What is the pH of a 0.001 M Ba(OH)2 solution, assuming complete dissociation?
Given: What is the pH of a 0.001 M Ba(OH)2 solution, assuming complete dissociation? These values define the system as per NCERT data. Formula: [OH-] = 2 × 0.001 = 0.002 M, pOH = -log(0.002) approx 2.7. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: pH = 14 - 2.7 = 11.3 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Concentration terms and colligative properties.