Equilibrium MCQ Quiz - Objective Question with Answer for Equilibrium - Download Free PDF

CONCEPT:

Solubility of Salt in a Common-Ion Solution

• The solubility product (K_sp) is an equilibrium constant that represents the solubility of a salt in water.
• In the presence of a common ion, the solubility of the salt is suppressed due to the common ion effect.
• The solubility product expression for CaSO₄ is:

  K_sp = [Ca²⁺] × [SO₄^2-]

• In the presence of CaCl₂, the concentration of Ca²⁺ is influenced by the CaCl₂ dissociation, so the total concentration of Ca²⁺ will be 0.2 + s , where s is the solubility of CaSO₄.
• We can use the solubility product equation to solve for s , the solubility of CaSO₄ in the solution.

EXPLANATION:

• Given:
  • K_sp = 4.0 × 10^-6
  • Concentration of CaCl₂ = 0.2 M
  • We approximate 0.2 + s \approx 0.2 because s is small.
• From the solubility product equation:

  4.0 × 10^-6 = (0.2 + s) × s ≈ 0.2s

• Solving for s :

  s = 2.0 × 10^-5 mol/L

• The value of x is 20, rounded to the nearest integer.

The value of x is 20.

CONCEPT:

Ionization Constants of Acids

• The ionization constant (Kₐ) represents the equilibrium constant for the dissociation of an acid in water.
• For a polyprotic acid like acetic acid, the ionization occurs in steps, each with its own ionization constant (Kₐ₁, Kₐ₂, etc.).
• The overall ionization constant (K) is related to the individual Kₐ values, and the acid strength typically decreases with each successive ionization step.

EXPLANATION:

• Option A: log K = log Kₐ₁ + log Kₐ₂ is not correct. The overall ionization constant K is not simply the sum of the individual ionization constants in terms of logarithms. The correct relationship would be derived from the equilibrium expressions, but it does not follow the form of log K = log Kₐ₁ + log Kₐ₂.
• Option B: CH₃COOH is a stronger acid than CH₃COO⁻. This statement is true because acetic acid (CH₃COOH) is a weak acid and dissociates to form acetate ion (CH₃COO⁻). The conjugate base CH₃COO⁻ is much weaker than the parent acid CH₃COOH.
• Option C: Kₐ₁ > Kₐ₂ is true. In polyprotic acids like acetic acid, the first ionization constant (Kₐ₁) is always greater than the second (Kₐ₂), indicating that the first dissociation step is stronger than the second.
• Option D: Kₐ = (Kₐ₂ + Kₐ₁) / 2 is not correct. The overall ionization constant K is not the average of the individual Kₐ values, but rather it involves the product of the equilibrium concentrations derived from the individual dissociation steps.

Therefore, the correct answer is: B and C only.

CONCEPT:

Precipitation of AY₂ from AB₂ and XY Solutions

• When two salt solutions AB₂ and XY are mixed, a precipitate of AY₂ will form if the ionic product (Qsp) exceeds the solubility product constant (Ksp).
• The ionic product is calculated as:

  Qsp = [A⁺][Y^-]²

• Since equal volumes are mixed, the concentrations of A⁺ and Y⁻ will be halved, leading to:
  • New concentration of ions after mixing: [A⁺]/2 and [Y^-]/2.
• For precipitation to occur, Qsp must be greater than Ksp.

EXPLANATION:

• For each combination, the ionic product Qsp is calculated using the diluted concentrations of A⁺ and Y^-:

Option 1:
Qsp = (2.5 × 10⁻³ / 2) × (2 × 10⁻⁴ / 2)²
Qsp = 1.25 × 10⁻³ × 1 × 10⁻⁸
Qsp = 1.25 × 10⁻¹¹
Since Qsp < Ksp, no precipitation occurs.

Option 2:
Qsp = (7.5 × 10⁻⁵ / 2) × (3 × 10⁻⁴ / 2)²
Qsp = 3.75 × 10⁻⁵ × 2.25 × 10⁻⁸
Qsp = 8.44 × 10⁻¹³
Since Qsp < Ksp, no precipitation occurs.

Option 3:
Qsp = (1.5 × 10⁻² / 2) × (1.5 × 10⁻² / 2)²
Qsp = 7.5 × 10⁻³ × 5.625 × 10⁻⁴
Qsp = 4.22 × 10⁻⁶
Since Qsp > Ksp, precipitation of AY₂ occurs.

Option 4:
Qsp = (1.0 × 10⁻⁴ / 2) × (1.25 × 10⁻³ / 2)² = 5.0 × 10⁻⁵ × 7.8125 × 10⁻⁷ = 3.91 × 10⁻¹²
Since Qsp < Ksp, no precipitation occurs.

• From the calculations, only Option 3 gives Qsp > Ksp, leading to the precipitation of AY₂.

Therefore, the correct answer is Option 3.

CONCEPT:

Le Chatelier’s Principle & Inert Gas Addition:

• When an inert gas is added to a reaction mixture at constant volume, the total pressure increases but the partial pressures of the reactants and products remain unchanged.
• Hence, equilibrium composition does not change.
• Ideal Gas Law: PV = nRT ⇒ P = (n/V)RT

EXPLANATION:

• Given:
  • Initial moles: H₂ = 16, N₂ = 4
  • Initial pressure = P, Volume = 1 L, T = constant
• The equilibrium reaction is:

  N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

• Let the extent of reaction be a, so:
  • N₂ at equilibrium = 4 - a
  • H₂ at equilibrium = 16 - 3a
  • NH₃ formed = 2a
  • Total moles at equilibrium = (4 - a) + (16 - 3a) + 2a = 20 - 2a
• Using ideal gas law:
  • Initial: P = 20RT (since total initial moles = 20)
  • New pressure at equilibrium: (9/10)P = (20 - 2a)RT
  • Substituting: (9/10) × 20RT = (20 - 2a)RT ⇒ a = 1
• Total moles at equilibrium = 20 - 2a = 18
• Now, adding x moles of inert gas at constant volume:
  • New total moles = 18 + x
  • Required pressure = P
  • Using P = (18 + x)RT = 20RT ⇒ x = 2

Therefore, the value of x is 2.

CONCEPT:

Buffer Solution and Henderson-Hasselbalch Equation

• A buffer solution consists of a weak base and its conjugate acid (or a weak acid and its conjugate base).
• The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:

  pH = pKa + log₁₀ ([Base]/[Acid])

• For a weak base like NH₄OH, the relationship between pKa and pKb is:

  pKa + pKb = 14

EXPLANATION:

• Given:
  • Moles of weak base (NH₄OH) = 0.20 mole/L
  • Moles of conjugate acid (NH₄Cl) = 0.25 mole/L
  • Dissociation constant of NH₄OH, Kb = 1.81 × 10^-5
• First, calculate pKb:

  pKb = -log₁₀(Kb)

  pKb = -log₁₀(1.81 × 10^-5)

  pKb = 4.74

• Next, calculate pKa using the relationship pKa + pKb = 14:

  pKa = 14 - pKb

  pKa = 14 - 4.74

  pKa = 9.26

• Now apply the Henderson-Hasselbalch equation:

  pH = pKa + log₁₀ ([Base]/[Acid])

  pH = 9.26 + log₁₀ (0.20/0.25)

  pH = 9.26 + log₁₀(0.8)

  pH = 9.26 + (-0.10)

  pH = 9.16

Therefore, the pH value of the solution is 9.16.

Concept:

The pH Scale

• The pH is the measure of the concentration of acid in the solution. 
• It is measured from the scale of 1 to 14. 
• If the solution is acidic or basic, depends upon the following parameters. 
• pH < 7 - Acidic
• pH > 7 - Basic
• pH = 7 - Neutral

​Conclusion:

• So, from the above explanation, it is clear that the Higher the pH, the weaker is the acid, and the Lower the pH, the weaker is the base.
• So, Statements 2 and 4 are correct. ​

Additional Information

• The pH of water at 25°c is 7. When it is heated to 100°C, the pH of water decreases to 6.14.
• On increasing the temperature of the water, the equilibrium would again move to lower the temperature.
• It would happen by absorbing the extra heat. 
• Hydrogen ions and hydroxide ions will remain at the same concentration in pure water and the water remains neutral even if its pH changes.

Concept:

pH of acidic buffer - 

• Buffer solution - Buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. It is a solution highly known for maintaining its pH on dilution with a small amount of strong acid or strong base by maintaining the H⁺ ion concentration inside it. 
• Two types of buffer solutions are there
• Acidic buffer - Highly specific for pH<7 and is a mixture of a weak acid and its conjugate base.
• Weak buffer - Highly specific for pH>7 and is a mixture of a weak base and its conjugate acid.
• pH of the acidic buffer is given by the formula -

pH = pKa + log ([salt]/[acid])

where pK_a = negative logarithms of the acid dissociation constant.

As acetic acid is a weak acid and sodium acetate is its conjugate salt with a strong base(NaOH), they form an acidic buffer whose pH is calculated by the formula -

pH = pKa + log ([salt]/[acid])

Chemical reaction - CH₃COOH + NaOH \(\rightarrow\) CH₃COONa +H₂O

Calculation -

Given,

• pKa = 4.57 (negative logarithms of the acid dissociation constant of acetic acid)
• [acid] = 0.01 M (concentration of acetic acid)
• [salt] = 0.10 M (concentration of salt)

Put all these values in the formula,

pH = pKa + log ([salt]/[acid])

=4.57 + log(0.10 / 0.01)

=4.57 + log(10)

= 4.57 + 1   -- (∵ log10 = 1)

=5.57

So, the pH of the buffer solution is = 5.57

Hence, the correct answer is option 2.

Chemical Equilibrium: Consider a general case of a reversible reaction

A + B ⇔ C + D

• With respect of time, the rate of forward reaction decreased, and the rate of backward reaction increased.
• With an instant of time, a stage is reached at which the rate of forward and reverse reaction become equal and the concentration of reactant and product becomes constant.
• The equilibrium is dynamic in 

The correct answer is Kolbe.

Key Points

• Synthesis of Acetic Acid
  • Acetic acid was first made or created from inorganic components in 1845 by German scientist Hermann Kolbe.
  • He created carbon disulfide with chlorine, and the end result was carbon tetrachloride.
  • After that, tetrachloroethylene was produced through thermal decomposition, trichloroacetic acid was created through aqueous chlorination, and acetic acid was finally produced through an electrolytic reduction.
  • He exclusively used inorganic materials made of hydrocarbons.
  • Carbon, hydrogen, and oxygen are inorganic elements that are used to create acetic acid while acetic acid is an organic one.
  • Acetic acid has a chemical formula and is a weak acid.
  • Kolbe's work was groundbreaking since he synthesised organic chemicals from inorganic ones.
  • In addition to bacterial fermentation, acetic acid is also made synthetically.
  • Some fruits naturally contain it.

Additional Information

• Kolbe's work was groundbreaking since he synthesised organic chemicals from inorganic ones. 
• In addition to bacterial fermentation, acetic acid is also made synthetically.
•  some fruits naturally contain it.

The correct answer is Less than 7

Key Points 

• When a salt is formed from the neutralization of a strong acid and a weak base, the resulting solution is acidic. This is because the strong acid completely dissociates in water, providing a high concentration of (H+) ions, while the weak base does not dissociate completely and therefore provides fewer (OH^-) ions to neutralize the (H+). As a result, the (H⁺) concentration in the solution is higher than the (OH-) concentration, leading to an acidic solution.

Therefore, the pH value of a salt made up of a strong acid and a weak base is Less than 7

• The cations of strong acids do not hydrolyze and therefore the solutions of salts formed by strong acids and strong bases are neutral, i.e. their pH is 7, whereas the solutions of salts formed by strong bases and weak acids are alkaline, i.e. their pH>7.
• Salts that are from strong bases and weak acids do hydrolyze, which gives them a pH greater than 7.
• The pH value range of a weak base is 7.1 to 10 whereas a strong base is 10.1 to 14.
• All substances can be classified as neutral (with a pH of about 7), basic (pH greater than 7), or acidic (pH less than 7), and the pH tells us how strong or weak that substance is as well. For example, a substance with pH = 8 is a very weak base, but a substance with pH = 3 is a strong acid.

The correct answer is II Only.

Key Points List of Acid and Natural Resources:

Explanation:

A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back. Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.

According to equilibrium constant, K_c

\({{\rm{K}}_{\rm{c}}} = \frac{{{{\left[ {{\rm{product}}} \right]}^{\rm{m}}}}}{{{{\left[ {{\rm{reactant}}} \right]}^{\rm{n}}}}}\)

Consider the given first equation:

\({A_2}\left( g \right)\; + \;{B_2}\;\left( g \right)\begin{array}{*{20}{c}} {{K_1}}\\ \rightleftharpoons \end{array}\;2AB\left( g \right)\)

\({K_1} = \frac{{{{\left[ {AB} \right]}^2}}}{{\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}\)

Consider the given second equation:

\(6AB\left( g \right)\begin{array}{*{20}{c}} {{K_2}}\\ \rightleftharpoons \end{array}\;3{A_2}\left( g \right) + 3{B_2}\left( g \right)\)

\({K_2} = \frac{{{{\left[ {{A_2}} \right]}^3}{{\left[ {{B_2}} \right]}^3}}}{{{{\left[ {AB} \right]}^6}}}\)

\(= \frac{1}{{{{\left( {\frac{{{{\left[ {AB} \right]}^2}}}{{\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}} \right)}^3}}}\)

\(= \frac{1}{{K_1^3}}\)

 → K2 = K1-3

 Concept:

Lewis concept of acids and bases is given as follows:

The interaction between Lewis acid and base- 

• Lewis acid has a vacant unoccupied orbital called LUMO.
• Lewis bases have fulfilled occupied orbital called HOMO.
• Electron density from HOMO is donated to LUMO of Lewis acid.
• A simple HOMO and LUMO interaction is shown below:

This donation of electron density forms a covalent coordinate between the acid and the base,

Explanation:

BCl₃ -

• BCl₃ molecule has an empty p orbital because its octet is not fulfilled.
• It can accept an electron pair inti this orbital

Thus it electron-deficient and a Lewis acid

FeCl3:

• Fe is a transition metal having empty d orbitals.
• It can easily accept a lone pair in the empty 4d orbital and can act as Lewis acid.

​

Hence, BCl3, FeCl3 is a pair of Lewis acids.

Additional Information

• AlCl3: AlCl3 also has empty p orbitals and thus is an electrophile.

• Proton acceptors are Bronsted-Lowry bases, while Proton donors are Bronsted-Lowry acids

Thus, HCl, HNO3 and H₂CO₃ NH4+ are Bronsted acids.

The correct answer is high pressure and low temperature.

Key Points

• This equation is an example of Equilibrium.
• Equilibrium is a condition that occurs when a chemical reaction is reversible, and the forward and reverse reactions occur simultaneously, at the same rate.
• Le Chatlier's Principle- "if a chemical system at equilibrium experiences a change in concentration, temperature or total pressure, the equilibrium will shift in order to minimize that change" 
• Any decrease in nitrogen or hydrogen pulls the reaction towards the left side.
• Any decrease in ammonia or temperature pulls the reaction towards the right side.