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Bottle X contains 10% acid and bottle Y contains 20% acid. In what ratio do I need to mix the liquid from the two bottles if I want to create a mixture that is 16% acid?

 Oct 9, 2020
 #1
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It is generally much easier to caculate the VOLUME of the final mixture, in this case 16% acid, if that volume is given. But, we can make up a theoretical volume of say, 10 c.c.

 

Let the amount of 10% acid needed = T

Then, the amount of 20% acid needed = 10 - T

 

0.10T + 0.20*[10 - T] =0.16 * 10, solve for T

 

T = 4 cc of 10% acid will be needed

10 - 4 = 6 cc of 20% acid will be needed.

 

So, the ratio needed to make a mixture of 16% acid will be:

4 : 6 = 2 : 3 . In other words, you will need to mix 2 parts of 10% acid with 3 parts of 20% acid to make a mixture that is 16% acid.

 Oct 9, 2020
 #2
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Acid of ingredients added together = acid in final solution

 

.10a      +    .20 ( f-a)   = .16 f               a = amount of X      f-a = amount of Y    f = final amount

.10a + .20f-.20a = .16f

-.1 a = - .04 f

a = .4 f        ( this is the amount of X)

then      Y = f - a = f - .4f = .6 f

 

then X:Y =    .4  :  .6 =   2 :3                         

 Oct 9, 2020
edited by ElectricPavlov  Oct 9, 2020

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