Question #f685a
2 Answers
Explanation:
The first important thing to notice here is that the concentration of the target solution is closer to
Right from the start, this tells you that the target solution will contain a higher volume of the
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
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#
Use this to calculate the number of moles of hydrochloric acid present in both solutions
#n_"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_"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_"total" = "0.15 moles" + "0.40 moles" = "0.55 moles HCl"#
The total volume of the solution will be
#V_"total" = "1 dm"^3 + "1 dm"^3 = "2 dm"^3#
The concentration of the solution would be
#c_"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,
The trick now is to realize that in order to reduce this concentration down to
I say "remove" because what you'll be doing is actually starting with
So, if you start with the
#V_"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
#n_x = "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_"total" = "0.15 moles" + (0.40x)color(white)(a)"moles"#
#n_"total" = (0.15 + 0.40x)color(white)(a)"moles"#
Use the known concentration of the target solution to find the value of
#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(1+x) = 0.15 + 0.40x#
#0.25 + 0.25x = 0.15 + 0.40x#
#0.15x = 0.10 implies x = 0.10/0.15 = "0.667"#
The maximum volume of
#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
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
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
In this example it will be the
By iteration it gives you the volume of B which you want -
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.