2 public questions tagged with this topic.
For HA H+ + A- , Ka = ([H+][A-]/[HA]) = (x²/0.1 - x) ≈ (x²/0.1) = 1.0 × 10⁻⁵ . Solving, x = sqrt1.0 × 10⁻⁶ = 1.0 × 10⁻³ , so pH = -log(1.0 × 10⁻³) = 3 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
1.3 accurately defines or describes the concept asked in this question. Within pH and Buffer, precise definitions and terminology are essential for clear scientific communication. The other options (0.7, 1, and 2) either describe related but distinct concepts, use incorrect terminology, or confuse similar-sounding terms that have different scientific meanings. A thorough understanding of exact definitions helps distinguish between closely related biological concepts and is crucial for competitive examinations.
Ref: Lehninger Principles of Biochemistry, Nelson & Cox, 8th Ed., Ch. 2
