Practice question

Given: The K_a of HF is 6.8 × 10⁻⁴. What is the pH of a 0.1 M HF solution? These values define the system as per NCERT data. Formula: K_a = frac[H+][F-][HF] approx x²/0.1, 6.8 × 10⁻⁴= x²/0.1, x² = 6.8 × 10⁻⁵, x approx 8.25 × 10⁻³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: pH = -log(8.25 × 10⁻³) approx 2.08 . 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.