What is the ratio of the concentrations of base to acid at each of the following levels of dissociation of the acid (1, 10, 20, 30, 40, 50, 60, 70, 80, 90, 99%), i.e. if the acid is 100 % dissociated (ionised) then all of the molecules are in the conjugate base form

For example, if acid is 1% dissociated, then the ratio of base to acid is 1/99 and if it is 20% dissociated, then the ratio is 20/80. However, we have to take into account of how many protons per acid?

@adianadiadi is right - let's say we have an acid (HA) which dissociates into its conjugate base (A-). The dissociation reaction is: HA --> A- + H+Let's say there is 0% dissociation. That gives us: HA --> A- + H+ 100% --> 0% Therefore [base]/[acid] or [A-]/[HA] is 0/100 Now let's say there is 10% dissociation. That gives us: HA --> A- + H+ 90% --> 10% Therefore [base]/[acid] or [A-]/[HA] is 10/90 Now let's say there is 50% dissociation. That gives us: HA --> A- + H+ 50% --> 50% Therefore [base]/[acid] or [A-]/[HA] is 50/50 It turns out that at 50% dissociation, we have a really intersting situation. The Henderson-Hasselbalch equation states: pH = pKa + log ([base]/[acid])What's really cool is that when half is in the acid form, and the other half is in the conjugate base form, the base:acid ratio cancels out! pH = pKa + log(50/50) pH = pKa + log(1) pH = pKa + 0 pH = pKaThe significance of this is the half-equivalence point: basically at this point on a titration curve (the flat parts of the curve), the acid is 50% dissociated. And, if we measure the pH, we can figure out experimentally what its pKa value is! So this is really important because it helps us measure how strong a given acid is, essentially. In other words, pH = pKa at the half-equivalence point. But I digress... Finally, let's say there is 100% dissociation. That gives us: HA --> A- + H+ 0% --> 100% Therefore [base]/[acid] or [A-]/[HA] is 100/0 (woops, that's not even a real number - that's probably why the question asks you to stop at 99% ).