0 .1 M NaOH is titrated with 0 .1 M, 20 ml HA till the end point, K a HA = 6×10 − 6 and degree of dissociation of HA is small as compared to unity. What is the pH of the resulting solution at the end point?

Step-by-step solution

NaOH 2 ⁡ + HA 2 ⁡ → Na A + H 2 O ⁡ At end point ≡ 0.1 × 20 = 2 ∵ 20 mL NaOH is required for the complete neutralization of HA NaA is a salt of strong base and weak acid Thus, will undergo hydrolysis and solution will becomes basic C   =   NaA   =   2 20   +   20   =   0 .05 M And pK a   =   − log 6   ×   10 − 6   =   5 .2 pH at the end point =   7   +   1 2 pK a