What is the molarity of a sample of phosphoric acid if 15.00 ml of 2 M #KOH# was required to titrate 12 ml of the #H_3PO_4# sample?

1 Answer
Jun 8, 2017

Answer:

#[H_3PO_4]=1.25*mol*L^-1#

Of course, there is a catch.............

Explanation:

This one is not so straightforward, and requires specialist knowledge.

It is a fact that phosphoric acid is a DIACID in aqueous solution, and reaches a stoichiometric endpoint at #HPO_4^(2-)#:

#H_3PO_4(aq) +2KOH(aq) rarr K_2HPO_4(aq) + 2H_2O#.

And thus moles of #KOH# #=# #15.00xx10^-3*Lxx2.0*mol*L^-1# #=0.030*mol#.

And thus the equivalent quantity of #H_3PO_4# was #0.015*mol# which were present in a #12*mL# volume of the acid........

#=((0.030*mol)/2)/(12xx10^-3*L)=1.25*mol*L^-1#.

Remember what I said with regard to the acid-base behaviour of phosphoric acid. It is NOT a triacid..........and you might win brownie points from your prof if you know this fact.