A 50.00 mL 1.5 M of #NaOH# titrated with 25.00 ml of #H_3PO_4# (aq). What is the concentration of the is #H_3PO_4# solution?

1 Answer
Aug 2, 2017

Answer:

#[H_3PO_4]-=1.50*mol*L^-1#

Explanation:

We follow the stoichiometric reaction......

#H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq) + 2H_2O(l)#

And it is a fact that phosphoric acid acts as DIACID in aqueous solution. #Na_3PO_4# WILL NOT BE ACCESSED even at high #pH#.

#"Moles of NaOH"=50.00xx10^-3Lxx1.5*mol*L^-1=0.075*mol#.....

And thus, with respect to #H_3PO_4#, there was a #(0.075*mol)/2# molar quantity.......

And #[H_3PO_4]=((0.075*mol)/2)/(25.00*mLxx10^-3*L*mL^-1)#

#=(0.075*mol)/2# of course, we might have been able to nut this result out directly, had we been on the ball in the first instance.