IB DP Chemistry - R3.1.9 pOH scale - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R3.1.9 pOH scale – Study Notes – New Syllabus
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Reactivity 3.1.9 — The pOH Scale
Reactivity 3.1.9 — The pOH Scale
The pOH scale provides a logarithmic measure of the hydroxide ion concentration, [OH⁻], in aqueous solutions. It is complementary to the pH scale and offers a convenient way to describe the basicity of a solution.
Definition of pOH:
- The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion concentration:
- \( \text{pOH} = -\log_{10} [\text{OH}^-] \)
- Conversely, if pOH is known, the [OH⁻] can be calculated as:
- \( [\text{OH}^-] = 10^{-\text{pOH}} \)
Key Characteristics of the pOH Scale:
- The pOH scale typically ranges from 0 to 14, similar to the pH scale.
- Low pOH values (close to 0) correspond to high hydroxide ion concentrations — these solutions are strongly basic.
- High pOH values (close to 14) indicate very low hydroxide ion concentrations — these solutions are acidic or near neutral.
- A pOH of 7 corresponds to a neutral solution at 25°C, where [OH⁻] = \( 1.0 \times 10^{-7} \, \text{mol dm}^{-3} \).
Relationship with Ion Concentration:
- The pOH value is directly related to the concentration of hydroxide ions, which increase as the solution becomes more basic.
- Because it is logarithmic, a change of 1 unit in pOH represents a 10-fold change in [OH⁻].
Why Use the pOH Scale?
- In strongly basic solutions, the concentration of [OH⁻] is high, and pH values may approach 14. Using pOH offers a more intuitive way to express how basic a solution is.
- It complements the pH scale and simplifies calculations in aqueous acid–base equilibria, especially when [OH⁻] is the directly measured or given quantity.
Interconverting [H⁺], [OH⁻], pH, and pOH
Acid–base calculations often involve converting between concentration of hydrogen ions \([ \text{H}^+ ]\), hydroxide ions \([ \text{OH}^- ]\), pH, and pOH. These values are mathematically related, and using logarithmic functions allows us to switch between them easily.
Key Relationships:
- \( \text{pH} = -\log_{10} [\text{H}^+] \)
Converts hydrogen ion concentration to the pH scale. A higher \([ \text{H}^+ ]\) means a lower pH (more acidic).
- \( \text{pOH} = -\log_{10} [\text{OH}^-] \)
Converts hydroxide ion concentration to the pOH scale. A higher \([ \text{OH}^- ]\) means a lower pOH (more basic).
- \( [\text{H}^+] = 10^{-\text{pH}} \)
Rearranged form of the first equation to find hydrogen ion concentration from pH.
- \( [\text{OH}^-] = 10^{-\text{pOH}} \)
Rearranged form of the second equation to find hydroxide ion concentration from pOH.
- \( \text{pH} + \text{pOH} = 14 \) (valid only at 25°C)
Used to find either pH or pOH if one is known. Only accurate at 25°C, as \( K_w \) is temperature-dependent.
- \( [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \) (from \( K_w \))
This equation, derived from the ionic product of water, allows interconversion of \([ \text{H}^+ ]\) and \([ \text{OH}^- ]\).
Step-by-Step Conversion Guide:
If You Know | To Find | Use Equation |
---|---|---|
\([ \text{H}^+ ]\) | pH | \( \text{pH} = -\log_{10} [\text{H}^+] \) |
pH | \([ \text{H}^+ ]\) | \( [\text{H}^+] = 10^{-\text{pH}} \) |
\([ \text{OH}^- ]\) | pOH | \( \text{pOH} = -\log_{10} [\text{OH}^-] \) |
pOH | \([ \text{OH}^- ]\) | \( [\text{OH}^-] = 10^{-\text{pOH}} \) |
pH | pOH | \( \text{pOH} = 14 – \text{pH} \) |
pOH | pH | \( \text{pH} = 14 – \text{pOH} \) |
\([ \text{H}^+ ]\) | \([ \text{OH}^- ]\) | \( [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{[\text{H}^+]} \) |
\([ \text{OH}^- ]\) | \([ \text{H}^+ ]\) | \( [\text{H}^+] = \frac{1.0 \times 10^{-14}}{[\text{OH}^-]} \) |
Important Notes:
- The value of pOH = 7 corresponds to neutrality only at 25°C, because it is based on the ion product of water, \( K_w = 1.0 \times 10^{-14} \).
- At different temperatures, the value of \( K_w \) changes, which affects the neutrality point on both the pH and pOH scales.
- Use at least 3 significant figures for all calculations unless stated otherwise.
- Always check the final units and ensure concentrations are in mol dm⁻³ before using logarithmic functions.
Example
A solution has a hydrogen ion concentration of \( [\text{H}^+] = 2.5 \times 10^{-3} \, \text{mol dm}^{-3} \). Calculate the pH of the solution.
▶️Answer/Explanation
Use the formula:
\( \text{pH} = -\log_{10}[ \text{H}^+ ] \)
\( \text{pH} = -\log_{10}(2.5 \times 10^{-3}) \)
\( \text{pH} \approx -(\log_{10}(2.5) + \log_{10}(10^{-3})) \)
\( \text{pH} \approx -(0.398 + (-3)) = 2.602 \)
Final Answer: pH ≈ 2.60
Example
At 50°C, the ion product of water is \( K_w = 5.5 \times 10^{-14} \). What is the pH of pure water at this temperature?
▶️Answer/Explanation
In pure water, the concentrations of \( [\text{H}^+] \) and \( [\text{OH}^-] \) are equal, so:
\( [\text{H}^+] = [\text{OH}^-] = \sqrt{K_w} \)
\( [\text{H}^+] = \sqrt{5.5 \times 10^{-14}} \approx 2.35 \times 10^{-7} \, \text{mol dm}^{-3} \)
Now calculate the pH:
\( \text{pH} = -\log_{10}(2.35 \times 10^{-7}) \)
\( \text{pH} \approx -(\log_{10}(2.35) + \log_{10}(10^{-7})) \)
\( \text{pH} \approx -(0.371 + (-7)) = 6.63 \)
Final Answer: pH ≈ 6.63
Important Insight: Although the pH is less than 7, the solution is still neutral because \( [\text{H}^+] = [\text{OH}^-] \). The lower pH is due to the increased ionization of water at higher temperature.
Example
A solution has a pH of 3.75. Calculate the hydroxide ion concentration \( [\text{OH}^-] \) in mol dm−3.
▶️Answer/Explanation
Step 1: Use the relationship \( \text{pOH} = 14 – \text{pH} \)
\( \text{pOH} = 14 – 3.75 = 10.25 \)
Step 2: Use the equation \( [\text{OH}^-] = 10^{-\text{pOH}} \)
\( [\text{OH}^-] = 10^{-10.25} \approx 5.62 \times 10^{-11} \, \text{mol dm}^{-3} \)
Final Answer: \( [\text{OH}^-] \approx 5.62 \times 10^{-11} \, \text{mol dm}^{-3} \)
Insight: Even though the pH is quite acidic (below 7), we can still calculate the small amount of \( \text{OH}^- \) present using pOH. This demonstrates the reciprocal relationship between \( [\text{H}^+] \) and \( [\text{OH}^-] \) via \( K_w \).