Thread: Mixture Type 2 - Problem

The question is:

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 ??

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

@Linalg123 You are absolutely correct. Thanks for pointing this out!!!.