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IB DP Chemistry – R3.1.9 pOH scale – Study Notes

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 KnowTo FindUse 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\([ \text{OH}^- ]\)\( [\text{OH}^-] = 10^{-\text{pOH}} \)
pHpOH\( \text{pOH} = 14 – \text{pH} \)
pOHpH\( \text{pH} = 14 – \text{pOH} \)
\([ \text{H}^+ ]\)\([ \text{OH}^- ]\)\( [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{[\text{H}^+]} \)
\([ \text{OH}^- ]\)\([ \text{H}^+ ]\)\( [\text{H}^+] = \frac{1.0 \times 10^{-14}}{[\text{OH}^-]} \)

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 \( K_w \).

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