Question

what is the ph of a 0.25M solution of HF if the Ka = 6.8x10^-4

Answer 1

pH = - log [H+].
So we know the concentration of HF initially used. We also know the Ka, acid dissociation constant, which tells us the ratio of [H+] and [F-] (products) per [HF] . We just need to determine the equilibrium concentration of [H+]. This is relatively easy using an ICE chart. There is one H+ for every F- dissociating, so [H+] = [F-] and Ka = [H+]^2 / [HF].