Question 280774: What quantity of 70 per cent acid solution must be mixed with a 30 solution to produce 1080 mL of a 50 per cent solution?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! A key part of solving these types of problems is keeping track of how much 'pure' stuff you need.
You need 1080 mL of 50% solution = .5*1080 = 540 mL of pure acid.
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x = volume of 70% acid you will use
y = volume of 30% acid you will use
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BUT since you know x+y = 1080, it is better to define the volumes in terms of 'x' only
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y = 1080-x
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Set up an equation that will describe the two parts and the final result
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.7x + .3(1080-x) = .5*1080
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multiply by 10 to get rid of decimals
7x + 3(1080-x) = 5*1080 = 5400
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expand, collect, simplify...
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7x + 3240 - 3x = 5400
4x = 2160
x = 540
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y = 1080 - 540 = 540
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So our proposed solution is 540 mL of 70% acid and 540 mL of 30% solution.
By inspection that makes sense because (.7+.3)/2 = .5
But we need to check our work...always...
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How much 'pure' acid will be in the final solution compared to how much you need.
You need 540 mL.
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.7*540 = 378
.3*540 = 162
378 + 162 = 540
Correct!
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Answer:
You need to mix 540 mL of 70% acid with 540 mL of 30% acid to produce 1080 mL of 50% acid.
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Done
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