Question

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

Answer 1

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

`"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 `"3.2 L acid"/"8 L soln"`

You want to add `x " L of water"` to get a 100 % solution.

Then

`3.2/(8+x) = 10/100`

`3.2 ×100=10(8+x)`

`320=80+10x`

`10x = 320-80`

`10x=240`

`x=24`

You would have to add 24 L of water.

Check:

`3.2/(8+24) = 10/100`

`3.2/32 = 10/100`

`1/10 = 1/10`

It checks!