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Acid-Base Equilibria

Titrations: curve shapes, salt pH and choosing an indicator

  • 1 thing to know
  • A2 Unit 3

Overview

Recall questions on curve shapes, salt pH and choosing an indicator.

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Key term

Why does the pH change so abruptly near the equivalence point of a strong acid and strong base titration?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    In a titration, what is the difference between the equivalence point and the end point?

    Model answer

    The equivalence point is the moment when the acid and base have reacted in exactly the stoichiometric ratio, which is a property of the chemistry. The end point is the moment the indicator changes colour, which is a property of the indicator; a well-chosen indicator makes the two coincide closely enough for accurate work.

  2. 2

    Why does the pH change so abruptly near the equivalence point of a strong acid and strong base titration?

    Model answer

    Very close to equivalence almost all the acid has been used up, so each further drop of alkali makes a very large fractional change in the tiny remaining [H⁺]. Because pH is a logarithmic scale, that fractional change appears as a jump of several pH units for a fraction of a cubic centimetre of titrant.

  3. 3

    What salt is formed when hydrochloric acid neutralises sodium hydroxide, and what is the pH of its solution?

    Model answer

    Sodium chloride is formed. Neither Na⁺ nor Cl⁻ reacts appreciably with water, because Cl⁻ is the conjugate base of a strong acid and so is an extremely poor proton acceptor. The solution is therefore neutral, with pH 7 at 298 K.

  4. 4

    Why is a solution of ammonium chloride acidic?

    Model answer

    It is the salt of a strong acid and a weak base, so the ammonium ion is the conjugate acid of a weak base and donates protons to water: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq). The chloride ion does nothing, so the extra hydrogen ions make the solution acidic, typically around pH 5.

  5. 5

    Why is a solution of sodium ethanoate alkaline?

    Model answer

    It is the salt of a weak acid and a strong base, so the ethanoate ion is the conjugate base of a weak acid and takes protons from water: CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq). The sodium ion has no effect, so the hydroxide ions released make the solution alkaline.

  6. 6

    Calculate the pH of 0.100 mol dm⁻³ sodium ethanoate at 298 K, given Ka for ethanoic acid = 1.74 × 10⁻⁵ mol dm⁻³ and Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

    Model answer

    For CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ the constant is Kw/Ka = 1.00 × 10⁻¹⁴ / (1.74 × 10⁻⁵) = 5.75 × 10⁻¹⁰ mol dm⁻³. Then [OH⁻] = √(5.75 × 10⁻¹⁰ × 0.100) = 7.58 × 10⁻⁶ mol dm⁻³, so [H⁺] = 1.00 × 10⁻¹⁴ / (7.58 × 10⁻⁶) = 1.32 × 10⁻⁹ mol dm⁻³ and pH = 8.88.

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