Question

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?

Answer 1

`[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.