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To solve this problem, we need to use the dilution formula:

\[
C_1 V_1 = C_2 V_2
\]

Where:
- \(C_1\) is the concentration of the stock solution (6.00 M),
- \(V_1\) is the volume of the stock solution we need to find,
- \(C_2\) is the concentration of the dilute solution (0.449 M),
- \(V_2\) is the volume of the dilute solution (75.0 mL).

Now, solving for \(V_1\):

\[
V_1 = \frac{C_2 V_2}{C_1}
\]

Substituting the known values:

\[
V_1 = \frac{0.449 \times 75.0}{6.00}
\]

Calculating this:

\[
V_1 = \frac{33.675}{6.00} = 5.61 \, \text{mL}
\]

So, you need to add **5.61 mL** of the concentrated hydriodic acid solution to obtain a total volume of 75.0 mL of the dilute solution.

Would you like more details or have any other questions?

Here are five related questions to expand your understanding:
1. What would happen if you added too much of the stock solution?
2. How does the molarity change when more solvent is added to a solution?
3. Can you calculate the final concentration if 10 mL of stock solution is used instead?
4. How would the volume change if the target molarity was 1.00 M instead of 0.449 M?
5. What is the significance of molarity in chemical reactions?

**Tip:** Always check your unit conversions in dilution problems to ensure consistency!

You wish to make a 0.449 M hydriodic acid solution from a stock solution of 6.00 M hydriodic acid. How much concentrated acid must you add to obtain a total volume of 75.0 mL of the dilute solution?

Learn how to calculate the volume of concentrated 6.00 M hydriodic acid needed to prepare 75.0 mL of a 0.449 M solution using the dilution formula. This step-by-step solution is suitable for high school or college students studying chemistry.

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