# What mass and volume of "25% orthophosphoric acid" is required for a 1.0*mol*L^-1 solution of H_3PO_4(aq)?

Jul 4, 2017

You need to quote the $\text{density}$, $\rho$, $\text{of 25% orthophosphoric acid}$.

#### Explanation:

And http://pubs.acs.org/doi/abs/10.1021/ie50546a062 quotes $\rho = 1.17 \cdot g \cdot m {L}^{-} 1$. This value should have been quoted with the question.

And so we can work out the concentration, by considering just a $\text{litre volume}$.

We need $\frac{\frac{97.99 \cdot g}{97.99 \cdot g \cdot m o {l}^{-} 1}}{1.0 \cdot L \text{ of solution}}$.

And so we need $97.99 \cdot g$ with respect to ${H}_{3} P {O}_{4}$.

=(97.99*g)/(25%xx1.17*g*mL^-1)=335.0*1/(1/(mL))=335.0*mL

And so we take a $335.0 \cdot m L$ volume of 25%(w/w) ${H}_{3} P {O}_{4}$, and dilute it up to a ONE LITRE volume. The order of addition is important, because $\text{if you SPIT in acid it spits back........}$