Acid solution problem

Question

You need 900 mL of a 30% alcohol solution. On hand, you have a 20% alcohol mixture and a 80% alcohol mixture. How much of each mixture will you need to obtain the desired solution?

Answer 1

For this problem, let's call the volume of 20% alcohol solution "a" and the volume of 80% alcohol mixture "b". We can generate two equations for our unknowns (volume and alcohol percentage).

a + b = 900 mL (volume)

.2a + .80b = .3 * 900 (alcohol percentage)

Once we have these two equations, we can substitute b = 900 - a into our second equation and solve. This will give us variable "a", which is the volume we need of 20% alcohol solution.

.2a + .8(900-a) = .3 * 900

.2a + 720 - .8a = 270

-.6a + 720 =270

-.6a = -440

a = 750

Now that we have solved for "a", we can plug this in to our first equation and solve for "b".

750 + b = 900

b = 150

Now that we have solved for both "a" and "b", we have completed the problem. To make 900 mL of 30% alcohol solution, you need 750 mL of 20% alcohol solution and 150 mL of 80% alcohol solution.