# What is the concentration of 85% H_3PO_4 in mol*L^-1?

Apr 15, 2017

You need to quote the density of the phosphoric acid.............and you need to add the conc. acid to water...........

#### Explanation:

This site quotes $\rho = 1.685 \cdot g \cdot m {L}^{-} 1$ for 85% ${H}_{3} P {O}_{4} \left(a q\right)$.

And thus $\text{mass of phosphoric acid}$ $=$ 85%xx1.685*g*mL^-1, with respect to a $1 \cdot m L$ volume.........

And we use these data to work out the concentration in $m o l \cdot {L}^{-} 1$.

$\rho = \text{Mass of phosphoric acid"/"Volume of solution}$

And thus $\text{mass of phosphoric acid"=rhoxx"volume of solution}$

((85%xx1.685*g*mL^-1xx1*mL)/(97.99*g*mol^-1))/(1xx10^-3*L)=14.6*mol*L^-1.

Given this, we can now work out the required volumes.........

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, where C_2=6.0*mol*L^-1, V_2=100*mL, V_1=??

${V}_{1} = \frac{{C}_{2} {V}_{2}}{C} _ 1 = \frac{6.0 \cdot m o l \cdot {L}^{-} 1 \times 100 \cdot m L}{14.6 \cdot m o l \cdot {L}^{-} 1} = 41.1 \cdot m L$.

So put approx. $50 \cdot m L$ of water in a flask, add the required volume of acid, and then make up to $100 \cdot m L$ volume. You find the EXACT concentration of your solution by titration with standard base, knowing that you have a solution of $N {a}_{2} H P {O}_{4} \left(a q\right)$ at the end point.