# 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

#### 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,

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 *every liter* of this solution will contain

In your case, the

#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 **per liter** of solution, you an say that you have

#1.5 color(red)(cancel(color(black)("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"color(white)(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

Rearrange to solve for

#c_1V_1 = c_2V_2 implies V_1 = c_2/c_1 * 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)|)))#