Topic 14 · page 15 of 18

Making a buffer of a given pH

Run the buffer calculation backwards. The target pH sets the ratio of conjugate base to acid, and the ratio sets the amounts you must mix.

13 min read14.20

Read first: pH of a buffer

Choose the acid, then fix the ratio 14.20

To make a buffer of a given pH you need two decisions.

  1. Choose a weak acid whose pKa\text{p}K_\text{a} is close to the target pH, within about one unit. Then the ratio you need is near 1 : 1, and the buffer has plenty of capacity in both directions. For pH 5 that means ethanoic acid (pKa\text{p}K_\text{a} 4.76), not methanoic acid (3.80) and not ammonium (9.25).
  2. Fix the ratio of [A⁻] to [HA] from the target pH and KaK_\text{a}. Rearrange the buffer equation:
[AX−][HA]=Ka[HX+] \frac{[\ce{A-}]}{[\ce{HA}]} = \frac{K_\text{a}}{[\ce{H+}]}

where [H⁺] = 10−pH10^{-\text{pH}}.

MethodPreparing a buffer

  1. Calculate the target [H⁺] = 10−pH10^{-\text{pH}}.
  2. Calculate the ratio [A⁻] ÷ [HA] = KaK_\text{a} ÷ [H⁺].
  3. Choose the concentration of the acid, then multiply by the ratio to get the concentration of the salt.
  4. Convert to a mass or volume with n = c × V and mass = n × M.

Worked examples 14.20

Worked exampleA buffer of pH 5.00

Describe how to make 1.00 dm³ of a buffer of pH 5.00 containing 0.100 mol dm⁻³ ethanoic acid, using sodium ethanoate (MrM_\text{r} = 82.0). KaK_\text{a} = 1.74 × 10⁻⁵ mol dm⁻³.

  1. [H⁺] = 10−5.0010^{-5.00} = 1.00 × 10⁻⁵ mol dm⁻³
  2. [A⁻] ÷ [HA] = 1.74 × 10⁻⁵ ÷ 1.00 × 10⁻⁵ = 1.74
  3. [CH₃COO⁻] = 1.74 × 0.100 = 0.174 mol dm⁻³, so 0.174 mol of sodium ethanoate in 1.00 dm³
  4. Mass = 0.174 × 82.0 = 14.3 g of sodium ethanoate, dissolved with 0.100 mol of ethanoic acid and made up to 1.00 dm³.

The ratio is above 1 because the target pH (5.00) is above the pKa\text{p}K_\text{a} (4.76): there must be more base than acid.

The same buffer can be made without any solid salt, by part-neutralising the acid with a strong base. The fraction of the acid to neutralise is the ratio ÷ (1 + ratio) = 1.74 ÷ 2.74 = 0.635, or 63.5%.

Worked exampleBy part-neutralisation

Take 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid. What volume of 0.100 mol dm⁻³ NaOH makes a buffer of pH 5.00?

  1. Moles of acid = 0.200 × 0.0500 = 1.00 × 10⁻² mol.
  2. 63.5% must become ethanoate: moles of NaOH = 0.635 × 1.00 × 10⁻² = 6.35 × 10⁻³ mol.
  3. Volume = 6.35 × 10⁻³ ÷ 0.100 = 0.0635 dm³ = 63.5 cm³.
  4. Check: moles of CH₃COO⁻ = 6.35 × 10⁻³ and moles of CH₃COOH = 3.65 × 10⁻³. The ratio is 1.74. ✓

Worked exampleTwo solutions of equal concentration

What volumes of 0.200 mol dm⁻³ ethanoic acid and 0.200 mol dm⁻³ sodium ethanoate give a buffer of pH 4.50? Give the volume of salt solution to add to 100 cm³ of acid.

  1. [H⁺] = 10−4.5010^{-4.50} = 3.16 × 10⁻⁵ mol dm⁻³
  2. [A⁻] ÷ [HA] = 1.74 × 10⁻⁵ ÷ 3.16 × 10⁻⁵ = 0.550
  3. Both solutions have the same concentration, so the volumes are in the same ratio: volume of salt = 0.550 × 100 cm³ = 55.0 cm³.

Check yourself

  1. What concentration of sodium ethanoate must be mixed with 0.250 mol dm⁻³ ethanoic acid to give a buffer of pH 4.30? KaK_\text{a} = 1.74 × 10⁻⁵ mol dm⁻³.
  2. Calculate the mass of sodium ethanoate (MrM_\text{r} = 82.0) to add to 500 cm³ of 0.200 mol dm⁻³ ethanoic acid to make a buffer of pH 4.50.
  3. A buffer of pH 9.50 is to contain 0.100 mol dm⁻³ ammonium chloride. What concentration of ammonia is needed? The pKa\text{p}K_\text{a} of NH₄⁺ is 9.25.
  4. What fraction of a sample of ethanoic acid must be neutralised with NaOH to give a buffer of pH 4.76? What volume of 0.100 mol dm⁻³ NaOH must be added to 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid?
  5. Which of methanoic acid (pKa\text{p}K_\text{a} 3.80), ethanoic acid (4.76) and the ammonium ion (9.25) would you choose to make a buffer of pH 3.9? Calculate the ratio [A⁻] ÷ [HA] needed. (KaK_\text{a} of methanoic acid = 1.6 × 10⁻⁴ mol dm⁻³.)
  6. A student wants a buffer of pH 5.00 from ethanoic acid and sodium ethanoate but makes up a mixture with [A⁻] ÷ [HA] = 0.575. Calculate its pH.

Answers cover these until you have tried the questions

  1. [H⁺] = 10−4.3010^{-4.30} = 5.01 × 10⁻⁵ mol dm⁻³. [A⁻] ÷ [HA] = 1.74 × 10⁻⁵ ÷ 5.01 × 10⁻⁵ = 0.347. [CH₃COO⁻] = 0.347 × 0.250 = 0.0868 mol dm⁻³.
  2. [H⁺] = 10−4.5010^{-4.50} = 3.16 × 10⁻⁵. Ratio = 1.74 × 10⁻⁵ ÷ 3.16 × 10⁻⁵ = 0.550. [A⁻] = 0.550 × 0.200 = 0.110 mol dm⁻³. Moles in 500 cm³ = 0.110 × 0.500 = 0.0550 mol. Mass = 0.0550 × 82.0 = 4.51 g.
  3. KaK_\text{a} = 10−9.2510^{-9.25} = 5.62 × 10⁻¹⁰. [H⁺] = 10−9.5010^{-9.50} = 3.16 × 10⁻¹⁰. [H⁺] = KaK_\text{a} × [NH₄⁺] ÷ [NH₃], so [NH₃] ÷ [NH₄⁺] = KaK_\text{a} ÷ [H⁺] = 5.62 × 10⁻¹⁰ ÷ 3.16 × 10⁻¹⁰ = 1.78. [NH₃] = 1.78 × 0.100 = 0.178 mol dm⁻³.
  4. At pH = pKa\text{p}K_\text{a} the ratio is 1, so half (50%) of the acid is neutralised. Moles of acid = 0.100 × 0.0250 = 2.50 × 10⁻³ mol, so moles of NaOH = 1.25 × 10⁻³ mol. Volume = 1.25 × 10⁻³ ÷ 0.100 = 12.5 cm³.
  5. Methanoic acid, because its pKa\text{p}K_\text{a} (3.80) is closest to 3.9. [H⁺] = 10−3.910^{-3.9} = 1.26 × 10⁻⁴. [A⁻] ÷ [HA] = 1.6 × 10⁻⁴ ÷ 1.26 × 10⁻⁴ = 1.27.
  6. pH = pKa\text{p}K_\text{a} + lg ([A⁻] ÷ [HA]) = 4.76 + lg 0.575 = 4.76 − 0.24 = 4.52.