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1.10.2Physical Chemistry

Definition and Determination of pH — Oxford AQA International A Level Chemistry Revision Notes & Topical Questions

5 min read 5 parts Lesson 52 of 63

📚 Oxford AQA International A Level Chemistry Hub / Topical Revision
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Curriculum Equation Key pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: pH + pOH = 14
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Part 1 of 5The pH Scale

1.10.2 Definition and Determination of pH

pH expresses hydrogen ion concentration on a logarithmic scale.

The pH Scale
Definition
  • pH = −log₁₀[H⁺], where [H⁺] is the hydrogen ion concentration in mol dm⁻³.
Meaning
  • pH below 7 is acidic, pH 7 is neutral at 298 K and pH above 7 is alkaline.
Logarithmic
  • A drop of one pH unit means [H⁺] has risen tenfold, and a drop of two units means a hundredfold increase.
Reverse
  • [H⁺] = 10⁻ᵖᴴ gives the concentration from a measured pH value.

Part 2 of 5Calculating pH of Strong Acids

Strong acids are fully ionised, so [H⁺] follows directly from the concentration.

1
For a monoprotic strong acid, [H⁺] equals the acid concentration, so 0.050 mol dm⁻³ HCl has [H⁺] = 0.050 mol dm⁻³.
2
Apply pH = −log₁₀[H⁺] and give the answer to two decimal places.
3
For a diprotic strong acid that ionises completely, double the acid concentration to find [H⁺].
4
To find the pH after dilution, calculate the new concentration using c₁V₁ = c₂V₂ before using the log formula.
5
To find the pH after mixing with alkali, work out the excess moles of H⁺ and divide by the total volume.

Part 3 of 5Measuring pH

pH meter
  • A glass electrode connected to a meter gives readings to two decimal places and must be calibrated with buffers of known pH.
Calibration
  • Use at least two buffers either side of the expected pH, then rinse the electrode with distilled water before every reading.
Universal indicator
  • Gives a rough pH by matching colours, but it is subjective and not suitable for deep colours or fine differences.
Stirring
  • Stir the solution gently before reading so that the electrode senses a uniform mixture and the reading is stable.

Part 4 of 5Worked Example

Question Calculate the pH of (a) 0.0250 mol dm⁻³ H₂SO₄ assuming complete ionisation to H⁺ and SO₄²⁻, and (b) the solution formed when 10.0 cm³ of 0.100 mol dm⁻³ HCl is diluted to 1000 cm³.

  1. (a) Each H₂SO₄ gives two H⁺, so [H⁺] = 2 × 0.0250 = 0.0500 mol dm⁻³.
  2. pH = −log₁₀(0.0500) = 1.30.
  3. (b) Moles HCl = 0.100 × 0.0100 = 1.00 × 10⁻³ mol, in 1.000 dm³.
  4. [H⁺] = 1.00 × 10⁻³ mol dm⁻³, so pH = 3.00.

Answer (a) pH 1.30 and (b) pH 3.00.

Part 5 of 5Exam Checklist & Tips

1.10.2 – Key Points for the Exam

  • pH The negative logarithm to base 10 of the hydrogen ion concentration.
  • [H⁺] The concentration of hydrogen ions in mol dm⁻³.
  • Monoprotic acid An acid that releases one proton per molecule.
  • Calibration Adjusting a meter using solutions of known pH.

Common Mistakes

  • Forgetting to double [H⁺] for a diprotic acid that ionises fully.
  • Using the volume of acid instead of the final volume when calculating the diluted concentration.
  • Giving a pH above 7 for a very dilute strong acid, which is impossible because the acid still adds H⁺ to water.
  • Saying that pH 3 is twice as acidic as pH 6, when it is a thousand times higher in [H⁺].

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