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"#?

I got #~~ 1.45xx10^3# #"mL"#, or #~~# #"1.45 L"#. Keep in mind that you won't be adding exactly #"50.0 mL"#---you'll probably add a bit less.

Just so you know, that initial concentration of #"HCl"# couldn't possibly be safe to use; it's about 10 times the concentration of the stock #"HCl"# that university labs let students use at all. But OK, let's see.

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, #M_i# is the concentration in molars (#"M"#) of solution #i#, and #V_i# is the volume of solution #i#.

For the volume, we can use milliliters (#"mL"#) for simplicity, since you will be 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, #V_2#:

#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:

#1500 - 50 = color(blue)("1450 mL")# water to begin with.

It's better, however, to start with the #"1450 mL"# of water and slowly add acid until you get to the #\mathbf("1500 mL")# mark, to account for the fact that not all solutions are 100% additive.

You won't necessarily transfer exactly #\mathbf("50 mL")#.