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")`.