<PRE><B>Question 1056637</B>
A lab technician wants to mix some amount of a 15% acid solution and another amount of a 25% acid solution so that his resultant mixture is 80 mL of a 22% acid solution. What volume of the 15% acid solution and 25% acid solution is required?
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acid + acid = acid
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0.15x + 0.25(80-x) = 0.22*80
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15x + 25*80 - 25x = 22*80
-10x = -3*80
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x = 24 mL (amt. of 15% solution needed)
80-x = 56 mL (amt. 25% solution needed)
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Cheers,
Stan H.
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