For a reaction,CH3COOH(aq)⇌H+(aq)+CH3COO-(aq)or HAc(aq)⇌H+(aq)+Ac-(aq)Evaluate the pH of the solution resulting on addition of 0.05 M acetate ion to 0.05 M acetic acid solution Ka=1.8×10-5

HAc(aq)⇌H+(aq)+Ac-(aq) Initial conc. (M)0.05           0      0.05 Equil.    0.05-x                x           0.05+xconc. (M)Ka=H+Ac-[HAc]={(x)(0.05+x)}(0.05-x)As K, is small for a very weak acid, .x << 0.05Hence, (0.05 + x) ≈ (0.05 - x) ≈ 0.05Thus, 1.8×10-5=(x)(0.05+x)(0.05-x)=x(0.05)(0.05)=x=H+=1.8×10-5MpH=-log1.8×10-5=4.74