The correct option is B 4.69
Given,
Ka=8.0×10−9, concentration (C) = 0.05 M
Let the weak monobasic acid is HA . The dissociation of weak acid is given by :
HA⇌ H++ A− C 0 0C−Cα Cα Cα
Ka=Cα2C(1−α)As,α<<1Ka=Cα2
α=√Kacα=√8×10−90.05=√16×10−8=4×10−4
Concentration of [H+]=Cα=(0.05×4×10−4)=2×10−5 mol L−1
pH=−log[2×10−5]=4.69