Question

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?

Answer 1

`"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, `"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 `"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 `"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 `"HCl"`. Since a concentration of `"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 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

`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_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)|)))`