Kyle M. answered • 09/19/12

Motivational tutoring in Science and Math

We should first try to think of a way to form an equation. We'll need to solve for the number of mL from the first solution and the number of mL from the second solution. Because there are two variables that need to be solved for, we'll need to form two equations. Two concepts we can use to form an equation for this problem are concervation of mass and concervation of volume. The concervation of volume equation is easy, so lets start with that one.

'x' will be the number of mL of the 70% acid solution

'y' will be the number of mL of the 20% solution

The question asks for a resulting mixture of 300mL, so the number of mL of both starting solutions has to add together to equal 300mL:

x + y = 300

Now, lets form the second equation using the concept of concervation of mass. There is no mass of these acids, just concentration and volume. By looking at the dimentional units for a solution, we can see that multiplying the concentration of a solution by the volume will produce the ammount of stuff in the solution

x% acid solution (concentration) = ammount of acid / volume of water

so the ammount of acid in the resulting solution from the solution with 70% acid would be the concenbtration, 70%, multiplied by the volume of solution added, x.

amount of acid from first solution: 0.7*x

amount of acid from second solution: 0.2*y

amount of acid in resulting mixture: 0.45*300 mL

0.7*x + 0.2*y = 0.45*300

Kyle M.

What confuses me about your explination is that we are not averaging the concentrations in the regular sense, but taking a weighted average of the concentrations to acheive a desired result, which can be significantly harder to conceptulaize. Although both solutions are weighted the same in this problem, I would not expect every word problem forming a system of equations to do the same. Also, a second equatiuon must always be formed, in this case expressing the relationship between the weights (x+y=1), forming the general equation Ax+B(1-x)=C for two solutions with concentrations A and B and resulting solution with concentration C. Furthermore, once the student has solved for the value of the weight, he or she would have to go back to solve for the ammount of liquid needed from each solution, which wouldn't be dificult, but would require again, an understanding of weighted averages. I beleive the explination I propose offers a little more insight into forming a system of equations. But I understand that my explination can be seen as 'removed' from the average person and I apreciate the simplicity in your explination.

10/29/12

Paul C.

Kyle, I like your solution, and I'd like to add a comment that might make it easier for some of us "regular folks" to understand.

Working with concentrations is similar to working with AVERAGES. We want the mix to average out to 45% concentration in the end. How many "parts" of 20 do we need, and how many "parts" of 70? When we AVERAGE, we add together, and divide by the number of inputs. For example, 20+70=90, which we then divide by 2, to get 45. If we add one part of 20% to one part of 70%, and mix them together, they "average" out to 45%. Since we want to have a total of 300mL in the end, we are going "half and half." We will use 150 mL of the 70% acid solution, and 150mL of the 20% acid solution.

10/29/12