A 50.0 mL sample of 0.2 M weak monoprotic acid HA is titrated with 0.2 M NaOH at 25^ C . The pH of the solution (a) when 25.0 mL of NaOH is added and (b) when 50.0 mL of NaOH is added, are respectively: (Given: K_a = 1.0 10⁻⁵ , 2 = 0.3 )

Step-by-step solution

For part (a), 25.0 mL of 0.2 M NaOH is added to 50.0 mL of 0.2 M HA . Initial millimoles of HA = 50.0 0.2 = 10.0 mmol Millimoles of NaOH added = 25.0 0.2 = 5.0 mmol This is exactly the half-equivalence point. The solution forms an acidic buffer where half the acid is converted to salt NaA . Millimoles of HA remaining = 5.0 mmol Millimoles of NaA formed = 5.0 mmol Using the Henderson-Hasselbalch equation: pH = pK_a + ( [ Salt ] [ Acid ] ) Since [ Salt ] = [ Acid ] , pH = pK_a = - (1.0 10⁻⁵) = 5.0 . For part (b), 50.