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

Answer:

#"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.

http://acidsandbasesfordummieschem.weebly.com/molarity.html

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)|)))#