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To solve this problem, we need to calculate the pH of a solution given the mass of hydroiodic acid (HI) and the volume of the solution.

**Step 1: Determine the molarity (M) of the hydroiodic acid solution.**

We are given:
- 827 mg of hydroiodic acid (HI)
- 280 mL of solution

First, convert the mass of HI from milligrams to grams:
\[
827 \, \text{mg} = 0.827 \, \text{g}
\]

Next, calculate the number of moles of HI. The molar mass of hydroiodic acid (HI) is approximately:
\[
M_{\text{HI}} = 1.008 \, (\text{H}) + 126.904 \, (\text{I}) = 127.912 \, \text{g/mol}
\]

So, the number of moles of HI is:
\[
\text{moles of HI} = \frac{0.827 \, \text{g}}{127.912 \, \text{g/mol}} = 6.467 \times 10^{-3} \, \text{mol}
\]

**Step 2: Calculate the molarity (M) of the solution.**

Convert the volume from mL to L:
\[
280 \, \text{mL} = 0.280 \, \text{L}
\]

Now, calculate the molarity of the HI solution:
\[
M = \frac{\text{moles of HI}}{\text{volume in liters}} = \frac{6.467 \times 10^{-3} \, \text{mol}}{0.280 \, \text{L}} = 0.0231 \, \text{M}
\]

**Step 3: Calculate the pH.**

Since HI is a strong acid, it dissociates completely in water:
\[
\text{HI} \rightarrow \text{H}^+ + \text{I}^-
\]
Thus, the concentration of \(\text{H}^+\) ions is equal to the concentration of the HI solution, which is 0.0231 M.

The pH is calculated as:
\[
\text{pH} = -\log[\text{H}^+] = -\log(0.0231) = 1.64
\]

So, the pH of the solution is approximately **1.64**.

Would you like further clarification on any step? Here are some follow-up questions:

1. What is the role of hydroiodic acid as a strong acid in determining the pH?
2. How would the calculation change if HI were a weak acid instead of a strong acid?
3. Why is it important to convert the volume to liters when calculating molarity?
4. Can you explain the significance of significant figures in this problem?
5. What would happen to the pH if the concentration of HI were doubled?

**Tip:** Always ensure that all units are converted correctly, especially when dealing with volume and mass in chemistry problems.

A chemist dissolves 827 mg of pure hydroiodic acid in enough water to make up 280 mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.

This problem walks through the step-by-step calculation of the pH of a hydroiodic acid solution. Starting with the mass of HI and the volume of the solution, learn how to calculate molarity and determine the pH using logarithms. Key concepts include molarity, pH calculation, and unit conversions. Suitable for high school or college chemistry students.

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