Question 522537
(How much pure acid should be mixed with 2 gallons of an 80% acid solution in order to get a 90% acid solution?) 
:
Let a = amt of 100% acid required
:
.80(2) + a = .90(a+2)
1.6 + a = .9a + 1.8
a - .9a = 1.8 - 1.6
.1a = .2
a = .2/.1
a = 2 gal of acid required