Question #26712

1 Answer
Jan 23, 2017

Answer:

Here's what I got.

Explanation:

The first thing to do here is to figure out exactly how much hydrochloric acid you get in #"25 mL"# of #15%# solution

#25 color(red)(cancel(color(black)("mL solution"))) * "15 g HCl"/(100color(red)(cancel(color(black)("mL solution")))) = "3.75 g HCl"#

Now, if you take #x# to be the number of grams of hydrochloric acid coming from the #5%# solution and #y# to be the number of grams of hydrochloric acid coming from the #20%# solution, you can say that

#x + y = 3.75" "color(orange)((1))#

For the #5%# solution, the volume that contains #x# #"g"# of hydrochloric acid is equal to

#x color(red)(cancel(color(black)("g HCl"))) * "100 mL solution"/(5color(red)(cancel(color(black)("g HCl")))) = (20x)" mL solution"#

For the #20%# solution, the volume that contains #y# #"g"# of hydrochloric acid is

#y color(red)(cancel(color(black)("g HCl"))) * "100 mL solution"/(20color(red)(cancel(color(black)("g HCl")))) = (5y)" mL solution"#

You can now say that

#20x + 5y = 25" "color(orange)((2))#

Use equation #color(orange)((1))# to find the value of #x# in terms of #y#

#x = 3.75 - y#

Plug this into equation #color(orange)((2))# to find the value of #y#

#20 * (3.75 - y) + 5y = 25#

#75 - 20y + 5y = 25#

#-15y = - 50 implies y = (-50)/(-15) = 10/3#

Use this to find the value of #x#

#x = 3.75 - 10/3 = 15/4 - 10/3 = 5/12#

Therefore, you can say that in order to get #"25 mL"# of #15%# hydrochloric acid solution, you must mix

#20 * x = 20 * 5/12 = color(darkgreen)(ul(color(black)("8.3 mL of 5% HCl solution")))#

and

#5 * y = 5 * 10/3 = color(darkgreen)(ul(color(black)("16.7 mL of 20% HCl solution")))#

I'll leave the answers rounded to two and three sig figs, respectively, but keep in mind that you only have one significant figure for the concentrations of the two stock solutions.