To determine how many milliliters of a 30% acid solution need to be added to 55 mL of a 55% acid solution to obtain a solution that is 40% acid, follow these steps:
1. Define Variables:
- Let [tex]\( x \)[/tex] be the amount of 30% acid solution to be added, in mL.
2. Amount of Acid in Each Solution:
- The 30% acid solution contains 30% acid, so the amount of pure acid in [tex]\( x \)[/tex] mL of this solution is [tex]\( 0.30x \)[/tex].
- The 55% acid solution contains 55% acid, so the amount of pure acid in 55 mL of this solution is [tex]\( 0.55 \times 55 \)[/tex].
3. Total Volume of the Final Solution:
- The total volume of the solution after adding [tex]\( x \)[/tex] mL of the 30% acid solution to the 55 mL of the 55% acid solution will be [tex]\( 55 + x \)[/tex] mL.
4. Amount of Acid in the Final Solution:
- The final solution should be 40% acid, so the amount of pure acid in the final solution should be 40% of the total volume, which is [tex]\( 0.40 \times (55 + x) \)[/tex].
5. Set Up the Equation:
- The total amount of acid from the initial solutions should equal the amount of acid in the final 40% solution.
- Therefore, we can set up the equation as:
[tex]\[
0.30x + 0.55 \times 55 = 0.40 \times (55 + x)
\][/tex]
6. Solve the Equation:
- First, simplify the equation step-by-step:
[tex]\[
0.30x + (0.55 \times 55) = 0.40 \times (55 + x)
\][/tex]
[tex]\[
0.30x + 30.25 = 0.40 \times (55 + x)
\][/tex]
- Expand the equation:
[tex]\[
0.30x + 30.25 = 0.40 \times 55 + 0.40x
\][/tex]
[tex]\[
0.30x + 30.25 = 22 + 0.40x
\][/tex]
- Rearrange to isolate [tex]\( x \)[/tex]:
[tex]\[
30.25 - 22 = 0.40x - 0.30x
\][/tex]
[tex]\[
8.25 = 0.10x
\][/tex]
[tex]\[
x = \frac{8.25}{0.10}
\][/tex]
[tex]\[
x = 82.5
\][/tex]
Therefore, 82.5 milliliters of the 30% acid solution need to be added to 55 mL of the 55% acid solution to obtain a 40% acid solution.
