# 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?

Aug 2, 2017

$\left[{H}_{3} P {O}_{4}\right] \equiv 1.50 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

${H}_{3} P {O}_{4} \left(a q\right) + 2 N a O H \left(a q\right) \rightarrow N {a}_{2} H P {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

And it is a fact that phosphoric acid acts as DIACID in aqueous solution. $N {a}_{3} P {O}_{4}$ WILL NOT BE ACCESSED even at high $p H$.

$\text{Moles of NaOH} = 50.00 \times {10}^{-} 3 L \times 1.5 \cdot m o l \cdot {L}^{-} 1 = 0.075 \cdot m o l$.....

And thus, with respect to ${H}_{3} P {O}_{4}$, there was a $\frac{0.075 \cdot m o l}{2}$ molar quantity.......

And $\left[{H}_{3} P {O}_{4}\right] = \frac{\frac{0.075 \cdot m o l}{2}}{25.00 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}$

$= \frac{0.075 \cdot m o l}{2}$ of course, we might have been able to nut this result out directly, had we been on the ball in the first instance.