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