# Question 216ab

Apr 14, 2017

$\text{0.083 L}$ of concentrated $12.0 M$ $H N {O}_{3}$

#### Explanation:

This is a dilution chemistry problem. For problems like this one, use the following equation

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$

Where
${M}_{1} = \text{molarity of solution 1}$
${V}_{1} = \text{volume of solution 1}$
${M}_{2} = \text{molarity of solution 2}$
${V}_{2} = \text{volume of solution 2}$

Whenever you dilute a concentrated stock solution, or for that matter, any solution, you are essentially changing the volume of that solution by adding water. If we look at what $\text{Molarity}$ exactly is,

"Molarity" = "moles"/(1" Liter solution")#

you can see when you increase the $\text{volume}$, you decrease the $\text{molarity}$ (concentration). In dilution cases, you are adding water so moles of the solute do not change, only the volume of the solution changes.

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$

Using our equation from above, we can find the number of moles of nitric acid, $H N {O}_{3}$, from what we are given.

We will let ${M}_{1}$ and ${V}_{1}$ be $8.0 M$ and $125 \text{ mL}$ respectively. Make sure to convert $\text{mL}$ to $\text{L}$ first. For ${M}_{2}$, use the $12.0 M$. Isolate ${V}_{2}$ and solve.

• ${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$

• $\left(0.125 L\right) \left(8.0 M\right) = \left(12.0 M\right) \left({V}_{2}\right)$

• $\frac{\left(0.125 L\right) \left(8.0 \cancel{M}\right)}{12.0 \cancel{M}} = {V}_{2}$

• $\text{0.083 L} = {V}_{2}$