How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution?

1 Answer
Sep 24, 2016

You must add 3 L of the 90 % acid.

Explanation:

For this problem, we can use the relation

color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = "amount of solute" color(white)(a/a)|)))" "

Let the "volume of 90 % acid" = xcolor(white)(l) "L"

Then, after mixing, we have (6 + x") L of 40 % acid".

This is made up of "6 L of 15 % acid" and xcolor(white)(l) "L of 90 % acid".

"Moles before = moles after"

c_1V_1 = c_2V_2 + c_3V_3

40 color(red)(cancel(color(black)(%))) × (6 + x) color(red)(cancel(color(black)("L"))) = 15 color(red)(cancel(color(black)(%))) × 6 color(red)(cancel(color(black)("L"))) + 90 color(red)(cancel(color(black)(%))) × x color(red)(cancel(color(black)("L")))

240 + 40x = 90 + 90x

50x = 150

x = 150/50 = 3

So, we add "3 L of 90 % acid" to "6 L of 15 % acid" and get "9 L of 40 % acid".

Check:

3 × 90 + 6 × 15 = 9 × 40

270 + 90 = 360

360 = 360

It checks!