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?

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 $\text{volume of 90 % acid" = xcolor(white)(l) "L}$

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

This is made up of $\text{6 L of 15 % acid}$ and $x \textcolor{w h i t e}{l} \text{L of 90 % acid}$.

$\text{Moles before = moles after}$

${c}_{1} {V}_{1} = {c}_{2} {V}_{2} + {c}_{3} {V}_{3}$

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")))

$240 + 40 x = 90 + 90 x$

$50 x = 150$

$x = \frac{150}{50} = 3$

So, we add $\text{3 L of 90 % acid}$ to $\text{6 L of 15 % acid}$ and get $\text{9 L of 40 % acid}$.

Check:

3 × 90 + 6 × 15 = 9 × 40

$270 + 90 = 360$

$360 = 360$

It checks!