# You have 67.0 mL of a 0.400 M stock solution that must be diluted to 0.100 M. Assuming the volumes are additive, how much water should you add?

Dec 30, 2016

$\text{201 mL}$

#### Explanation:

The thing to remember about dilutions is that the ratio that exists between the concentration of the concentrated solution and that of the diluted solution is equal to the ratio that exists between the volume of the diluted solution and that of the concentrated solution.

This ratio is called the dilution factor. You thus have

$\text{DF" = "concentration of the concentrated solution"/"concentration of the diluted solution}$

and

$\text{DF" = "volume of the diluted solution"/"volume of the concentrated solution}$

In your case, the solution must be diluted from an initial concentration of $\text{0.400 M}$ to a final concentration of $\text{0.100 M}$, the equivalent of a dilution factor equal to

"DF" = (0.400 color(red)(cancel(color(black)("M"))))/(0.100color(red)(cancel(color(black)("M")))) = color(blue)(4)

You can now say that the volume of the diluted solution must be $4$ times greater than the volume of the concentrated solution, since

$\text{DF" = "volume of diluted solution"/"67.0 mL}$

$\text{volume of diluted solution" = "DF" xx "67.0 mL}$

and therefore

$\text{volume of diluted solution" = color(blue)(4) xx "67.0 mL" = "268 mL}$

Assuming that the volumes are addictive, you will have to add

"volume of water added" = "268 mL" - "67.0 mL" = color(darkgreen)(ul(color(black)("201 mL")))

of water to your concentrated solution in order to go from $\text{67.0 mL}$ of $\text{0.400 M}$ solution to $\text{268 mL}$ of $\text{0.100 M}$ solution.

The answer is rounded to three sig figs.