A 20*g mass of phosphoric acid dissolved in 250.0*mL aqueous solution, reaches a stoichiometric end-point with WHAT volume of NaOH(aq) at 0.120*mol*L^-1 concentration?

Jul 31, 2017

We need (i) a stoichiometric equation.....

Explanation:

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

Note a stoichiometric endpoint is reached upon the addition of 2 equiv of base.

$\text{Concentration}$ $\left[{H}_{3} P {O}_{4}\right] = \text{Moles of phosphoric acid"/"Volume of solution}$

$= \frac{\frac{20.0 \cdot g}{98.08 \cdot g \cdot m o {l}^{-} 1}}{250 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1} = 0.816 \cdot m o l \cdot {L}^{-} 1$

And (ii) we need equivalent quantitites of $N a O H$ when $20 \cdot m L$ is tritrated.

$= \frac{0.816 \cdot m o l \cdot {L}^{-} 1 \times 20 \times {10}^{-} 3 \cdot L}{\frac{1}{2} \times 0.120 \cdot m o l \cdot {L}^{-} 1} = 0.272 \cdot L$

$= 27.2 \cdot m L$.

Phosphoric acid is a diacid in aqueous solution.