1. ## need help fast!!!

i hav gotten different ansers for tis poblem "Mow much pure hydrochloric acid must be added to 40 mililiters of a 50% solution to obtain a 75% acid solution?" can anyone clarify this problem for me?

2. Originally Posted by reeree1987
i have gotten different ansers for tis poblem "Mow much pure hydrochloric acid must be added to 40 mililiters of a 50% solution to obtain a 75% acid solution?" can anyone clarify this problem for me?
I am pretty sure it's 40 ml of pure hydrochloric acid. If you take the average of 100% and 50% (assuming equal weights of 40 ml each), you will get a 75% acid solution.

3. i hav gotten different ansers for tis poblem
what "answers" did you get, and how did you get them?

here is the set-up I would use ...

let x = amount of pure HCl in mL

x(100%) + 40(50%) = (x+40)(75%)

solve for x.

4. Originally Posted by reeree1987
i have gotten different ansers for tis poblem "Mow much pure hydrochloric acid must be added to 40 mililiters of a 50% solution to obtain a 75% acid solution?" can anyone clarify this problem for me?

let:
x = the amount of pure hydrochloric acid (100%=1)
(40+x) =total amount

x(1)+ 40(0.5) = 40+x(0.75)
x(100%) amount of pure acid + 40ml(50%) solution = 40 + x(75%)
x+ 20 = 30 + 0.75x
x - 0.75x = 30 -20
0.25x = 10
x= 40 ml

let's check it
40(1) + 40(.5) = 40+40(.75)
40+20= 80(.75)
60=60