SOLUTION: please help me find the answer! suppose you have 80 ml of a solution that is 60% acid and 40% water. how much acid do you need to add to make a solution that is 75% acid.

Question 750912: please help me find the answer!
suppose you have 80 ml of a solution that is 60% acid and 40% water. how much acid do you need to add to make a solution that is 75% acid.

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
you have 80 ml of a solution that is 60% acid

so you have 80*0.6 = 48 mL of pure acid

Let x = amount of pure acid

If you add x mL of pure acid to 48 mL of pure acid, then you'll have 48+x mL of pure acid total. This is out of 80 + x mL of new total solution.

This ratio (48+x)/(80+x) must be equal to 0.75 since we want this new percentage to be 75%, so

(48+x)/(80+x) = 0.75

now solve for x

(48+x)/(80+x) = 0.75

48+x = 0.75(80+x)

48+x = 60+0.75x

x = 60+0.75x - 48

x-0.75x = 60 - 48

0.25x = 12

x = 12/0.25

x = 48

So you need to add 48 mL of pure acid.

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