A lab technician needs 60mL of 25% acid solution for a certain experiment , but he has only 10% solution and 40% solution. How many milliliters of the 10% and the 40% solutions should he mix to get what he needs?

Step-by-step explanation:

The needed concentration is exactly halfway between the available concentrations, hence they must be mixed equally.

He should mix 30 mL each of the 10% and 40% solutions.

The first step in solving any problem is to look at the given information, and at what is being asked for. Here, the mix that is being asked for is exactly halfway between the given concentrations, so you know without any further contemplation that the mix will be 50%/50% of each.

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If the mix were something else, there are several ways the problem can be solved. I like to use a diagram that puts the available concentrations on the left in a column with the highest on top, the needed concentration in the next column in the middle, and the differences of these numbers on the diagonals in the right column.

Example: if the needed mix is 15%, the diagram would look like ...

   40            5

          15

   10           25

The numbers on the right tell you the proportions required (5:25 = 1:5). For this example, you need 1 part 40% solution and 5 parts 10% solution to make a mix that is 15%. That's a total of 6 parts, so each "part" is 10 mL for 60 mL of solution.

We chose this example so every number in the diagram is different, so you could see how the instructions for use apply.

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Another way to solve this is to let a variable represent the amount needed of the highest concentration. If we let that variable be x, for the given problem, we can write the equation for the amount of acid in the mix as ...

  40%(x) + 10%(60 -x) = 25%(60)

  0.30x = 9 . . . . . . . . subtract 6, simplify

  x = 9/0.30 = 30 . . . . . mL of 40% solution