At the equivalence point of a khp/naoh titration, you have added enough oh- to react with all of the hp- such that the only species left in solution are p2- (the conjugate base of hp-) and the spectator ions na+ and k+. if a titration of khp with naoh resulted in [p2-] = 0.042 m at the equivalence point, what is the ph of the solution? ka for hp- = 3.9 x 10-6 3.40 4.98 7.00 9.02 10.61 flag this question