Bottle X contains 10% acid and bottle Y contains 20% acid. In what ratio do I need to mix the liquid from the two bottles if I want to create a mixture that is 16% acid?
It is generally much easier to caculate the VOLUME of the final mixture, in this case 16% acid, if that volume is given. But, we can make up a theoretical volume of say, 10 c.c.
Let the amount of 10% acid needed = T
Then, the amount of 20% acid needed = 10 - T
0.10T + 0.20*[10 - T] =0.16 * 10, solve for T
T = 4 cc of 10% acid will be needed
10 - 4 = 6 cc of 20% acid will be needed.
So, the ratio needed to make a mixture of 16% acid will be:
4 : 6 = 2 : 3 . In other words, you will need to mix 2 parts of 10% acid with 3 parts of 20% acid to make a mixture that is 16% acid.
Acid of ingredients added together = acid in final solution
.10a + .20 ( f-a) = .16 f a = amount of X f-a = amount of Y f = final amount
.10a + .20f-.20a = .16f
-.1 a = - .04 f
a = .4 f ( this is the amount of X)
then Y = f - a = f - .4f = .6 f
then X:Y = .4 : .6 = 2 :3