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Acid-Base Equilibria

pH calculations for acids and alkalis: how to use pH

  • 3 things to know
  • A2 Unit 3

Overview

Fourteen recall questions on pH calculations for acids and alkalis: how to use pH.

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Answer each one before you reveal it.

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Numbers

Calculate the pH of 0.0100 mol dm⁻³ benzoic acid, Ka = 6.3 × 10⁻⁵ mol dm⁻³.

Numbers

Hydrocyanic acid has Ka = 4.9 × 10⁻¹⁰ mol dm⁻³. Calculate the pH of a 0.100 mol dm⁻³ solution.

Numbers

Calculate the pH of 0.100 mol dm⁻³ chloroethanoic acid, Ka = 1.38 × 10⁻³ mol dm⁻³, and comment on the reliability of the usual approximation.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Calculate the pH of 0.0250 mol dm⁻³ nitric acid.

    Model answer

    Nitric acid is a strong monobasic acid, so it is fully dissociated and [H⁺] = 0.0250 mol dm⁻³. pH = −log₁₀(0.0250) = 1.60.

  2. 2

    Calculate the pH of 2.50 × 10⁻³ mol dm⁻³ hydrochloric acid.

    Model answer

    Hydrochloric acid is strong and monobasic, so [H⁺] = 2.50 × 10⁻³ mol dm⁻³. pH = −log₁₀(2.50 × 10⁻³) = 2.60.

  3. 3

    A solution of sulfuric acid is 0.0500 mol dm⁻³. Treating both ionisations as complete, calculate its pH.

    Model answer

    Sulfuric acid is dibasic, so each mole releases two moles of H⁺ and [H⁺] = 2 × 0.0500 = 0.100 mol dm⁻³. pH = −log₁₀(0.100) = 1.00, which is 0.30 units lower than the pH 1.30 a monobasic acid of the same concentration would give, because doubling the hydrogen ion concentration lowers the pH by log₁₀ 2.

  4. 4

    25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid is diluted with water to 250 cm³. What is the pH of the diluted solution?

    Model answer

    Moles of HCl = 0.100 × 0.0250 = 2.50 × 10⁻³ mol, and this is now in 0.250 dm³, so the concentration is 2.50 × 10⁻³ / 0.250 = 0.0100 mol dm⁻³. The acid is strong, so [H⁺] = 0.0100 mol dm⁻³ and pH = 2.00. A tenfold dilution of a strong acid raises the pH by exactly one unit.

  5. 5

    A strong monobasic acid has pH 1.40. Calculate its concentration.

    Model answer

    [H⁺] = 10⁻ᵖᴴ = 10⁻¹·⁴⁰ = 0.0398 mol dm⁻³. Because the acid is strong and monobasic, the acid concentration is the same, 0.0398 mol dm⁻³.

  6. 6

    Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, for which Ka = 1.74 × 10⁻⁵ mol dm⁻³.

    Model answer

    For a weak acid, Ka = [H⁺][A⁻]/[HA]. Since [H⁺] = [A⁻] and very little acid dissociates, [HA] is taken as 0.100 mol dm⁻³, giving [H⁺] = √(Ka × c) = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³. pH = −log₁₀(1.32 × 10⁻³) = 2.88.

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