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

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Answer 1 · 2015-08-24T21:03:20

You would add 24 L of water.

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!