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CBSE Class 11 Online Test: Chemical and Ionic Equilibrium MCQs with Solutions
Introduction
The topic “Ionic and Chemical Equilibrium” forms a vital part of Physical Chemistry (Chapter 7) in the CBSE Class 11 Chemistry syllabus. Mastering this chapter is essential for understanding the balance between forward and reverse reactions, ionic dissociation, and the role of equilibrium constants (Kc, Kp, and Ksp) in chemical systems.
This online MCQ practice test is designed strictly as per the NCERT syllabus, making it perfect for CBSE Class 11 Board Exam preparation. Each question focuses on the conceptual understanding of acid-base equilibrium, solubility product, common-ion effect, and buffer systems — topics that often appear in board and entrance exams like NEET and JEE (Main).
With automatic scoring and per-question feedback, this quiz helps students assess their knowledge in real time, strengthen their conceptual base, and improve problem-solving accuracy.
Highlight: These MCQs follow the NCERT Class 11 pattern, ensuring the best preparation for CBSE Board standards.
Sample MCQs (with Answers and Explanations)
Q1. In the equilibrium HA ⇌ H⁺ + A⁻, the value of Ka indicates —
A) The strength of the acid
B) The pH of the solution
C) The concentration of A⁻
D) The temperature of the system
Answer: A) The strength of the acid
Explanation: A higher Ka value means greater dissociation and hence a stronger acid.
Q2. For the salt NH₄Cl dissolved in water, the resulting solution is —
A) Basic
B) Acidic
C) Neutral
D) Amphoteric
Answer: B) Acidic
Explanation: NH₄⁺ acts as a weak acid, releasing H⁺ ions, making the solution acidic.
Q3. The solubility product (Ksp) of AgCl at 25°C is 1.8 × 10⁻¹⁰. Its molar solubility is —
A) 1.34 × 10⁻⁵ M
B) 1.34 × 10⁻⁴ M
C) 3.6 × 10⁻⁵ M
D) 1.8 × 10⁻¹⁰ M
Answer: A) 1.34 × 10⁻⁵ M
Explanation: AgCl ⇌ Ag⁺ + Cl⁻; Ksp = s² → s = √(1.8×10⁻¹⁰) = 1.34×10⁻⁵ M.
Q4. The pH of a buffer containing equal concentrations of acetic acid and sodium acetate is —
A) 7
B) pKa
C) 14
D) Depends on [H⁺]
Answer: B) pKa
Explanation: For equal [acid] and [salt], pH = pKa + log(1) = pKa (Henderson-Hasselbalch equation).
Q5. Which of the following decreases the solubility of a sparingly soluble salt?
A) Common-ion effect
B) Temperature increase
C) Removal of ions
D) Addition of complexing agent
Answer: A) Common-ion effect
Explanation: Adding a common ion shifts the equilibrium toward undissociated solid, reducing solubility.