# Question #440a6

##### 1 Answer

Here's what I got.

#### Explanation:

The trick here is to realize that when you're diluting a solution, the *increase* in volume must **match** the *decrease* in concentration **remains constant**.

In other words, the volume will increase by a factor and the concentration will decrease **by the same factor**.

#color(blue)(ul(color(black)("DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted"))) -># thedilution factor

In your case, the concentration of the solution must decrease by a factor of

#"DF" = (3.02 color(red)(cancel(color(black)("M"))))/(0.150 color(red)(cancel(color(black)("M")))) = color(blue)(20.133)#

This means that the volume must have increased by the same dilution factor.

You will thus have

#color(blue)(20.133) = V_"diluted"/V_"concentrated" implies V_"concentrated" = V_"diluted"/color(blue)(20.133)#

Therefore, the student will need

#V_"concentrated" = "125 mL"/color(blue)(20.133) = color(darkgreen)(ul(color(black)("6.21 mL")))#

of

The answer is rounded to three **sig figs**.

The difference between the volume of the stock solution and the volume of the diluted solution represents the **volume of water** needed for the dilution.

Also, keep in mind that when you're dealing with **strong acids** and **strong bases**, you must add the acid or the base to water, not the other way around!