The correct option is C 10−4
According to the question, 12 mL of 0.1 M strong base is used. Thus, at the end point, the volume of same base required is 13.2 mL
HA+BOH⇌BA+H2Ot=0 a 12×0.1 0 0t=eq (a−1.2) 0 1.2 1.2
Given, at the end point, the volume of same base required is 13.2 ml
Thus, Meq of HA = Meq of BOH
Meq of BOH at the end point =13.2×0.1=1.32
∴ Meq of HA left =1.32−1.2=0.12
Now, pH=pKa+log[Salt][acid]
5=−logKa+log1.2(0.12)⇒5=−logKa+1⇒Ka=10−4