Topic 8 · page 19 of 29

Titration calculations

Write the equation, find the amount of the solution you know, use the ratio to get the amount of the other, then divide by its volume. Give the answer in mol dm⁻³ and g dm⁻³ to the right number of significant figures.

14 min read8.20

Read first: Concentration and titration

The five-step method 8.20

MethodTitration calculation

  1. Write the balanced equation, and read off the mole ratio.
  2. Calculate the amount (mol) of the solution whose concentration and volume you know: n = c × V, with V in dm³.
  3. Use the ratio from the equation to find the amount of the other substance.
  4. Calculate the concentration of the unknown: c = n ÷ V, with V the volume of that solution (in dm³).
  5. If asked, convert to g dm⁻³ by multiplying by the molar mass.

Use the mean of the concordant titres for the volume from the burette, and keep unrounded values until the final answer.

Worked exampleHydrochloric acid and sodium hydroxide (1 : 1)

25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide is neutralised by 24.60 cm³ of hydrochloric acid. Find the concentration of the acid in mol dm⁻³ and g dm⁻³.

  1. HCl+NaOH→NaCl+HX2O\ce{HCl + NaOH -> NaCl + H2O} (ratio 1 : 1)
  2. n(NaOH) = 0.100 × 25.0 ÷ 1000 = 2.50 × 10⁻³ mol
  3. n(HCl) = 2.50 × 10⁻³ mol (ratio 1 : 1)
  4. c(HCl) = 2.50 × 10⁻³ ÷ 0.02460 = 0.1016 = 0.102 mol dm⁻³ (3 s.f.)
  5. In g dm⁻³: 0.1016 × 36.5 = 3.71 g dm⁻³ (using the unrounded 0.10163)

Worked exampleSulfuric acid and sodium hydroxide (1 : 2)

25.0 cm³ of sodium hydroxide solution needs 20.40 cm³ of 0.0500 mol dm⁻³ sulfuric acid. Find the concentration of the sodium hydroxide.

  1. HX2SOX4+2 NaOH→NaX2SOX4+2 HX2O\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O} (ratio 1 : 2)
  2. n(H₂SO₄) = 0.0500 × 20.40 ÷ 1000 = 1.02 × 10⁻³ mol
  3. n(NaOH) = 2 × 1.02 × 10⁻³ = 2.04 × 10⁻³ mol
  4. c(NaOH) = 2.04 × 10⁻³ ÷ 0.0250 = 0.0816 mol dm⁻³

Worked exampleSodium carbonate and hydrochloric acid (1 : 2)

25.0 cm³ of sodium carbonate solution reacts with 21.30 cm³ of 0.100 mol dm⁻³ hydrochloric acid, using methyl orange. (Illustrative data.) Find the concentration of the carbonate in mol dm⁻³ and g dm⁻³.

  1. NaX2COX3+2 HCl→2 NaCl+HX2O+COX2\ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2} (ratio 1 : 2)
  2. n(HCl) = 0.100 × 21.30 ÷ 1000 = 2.130 × 10⁻³ mol
  3. n(Na₂CO₃) = 2.130 × 10⁻³ ÷ 2 = 1.065 × 10⁻³ mol
  4. c = 1.065 × 10⁻³ ÷ 0.0250 = 0.0426 mol dm⁻³
  5. MrM_\text{r} = 106.0, so 0.0426 × 106.0 = 4.52 g dm⁻³

When the sample has been diluted 8.20

Sometimes a solution is too concentrated to titrate directly. It is diluted first, and the answer must be scaled back up.

Worked exampleA diluted sample

25.0 cm³ of a sodium hydroxide solution is made up to 250 cm³ in a volumetric flask. 25.0 cm³ of the diluted solution is neutralised by 23.80 cm³ of 0.0500 mol dm⁻³ hydrochloric acid. Find the concentration of the original sodium hydroxide solution.

  1. n(HCl) = 0.0500 × 23.80 ÷ 1000 = 1.19 × 10⁻³ mol
  2. NaOH+HCl→NaCl+HX2O\ce{NaOH + HCl -> NaCl + H2O}, ratio 1 : 1, so n(NaOH) in the 25.0 cm³ sample = 1.19 × 10⁻³ mol
  3. c(diluted) = 1.19 × 10⁻³ ÷ 0.0250 = 0.0476 mol dm⁻³
  4. The dilution was 25.0 → 250 cm³, a factor of 10, so c(original) = 0.0476 × 10 = 0.476 mol dm⁻³

Finding a molar mass 8.20

A titration with a solid gives its molar mass.

Worked exampleMolar mass of a solid acid

0.250 g of a solid monoprotic acid, HA, is dissolved in water and titrated with 0.100 mol dm⁻³ sodium hydroxide. The titre is 20.50 cm³. Find the molar mass of the acid.

  1. HA+NaOH→NaA+HX2O\ce{HA + NaOH -> NaA + H2O} (1 : 1)
  2. n(NaOH) = 0.100 × 20.50 ÷ 1000 = 2.050 × 10⁻³ mol = n(HA)
  3. M = m ÷ n = 0.250 ÷ 2.050 × 10⁻³ = 122 g mol⁻¹ (3 s.f.)

Check yourself

  1. 25.0 cm³ of 0.0800 mol dm⁻³ potassium hydroxide needs 20.00 cm³ of hydrochloric acid for neutralisation. Calculate the concentration of the acid.
  2. 25.0 cm³ of sulfuric acid is neutralised by 30.50 cm³ of 0.100 mol dm⁻³ sodium hydroxide. Calculate the concentration of the acid in mol dm⁻³.
  3. 25.0 cm³ of a sodium carbonate solution needs 21.30 cm³ of 0.100 mol dm⁻³ hydrochloric acid. Show that the concentration of the carbonate is 4.52 g dm⁻³.
  4. Calculate the volume of 0.100 mol dm⁻³ hydrochloric acid needed to neutralise 25.0 cm³ of 0.0500 mol dm⁻³ barium hydroxide solution.
  5. 0.250 g of a monoprotic acid needed 20.50 cm³ of 0.100 mol dm⁻³ NaOH. Suggest the identity of the acid given that its molar mass is close to 122 g mol⁻¹ and it is a white solid containing C₇H₆O₂.
  6. A solution of ethanoic acid is diluted by taking 10.0 cm³ and making it up to 250 cm³. 25.0 cm³ of the diluted solution needs 18.40 cm³ of 0.100 mol dm⁻³ NaOH. Calculate the concentration of the original acid.

Answers cover these until you have tried the questions

  1. n(KOH) = 0.0800 × 0.0250 = 2.00 × 10⁻³ mol. HCl+KOH→KCl+HX2O\ce{HCl + KOH -> KCl + H2O} (1 : 1). c(HCl) = 2.00 × 10⁻³ ÷ 0.02000 = 0.100 mol dm⁻³.
  2. n(NaOH) = 0.100 × 0.03050 = 3.050 × 10⁻³ mol. HX2SOX4+2 NaOH→NaX2SOX4+2 HX2O\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}, so n(H₂SO₄) = 1.525 × 10⁻³ mol. c = 1.525 × 10⁻³ ÷ 0.0250 = 0.0610 mol dm⁻³.
  3. n(HCl) = 0.100 × 0.02130 = 2.130 × 10⁻³ mol. NaX2COX3+2 HCl→2 NaCl+HX2O+COX2\ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2}, so n(Na₂CO₃) = 1.065 × 10⁻³ mol. c = 1.065 × 10⁻³ ÷ 0.0250 = 0.0426 mol dm⁻³. × 106.0 = 4.516 = 4.52 g dm⁻³. ✓
  4. Ba(OH)X2+2 HCl→BaClX2+2 HX2O\ce{Ba(OH)2 + 2HCl -> BaCl2 + 2H2O}. n(Ba(OH)₂) = 0.0500 × 0.0250 = 1.25 × 10⁻³ mol, so n(HCl) = 2.50 × 10⁻³ mol. V = 2.50 × 10⁻³ ÷ 0.100 = 0.0250 dm³ = 25.0 cm³.
  5. M = 0.250 ÷ 2.050 × 10⁻³ = 122. C₇H₆O₂: 7 × 12.0 + 6 × 1.0 + 2 × 16.0 = 122.0. The acid is consistent with benzoic acid, C₆H₅COOH.
  6. n(NaOH) = 0.100 × 0.01840 = 1.840 × 10⁻³ mol = n(CH₃COOH) (1 : 1). c(diluted) = 1.840 × 10⁻³ ÷ 0.0250 = 0.0736 mol dm⁻³. Dilution factor 250 ÷ 10.0 = 25, so c(original) = 0.0736 × 25 = 1.84 mol dm⁻³.