# If somehow, #"50.0 mL"# of a stock solution of #"120. M"# #"HCl"# is available, what volume of water should the appropriate amount of #"HCl"# be added to, in order to reduce its concentration to #"4.00 M"#?

##### 1 Answer

I got

Just so you know, that initial concentration of **couldn't possibly be safe to use**; it's about 10 times the concentration of the stock

I assume you mean the "before/after" formula:

#\mathbf(M_1V_1 = M_2V_2)#

Basically, you have a relationship of concentration to volume:

- If concentration is to
*increase*, then volume has to*decrease*. - So, if volume
*increases*, then concentration must*decrease*(the contrapositive).

Here,

For the volume, we can use milliliters (**dividing two concentrations** and *canceling out* their units.

What you already have are:

#V_1 = "50.0 mL"# #M_1 = "120. M"# #M_2 = "4.00 M"#

So, you are solving for the **final volume**,

#color(green)(V_2) = ((M_1)/(M_2)) V_1#

#= (("120." cancel"M")/(4.00 cancel"M")) ("50.0 mL")#

#=# #"1500 mL"#

Or, to three sig figs:

#= color(green)(1.50xx10^3)# #color(green)("mL")#

Remember, that's your **FINAL volume**, NOT the amount of water you might start with, so you should subtract to get:

It's better, however, to start with the *slowly* add acid **until you get to the** **mark**, to account for the fact that not all solutions are 100% additive.

**You won't necessarily transfer exactly**