# Suppose you wanted a 2:25 dilution of a stock solution, but you wanted 200 ml of it. How would you make this?

Apr 8, 2018

Here's what I got.

#### Explanation:

Your tool of choice here is the dilution factor, $\text{DF}$, which tells you

1. The ratio that exists between the volume of the diluted solution and the volume of the stock solution
2. The ratio that exists between the concentration of the stock solution and the concentration of the diluted solution

So for any dilution, you have

"DF" = color(white)(overbrace(color(black)(V_"diluted"/V_"stock"))^(color(blue)("volume ratio: diluted/stock"))) = color(white)(overbrace(color(black)(c_"stock"/c_"diluted"))^(color(blue)("concentration ratio: stock/diluted")))

Now, you know that you're performing a

$2 : 25 = 1 : \textcolor{b l u e}{12.5}$

dilution that will have

$\text{DF} = \textcolor{b l u e}{12.5}$

This tells you that the stock solution was $\textcolor{b l u e}{12.5}$ more concentrated than the diluted solution and that the volume of the diluted solution is $\textcolor{b l u e}{12.5}$ times the volume of the stock solution.

If you want the volume of the diluted solution to be equal to $\text{200 mL}$, you can use the dilution factor to calculate the volume of the stock solution needed to make this dilution.

${V}_{\text{stock" = V_"diluted"/"DF}}$

Plug in your values to find

${V}_{\text{stock" = "200 mL"/color(blue)(12.5) = "16 mL}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the volume of the diluted solution.

So, you can perform this dilution by taking $\text{16 mL}$ of your stock solution and adding enough water to get the volume of the diluted solution to $\text{200 mL}$.