How much pure acid should be mixed with 3 gallons of 40% acid solution in order to get a 90% acid solution?

Guest Aug 29, 2014

#2**+5 **

I set this type of problem up a little differently from Alan. Let 3 gallons of the 40% solution be reperesented by 3(.40). And let the amount of added 100% solution be represented by 1x. And let the amount of the resulting 90% solution be represented by (3 + x)(.90)....so we have......

3(.40) + 1x = (3 + x)(.90) simplify

1.2 + x = 2.7 +.9x combine like terms

.1x = 1.5 divide both sides by .1

x = 15 gallons

And that's the amount of 100% solution that should be added.

CPhill Aug 30, 2014

#1**+5 **

Amount of acid in the 40% solution is 40% of 3 gallons = 1.2 gallons.

Amount of non-acid in the 40% solution is 60% of 3 gallons = 1.8 gallons.

Let x be the total amount of acid in a 90% acid solution. Then we must have 0.9 = x/(x+1.8)

0.9*x + 0.9*1.8 = x

0.1x = 1.62

x = 16.2 gallons

This is the *total* amount, so the amount that must be *added* is this minus the original 1.2 gallons, or 15 gallons.

Alan Aug 29, 2014

#2**+5 **

Best Answer

I set this type of problem up a little differently from Alan. Let 3 gallons of the 40% solution be reperesented by 3(.40). And let the amount of added 100% solution be represented by 1x. And let the amount of the resulting 90% solution be represented by (3 + x)(.90)....so we have......

3(.40) + 1x = (3 + x)(.90) simplify

1.2 + x = 2.7 +.9x combine like terms

.1x = 1.5 divide both sides by .1

x = 15 gallons

And that's the amount of 100% solution that should be added.

CPhill Aug 30, 2014