All subjects · Chemistry · WCH14 (Unit 4)

Buffer solutions

A buffer keeps pH almost constant when small amounts of acid or alkali are added. You need to explain how it works and calculate its pH.

What a buffer is (spec 14.17, 14.18)

A buffer solution resists changes in pH when small amounts of acid or alkali are added. An acidic buffer contains a weak acid and its conjugate base in similar, large amounts, for example ethanoic acid and sodium ethanoate:

CH3COOH ⇌ CH3COO- + H+

Because [HA] and [A-] are both large, their ratio hardly changes, so [H+] hardly changes. A basic buffer uses a weak base and its salt, such as NH3 and NH4Cl.

Buffer pH (spec 14.19)

Assume [HA] = acid concentration and [A-] = salt concentration (the salt is fully dissociated and the acid hardly at all):

[H+] = Ka x [HA] / [A-], then pH = -log10[H+].

Equivalently pH = pKa + log10([A-]/[HA]). When [HA] = [A-], pH = pKa.

Making a buffer (spec 14.20, 14.21)

Rearrange: [A-] = Ka[HA] / [H+] to find how much salt to add for a chosen pH. A buffer can also be made by partly neutralising a weak acid with a strong base: half-neutralised ethanoic acid has pH = pKa. The flat region of a weak acid-strong base titration curve is this buffer action.

Buffers in life (spec 14.22)

Blood is kept at pH 7.35 to 7.45 mainly by the H2CO3/HCO3- buffer: H2CO3 ⇌ H+ + HCO3-. Buffers are also used in foods and in shampoos and biological solutions where a fixed pH matters.

Worked example

Calculate the pH of a buffer containing 0.100 mol dm-3 ethanoic acid and 0.200 mol dm-3 sodium ethanoate. Ka = 1.7 x 10-5 mol dm-3.

  1. [H+] = Ka x [HA] / [A-] = 1.7 x 10-5 x 0.100 / 0.200
  2. [H+] = 8.5 x 10-6 mol dm-3

Answer: pH = 5.07

Key idea: Buffer: weak acid + conjugate base in large amounts. [H+] = Ka[HA]/[A-]. Added H+ is removed by A-, added OH- by HA. Blood uses H2CO3/HCO3-.

Check yourself

  1. What is a buffer solution?
    • A solution that resists changes in pH when small amounts of acid or alkali are added
    • A solution with a pH of exactly 7
    • A solution of a strong acid and a strong base
    • A solution whose pH never changes at all
    Show answer

    A solution that resists changes in pH when small amounts of acid or alkali are added. It resists change; it does not stop all change, and only for small additions.

  2. Calculate the pH of a buffer that is 0.150 mol dm-3 in ethanoic acid and 0.100 mol dm-3 in sodium ethanoate. Ka = 1.7 x 10-5 mol dm-3.
    Show answer

    4.59. [H+] = 1.7 x 10-5 x 0.150 / 0.100 = 2.55 x 10-5; pH = 4.59.

  3. In an ethanoic acid / sodium ethanoate buffer, what removes a small amount of added H+?
    • Ethanoate ions, forming ethanoic acid
    • Ethanoic acid molecules, forming ethanoate ions
    • Sodium ions
    • Water molecules, forming OH-
    Show answer

    Ethanoate ions, forming ethanoic acid. CH3COO- + H+ -> CH3COOH; there is a large reservoir of ethanoate.

  4. Ethanoic acid has pKa 4.76. What ratio [CH3COO-] / [CH3COOH] gives a buffer of pH 5.00?
    Show answer

    1.74. log([A-]/[HA]) = 5.00 - 4.76 = 0.24, so the ratio = 100.24 = 1.74.

  5. What mass of sodium ethanoate (CH3COONa, Mr 82.0) must be dissolved in 500 cm3 of 0.100 mol dm-3 ethanoic acid to make a buffer of pH 5.00? (Ka = 1.7 x 10-5 mol dm-3; assume no volume change)
    Show answer

    6.97 g. [H+] = 1.0 x 10-5; [A-] = 1.7 x 10-5 x 0.100 / 1.0 x 10-5 = 0.170 mol dm-3; n = 0.170 x 0.500 = 0.0850 mol; mass = 0.0850 x 82.0 = 6.97 g.

  6. 50.0 cm3 of 0.200 mol dm-3 ethanoic acid is mixed with 25.0 cm3 of 0.200 mol dm-3 sodium hydroxide. Ka = 1.7 x 10-5 mol dm-3. Calculate the pH.
    Show answer

    4.77. Half the acid is neutralised, so [HA] = [A-] and pH = pKa = 4.77.

  7. Which conjugate pair is mainly responsible for buffering blood?
    • HCl and Cl-
    • H2CO3 and HCO3-
    • NH4+ and NH3
    • H2SO4 and HSO4-
    Show answer

    H2CO3 and HCO3-. Carbonic acid and hydrogencarbonate keep blood at about pH 7.4.

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Last updated 7 November 2026. Spotted a mistake? Tell us.