Question 295910
1. A chemical engineer mixed 40ml of 8% hydrochloric acid with 60ml of 12% hydrochloric acid solution. He used a portion of this solution and replaced it with distilled water. If the new solution tested 5.2% hydrochloric acid, how much of the original mixture did he use. 
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From:A chemical engineer mixed 40ml of 8% hydrochloric acid with 60ml of 12% hydrochloric acid solution. 
amount of acid = .08(40) + .12(60)
amount of acid = 3.2 + 7.2
amount of acid = 10.4 ml
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total amount of solution = 40+60 = 100 ml
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concentration of solution = 10.4/100 = .104
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Let x = amount of solution used and replaced with distilled water
then
.104(100-x) = .052(100+x)
10.4 - .104x = 5.2 + .052x
10.4 = 5.2 + .156x
5.2 = .156x
33.33 ml = x