A solution of 24 gallons is 50% acidic. It was combined using a solution that is 20% acidic and another that is 70% acidic. How much is used of each solution?

Question

Kelly, a chemist, was asked to prepare 24 gallons of a solution that is 50% acidic by mixing a solution that is 20% acidic with another solution that is 70% acidic. How much of each solution should she use?

Answer 1

Let x = gallons of the 20% acidic solution, then 0.2x = amount of acid in the 20% solution

     y = gallons of the 70% acidic solution, then 0.7y = amount of acid in the 70% solution

    24 = gallons of the 50% acidic solution, then 0.5(24) = amount of acid in the 50% solution

 

Thus,         x + y = 24

               0.2x + 0.7y = 0.5(24) 

 

Solving by substitution:    x = 24 - y

                                     0.2(24 - y) + 0.7y = 0.5(24)

                                        4.8 - 0.2y + 0.7y = 12

                                                       0.5y = 7.2

                                                           y = 14.4 gallons of the 70% acidic solution

                                                           x = 9.6 gallons of the 20% acidic solution 

Answer 2

There are many ways of doing this.

 

Since you need a total of 24 gallons, I'm going to call x the number of 20% gallons and 24-x the number of 70% gallons.

 

Since you need the solution to be 50% acidic, thats 12 gallons of acid and 12 gallons of other liquid.

 

.20x is the amount of acid in the first solution and .70(24-x) is the amount of acid in the second solution.

 

So .2x+.7(24-x)=12

    .2x + 16.8-.7x =12

     -.5x              = - 4.8

         x              =  9.6

 

So 9.6 gallons of 20% acid and 24-9.6 = 14.4 gallons of the 70% acid shoud work.

 

Let's check

 

9.6 * 20% = 1.92 gallons of acid.

14.4*70% = 10.08 gallons of acid

 

Total of 12 gallons of acid in 24 gallons of solution equaling 50% acid.

 

Hope this helped.