibrium Constant:

1. Constant at a given temperature : The value of equilibrium constant, K depends only on temperature, not on initial concentrations or pressures. Changing temperature shifts equilibrium, changing the value of K.
2. Predicts direction of reaction :
   • If K > 1 products are favoured (reaction proceeds forward).
   • If K < 1, reactants are favoured (reaction tends to go backward).
3. Independent of catalyst: The value of equilibrium constant, K is not affected by the presence of a catalyst. Catalysts only affect the speed of reaching equilibriilm, not its position.
4. Affected by temperature: Changing temperature changes the value of K, depending on whether the reaction is exothermic or endothermic.
5. Independent of initial concentrations : No matter the starting amounts of reactants/products, K remains the same if temperature is constant.
6. Reverse reaction: For the reverse reaction, K is the reciprocal of the K for the forward reaction.

Applications of Equilibrium Constant:

1. Predicting the Extent of a Reaction:
   • The magnitude of K tells how far a reaction goes before reaching equilibrium.
   • Large K (>>1): Reaction goes almost to completion (more products).
   • Small K (<<1): Reaction hardly proceeds (more reactants).
2. Predicting direction of reaction: Comparing the reaction quotient (Q) to the equilibrium constant (K) can predict the direction a reaction will shift to reach equilibrium.
   • If Q < K, it indicates that the reaction will proceed in the direction of products to reach equilibrium.
   • If Q > K, it indicates that the reaction will proceed in the direction of reactants to reach equilibrium.
   • If Q = K, it indicates that the reaction mixture is already at equilibrium.

Question 4.
What is Lechatelier’s principle ? Discuss briefly the factors which can influence the equilibrium.
Answer:
Statement: “If a system at equilibrium is subjected to the change of pressure, temperature (or) concentration, the system is shifted in such a way as to nullify the effect of change”.

Factors influencing the equilibrium:

1) Concentration: Increase of reactant concentrations pushes the equilibrium state to the products side and increase of the products concentrations pushes the equilibrium to the reactants side. For example in the chemical equilibrium
H₂(g) + I₂(g) ⇌ 2HI(g)
Increase of H₂ or I₂ concentrations pushes the equilibrium in the forward direction and similarly the increase of HI concentrations pushes the equilibrium in the reverse direction.

2) Pressure : Increase in pressure shifts the equilibrium toward the side with fewer gas molecules (to reduce pressure). Decrease in pressure shifts equilibrium toward the side with more gas molecules (to increase pressure).
Example:

For the above reaction, increasing pressure favours the formation of NH₃ (fewer gas molecules) i.e., the equilibrium shifts to products side and forward reaction is favourable. Similarly, decreasing pressure favours the formation of N₂ and H₂ (more gas molecules) i.e., the equilibrium shifts to reactants side and reverse reaction is favourable. If moles of gases are equal on both sides, pressure changes have no effect.

3) Temperature: Effect of change of temperature can be explained below.
For exothermic reactions : Increasing of temperature shifts the equilibrium towards reactants and decreasing temperature shifts the equilibrium toward products.
For endothermic reactions : Increasing of temperature shifts the equilibrium towards products and decreasing temperature shifts the equilibrium toward reactants.

Question 5.
Describe the effect of :
(a) additioñ of H₂
(b) addition of CH₃OH
(c) removal of CO
(d) removal of CH₃OH
On the equilibrium of the reaction, 2H₂(g) + CO(g) ⇌ CH₃OH(g).
Answer:
Using Le Chatelier’s Principle, the equilibrium will shift to counteract any imposed changes. Below is the effect of each change:

Addition of H₂ (Reactant): Addition of Hydrogen increases the concentration of a reactant. Then equilibrium shifts toward the products (right) to consume the excess H₂ and produce more CH₃OH. Adding H₂ increases the forward reaction rate, forming more CH₃OH until equilibrium is re-established.

Addition of CH₃OH (Product): Addition of CH₃OH increases the concentration of a product. Then equilibrium shifts toward the reactants (left) to consume the excess CH₂OH and produce more H₂ and CO. Adding CH₃OH increases the reverse reaction rate, breaking it back into H₂ and CO.

Removal of CO (Reactant): Removal of CO decreases the concentration of a reactant. Then equilibrium shifts toward the reactants (left) to replenish the lost CO by decomposing CH₃OH. Removing CO reduces the forward reaction rate, causing the reverse reaction to dominate, producing more H₂ and CO.

Removal of CH₃OH (Product): Removal of CH₃OH decreases the concentration of a product. Then equilibrium shifts toward the products (right) to replace the removed CH₃OH by consuming more H₂ and CO. Removing CH₃OH reduces the reverse reaction rate, favouring the forward reaction to produce more methanol.

Question 6.
What is, degree of ionization in respect of weak acids and weak bases ? Derive the relationship betw een degree of ionization (α) and ionization constant (K_a) for the w eak acid HX.
Answer:
Degree of ionization (α) refers to the fraction of the total number of acid or base molecules that ionize in solution.
For weak acids and weak bases, ionization is partial, so α < 1, Strong electrolytes (like strong acids/bases) ionize completely and have α = 1, but weak acids and bases ionize only slightly, typically having α << 1.
Consider a weak acid HX in aqueous solution
HX ⇌ H⁺ + X^–
Let initial concentration of HX = C mol/lit
Degree of ionization = α
At equilibrium, [HX] = C (1 – α) and
[H⁺] = [X^–] = Cα
We know that ionization constant, K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{X}^{-}\right]}{[\mathrm{HX}]}\)
Substituting the equilibrium concentrations,
K_a = \(\frac{(C \alpha)(C \alpha)}{C(1-\alpha)}\) = \(\frac{C \alpha^2}{1-\alpha}\)
Since α is very small for weak acids, (1 – α) ~ 1
From this equation, the degree of ionization (α ) increases with decreasing con-centration(C) and increasing ionization constant, K_a.

Question 7.
Define pH. What is buffer solution ? Derive Henderson – Hasselbalch equation for Calculating the pH of an acid buffer solution
Answer:
p^H is defined as, “The negative logarithm of the hydrogen ion concentration expressed in moles / lit is called p^H of the solution”.
∴ p^H = -log [H⁺]
p^H scale is in the range of 0 – 14.
For acidic solutions, p^H < 7
For bašic solutions, p^H > 7
For neutral solutions, p^H = 7.
The lower the p^H, more acidic is the solution. Conversely, higher the p^H, more basic iš the solution.

BUFFER SOLUTION : It can be defined as “A solution which resists any change in its p^H value on addition of small amount of strong acid or strong baše or on dilution is called as Buffer solution”.

Types of buffer solútion : The buffer solutions are of two types.

1. Acidic buffer solution An acidic buffer solution is a mixture of weak acid and its salt with strong base. E.g.: CH₃CÖOH and CH₃COONa
2. Basic buffer solution : A basic buffer solution is mixture of weak base and its salt with strong acid.
   E.g. : NH₄OH and NH₄Cl

Derivation of Henderson-Hasselbaich Equation for Acid Buffer:
Consider a weak acid HA and itš salt, which provides the conjugate base A^–
HA ⇌ H⁺ + A^–
For this reaction, ionization constant, K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{\mathrm{HA}}\)
⇒ [H⁺] = K_a
⇒ [H⁺] = K_a . \(\frac{[\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}\)

Question 8.
Write notes on
(i) Common ion effect.
(ii) The relation between K_SP and solubility (S) of a sparingly soluble salt BaSO₄
Answer:
i) Common ion effect: The decrease in the ionisation or dissociation of a weak electrolyte, by the addition of a strong electrolyte having an ion common with the weak electrolyte, is known as common ion effect.
Example: The dissociation of NH₄OH is diminished by the addition of NH₄Cl due to the common ion, N\(\mathrm{H}_4^{+}\) ion
NH₄ ⇌ N\(\mathrm{H}_4^{+}\) + OH^–

Importance of Common ion effect in qualitative analysis:

1. This principle is used in the systematic qualitative analysis of cations.
2. Selective precipitation of ions: In group separation, adding a common ion helps precipitate specific ions while keeping others in solution.
3. Confirmation Tests : The common ion effect is used to confirm the presence of ions by enhancing precipitation.
4. The common ion effect helps control the pH and ionic concentrations to ensure only specific ions precipitate at a given time. This prevents false positives or co-precipitation of unwanted ions.
5. Separation of chemically similar ions : Ions with similar properties (e.g., Ag⁺ and Pb²⁺) can be separated using a common ion.
6. The common ion effect principle is also used in controlling the H⁺ ion concentration in buffer solutions.
7. It is also used in the purification of common salts by passing dry HCl gas into impure salt solution.

ii) The relation between K_sp and solubility (S) of a sparingly soluble salt BaSO₄:
Solubility product: It is the.product of the molar concentrations of its ions at saturation point, each raised to the power of their stoichiometric coefficients. It represents the extent to which a salt dissolves in water to form its constituent ions. It can be denoted by K_sp.

Consider dissociation of a salt AB
A_xB_y(s) ⇌ xA^y+(aq) + y Bx^{– (aq)}

For this reaction, the solubility product can be written as, K_sp = [A^y+]^x [B^x-]^y
Solubility: The amount of salt in mol/L that dissolves to form a saturated solution is called as solubility. It can be denoted by S.
BaSO₄ a sparingly soluble salt, can undergo ionization as follows.
BaSO₄(s) ⇌ Ba²⁺ (aq) + S\(O_4^{2-}\) (aq)
For this reaction, K_sp = [Ba²⁺] [S\(\mathrm{O}_4^{2-}\)]
Let S is the solubility of BaSO₄. At equilibrium, [Ba²⁺] = S, [S\(\mathrm{O}_4^{2-}\)] = S
K_sp = S × S = S² = \(\sqrt{\mathrm{K}_{\mathrm{sp}}}\).