Question

What is the molarity of a solution of `H_3PO_4`, if 15.0 mL is neutralized by 38.5 mL of 0.150 M NaOH#?

Answer 1

We need a stoichiometric equation.........and of course there is a catch.

Explanation:

You will have to liaise with your teacher to confirm that phosphoric acid behaves as a DIACID in water. And thus we investigate the reaction:

`H_3PO_4(aq) + 2NaOH(aq) rarr Na_2^(+)""^(-)HPO_4(aq) + 2H_2O(l)`

And thus..........................

`"moles of NaOH"=38.5*mLxx10^-3L*mL^-1xx0.150*mol*L^-1=5.78xx10^-3*mol.`

And given the stoichiometry, there was thus a concentration of..........`(5.78xx10^-3*molxx1/2)/(15.0*mLxx10^-3L*mL^-1)=0.193*mol*L^-1..................`

with respect to `H_3PO_4`.