# A 17.15*mL volume of phosphoric acid of 0.077*mol*L^-1 concentration reacts with how many grams of sodium hydroxide to reach stoichiometric equivalence?

May 13, 2017

A tricky one ..........I gets a volume of approx. $5 \cdot m L$.

#### Explanation:

Phosphoric acid acts as a diacid in water, and thus we interrogate the stoichiometric reaction:

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

And so at the endpoint we have $\text{sodium biphosphate}$......

And thus moles of ${n}_{{H}_{3} P {O}_{4}} \equiv 17.15 \times {10}^{-} 3 \cdot L \times 0.077 \cdot m o l \cdot {L}^{-} 1 = 1.32 \times {10}^{-} 3 \cdot m o l$.

And this reacts with TWICE this molar quantity of $\text{sodium hydroxide}$ as per the stoichiometry:

$\frac{2 \times 1.32 \times {10}^{-} 3 \cdot m o l}{0.508 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 = 5.20 \cdot m L$.