Practice question

Given: The K_a of propanoic acid is 1.32 × 10⁻⁵. What is the pH of a 0.01 M solution? These values define the system as per NCERT data. Formula: K_a = x²/0.01, 1.32 × 10⁻⁵= x²/0.01, x² = 1.32 × 10⁻⁷. This is standard NCERT relation. Substitution & Calculation: x = sqrt1.32 × 10⁻⁷approx 3.63 × 10⁻⁴, pH = -log(3.63 × 10⁻⁴) approx 3.44 . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.