# If you need to prepare 1.5 L of a 1.5 M solution of nitric acid, how many milliliters of 9.0 M nitric acid will you have to dilute to 1.5 L?

Jan 8, 2018

Well, we must remember to add acid to water....we need $250 \cdot m L$ of the starting acid.

#### Explanation:

We want a solution whose volume is $1.50 \cdot L$, and whose concentration is $1.50 \cdot m o l \cdot {L}^{-} 1$....

And this represents a molar quantity of $1.50 \cdot L \times 1.50 \cdot m o l \cdot {L}^{-} 1 = 2.25 \cdot m o l$ with respect to nitric acid...

We have $9.0 \cdot m o l \cdot {L}^{-} 1$ nitric available.....and so we take the quotient....

$\frac{2.25 \cdot m o l}{9.0 \cdot m o l \cdot {L}^{-} 1} \times 1000 \cdot m L \cdot {L}^{-} 1 \equiv 250 \cdot m L$...

And so you add $250 \cdot m L$ nitric acid to $1250 \cdot m L$ of distilled water.

We remember that if you spit in acid, it spits back.