How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution?

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Answer 1 · 2017-05-30T16:57:42

15 litres of added water

We need to reduce 25% to 10%

Let the amount of acid in the mix be $x/100$

This gives a final acid content of $x/100xx25%$

The target concentration is 10%

Set $x/100xx25%=10%$

$x=(10%xx100)/(25%)$

$x=40$

Thus $40/100$ of the final blend is the acid
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Let the total volume of the blend be $v$

We have 10 litres of acid so

$40/100v=10$

$v=(10xx100)/40 =100/4=25$ litres all together

Thus the added water is 25-10=15 litres