Question

What amount of a 80% acid solution must be mixed with a 30% solution to produce 200 mL of a 50% solution?

Answer 1

you need tomix 80 mL of the 80% acid solution with 120 mL of the 30% solution to produce 200 mL of a 50% solution.

Explanation:

Let x be the amount (in mL) of the 80% solution to be mixed with the 30% solution.

Here's the equation based on the amount of acid in each solution:

0.80x (acid from the 80% solution) + 0.30(200 - x) (acid from the 30% solution) = 0.50(200) (acid in the final 50% solution)

Now, let's solve for x:

0.80x + 60 - 0.30x = 100

Combine like terms:

0.50x + 60 = 100

Subtract 60 from both sides:

0.50x = 100 - 60

0.50x = 40

Now, divide by 0.50 to isolate x:

x = 40 / 0.50

x = 80