The answer for this issue is: The chemical equation is: HBz + H2O <- - > H3O+ + Bz- Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz] Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x. Accept that x is little contrasted with 0.5 M. At that point, Ka = 6.4X10^-5 = x^2/0.5 x = [H3O+] = 5.6X10^-3 M pH = 2.25 (x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)
