The key to this question is understanding what we mean by #pH# and #pK_a#, which are logarithmic functions. Students tend to have problems with the logarithmic function. I will introduce the subject briefly.

When I write #log_(a)b=c#, I am asking to what power I raise the base #a#, to get #b#. Here #a^c=b#. Now, usually we use the bases #10# or #e#. So #log_(10)100=2#, #log_(10)1000=3#, #log_(10)1000000=6#. Likewise #log_(10)0.1=log_(10)10^(-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# #=# #-log_(10)[H_3O^+]#, and #pK_a# #=# #-log_(10)K_a#. 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=pK_a + log_(10){[[A^-]]/[[HA]]}#.

It is clear that when #[HA]# #=# #[A^-]#, then #pH=pK_a# because #log_(10){[[A^-]]/[[HA]]}# #=# #log_(10)1 =0#

So we want an acid whose #pK_a# #~=# #pH#. To get #pK_a# I simply perform the function #-log_(10)K_a#, on each of the acid dissociation constants.

#pK_a# #HA# #=# #2.57#;

#pK_a# #HB# #=# #5.36#;

#pK_a# #HC# #=# #8.58#;

We want to maintain #pH# at #8.6#, so it is clear that acid #HC# is the acid of choice.