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6.1.10Additional lessons

The pOH Scale (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 124 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 5Definition

The pOH scale expresses alkalinity in the same style as pH expresses acidity.

pOH formula
  • pOH formula pOH = −log₁₀[OH⁻], in the same way that pH is the negative logarithm of [H⁺].
Reverse
  • Reverse The hydroxide ion concentration is found with [OH⁻] = 10⁻ᵖᴼᴴ mol dm⁻³.
Sum rule
  • Sum rule At 298 K, pH + pOH = 14.00, which follows from Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴.
Neutral
  • Neutral A neutral solution at 298 K has pOH = 7, equal to its pH.

Part 2 of 5Using pOH

Using pOH
Strong bases
  • Strong bases For NaOH and KOH, [OH⁻] equals the concentration of the base, so pOH is calculated directly.
Group 2 hydroxides
  • Group 2 hydroxides Ba(OH)₂ releases two moles of OH⁻ per mole, so [OH⁻] is twice the concentration of the base.
Convert to pH
  • Convert to pH Subtract pOH from 14.00 to obtain the pH of the alkali.
Temperature
  • Temperature The sum 14.00 applies at 298 K only, because Kw changes with temperature.
pOH[OH⁻] / mol dm⁻³pH at 298 K
11 × 10⁻¹13
71 × 10⁻⁷7
121 × 10⁻¹²2

Part 3 of 5Calculation Route

1
Write the equation for the dissociation of the base to see how many OH⁻ ions are released.
2
Calculate [OH⁻] from the concentration and the stoichiometry.
3
Find the pOH by taking −log₁₀ of the hydroxide ion concentration.
4
Subtract from 14.00 to get the pH, or use Kw to find [H⁺] and then −log₁₀[H⁺].
5
Check that the pH is above 7 for a basic solution, and quote the answer to two decimal places.

Part 4 of 5Worked Example

Question Calculate the pH of 0.020 mol dm⁻³ barium hydroxide at 298 K.

  1. Ba(OH)₂ → Ba²⁺ + 2OH⁻, so [OH⁻] = 2 × 0.020 = 0.040 mol dm⁻³.
  2. pOH = −log₁₀(0.040) = 1.40.
  3. pH = 14.00 − 1.40 = 12.60.

Answer pH = 12.60

Part 5 of 5Exam Checklist & Tips

6.1.10 – Key Points for the Exam

  • pOH The negative logarithm of the hydroxide ion concentration.
  • pH + pOH = 14 The relationship valid at 298 K.
  • Hydroxide ion OH⁻, the species responsible for the alkalinity of a solution.
  • Strong base A base that is fully dissociated in solution.

Common Mistakes

  • Using the concentration of Ba(OH)₂ as [OH⁻] without multiplying by two.
  • Applying pH + pOH = 14 at a temperature other than 298 K without being told.
  • Forgetting the minus sign in the pOH formula.
  • Giving a pH below 7 for a basic solution after subtracting incorrectly.

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