Question

A quantity of 22% acid solution is being diluted with some 8% acid solution. The final solution must be 12% acid and contain 101 mL. How much of each solution must be added to achieve the result? ROUND TO THE NEAREST mL.

Answer 1

Amount of 22% acid solution needed = 29 ml

Amount of 8% acid solution needed = 72 ml

Explanation:

Let x=amount of 22% solution needed

Then 101 - x= amount of 8% solution needed

The pure acid in the 22% solution (0.22x) plus the amount of pure acid in the 8% solution 0.08(101 - x) has to equal the amount of pure acid in the final mixture (0.12*101).

The equation to solve is:

0.22x + 0.08(101 - x) = (0.12*101)

0.22x + 8.08 - 0.08x = 12.12

0.14x = 12.12 - 8.08

0.14x = 4.04

x = 4.04 / 0.14

= 28.86

x = 29 ml to the nearest ml

Amount of 22% acid solution needed = 29 ml

Amount of 8% acid solution needed = 101 - x

= 101 - 29

= 72ml