#pH = pK_a + log_(10){[A^-]/[HA]}#. Clearly, we need #pK_a# for #H_2CO_3#, i.e. #pK_a = 6.35.#
Explanation:
If #pH= 7.40#, then #log_(10){[HC(=O)O^-]/[HC(=O)OH]}##=##1.05#. This reasonable because the #pH# is higher than the #pK_a# and there should be more carbonate than carbonic acid.