Question

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?

Answer 1

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!