Question #03d29

1 Answer
Apr 11, 2016

Answer:

Here's what I got.

Explanation:

The first thing to do here is use the volume by volume percent concentration, #"%v/v"#, of the target solution to determine how many liters of solute, which in your case is sulfuric acid, #"H"_2"SO"_4#, it must contain.

A #"29% v/v"# sulfuric acid solution will contain #"29 L"# of sulfuric acid for every #"100 L"# of solution, which means that your solution must contain

#600 color(red)(cancel(color(black)("L solution"))) * ("29 L H"_2"SO"_4)/(100color(red)(cancel(color(black)("L solution")))) = "174 L H"_2"SO"_4#

Now, let's assume the #x# represents the volume of the #"70% v/v"# sulfuric acid solution and #y# represents the volume of the #"25% v/v"# sulfuric acid solution.

The first equation that you can write here will be

#x + y = "600 L"" " " "color(orange)((1))#

This simply uses the fact that the two solutions must be mixed together to form a total volume of #"600 L"#.

Now, use the given percent concentrations to figure out how many liters of sulfuric acid you'd get in those two solutions

#xcolor(white)(a)color(red)(cancel(color(black)("L solution"))) * ("70 L H"_2"SO"_4)/(100color(red)(cancel(color(black)("L solution")))) = 7/10xcolor(white)(a)"L H"_2"SO"_4#

#ycolor(white)(a)color(red)(cancel(color(black)("L solution"))) * ("25 L H"_2"SO"_4)/(100color(red)(cancel(color(black)("L solution")))) = 1/4ycolor(white)(a)"L H"_2"SO"_4#

This means that you can write

#7/10x + 1/4y = "174 L"" " " "color(orange)((2))#

This equation describes the fact that the amount of sulfuric acid you get from the two solutions you're mixing must add up to give #"174 L"#.

Use equation #color(orange)((1))# to get

#x = 600 - y#

Plug this into the second equation to get

#7/10(600 -y) + y/4 = 174#

#420 - 7/10y + y/4 = 174#

#-9/20y = -246 implies y = 547#

This means that you have

#x = 600 - 547 = 53#

I'll leave the answers as

#"volume of 70% solution" = color(green)(|bar(ul(color(white)(a/a)"50 L"color(white)(a/a)|)))#

#"volume of 25% solution" = color(green)(|bar(ul(color(white)(a/a)"550 L"color(white)(a/a)|)))#

Finally, the result makes sense because the concentration of the target solution is much closer to the concentration of the #25%# solution, which implies that you'd need a lot more of this solution than of the #70%# one.