# How many milliliters of 12.0 M HCl(aq) must be diluted with water to make exactly 500. mL of 3.00 M hydrochloric acid?

##### 1 Answer
Apr 10, 2016

$\text{125 mL}$

#### Explanation:

Your strategy here will be to use the molarity and volume of the diluted solution to determine how many moles of solute, which in your case is hydrochloric acid, $\text{HCl}$, it must contain..

Once you know this value, you an sue the molarity of the stock solution as a conversion factor to see how many milliliters would contain this many moles.

The underlying principle of a dilution is the fact that the number of moles of solute must remain constant. Basically, a dilution decreases the concentration of a solution by increasing its volume.

As you know, a solution's molarity tells you how many moles of solute you get per liter of solution. The diluted solution has a molarity of ${\text{3.00 mol L}}^{- 1}$, which means that every liter of this solution will contain $3.00$ moles of hydrochloric acid.

In your case, the $\text{500.-mL}$ sample will contain

500. color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "3.00 moles HCl"/(1color(red)(cancel(color(black)("L")))) = "1.5 moles HCl"

Now use the molarity of the stock solution to determine how many milliliters would contain $1.5$ moles of $\text{HCl}$. Since a concentration of ${\text{12.0 mol L}}^{- 1}$ means that you get $12.0$ moles of hydrochloric acid per liter of solution, you an say that you have

$1.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * "1 L"/(12.0color(red)(cancel(color(black)("moles")))) * (10^3color(white)(a)"mL")/(1color(red)(cancel(color(black)("L")))) = color(green)(|bar(ul(color(white)(a/a)"125 mL} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.

ALTERNATIVE APPROACH

You can get the same result by using the equation for dilution calculations, which looks like this

color(blue)(overbrace(c_1 xx V_1)^(color(red)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(red)("moles of solute in diluted solution"))

Here you have

${c}_{1}$, ${V}_{1}$ - the molarity and volume of the concentrated solution
${c}_{2}$, ${V}_{2}$ - the molarity and volume of the diluted solution

Rearrange to solve for ${V}_{1}$ and plug in your values to find

${c}_{1} {V}_{1} = {c}_{2} {V}_{2} \implies {V}_{1} = {c}_{2} / {c}_{1} \cdot {V}_{2}$

V_1 = (3.00 color(red)(cancel(color(black)("M"))))/(12.0color(red)(cancel(color(black)("M")))) * "500. mL" = color(green)(|bar(ul(color(white)(a/a)"125 mL"color(white)(a/a)|)))