The dissociation constant of ethanoic acid is 1.8x10-5 . Find the pH of 0.02M solution of the acid. pls can i get the answer by today .**** Share with your friends Share 0 Akanksha C. answered this Hi, CH3COOH ↔ CH3COO- + H+ t=0 C 0 0 teq C(1-α) Cα Cα Here, C= 0.02 M Ka = 1.8×10-5Therefore, α = KaC = 1.8×10-50.02 =3 ×10-2pH= -log[H+ ]=-log[Cα]=-log(0.02×3×10-2)=-log(0.0006)= 3.22 =3.22Therefore, pH of 0.02 M solution of acetic acid is 3.22Regards 1 View Full Answer Mohanish Venkatesh answered this pls experts i need this answer 1
