Question

How much pure acid must be added to a 30% acid solution to make a 60L solution which is 50% acid?

Answer 1

amount of pure acid to be added -- x L

1(x) + .3(60-x) = .5(60)
10x + 3(60-x) = 5(60)
10x + 180 - 3x = 300
7x = 120
x = 120/7 or appr 17.14 L