Question

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?

Answer 1

`[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.