To solve this problem, we need to determine the quantity of the 60% acid solution that must be mixed with the 25% acid solution to produce a final solution that is 50% acid, with a total volume of 588 mL. Here is the detailed step-by-step solution:
1. Define variables:
- Let [tex]\( x \)[/tex] be the quantity (in mL) of the 60% acid solution.
- Therefore, [tex]\( 588 - x \)[/tex] will be the quantity of the 25% acid solution since the total volume is 588 mL.
2. Set up the equation using the concept of mass balance (weighted average):
- The equation can be formulated from the percentages of acid in each solution.
- The amount of pure acid in the 60% solution is [tex]\( 0.60x \)[/tex] mL.
- The amount of pure acid in the 25% solution is [tex]\( 0.25(588 - x) \)[/tex] mL.
- The amount of pure acid in the final 50% solution is [tex]\( 0.50 \times 588 \)[/tex] mL.
3. Combine the amounts of pure acid from both solutions to form the final solution:
[tex]\[
0.60x + 0.25(588 - x) = 0.50 \times 588
\][/tex]
4. Solve the equation step-by-step:
- Simplify the equation:
[tex]\[
0.60x + 0.25(588 - x) = 294
\][/tex]
- Distribute the 0.25:
[tex]\[
0.60x + 147 - 0.25x = 294
\][/tex]
- Combine like terms:
[tex]\[
0.35x + 147 = 294
\][/tex]
- Isolate [tex]\( x \)[/tex]:
[tex]\[
0.35x = 294 - 147
\][/tex]
[tex]\[
0.35x = 147
\][/tex]
[tex]\[
x = \frac{147}{0.35}
\][/tex]
[tex]\[
x = 420 \, \text{mL}
\][/tex]
5. Calculate the quantity of the 25% acid solution:
[tex]\[
588 - 420 = 168 \, \text{mL}
\][/tex]
Therefore, to produce 588 mL of a 50% acid solution, you need to mix 420 mL of the 60% acid solution with 168 mL of the 25% acid solution.
