The key to this question is understanding what we mean by pH and pKa, which are logarithmic functions. Students tend to have problems with the logarithmic function. I will introduce the subject briefly.
When I write logab=c, I am asking to what power I raise the base a, to get b. Here ac=b. Now, usually we use the bases 10 or e. So log10100=2, log101000=3, log101000000=6. Likewise log100.1=log1010−1 = −1. In the days before electronic calculators (approx. 30-40 years), students would be issued log tables so that complicated calculations could be performed.
From the above pH = −log10[H3O+], and pKa = −log10Ka. These are simple functions that have been widely used in chemistry.
Now it is fact, that when a weak acid is mixed with its conjugate base in appreciable concentrations, a buffer solution is formed that tends to resist gross changes in pH.
We can write, pH=pKa+log10{[A−][HA]}.
It is clear that when [HA] = [A−], then pH=pKa because log10{[A−][HA]} = log101=0
So we want an acid whose pKa ≅ pH. To get pKa I simply perform the function −log10Ka, on each of the acid dissociation constants.
pKa HA = 2.57;
pKa HB = 5.36;
pKa HC = 8.58;
We want to maintain pH at 8.6, so it is clear that acid HC is the acid of choice.
