# A 31.27*mL volume of sodium hydroxide at 0.167*mol*L^-1 concentration was used to neutralize a 22.5*mL volume of AQUEOUS H_3PO_4. What is [H_3PO_4](aq)?

Aug 26, 2016

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

#### Explanation:

Note the stoichiometry. You reach a stoichiometric endpoint at biphosphate, $H P {O}_{4}^{2 -}$, NOT $P {O}_{4}^{3 -}$.

$\text{Moles of sodium hydroxide}$ $=$ $31.27 \times {10}^{-} 3 L \times 0.167 \cdot m o l \cdot {L}^{-} 1$ $=$ $5.22 \times {10}^{-} 3 \cdot m o l$.

There were thus $\frac{5.22 \times {10}^{-} 3 \cdot m o l}{2}$ with respect to ${H}_{3} P {O}_{4}$ in the volume of $22.5 \cdot m L$.

$\text{Molarity"_"phosphoric acid}$ $=$ $\frac{5.22 \times {10}^{-} 3 \cdot m o l}{2} \times \frac{1}{22.5 \times {10}^{-} 3 L}$ $\cong$ $0.1 \cdot m o l \cdot {L}^{-} 1$ with respect to phosphoric acid.