This chemistry question is driving me nuts. To make a solution, 7.38 mmol of a diprotic acid (quinolinic acid), is dissolved in 162.5 mL of water. The pH of the resulting solution is measured to be 1.99. From the equation, what is the pKa of the acid? i am using henderson hasselbach pka = ph - log([A-]/[HA]) [H2A] = 7.38/162.5 = .0454 ph = 1.99 [H+]/2 = [A-] (10^-1.99)/2 = .005116 plug and chug... i get 2.93 the answer is 2.53. what am i doing wrongggggggg?