20 mL of a weak monobasic acid (HA) requires 20 mL 0.2 M NaOH for complete titration. If pH of solution upon addition of 10 mL of this alkali to 25 mL of the above solution of HA is 5.8. The pKa of the weak acid is

meq. of acid = meq of base ⇒  20 × M = 20 × 0.2 = 4Molarity of HA = 0.2 MWhen 25ml,0.2M acid reacts with 10ml 0.2M NaOH         HA    +      OH-      →       A-      +         H2O    5 (=25×.2)     2 (10×.2)              0      3(=5-2)              -                 2 mmole    ∴   pH = pKa + log   A-HA ⇒    5.8 =  pKa +  log 23 ⇒     pKa = 5.98