To find the maximum volume of 0.243 M HCl that can be made from 1 liter of 0.183 M HCl and 1 liter of 0.381 M HCl, we can use the concept of molarity and the conservation of moles.
### Step-by-Step Solution:
1. **Identify the given information:**
- Volume of HCl solution 1 (V1) = 1 L
- Molarity of HCl solution 1 (M1) = 0.183 M
- Volume of HCl solution 2 (V2) = 1 L
- Molarity of HCl solution 2 (M2) = 0.381 M
- Desired molarity of the final solution (M3) = 0.243 M
2. **Calculate the total moles of HCl in each solution:**
- Moles of HCl from solution 1:
\[
\text{Moles from solution 1} = M1 \times V1 = 0.183 \, \text{mol/L} \times 1 \, \text{L} = 0.183 \, \text{mol}
\]
- Moles of HCl from solution 2:
\[
\text{Moles from solution 2} = M2 \times V2 = 0.381 \, \text{mol/L} \times 1 \, \text{L} = 0.381 \, \text{mol}
\]
3. **Calculate the total moles of HCl available:**
\[
\text{Total moles} = 0.183 \, \text{mol} + 0.381 \, \text{mol} = 0.564 \, \text{mol}
\]
4. **Set up the equation for the final solution:**
- Let \( x \) be the volume of the 0.381 M solution that we will use. The total volume of the final solution will then be:
\[
V3 = 1 \, \text{L} + x \, \text{L}
\]
- The moles of HCl in the final solution can be expressed as:
\[
\text{Moles in final solution} = 0.183 \, \text{mol} + 0.381 \, \text{mol} \times x
\]
- The molarity of the final solution is given by:
\[
M3 = \frac{\text{Total moles}}{V3}
\]
- Therefore, we can write:
\[
0.243 = \frac{0.183 + 0.381x}{1 + x}
\]
5. **Cross-multiply to solve for \( x \):**
\[
0.243(1 + x) = 0.183 + 0.381x
\]
\[
0.243 + 0.243x = 0.183 + 0.381x
\]
6. **Rearranging the equation:**
\[
0.243 - 0.183 = 0.381x - 0.243x
\]
\[
0.060 = 0.138x
\]
\[
x = \frac{0.060}{0.138} \approx 0.435 \, \text{L}
\]
7. **Convert \( x \) to milliliters:**
\[
x \approx 0.435 \, \text{L} \times 1000 \, \text{mL/L} = 435 \, \text{mL}
\]
8. **Calculate the total volume of the final solution:**
\[
V3 = 1 \, \text{L} + 0.435 \, \text{L} = 1.435 \, \text{L} = 1435 \, \text{mL}
\]
### Final Answer:
The maximum volume of 0.243 M HCl that can be made from the two solutions is **1435 mL**.
