Word problem

Question

Pure acid is added to a 10% solution to obtain 90L of 51% solution. What amounts of each should be used? How many liters of 100% pure acid should be used to make a solution?

Answer 1

The trick to this type of mixture problem is calculating the amount of pure acid in the components and the mixture.

 

The final mixture is 90L of 51% solution.

(0.51)(90) = 45.9 L of acid

 

A = volume of pure acid

B = volume of 10% acid

 

1.00A + 0.10B = 0.51(90)   

A + 0.1B = 45.9                  volume of pure acid in A, B and the final mixture,

 

A + B = 90                         total volume of the final mixture.

B = 90-A                            Subsititute this in the first equation

 

A + (0.1)((90-A)= 45.9

A + 9 -0.1A =45.9

0.9A = 36.9

A = 41 L                       Pure acid

B = 90-41 = 49 L          10% acid

 

Check:

A + 0.1B =? 45.9

41 + (0.1)(49) =? 45.9

41 + 4.9 =? 45.9

45.9 = 45.9