WHAT QUANTITY OF A 70% ACID SOLUTION HAS TO BE MIXED WITH A 20% ACID SOLUTION TO GET 300ML OF A 45% SOLUTION?
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