The acid dissociation constant, Ka, of ethanoic acid is 1.78 x 10^-5 at 298K. Given that the concentration of a sample of ethanoic acid is 0.4moldm^-3, calculate its pH at 298K.

Using the acid dissociation equation, Ka = [H+][A-]/[HA]. (where Substitute the known values into the concentration to give 1.78x10^-5= [H+][A-]/0.4 . Because the acid is dissociating in solution the acid dissociates in water which is neutral, then [H+] and [A-] must be equal. So we can write: [H+]²/0.4=1.78x10^-5. So [H+]2 =7.12x10^-6 and [H+] = 2.67x10^-3.Substitute [H+] into the pH equation: pH =-log[H+] = -log[2.67x10^-3] = 2.57