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To calculate the pOH of the 0.0184 M \( \text{HClO}_4 \) solution at each specified temperature, let's go through the steps:

### Step 1: Determine the pH of the Solution
Since \( \text{HClO}_4 \) is a strong acid, it will fully dissociate in solution, so:
\[
[\text{H}^+] = 0.0184 \, \text{M}
\]
The pH is calculated as:
\[
\text{pH} = -\log[\text{H}^+]
\]
Substitute \( [\text{H}^+] = 0.0184 \):
\[
\text{pH} = -\log(0.0184) \approx 1.735
\]

### Step 2: Calculate pOH Using \( \text{pH} + \text{pOH} = \text{pKw} \)
The relationship between pH, pOH, and \( K_w \) (the ion-product constant of water) is:
\[
\text{pH} + \text{pOH} = \text{pKw}
\]
where \( \text{pKw} = -\log K_w \) for the given temperature.

### Step 3: Substitute Values for Each Temperature

#### At 15°C:
1. Find \( K_w \) at 15°C: \( K_w = 4.57 \times 10^{-15} \).
2. Calculate \( \text{pKw} = -\log(4.57 \times 10^{-15}) \approx 14.34 \).
3. Use \( \text{pOH} = \text{pKw} - \text{pH} \):
   \[
   \text{pOH} = 14.34 - 1.735 \approx 12.605
   \]

#### At 45°C:
1. Find \( K_w \) at 45°C: \( K_w = 3.94 \times 10^{-14} \).
2. Calculate \( \text{pKw} = -\log(3.94 \times 10^{-14}) \approx 13.40 \).
3. Use \( \text{pOH} = \text{pKw} - \text{pH} \):
   \[
   \text{pOH} = 13.40 - 1.735 \approx 11.665
   \]

### Final Answers
- **At 15°C, pOH ≈ 12.605**
- **At 45°C, pOH ≈ 11.665**

Would you like additional details on any part of the process? 

Here are some further related questions to consider:

1. How would the pOH change if the concentration of \( \text{HClO}_4 \) were different?
2. Why does \( K_w \) change with temperature, and how does this affect pH and pOH?
3. How would you find the pH and pOH of a weak acid solution using similar data?
4. What is the significance of \( K_w \) being different at various temperatures?
5. Can you calculate the pOH for another temperature using similar steps?

**Tip:** Always remember that temperature affects the ionization constant \( K_w \), which in turn affects pH and pOH calculations at different temperatures.

Calculate the pOH of a 0.0184 M HClO4 solution at 15°C and 45°C based on the provided Kw values.

Learn how to calculate the pOH of a 0.0184 M HClO4 solution at 15°C and 45°C. This involves key chemistry concepts like Kw, pH, and pOH. The calculations are based on temperature-specific Kw values, and formulas such as pH = -log[H+]. Suitable for high school chemistry students.

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