Question

Question #26712

Answer 1

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.