# 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?

May 24, 2017

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..........................

$\text{moles of NaOH} = 38.5 \cdot m L \times {10}^{-} 3 L \cdot m {L}^{-} 1 \times 0.150 \cdot m o l \cdot {L}^{-} 1 = 5.78 \times {10}^{-} 3 \cdot m o l .$

And given the stoichiometry, there was thus a concentration of..........$\frac{5.78 \times {10}^{-} 3 \cdot m o l \times \frac{1}{2}}{15.0 \cdot m L \times {10}^{-} 3 L \cdot m {L}^{-} 1} = 0.193 \cdot m o l \cdot {L}^{-} 1. \ldots \ldots \ldots \ldots \ldots . .$

with respect to ${H}_{3} P {O}_{4}$.