To determine the quantities of 80% acid solution and 20% acid solution required to produce 240 mL of a 50% acid solution, we can follow these steps:
1. Identify the variables:
- Let [tex]\( x \)[/tex] be the volume of the 80% acid solution.
- Therefore, the volume of the 20% acid solution will be [tex]\( 240 - x \)[/tex] mL since the total volume must be 240 mL.
2. Set up the equation:
- The acid content in the 80% solution will be [tex]\( 0.80x \)[/tex].
- The acid content in the 20% solution will be [tex]\( 0.20(240 - x) \)[/tex].
- The total acid content needed in the final 50% solution will be [tex]\( 0.50 \times 240 \)[/tex] mL.
3. Write the mixture equation:
[tex]\[
(0.80x) + (0.20(240 - x)) = 0.50 \times 240
\][/tex]
4. Simplify the equation:
[tex]\[
0.80x + 48 - 0.20x = 120
\][/tex]
[tex]\[
0.60x + 48 = 120
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
- Subtract 48 from both sides:
[tex]\[
0.60x = 72
\][/tex]
- Divide both sides by 0.60:
[tex]\[
x = 120
\][/tex]
6. Finding the volume of the 20% solution:
- Since the volume of the 80% solution is [tex]\( x = 120 \)[/tex] mL, the volume of the 20% solution will be:
[tex]\[
240 - 120 = 120 \text{ mL}
\][/tex]
Therefore, 120 mL of the 80% acid solution and 120 mL of the 20% acid solution must be mixed to produce 240 mL of a 50% acid solution.
