# 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"?

Jul 23, 2016

I got $\approx 1.45 \times {10}^{3}$ $\text{mL}$, or $\approx$ $\text{1.45 L}$. Keep in mind that you won't be adding exactly $\text{50.0 mL}$---you'll probably add a bit less.

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

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

$\setminus m a t h b f \left({M}_{1} {V}_{1} = {M}_{2} {V}_{2}\right)$

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 ($\text{M}$) of solution $i$, and ${V}_{i}$ is the volume of solution $i$.

For the volume, we can use milliliters ($\text{mL}$) for simplicity, since you will be dividing two concentrations and canceling out their units.

• ${V}_{1} = \text{50.0 mL}$
• ${M}_{1} = \text{120. M}$
• ${M}_{2} = \text{4.00 M}$

So, you are solving for the final volume, ${V}_{2}$:

$\textcolor{g r e e n}{{V}_{2}} = \left(\frac{{M}_{1}}{{M}_{2}}\right) {V}_{1}$

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

$=$ $\text{1500 mL}$

Or, to three sig figs:

$= \textcolor{g r e e n}{1.50 \times {10}^{3}}$ $\textcolor{g r e e n}{\text{mL}}$

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

$1500 - 50 = \textcolor{b l u e}{\text{1450 mL}}$ water to begin with.

It's better, however, to start with the $\text{1450 mL}$ of water and slowly add acid until you get to the $\setminus m a t h b f \left(\text{1500 mL}\right)$ mark, to account for the fact that not all solutions are 100% additive.

You won't necessarily transfer exactly $\setminus m a t h b f \left(\text{50 mL}\right)$.