To solve this problem, we can use the concept of the mass (or volume) balance equation, focusing on the acid content.

Let $x$ represent the amount (in liters) of the 25% acid solution that needs to be added.

Step 1: Write the equation for the acid content

• The total amount of acid in the 25% solution is $0.25x$ (since 25% of $x$ is acid).
• The total amount of acid in the 30 liters of the 80% solution is $0.80 \times 30 = 24$ liters (since 80% of 30 liters is acid).
• After adding $x$ liters of the 25% solution to the 30 liters of the 80% solution, the total volume becomes $30 + x$ liters.
• We want the resulting mixture to be 50% acid, so the total amount of acid in the mixture should be $0.50 \times (30 + x)$.

Step 2: Set up the equation

The total amount of acid after mixing must equal the sum of the acid in both solutions:

$0.25x + 24 = 0.50(30 + x)$

Step 3: Solve for $x$

First, expand both sides of the equation:

$0.25x + 24 = 15 + 0.50x$

Now, move the terms involving $x$ to one side and the constants to the other side:

$24 - 15 = 0.50x - 0.25x$

Simplify:

$9 = 0.25x$

Now, solve for $x$:

$x = \frac{9}{0.25} = 36$

Conclusion

You need to add 36 liters of the 25% acid solution to the 30 liters of the 80% acid solution to create a 50% acid solution.

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Would you like a further breakdown of the steps or have additional questions? Here are some related questions:

1. What would happen if the concentrations were reversed (adding a 80% solution to a 25% solution)?
2. How can this method be applied to different concentration mixtures?
3. How does the total volume affect the final acid concentration in this type of problem?
4. Can you solve similar problems using a graphical approach?
5. What real-world scenarios might require this type of calculation?

Tip: Always make sure to balance both the acid content and the total volume in mixture problems!