Chemical and Ionic Equilibrium Previous Year Questions with Solutions

1.0Chemical Equilibrium (Introduction)

Like a perfectly balanced seesaw – rate of forward reaction = rate of backward reaction.
A state in a reversible reaction where concentrations of reactants and products remain constant over time.

Dynamic Nature

• Reactions continue in both directions.
• No net change in concentration.

2.0Key Concepts

General Reaction & Rate Equality

aA+bB⇌cC+dD

Rate_f=Rate_b⇒k_f[A]^a[B]^b=k_b[C]^c[D]^d

Equilibrium Constant (Keq)

$K_{eq}=\frac{[C{]}^c[D{]}^d}{[A]a[B]b}$

• Depends on temperature.
• Expressed as Kc (concentration) or Kp (pressure for gases).

3.0Types of Equilibrium

• Homogeneous: All reactants/products in same phase.

Ex: N2(g)+3H2(g)⇌2NH3(g)

${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)N2(g)+3H2(g)\rightleftharpoons 2NH3(g)$

• Heterogeneous: Reactants/products in different phases.
  Ex: H2O(g)+C(s)⇌H2(g)+CO(g)

${\text{H}}_2\text{O}(g)+\text{C}(s)\rightleftharpoons {\text{H}}_2(g)+\text{CO}(g)H2O(g)+C(s)\rightleftharpoons H2(g)+CO(g)$

Factors Affecting Equilibrium (Le Chatelier’s Principle)

4.0Ionic Equilibrium

Ionic equilibrium occurs when the rate of ion formation equals the rate of recombination, leading to constant ion concentration in solution.
Example: Acetic acid (CH₃COOH) in water establishes equilibrium:
CH₃COOH ⇌ CH₃COO⁻ + H⁺

5.0Key concepts

Electrolytes vs. Nonelectrolytes

Degree of Dissociation (α)

Fraction of total molecules that dissociate into ions.
α=No. of molecules dissociated Total no. of molecules taken

$\alpha =\frac{\text{No. of molecules dissociated}}{\text{Total no. of molecules taken}}$

Strong Electrolytes

• α ≈ 1
• Irreversible ionization
• High conductivity

Examples:

• Acids: HCl, HNO₃, H₂SO₄
• Bases: NaOH, KOH, Ba(OH)₂
• Salts: NaCl, KNO₃
  Weak Electrolytes
• α < 1
• Reversible ionization
• Low conductivity

Examples:

• Acids: CH₃COOH, HF, H₂CO₃
• Bases: NH₄OH, Aniline (C₆H₅NH₂)

Ostwald’s Dilution Law

For weak electrolytes:

α = KC

$\alpha =\sqrt{\frac{K}{C}}$

Where:

• α = degree of dissociation
• K = dissociation constant
• C = molar concentration

When α≪1

K≈Cα²

K_a and K_b Relationship

K_a×K_b=K_w=1.0×10⁻¹⁴

• K_a: Dissociation constant of an acid
  K_b: Dissociation constant of a base
  K_w: Ionization constant of water

6.0Concepts of Acids and Bases

Arrhenius Concept

• Acid: Produces H⁺ ions (e.g., HCl → H⁺ + Cl⁻)
• Base: Produces OH⁻ ions (e.g., NaOH → Na⁺ + OH⁻)

Bronsted-Lowry Concept

• Acid: Proton (H⁺) donor
• Base: Proton acceptor

Example:
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

• NH₃ = base, H₂O = acid (here)

Conjugate Pairs:

• HCl / Cl⁻
• NH₃ / NH₄⁺

Lewis Concept

• Acid: Electron pair acceptor
• Base: Electron pair donor

Examples:

• BF₃ + NH₃ → H₃N→BF₃ (coordinate bond)
• Ag⁺ + 2CN⁻ → [Ag(CN)₂]⁻

Acid & Base Strength

• pH = −log[H⁺]
• pOH = −log[OH⁻]
• pH + pOH = 14

Buffer Solutions

• Resist change in pH on addition of small amounts of acid or base.
• Consist of a weak acid + conjugate base or weak base + conjugate acid.

Henderson-Hasselbalch Equation

• For acid buffer: $\mathrm{pH}=\mathrm{p}{K}_a+\log \left(\frac{[A^{-}]}{[HA]}\right)pH=pKa+log([HA][A-])$
• For base buffer: $\mathrm{pOH}=\mathrm{p}{K}_b+\log \left(\frac{[B{H}^{+}]}{[B]}\right)pOH=pKb+log([B][BH+])$

Salt Hydrolysis & pH

7.0Chemical and Ionic Equilibrium Important PYQs

Q.1 For the reaction ,  N₂O₂ (g)→2NO₂ (g)    K_p= 0.492 atm at 300K. K_c for the reaction at the same temperature is ___ × 10^–2 .

(Given : R = 0.082 L atm mol^–1 K^–1)]

Ans. (2)

Solution.  

$$\begin{gathered} K_p=K_c(RT{)}^{\Delta n_g} \\ \Delta n_g=1 \\ \Rightarrow K_c=\frac{K_p}{RT}=\frac{0.492}{0.082\times 300}=2\times 10^{-2} \end{gathered}$$

Q.2   $A(g)\to B(g)+C2\frac{C}{2}(g)$The correct relationship between K_P​, α, and equilibrium pressure P is:      

$$\begin{gathered} (1)K_p=\frac{a^{1/2}{P}^{1/2}}{(2+\alpha {)}^{1/2}} \\ (2)K_p=\frac{a^{3/2}{P}^{1/2}}{(2+\alpha {)}^{1/2}(1-\alpha )} \\ \text{(3)}{K}_p=\frac{a^{3/2}{P}^{1/2}}{(2+\alpha {)}^{1/2}(1-\alpha )} \\ \text{(4)}{K}_p=\frac{a^{3/2}{P}^{1/2}}{(2+\alpha {)}^{3/2}} \end{gathered}$$

Ans. (2)      A(g)→B(g) + C2 (g)

t = t_eq         (1-ɑ)       ɑ       $\frac{a}{2}$

P_B=   ɑ 1+   ɑ2.P

P_B=   1-ɑ 1+   ɑ2.P

P_C=   ɑ21+   ɑ2.P

KP= PB.PC.PA.

$$\begin{gathered} \text{A(g)}\to \text{B(g)}+\frac{C}{2}\text{(g)}t=t_{\text{eq}}\begin{array}{ccc}\text{A}& \text{B}& \text{C}\\ 1-\alpha & \alpha & \frac{\alpha }{2}\end{array} \\ {P}_B=\frac{\alpha }{\left(1+\frac{\alpha }{2}\right)}P,P_A=\frac{1-\alpha }{\left(1+\frac{\alpha }{2}\right)}P,P_C=\frac{\frac{\alpha }{2}}{\left(1+\frac{\alpha }{2}\right)}P \\ {K}_p=\frac{P_B\cdot P_C^{1/2}}{P_A} \\ =\frac{\left(\frac{\alpha }{1+\frac{\alpha }{2}}P\right)\cdot {\left(\frac{\alpha }{2(1+\frac{\alpha }{2})}P\right)}^{1/2}}{\left(\frac{1-\alpha }{1+\frac{\alpha }{2}}P\right)} \\ =\frac{{\alpha }^{3/2}{P}^{1/2}}{(1-\alpha )(2+\alpha {)}^{1/2}} \\ {P}_B=\frac{\alpha }{\left(1+\frac{\alpha }{2}\right)}P \\ {P}_B=\frac{1-\alpha }{\left(1+\frac{\alpha }{2}\right)}P \\ {P}_C=\frac{\frac{\alpha }{2}}{\left(1+\frac{\alpha }{2}\right)}P \\ {K}_p=\frac{P_B\cdot P_C^{1/2}}{P_A} \end{gathered}$$

Q.3The equilibrium constant for the reaction SO₃ →SO₂(g) + 12 O₂(g) is K_C = 4.9 × 10^–2. The value of K_C for the reaction given below is 

2SO₂ (g)+ O₂(g) →   2SO₂ (g) is

(1) 4.9                               

(2) 41.6

(3) 49                           

(4) 416

Ans.   (4)

Sol.

${\mathrm{K}}_{\mathrm{c}}^{\prime }={\left\{\frac{1}{K_c}\right\}}^2={\left\{\frac{1}{4.9\times 10^{-2}}\right\}}^2$

K'_C = 416.49

Q.4  The effect of addition of helium gas to the following reaction in equilibrium state, is :

PCl₅(g) →PCl₃(g)  +Cl₂(g)

(1) the equilibrium will shift in the forward direction and more of Cl₂ and PCl₃ gases will be produced.

(2) the equilibrium will go backward due to suppression of dissociation of PCl₅.

(3) helium will deactivate PCl₅ and reaction will stop.

(4) addition of helium will not affect the equilibrium.

Ans. (1 & 4)

Solution. PCl₅(g) →PCl₃(g)  +Cl₂(g)

Case 1 : At constant P – volume will increase so reaction will shift in forward direction then answer will be A

Case 2 : At constant volume no change in active mass so reaction will not shift in any direction then answer will be D.

Q.5    For the given reaction  A → 2B(g)  +C(g), if the initial pressure is 450 mm Hg and the pressure at time t is 720 mm Hg at a constant temperature T and constant volume V. The fraction of A(g) decomposed under these conditions is x × 10^­–1. The value of x is ________. (Nearest integer)

Ans.(3)  

Solution:                       

A →    2B(g)  +  C(g)

 t=0        450

time t    450-x     2x         x

P_T =P_A + P_B +P_C

720 = 450 – x + 2x + x

2x = 270

x = 135

Fraction of A decomposed =135450 \frac{135}{450}= 0.3 = 3×10^–1

So, x = 3

Q.6  Rate law for a reaction between A and B is given by R = k [A]ⁿ[B]^m  If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction is

(1) 2^(n–m)                    

(2) (n–m)

(3) (m + n)                

(4) $\frac{1}{2^{m+n}}$

Ans.   (1)

Solution.

r₁ = k[A]ⁿ [B]^m

Now

 A is doubled & B is halved in concentration

r₂ = k2ⁿ[A]ⁿ.$\frac{B^m}{2^m}$[B]^m

$\frac{r_2}{r_1}=2^{n-m}$

Q.7 4.0 moles of argon and 5.0 moles of PCI₅ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K_p for the reaction is [Given : R = 0.082L atm K^–1 mol^–1]

(1) 2.25              

(2) 6.24                           

(3) 12.13                           

(4) 15.24 

Ans. 1

Solution 

$$\begin{gathered} PC{l}_5=5mole \\ Ar=4mole \\ {P}_{\text{total}}=\frac{9\times 0.82\times 610}{100}=4.5,\text{atm} \\ {P}_{{\text{PCl}}_5}=\frac{9\times 4.5}{9}=2.5 \\ {P}_{\text{Ar}}=\frac{4\times 4.5}{9}=2 \end{gathered}$$

PCl₅(g) →PCl₃(g)  +Cl₂(g)

 2.5 – P           P          P

$$\begin{gathered} P_{\text{total}}=2.5-P+P+P+P_{\text{Ar}}=6 \\ P=1.5 \\ {K}_p=\frac{1.5\times 1.5}{1}=2.25\text{atm} \end{gathered}$$

Q.8 In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30°C is ______ × 10^–4.

2NOCl(g)→2NO(g) + Cl₂(g)

Ans. by NTA (125)

Sol.      

2NOCl(g) → 2NO(g) + Cl₂(g)

t = 0        2M                       ----

t=t_eq       (2–x)M          x M        12M

∵ x= 0.4M

[NOCl]_eq = 1.6 M

[NO]_eq = 0.4 M

[Cl₂]_eq = 0.2 M

$$\begin{gathered} K_c=\frac{[\mathrm{ch}NO{]}^2[\mathrm{ch}{Cl}_2]}{[\mathrm{ch}NOCl{]}^2}=\frac{[0.4{]}^2[0.2]}{[1.6{]}^2} \\ {K}_c=\frac{32}{2.56}\times 10^{-3} \end{gathered}$$

K_c = 12.5 × 10^-3

K_c = 125 × 10^-4

Integer answer is 125

Q.9 Two solutions A and B, each of 100 L was made by dissolving 4g of NaOH and 9.8 g of H₂SO₄ in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is_______.

 Ans. (10.60)

Solution . 

4 gm of NaOH in 100 L sol. 10^–3 M sol.

9.8 gm of H₂SO₄  in 100 L sol.10^–3 M sol.

Mixture : 40L of 10–3 M NaOH and 10 L of 10^–3 M H₂SO₄  sol.

Final Conc. of OH^– 

$=\frac{10^{-3}\left(40\times 1-10\times 1\times 2\right)}{40+10}=6\times 10^{-4}\text{M}$

pOH = – log (6 ×10^–4)

= 4 – log 6 = 4 – 0.60 = 3.40

pH = 14 – 3.40 = 10.60

Q.10  100 mL of 0.1 M HCl is taken in a beaker and to it 100 mL of 0.1 M NaOH is added in steps of 2 mL and the pH is continuously measured. Which of the following graphs correctly depicts the change in pH?

(1) 7pH vol. of NaOH                        

(2) pH 7vol. of NaOH

(3) pH 7vol. of NaOH                       

(4) pH7vol. of NaO

Ans. by NTA (3)

Solution. 

Steep rise in pH around the equivalence point  for titration of strong acid with strong base. 

Q.11 A solution is 0.1 M in Cl^– and 0.001 M in $\mathrm{ch}CrO{4}^{2-}$.  Solid AgNO₃ is gradually added to it. Assuming that the addition does not change in volume and K_sp(AgCl) = 1.7 × 10^–10 M² and K_sp(Ag₂CrO₄) = 1.9 × 10^–12 M³. Select correct statement from the following :

(1)  AgCl precipitates first because its K_sp is high.

(2) Ag₂CrO₄ precipitates first as its K_sp is low.

(3) Ag₂CrO₄ precipitates first because the amount of Ag⁺ needed is low.

(4) AgCl will precipitate first as the amount of Ag⁺ needed to precipitate is low.

Ans. (4)

Solution.

(i)  [Ag⁺] required to ppt AgCl(s)

Ksp = IP = [Ag⁺] [Cl^–] = 1.7 × 10^–10

[Ag⁺] = 1.7 × 10^–9

(ii) [Ag⁺] required to ppt Ag₂CrO₄(s)

Ksp = IP = [Ag+]² [CrO₄^–2] = 1.9 × 10^–12

[Ag⁺] = 4.3 × 10^–5     

[Ag⁺] required to ppt AgCl is low so AgCl will ppt First.

Q.12 The pH of ammonium phosphate solution, if pKa of phosphoric acid and pkb of ammonium hydroxide are 5.23 and 4.75 respectively, is ______.

Ans  (7)

Solution . 

Since (NH4)3PO4 is salt of weak acid (H3PO4) & weak base (NH4OH).

$$\begin{gathered} \text{pH}=7+\frac{1}{2}(\text{p}{K}_a-\text{p}{K}_b) \\ =7+\frac{1}{2}(5.23-4.75) \\ =7.24\approx 7 \end{gathered}$$

Q.13 If the solubility product of PbS is 8 × 10^–28, then the solubility of PbS in pure water at 298 K is x × 10^–l6 mol L^–1. The value of x is ________. (Nearest Integer)  [Given2 = 1.41]

Ans  (282)

Solution 

K_sp = S²

$S=\sqrt{K_{\text{sp}}}=\sqrt{8\times 10^{-28}}=\sqrt{2}\times 10^{-14}$

= 2.82 × 10^–14

= 282 ×10^–16          

Ans. = 282

Q.14 Given below are two statements :

Statement (I) : A Buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities.

Statement (II) : Blood is naturally occurring buffer solution whose pH is maintained by H₂CO₃ concentrations.

Q.15 In the light of the above statements, choose the correct answer from the options given below.

(1) Statement I is false but Statement II is true

(2) Both Statement I and Statement II is true

(3) Both Statement I and Statement II is false

(4) Statement I is true but Statement II is false

Ans.  

Sol.

Buffer solution is a mixture of either weak acid / weak base and its respective conjugate.

Blood is a buffer solution of carbonic acid H₂CO₃ and bicarbonate HCO₃^- 

Statement 1 is false but Statement II is true.