# How many liters of water must be added to 8 liters of a 40% acid solution to obtain a 10% acid solution?

Aug 24, 2015

You would add 24 L of water.

#### Explanation:

We will assume that the density of the solution remains constant (even though the density changes with concentration!).

A 40 % concentration means

40 % = "40 L acid"/"100 L solution"

So the volume of acid in 8 L of solution is

$\text{Volume of acid" = 8 color(red)(cancel(color(black)("L soln"))) × "40 L acid"/(100 color(red)(cancel(color(black)("L soln")))) = "3.2 L acid}$

So you have $\text{3.2 L acid"/"8 L soln}$

You want to add $x \text{ L of water}$ to get a 100 % solution.

Then

$\frac{3.2}{8 + x} = \frac{10}{100}$

3.2 ×100=10(8+x)

$320 = 80 + 10 x$

$10 x = 320 - 80$

$10 x = 240$

$x = 24$

You would have to add 24 L of water.

Check:

$\frac{3.2}{8 + 24} = \frac{10}{100}$

$\frac{3.2}{32} = \frac{10}{100}$

$\frac{1}{10} = \frac{1}{10}$

It checks!