SOLUTION: If you wish to increase the percent of acid in 60 mL of a 25% acid solution in water to 35% acid, how much pure acid must you add?

Question 934141: If you wish to increase the percent of acid in 60 mL of a 25% acid solution in water to 35% acid, how much pure acid must you add?
Found 2 solutions by mananth, Theo:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
--- percent ---------------- Quantity
Acid I 25 ---------------- 60 ml
Pure acid 100 ---------------- x ml
Mixture 35 ---------------- 60 + x ml

25 * 60 + 100 x = 35 ( 60 + x )
1500 + 100 x = 2100 + 35 x
100 x -35 x = -1500 + 2100
65 x = 600
/ 65
x = 9.23 ml Pure acid

m.ananth@hotmail.ca

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let x = amount of acid you need to add to your solution to make it 35% acid.

your new solution will be equal to 60 + x.

you currently have 15 ml of acid because .25 * 60 = 15.

if you add x to that, you will get x + 15 ml of acid.

if you add x to 60, you will get x + 60 of total solution.

x + 15 must be equal to 35% of 60 + x in your new solution.

your equation is x + 15 = .35 * (60 + x)

simplify this equation to get x + 15 = .35 * 60 + .35 * x

subtract .35 * x from both sides of the equation and subtract 15 from both sides of the equation and you get x - .35 * x = .35 * 60 - 15.

simplify that equation to get .65 * x = 6.

divide both sides of that equation by .65 to get x = 6/.65 = 9.l230769231.

that's your solution.

you start with 15 ml of acid in 60 ml of solution.

15 / 60 = .25 which is the current ratio.

add 9.1230769231 of acid and you get 15 + 9.1230769231 ml of acid in a total solution of 60 + 9.1230769231 ml.

that simplifies to 24.1230769231 ml of acid in a total solution of 69.1230769231 ml.

24.1230769231 / 69.1230769231 = .35

your solution is now 35% acid.

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