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?

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- January 21st 2009, 12:25 PMreeree1987need 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?

- January 21st 2009, 12:48 PMLast_Singularity
- January 21st 2009, 02:18 PMskeeterQuote:

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. - January 21st 2009, 11:08 PMprincess_21
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