Question

A solution containing an 8% concentration of acid is mixed with a solution containing a 5% concentration of acid to get 1200 mL of solution containing a 7% concentration of acid. How many mL of each solution is needed?

Answer 1

You would need 400 mL of the 8% acid solution and 800 mL of the 5% acid solution.

Step-by-step explanation:

To solve this problem, we can use the method of mixtures. Let's assume that x mL of the 8% acid solution is mixed with (1200 - x) mL of the 5% acid solution to get 1200 mL of a 7% acid solution.

Using the equation for the concentration of the final solution: 0.08x + 0.05(1200 - x) = 0.07 * 1200

Simplifying and solving for x, we find that x = 400 mL of the 8% acid solution is needed, and (1200 - x) = 800 mL of the 5% acid solution is needed.