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To solve this problem, let’s go through each part step-by-step.

Given:
- pH = 9.49 at 25 °C.

### 1. Calculate \([ \text{H}^+ ]\)
The relationship between pH and hydrogen ion concentration \([ \text{H}^+ ]\) is:
\[
[ \text{H}^+ ] = 10^{-\text{pH}}
\]
Substitute pH = 9.49:
\[
[ \text{H}^+ ] = 10^{-9.49} \approx 3.24 \times 10^{-10} \, \text{M}
\]

### 2. Calculate pOH
Using the relationship between pH and pOH:
\[
\text{pH} + \text{pOH} = 14
\]
So,
\[
\text{pOH} = 14 - \text{pH} = 14 - 9.49 = 4.51
\]

### 3. Calculate \([ \text{OH}^- ]\)
The relationship between pOH and hydroxide ion concentration \([ \text{OH}^- ]\) is:
\[
[ \text{OH}^- ] = 10^{-\text{pOH}}
\]
Substitute pOH = 4.51:
\[
[ \text{OH}^- ] = 10^{-4.51} \approx 3.09 \times 10^{-5} \, \text{M}
\]

### Summary of Answers:
- \([ \text{H}^+ ] \approx 3.24 \times 10^{-10} \, \text{M}\)
- \([ \text{OH}^- ] \approx 3.09 \times 10^{-5} \, \text{M}\)
- \(\text{pOH} \approx 4.51\)

Would you like further details or have any questions?

---

1. What is the relationship between pH, pOH, and the ion product of water?
2. How does temperature affect pH and pOH calculations?
3. What methods can we use to verify pH or pOH experimentally?
4. Why is pH a logarithmic scale rather than linear?
5. How do strong acids and bases differ in their pH and pOH calculations?

**Tip:** Remember that in water at 25 °C, \(\text{pH} + \text{pOH} = 14\) due to the constant ion product of water (\(K_w = 10^{-14}\) M²).

Determine the [H+], [OH-], and pOH of a solution with a pH of 9.49 at 25 °C.

Step-by-step guide to calculate [H+], [OH-], and pOH of a solution with a pH of 9.49 at 25 °C. Learn essential acid-base equilibrium concepts with key formulas for chemistry students in Grades 10-12.

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