In a chemistry lab, there is some quantity of 50% acid solution and some quantity of 25% acid solution . How much of each should be mixed to make 10 liters of 40% acid solution ?+

To solve this, let’s set up variables to represent the unknowns:

Let:
- \( x \) be the amount of the 50% acid solution (in liters),
- \( y \) be the amount of the 25% acid solution (in liters).

We need to mix these two solutions to get 10 liters of 40% acid solution.

### 1. Total volume equation:
Since the total volume of the mixture is 10 liters, we can write the equation:
\[
x + y = 10
\]

### 2. Concentration equation:
The amount of acid in each solution is the product of the volume and the concentration. For the 50% acid solution, the amount of acid is \( 0.50x \), and for the 25% acid solution, the amount of acid is \( 0.25y \). In the final mixture, the amount of acid should be 40% of 10 liters, which is \( 0.40 \times 10 = 4 \) liters of acid.

Thus, the second equation is:
\[
0.50x + 0.25y = 4
\]

### Solving the system of equations:
You now have the system of two equations:
1. \( x + y = 10 \)
2. \( 0.50x + 0.25y = 4 \)

From the first equation, solve for \( y \) in terms of \( x \):
\[
y = 10 - x
\]

Substitute this expression for \( y \) into the second equation:
\[
0.50x + 0.25(10 - x) = 4
\]

Now, distribute the \( 0.25 \):
\[
0.50x + 2.5 - 0.25x = 4
\]

Simplify:
\[
0.25x + 2.5 = 4
\]

Subtract 2.5 from both sides:
\[
0.25x = 1.5
\]

Solve for \( x \):
\[
x = \frac{1.5}{0.25} = 6
\]

Now, substitute \( x = 6 \) into the equation \( y = 10 - x \):
\[
y = 10 - 6 = 4
\]

### Conclusion:
- You need 6 liters of the 50% acid solution.
- You need 4 liters of the 25% acid solution.

Would you like more details or clarifications? 

Here are five related questions to expand your understanding:
1. How would the solution change if the required acid concentration was 35% instead of 40%?
2. What happens if the total volume needed is 15 liters instead of 10 liters?
3. Can this method be applied to create a solution with a 60% concentration?
4. How would you approach the problem if the concentrations of the starting solutions were reversed?
5. How would the system of equations look if one of the solutions was pure acid?

**Tip:** Always check your final answer by verifying both the volume and concentration equations are satisfied.

In a chemistry lab, there is some quantity of 50% acid solution and some quantity of 25% acid solution. How much of each should be mixed to make 10 liters of 40% acid solution?

Learn how to solve a mixture problem by calculating how much of a 50% acid solution and a 25% acid solution you need to create 10 liters of a 40% acid solution. This algebraic problem involves solving systems of linear equations. Suitable for students in Grades 9-10.

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