Relevant Titles
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Buffer Solution and Solubility Product MCQs for Class 11 Chemistry
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CBSE Class 11 Chapter 7 Equilibrium – Online MCQ Practice Test
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Class 11 Chemistry MCQs on Solubility Product and Buffer Concept
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Physical Chemistry Chapter 7 Equilibrium Quiz – CBSE Class 11
Introduction
Understanding the Buffer Solution and Solubility Product concepts is crucial for mastering Chapter 7: Equilibrium in CBSE Class 11 Chemistry. These two topics explain how chemical systems maintain stability — buffer solutions resist pH changes, while solubility product (Ksp) governs the saturation and precipitation of ionic compounds.
This online MCQ practice test is based strictly on the NCERT Class 11 Chemistry syllabus, ensuring alignment with CBSE Board Exam standards. It features 30 conceptual and numerical questions covering the Henderson–Hasselbalch equation, buffer capacity, common-ion effect, and Ksp calculations, each with instant feedback and detailed solutions.
Ideal for students preparing for CBSE Class 11 exams, as well as NEET and JEE (Main) aspirants, this quiz helps in revising core equilibrium concepts effectively. With automatic scoring, timed assessment, and per-question feedback, it builds accuracy, speed, and confidence in physical chemistry fundamentals.
Highlight: These MCQs are designed strictly according to the NCERT Class 11 syllabus, making them ideal for CBSE Board standards and entrance preparation.
Sample MCQs (with Answers and Explanations)
Q1. Which of the following pairs can form a buffer solution?
A) HCl and NaCl B) CH₃COOH and CH₃COONa C) NaOH and NaCl D) H₂SO₄ and Na₂SO₄
Answer: B) CH₃COOH and CH₃COONa
Explanation: A buffer is made from a weak acid and its conjugate base, like acetic acid and sodium acetate.
Q2. The pH of a buffer is given by:
A) pH = pKa + log([A⁻]/[HA]) B) pH = pKa – log([A⁻]/[HA]) C) pH = [H⁺] D) pH = –log Kb
Answer: A) pH = pKa + log([A⁻]/[HA])
Explanation: This is the Henderson–Hasselbalch equation for acid buffer solutions.
Q3. Common-ion effect on solubility causes —
A) Increase in solubility B) Decrease in solubility C) No change D) Increase in Ksp
Answer: B) Decrease in solubility
Explanation: Adding a common ion shifts equilibrium toward the undissociated solid, reducing solubility.
Q4. If Ksp of AgCl = 1.8 × 10⁻¹⁰, its solubility in mol/L is —
A) 1.34 × 10⁻⁵ B) 1.8 × 10⁻¹⁰ C) 3.6 × 10⁻⁵ D) 1.34 × 10⁻⁴
Answer: A) 1.34 × 10⁻⁵
Explanation: For AgCl ⇌ Ag⁺ + Cl⁻, Ksp = s² ⇒ s = √(1.8×10⁻¹⁰) = 1.34×10⁻⁵ M.
Q5. Which of the following will increase solubility of AgCl in water?
A) Adding NH₃ B) Adding NaCl C) Adding HCl D) Adding AgNO₃
Answer: A) Adding NH₃
Explanation: NH₃ forms a soluble complex [Ag(NH₃)₂]⁺, increasing solubility through complexation.