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CBSE Class 11 Chemistry MCQs: Law of Mass Action and Ka, Kb Practice Test
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Introduction
Understanding the Law of Mass Action and the Ka–Kb concepts is a vital part of the CBSE Class 11 Chemistry Chapter 7 – Equilibrium syllabus. These concepts form the foundation for analyzing chemical reactions, determining equilibrium constants, and understanding the strength of acids and bases.
This online practice test presents a collection of CBSE Class 11 Chemistry MCQs designed strictly as per the latest NCERT syllabus and CBSE board exam pattern. The questions are crafted to strengthen conceptual clarity on equilibrium constant (Kc, Kp), ionization constants (Ka and Kb), and their relationships through Kw.
Each question comes with detailed explanations and automatic scoring — perfect for self-assessment. These Equilibrium MCQs also help students preparing for competitive exams like NEET and JEE, where mastery of acid-base equilibrium is crucial.
Take this interactive Class 11 Chemistry online quiz to enhance your understanding of equilibrium laws, reaction dynamics, and the mathematical relationships between various equilibrium constants.
Sample MCQs (with Answers and Explanations)
Q1. According to the Law of Mass Action, the rate of a chemical reaction is proportional to:
A) The total pressure B) The product of reactant concentrations raised to their coefficients C) The molecular weight D) The sum of reactant concentrations
Answer: B) The product of reactant concentrations raised to their coefficients
Explanation: The Law of Mass Action states that for a reaction aA + bB ⇌ cC + dD, the equilibrium constant Kc = [C]^c[D]^d / [A]^a[B]^b.
Q2. For the equilibrium: N₂O₄ ⇌ 2NO₂, if Kc = 0.0008, doubling the concentration of N₂O₄ will:
A) Double Kc B) Halve Kc C) Not change Kc D) Increase Kc fourfold
Answer: C) Not change Kc
Explanation: The equilibrium constant Kc depends only on temperature, not on the initial concentrations of reactants or products.
Q3. The relationship between Ka and Kb for a conjugate acid-base pair is:
A) Ka + Kb = Kw B) Ka × Kb = Kw C) Ka / Kb = Kw D) Ka = Kb
Answer: B) Ka × Kb = Kw
Explanation: For conjugate acid-base pairs like NH₄⁺/NH₃, the product of Ka and Kb equals the ionic product of water, Kw = 1 × 10⁻¹⁴ at 25°C.
Q4. The value of Ka for acetic acid is 1.8 × 10⁻⁵. The pKa is approximately:
A) 5.74 B) 4.74 C) 6.00 D) 3.74
Answer: B) 4.74
Explanation: pKa = –log(Ka) = –log(1.8 × 10⁻⁵) ≈ 4.74, indicating that acetic acid is a weak acid.
Q5. A weak acid HA has Ka = 1 × 10⁻⁴. The conjugate base’s Kb is:
A) 1 × 10⁻¹⁰ B) 1 × 10⁻⁵ C) 1 × 10⁻¹⁴ D) 1 × 10⁻⁸
Answer: A) 1 × 10⁻¹⁰
Explanation: Since Ka × Kb = Kw = 1 × 10⁻¹⁴, Kb = Kw / Ka = 1 × 10⁻¹⁴ / 1 × 10⁻⁴ = 1 × 10⁻¹⁰.