A student is asked to prepare a buffer solution with pH = 8.6, using one of the following weak acids: HA (#Ka = 2.7 × 10^-3#), HB (#Ka = 4.4 × 10^-6#) or HC (#Ka = 2.6 × 10^-9#). Which acid should she choose?

Question

Henry Antrim · Accepted Answer

#HC# should be the acid of choice.

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. It is a known fact that a buffer solution that is resistant to significant changes in pH is created when a weak acid and its conjugate base are combined in noticeable concentrations. 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.

Jeremiah Adams · Answer

undefined The student should choose acid HA to prepare the buffer solution with pH = 8.6.