# If a student needs to make exactly 2.5 liters of a 1.25 M solution of acetic acid from the 12.0 M stock solution in the chemistry closet, what volume of the stock solution should be used?

Jul 1, 2016

$\text{260 mL}$

#### Explanation:

The idea here is that you can use the dilution factor to figure out the volume of stock solution needed to make $\text{2.5 L}$ of diluted solution.

As you know, you can decrease the concentration of a solution by increasing its volume while keeping the number of moles of solute constant, i.e. by diluting it.

The important thing to remember here is that if you increase the volume of a solution by a specific factor, the concentration must decrease by the same factor.

This factor is called the dilution factor

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{D.F." = V_"diluted"/V_"stock" = c_"stock"/c_"diluted} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, the concentration of the solution must go from $\text{12.0 M}$, which is what you have for the stock solution, to $\text{1.25 M}$.

The dilution factor will thus be

"D.F." = (12.0 color(red)(cancel(color(black)("M"))))/(1.25color(red)(cancel(color(black)("M")))) = 9.6

This means that the volume of the stock solution used for the dilution was

$\text{D.F." = V_"diluted"/V_"stock" implies V_"stock" = V_"diluted"/"D.F.}$

${V}_{\text{stock" = "2.5 L"/9.6 = "0.26 L}}$

Expressed in milliliters and rounded to two sig figs, the answer will be

V_"stock" = color(green)(|bar(ul(color(white)(a/a)color(black)("260 mL")color(white)(a/a)|)))