# Mixture Type 2 - Problem

• May 28th 2012, 08:56 PM
MikeNoob
Mixture Type 2 - Problem
The question is:
Quote:

A solution of 27 gallons of acid contains 9 gallons of pure acid.How much water in gallons should be added to produce a 25% solution of this acid ?
The answer is 9. However here is how i approached it.

 Quantity %age Cal water x 100% x acid solution 27 9/27 or (33.33%) 0.33 x 27 = 9 Required Solution x+27 25% Equation
Equation is :
x+9 = 0.25(x+27)
x+9 = 0.25x + 6.75

Could anyone tell me what am i doing wrong i am getting a negative value ??
• May 29th 2012, 12:26 AM
linalg123
Re: Mixture Type 2 - Problem
Wouldn't it just be:
you are adding water which has a 0% acid concentration
since you have 9gallons of acid, you want the total solution to be 36 gallons.
Add 36 - 27 gallons of water = 9 gallons.

Doing it in your table you have conc of water is 100%, but actually it's 0% as you're talking about the acid concentration.
so it would be
 Quantity %age Cal Initially 27 1/3 9 addWater x 0 0 Required Solution x+27 25% 9
so (27+x)/4 = 9
x + 27 = 36
x= 9
• May 29th 2012, 12:40 AM
MikeNoob
Re: Mixture Type 2 - Problem
@Linalg123 You are absolutely correct. Thanks for pointing this out!!!.