Topic 14 · page 6 of 18

Ka from a measured pH

Run the weak-acid calculation backwards. A known mass of acid and a measured pH give the concentration, then [H⁺], then Ka.

13 min read14.14

Read first: pH of a weak acid

Working backwards 14.14

On the last page you were given KaK_\text{a} and found the pH. In the laboratory you usually have it the other way round. You weigh a sample of acid, dissolve it in a known volume, and measure the pH with a meter. The pH tells you [H⁺]. The same two assumptions then give you KaK_\text{a}.

Ka=[HX+]2c K_\text{a} = \frac{[\ce{H+}]^2}{c}

Here [H⁺] = 10−pH10^{-\text{pH}} and c is the concentration of the acid you made up.

MethodKa from a mass and a pH

  1. Moles of acid n = mass ÷ MrM_\text{r}.
  2. Concentration c = n ÷ V, with V in dm³.
  3. [H⁺] = 10−pH10^{-\text{pH}}.
  4. State the assumptions: [A⁻] = [H⁺] and [HA] = c.
  5. KaK_\text{a} = [H⁺]² ÷ c, with units mol dm⁻³.
  6. If asked, pKa\text{p}K_\text{a} = −lg KaK_\text{a}.

Worked examples 14.14

Worked exampleEthanoic acid

0.600 g of ethanoic acid (MrM_\text{r} = 60.0) is dissolved in water and made up to 250 cm³. The pH is 3.08. Calculate KaK_\text{a}.

  1. n = 0.600 ÷ 60.0 = 0.0100 mol
  2. c = 0.0100 ÷ 0.250 = 0.0400 mol dm⁻³
  3. [H⁺] = 10−3.0810^{-3.08} = 8.32 × 10⁻⁴ mol dm⁻³
  4. Assume [CH₃COO⁻] = [H⁺] and [CH₃COOH] = 0.0400 mol dm⁻³.
  5. KaK_\text{a} = (8.32 × 10⁻⁴)² ÷ 0.0400 = 6.92 × 10⁻⁷ ÷ 0.0400 = 1.73 × 10⁻⁵, so 1.7 × 10⁻⁵ mol dm⁻³
  6. Check: 8.32 × 10⁻⁴ ÷ 0.0400 = 2.1% dissociated, so assumption 2 is sound.

Worked exampleBenzoic acid

1.22 g of benzoic acid, C₆H₅COOH, is made up to 500 cm³ of solution. The pH is 2.95. Calculate KaK_\text{a} and pKa\text{p}K_\text{a}.

  1. MrM_\text{r} = 7 × 12.0 + 6 × 1.0 + 2 × 16.0 = 122.0, so n = 1.22 ÷ 122.0 = 0.0100 mol
  2. c = 0.0100 ÷ 0.500 = 0.0200 mol dm⁻³
  3. [H⁺] = 10−2.9510^{-2.95} = 1.12 × 10⁻³ mol dm⁻³
  4. KaK_\text{a} = (1.12 × 10⁻³)² ÷ 0.0200 = 1.26 × 10⁻⁶ ÷ 0.0200 = 6.3 × 10⁻⁵, so 6.3 × 10⁻⁵ mol dm⁻³
  5. pKa\text{p}K_\text{a} = −lg (6.3 × 10⁻⁵) = 4.20

When the question gives the concentration directly, skip steps 1 and 2. For 0.100 mol dm⁻³ of an acid HX with pH 2.85, [H⁺] = 1.41 × 10⁻³, so KaK_\text{a} = (1.41 × 10⁻³)² ÷ 0.100 = 2.0 × 10⁻⁵ mol dm⁻³.

How reliable is the answer? 14.14

Because [H⁺] is squared, an error in pH is doubled in KaK_\text{a}.

A pH meter reading to ±0.01 means [H⁺] is uncertain by a factor of 100.0110^{0.01}, about 2.3%. Squaring doubles this to about ±5% in KaK_\text{a}. If the meter reads to only ±0.02, the uncertainty in KaK_\text{a} is about ±10%.

Two other sources of error matter. The acid may not be at the concentration you worked out, if the mass or volume was measured carelessly. And the meter must be calibrated, or every pH reading shifts by the same amount.

Check yourself

  1. A 0.0500 mol dm⁻³ solution of a weak acid HX has pH 3.10. Calculate KaK_\text{a} and pKa\text{p}K_\text{a}.
  2. 0.370 g of propanoic acid, CH₃CH₂COOH (MrM_\text{r} = 74.0), is made up to 100 cm³ of solution. The pH is 3.09. Calculate KaK_\text{a}.
  3. 0.220 g of butanoic acid (MrM_\text{r} = 88.0) is made up to 250 cm³ of solution. The pH is 3.40. Calculate KaK_\text{a}.
  4. Show that the assumption [HA] = c is reasonable for the solution in question 1.
  5. The pH in question 1 is measured to ±0.02. Estimate the percentage uncertainty in KaK_\text{a}.
  6. Explain why [A⁻] can be taken as equal to [H⁺] when calculating KaK_\text{a} from a measured pH.

Answers cover these until you have tried the questions

  1. [H⁺] = 10−3.1010^{-3.10} = 7.94 × 10⁻⁴ mol dm⁻³. KaK_\text{a} = (7.94 × 10⁻⁴)² ÷ 0.0500 = 6.31 × 10⁻⁷ ÷ 0.0500 = 1.26 × 10⁻⁵, so 1.3 × 10⁻⁵ mol dm⁻³. pKa\text{p}K_\text{a} = 4.90
  2. n = 0.370 ÷ 74.0 = 0.00500 mol; c = 0.00500 ÷ 0.100 = 0.0500 mol dm⁻³. [H⁺] = 10−3.0910^{-3.09} = 8.13 × 10⁻⁴. KaK_\text{a} = (8.13 × 10⁻⁴)² ÷ 0.0500 = 6.61 × 10⁻⁷ ÷ 0.0500 = 1.32 × 10⁻⁵, so 1.3 × 10⁻⁵ mol dm⁻³
  3. n = 0.220 ÷ 88.0 = 0.00250 mol; c = 0.00250 ÷ 0.250 = 0.0100 mol dm⁻³. [H⁺] = 10−3.4010^{-3.40} = 3.98 × 10⁻⁴. KaK_\text{a} = (3.98 × 10⁻⁴)² ÷ 0.0100 = 1.58 × 10⁻⁷ ÷ 0.0100 = 1.58 × 10⁻⁵, so 1.6 × 10⁻⁵ mol dm⁻³
  4. Percentage dissociated = 7.94 × 10⁻⁴ ÷ 0.0500 × 100 = 1.6%. This is small, so the equilibrium concentration of HA is almost the same as the starting concentration.
  5. [H⁺] is uncertain by a factor of 100.0210^{0.02} = 1.047, about 4.7%. KaK_\text{a} depends on [H⁺]², so the uncertainty doubles to about ±9% to 10%.
  6. Each HA that dissociates gives one H⁺ and one A⁻, in equal numbers. The H⁺ from the ionisation of water is about 10⁻⁷ mol dm⁻³, far smaller than the H⁺ from the acid, so [A⁻] ≈ [H⁺].