How would you describe the preparation of 900 mL of 3.00 M HNO_3 from the commercial reagent that is 70.5% HNO_3 (w/w) and has a specific gravity of 1.42.?

Apr 14, 2017

Well first, we determine the molar concentration of the mother acid.

...........and finally we get a required volume of $170 \cdot m L$ with respect to the starting acid.

Explanation:

$\text{Concentration of nitric acid}$ $=$ $\text{Moles of nitric acid"/"Volume of solution}$

And if we specify a $1 \cdot m L$ volume then..........

We have ((1.42*g*mL^-1xx1*mLxx70.5%)/(63.01*g*mol^-1))/(1xx10^-3*L)=15.9*mol*L^-1

So that is our mother acid. And as you know WE ALWAYS ADD ACID TO WATER AND NEVER THE REVERSE. Because if you spit in acid it will spit back.............

And thus, we want a molar quantity of $900 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 3.00 \cdot m o l \cdot {L}^{-} 1 = 2.70 \cdot m o l$.

And so we divide this molar quantity by the concentration of the conc. nitric..........

$= \frac{2.70 \cdot m o l}{15.9 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 = 169.8 \cdot m L$

And so we dilute (IN THIS ORDER), $170 \cdot m L$ of the conc. acid TO a volume of $900 \cdot m L$.