Question
How much pure acid must be added to a 30% acid solution to make a 60L solution which is 50% acid?
This was the answer I got from Reiny.
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
I was just wondering how did you come up 10x on the second part? Did you multiply everything by 10? Why so? Why not 100?
This was the answer I got from Reiny.
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
I was just wondering how did you come up 10x on the second part? Did you multiply everything by 10? Why so? Why not 100?
Answers
Brittany Jones
Oh I think I got it now. Never mind. Thanks much for the help.