Practice question

Given: What is the pH of a 0.003 M NaOH solution, assuming complete dissociation? These values define the system as per NCERT data. Formula: [OH-] = 0.003 M, pOH = -log(0.003) approx 2.52. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: pH = 14 - 2.52 = 11.48 approx 11.5 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.