# Question #63a58

##### 1 Answer

#### Answer:

You can use a simple dilution calculation.

#### Explanation:

I assume that by "concentration and volume of the concentration acid" you mean *molar concentration and volume of the concentrated acid solution*, right?

If you know the molar concentration and volume of the concentrated acid solution, and assuming that you know what the volume of the target solution must be, then you can say that

#C_1 * V_1 = C_2 * V_2" "# , where

So, if your concentrated acid solution has amolarity of

#C_2 = V_1/V_2 * C_1#

Now, if you know the volume and *percent concentration by mass* of the concentrated acid solution, then you must use its density to find the molarity of the stock solution.

Once you know the volume and molarity of the stock solution, you can use the formula for dilution calculations again.

For example, if you have one liter of a concentrated hydrochloric acid solution, which is about

#1 * 10^3color(red)(cancel(color(black)("mL"))) * "1.18 g"/(1color(red)(cancel(color(black)("mL")))) = "1180 g"#

The mass of hydrochloric acid in this solution will be

#1180color(red)(cancel(color(black)("g solution"))) * "37 g HCl"/(100color(red)(cancel(color(black)("g solution")))) = "436.6 g"#

Now use the compound's molar mass to get the number of moles you have in one liter of solution

#436.6color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) ~= "12.0 moles HCl"#

The molarity of the stock solution will be

#C_1 = "12 moles"/"1 L" = "12 M"#

So, if you dilute a

#C_2 = V_1/V_2 * C_1#

#C_2 = (100color(red)(cancel(color(black)("mL"))))/(500color(red)(cancel(color(black)("mL")))) * "12 M" = "2.4 M"#