Mixture problems

Question

500 ml of 75% acid solution is to be diluted with pure water to bring the concentration down to 20% acid. How much pure water (pure water is 0% acid) should be added?

Answer 1

Percentage*Volume = Percentage * Volume 

(75%)(500 mL) = (20%)(x mL)

Solving for x gives 1875 mL.  However, this is the total volume of the solution, and the question asks us how much pure water we have to add.

 

Since we started with 500 mL, we subtract that from our total calculated volume: 1875 mL - 500 mL = 1376 mL H2O to be added.