# Question f685a

May 6, 2016

${\text{1.67 dm}}^{3}$

#### Explanation:

The first important thing to notice here is that the concentration of the target solution is closer to $\text{0.15 M}$ than it is to $\text{0.40 M}$.

Right from the start, this tells you that the target solution will contain a higher volume of the $\text{0.15 M}$ solution than of the $\text{0.40 M}$ solution.

As you know, molarity is defined as the number of moles of solute per cubic decimeter of solution. This means that you can express the number of moles of solute in terms of the solution's molarity and volume

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Use this to calculate the number of moles of hydrochloric acid present in both solutions

${n}_{\text{0.15 M" = "0.15 mol" color(red)(cancel(color(black)("dm"^(-3)))) * 1 color(red)(cancel(color(black)("dm"^3))) = "0.15 moles HCl}}$

${n}_{\text{0.40 M" = "0.40 mol" color(red)(cancel(color(black)("dm"^(-3)))) * 1color(red)(cancel(color(black)("dm"^3))) = "0.40 moles HCl}}$

Now, notice what you get when you mix these solutions completely. The total number of moles of hydrochloric acid will be

${n}_{\text{total" = "0.15 moles" + "0.40 moles" = "0.55 moles HCl}}$

The total volume of the solution will be

${V}_{\text{total" = "1 dm"^3 + "1 dm"^3 = "2 dm}}^{3}$

The concentration of the solution would be

${c}_{\text{target" = "0.55 moles"/"2 dm"^3 = "0.275 moles HCl}}$

As you can see, this solution is slightly more concentrated than what you need, $\text{0.25 M}$.

The trick now is to realize that in order to reduce this concentration down to $\text{0.25 M}$ and still get a maximum volume, you must "remove" some of the solution that contains more moles per unit of volume, i.e. the more concentrated one.

I say "remove" because what you'll be doing is actually starting with ${\text{1 dm}}^{3}$ of the less concentrated solution and adding just enough of the more concentrated solution to get the concentration to $\text{0.25 M}$.

So, if you start with the $\text{0.15 M}$ solution and add $x$ ${\text{dm}}^{3}$ of $\text{0.40 M}$ solution, you can say that the total volume of the resulting solution will be

${V}_{\text{total" = "1 dm"^3 + xcolor(white)(a)"dm"^3 = (1+x)color(white)(a)"dm}}^{3}$

The number of moles of hydrochloric acid present in $x$ ${\text{dm}}^{3}$ of the $\text{0.40 M}$ solution will be

${n}_{x} = \text{0.40 mol" color(red)(cancel(color(black)("dm"^(-3)))) * xcolor(red)(cancel(color(black)("dm"^3))) = (0.40x)color(white)(a)"moles HCl}$

The total number of moles of hydrochloric acid present in the target solution will be

${n}_{\text{total" = "0.15 moles" + (0.40x)color(white)(a)"moles}}$

${n}_{\text{total" = (0.15 + 0.40x)color(white)(a)"moles}}$

Use the known concentration of the target solution to find the value of $x$

0.25 color(red)(cancel(color(black)("mol")))/color(red)(cancel(color(black)("dm"^3))) = ( (0.15 + 0.40x)color(red)(cancel(color(black)("moles"))))/((1+x)color(red)(cancel(color(black)("dm"^3))))

Rearrange to get

$0.25 \left(1 + x\right) = 0.15 + 0.40 x$

$0.25 + 0.25 x = 0.15 + 0.40 x$

$0.15 x = 0.10 \implies x = \frac{0.10}{0.15} = \text{0.667}$

The maximum volume of $\text{0.25 M}$ hydrochloric acid solution that can be made from ${\text{1 dm}}^{3}$ of $\text{0.15 M}$ solution and ${\text{1 dm}}^{3}$ of $\text{0.40 M}$ solution will be

V_"max" = "1 dm"^3 + "0.667 dm"^3 = color(green)(|bar(ul(color(white)(a/a)"1.67 dm"^3color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs.

So, in order to make your target solution, you must mix ${\text{1 dm}}^{3}$ of $\text{0.15 M}$ solution and ${\text{0.67 dm}}^{3}$ of $\text{0.40 M}$ solution.

May 10, 2016

$1.67 \text{L}$

#### Explanation:

An alternative method which is useful if you have a lot of these calculations to do is to set up a spreadsheet:

In this example you can input the original concentrations into the 2 cells above the table.

These are linked to the table so if you change them, the values in the table will reflect that change automatically.

These are solutions A and B.

Because the target concentration is closer to A I have set the volume of A to be $1 \text{L}$. The no. of moles of HCl is automatically calculated in the next column using $n = c \times v$.

In the next column I have gradually increased the volume of B. The next 2 columns give the total moles of HCl and the total volume.

The last column gives the final concentration of the mixture.

If you go onto the "data" menu bar then select "what if scenarios" then "goal seek" you can set the target cell to be the value you want, in this case 0.25, by altering any of the precedent cells.

(I am using $\text{Excel}$ - other spreadsheets are available)

In this example it will be the ${V}_{b}$ column.

By iteration it gives you the volume of B which you want - $0.66 \text{L}$ which gives a maximum volume of $1.67 \text{L}$ rounded up.

I wouldn't bother doing this for a single calculation but if, say, you are doing a lab which involves lots of multiple trials an automated table like this can save you a lot of time number crunching with a calculator.