SOLUTION: How much pure acid must be added to 60 ml of a 35% solution to produce a mixture that is 80% acid?
ml of pure acid/ml of mixture = concentration of acid
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ml of pure acid/ml of mixture = concentration of acid
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Question 489271: How much pure acid must be added to 60 ml of a 35% solution to produce a mixture that is 80% acid?
ml of pure acid/ml of mixture = concentration of acid Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! How much pure acid must be added to 60 ml of a 35% solution to produce a mixture that is 80% acid?
ml of pure acid/ml of mixture = concentration of acid
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Equation:
acid + acid = acid
0.35*60 + x = 0.80(60+x)
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Multiply thru by 100 to get:
35*60 + 100x = 80*60 + 80x
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20x = 45*60
x = 15 ml (amt. of pure acid needed)
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Cheers,
Stan H.
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