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Class 11 Chemistry MCQs on pH, pOH, and Hydrolysis Calculations
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CBSE Class 11 Equilibrium Chapter – pH, pOH, Hydrolysis MCQs with Answers
Introduction
Understanding pH, pOH, and Hydrolysis Calculations is essential for mastering Chapter 7: Equilibrium in CBSE Class 11 Physical Chemistry. These concepts explain how acids, bases, and salts behave in aqueous solutions and how their ionization levels determine the acidity or basicity of a solution.
This online MCQ practice test covers the fundamentals of pH and pOH relations, ionic product of water (Kw), weak acid–base equilibria, buffer solutions, and hydrolysis of salts — all strictly as per the NCERT syllabus. Each question includes instant feedback and detailed explanations, helping students correct conceptual errors on the spot.
Perfect for CBSE Class 11 Board Exam preparation, this quiz also builds the foundation for NEET and JEE (Main) aspirants, who frequently encounter these numerical and conceptual problems in competitive exams.
Highlight: These MCQs are designed strictly according to the NCERT Class 11 Chemistry syllabus, ensuring clarity, accuracy, and board-level relevance.
Sample MCQs (with Answers and Explanations)
Q1. What is the pH of a 0.01 M HCl solution?
A) 1 B) 2 C) 3 D) 4
Answer: B) 2
Explanation: For strong acid HCl, [H⁺] = 0.01 M. Thus, pH = -log[H⁺] = 2.
Q2. If the pOH of a solution is 5, then the pH at 25°C is —
A) 5 B) 9 C) 7 D) 14
Answer: B) 9
Explanation: pH + pOH = 14 ⇒ pH = 14 – 5 = 9.
Q3. The ionization constant (Ka) of acetic acid is 1.8 × 10⁻⁵. The pKa is —
A) 5.74 B) 4.74 C) 6.74 D) 7.00
Answer: B) 4.74
Explanation: pKa = -log(1.8 × 10⁻⁵) ≈ 4.74.
Q4. A buffer solution is formed by mixing —
A) HCl and NaCl B) CH₃COOH and CH₃COONa C) NaOH and NaCl D) H₂SO₄ and Na₂SO₄
Answer: B) CH₃COOH and CH₃COONa
Explanation: Buffer solutions consist of a weak acid and its conjugate base (salt) to resist pH changes.
Q5. The solubility of a weak acid increases in the presence of a strong base due to —
A) Hydrolysis B) Common-ion effect C) Neutralization D) Salt formation
Answer: D) Salt formation
Explanation: The strong base neutralizes the weak acid, forming salt and increasing solubility.