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

1 Answer
Aug 24, 2015

Answer:

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!