Topic 8 · page 18 of 29

Concentration and titration

Concentration is the amount of solute in 1 dm³ of solution, in mol dm⁻³ or g dm⁻³. A titration measures exactly how much of one solution reacts with a known volume of another, and an indicator shows the end point.

14 min read8.20

Concentration in mol dm⁻³ and g dm⁻³ 8.20

Concentration was introduced in Topic 1. The two forms you need here are:

c=nVconcentration / g dm−3=concentration / mol dm−3×Mr c = \frac{n}{V} \qquad \text{concentration / g dm}^{-3} = \text{concentration / mol dm}^{-3} \times M_\text{r}

where c is in mol dm⁻³, n is the amount in mol and V is the volume of solution in dm³ (cm³ ÷ 1000). The amount of solute is n=m÷Mrn = m \div M_\text{r}.

Worked exampleTwo units for one solution

4.00 g of sodium hydroxide, NaOH, is dissolved and made up to 250 cm³ of solution. Find its concentration in mol dm⁻³ and in g dm⁻³.

  1. MrM_\text{r}(NaOH) = 23.0 + 16.0 + 1.0 = 40.0
  2. n = 4.00 ÷ 40.0 = 0.100 mol
  3. V = 250 ÷ 1000 = 0.250 dm³
  4. c = 0.100 ÷ 0.250 = 0.400 mol dm⁻³
  5. In g dm⁻³: 0.400 × 40.0 = 16.0 g dm⁻³ (check: 4.00 g ÷ 0.250 dm³ = 16.0 ✓)

Worked exampleFrom g dm⁻³ to mol dm⁻³

A hydrochloric acid solution contains 3.65 g dm⁻³ of HCl. What is its concentration in mol dm⁻³?

  1. MrM_\text{r}(HCl) = 1.0 + 35.5 = 36.5
  2. c = 3.65 ÷ 36.5 = 0.100 mol dm⁻³

What a titration does 8.20

A burette on a stand, filled with solution, with its tap above a conical flask standing on a white tile. The flask holds a pipetted volume of the other solution and a few drops of indicator. A pipette is drawn to the side with a 25.0 cm3 mark.
The burette delivers the solution drop by drop into the flask, which holds a measured volume from the pipette.

The apparatus has jobs:

The point where the indicator changes colour is the end point. For a well-chosen indicator it coincides with the equivalence point, where the acid and alkali have reacted in exactly the ratio in the equation.

Indicators 8.20

An indicator is a weak acid or base that has different colours in acid and in alkali. You need two, as the specification says.

Two colour strips. Methyl orange: red in acid, orange at the end point, yellow in alkali. Phenolphthalein: colourless in acid, pale pink at the end point, pink in alkali. Arrows show the change in colour in each direction.
The colour at the end point is between the two extremes. It depends on which solution is in the burette.
Indicator In acid End point In alkali
Methyl orange red orange yellow
Phenolphthalein colourless pale pink pink

The end-point colour is between the two. Which way you see the colour change depends on which solution is in the flask.

Indicator Acid in the flask, alkali added Alkali in the flask, acid added
Methyl orange red → orange yellow → orange
Phenolphthalein colourless → pale pink pink → colourless

In an exam the question names the indicator. As a guide, methyl orange works when the acid is strong (hydrochloric or sulfuric acid, even with a weak alkali such as sodium carbonate). Phenolphthalein works when the alkali is strong (sodium hydroxide), even with a weak acid such as ethanedioic acid in Core Practical 4. With a strong acid and a strong alkali either works.

Good technique 8.20

MethodA titration

  1. Rinse the burette with the solution it will hold, fill it (using a funnel, then remove the funnel) and run solution through the tip to remove air. Read the starting volume.
  2. Pipette a measured volume into a conical flask and add 2–3 drops of indicator. Stand it on a white tile.
  3. Run in solution from the burette, swirling. Do a rough titration first, to find the approximate end point.
  4. Repeat, running in solution quickly to within about 1 cm³ of the rough value, then drop by drop. Stop at the first permanent colour change.
  5. Record the burette reading to 0.05 cm³ (two decimal places, ending in 0 or 5). The titre is final minus initial reading.
  6. Repeat until you have two or more concordant titres: titres within 0.10 cm³ of each other. Take the mean of the concordant titres and ignore the rough one.

Check yourself

  1. Calculate the concentration, in mol dm⁻³ and g dm⁻³, of a solution of 5.30 g of sodium carbonate, Na₂CO₃, in 250 cm³ of solution.
  2. Convert a 0.0400 mol dm⁻³ solution of sulfuric acid, H₂SO₄ (MrM_\text{r} = 98.1), to g dm⁻³.
  3. State the colour of methyl orange (a) in dilute acid, (b) at the end point, (c) in dilute alkali.
  4. In a titration, hydrochloric acid is added from a burette to sodium hydroxide solution in a flask with phenolphthalein. Describe the colour change at the end point.
  5. A student's titres are 26.20 (rough), 25.40, 25.50 and 25.70 cm³. State which titres are concordant and give the mean titre.
  6. State two reasons for doing a rough titration first.

Answers cover these until you have tried the questions

  1. MrM_\text{r}(Na₂CO₃) = 2 × 23.0 + 12.0 + 3 × 16.0 = 106.0. n = 5.30 ÷ 106.0 = 0.0500 mol. c = 0.0500 ÷ 0.250 = 0.200 mol dm⁻³. In g dm⁻³: 0.200 × 106.0 = 21.2 g dm⁻³.
  2. 0.0400 × 98.1 = 3.924 = 3.92 g dm⁻³.
  3. (a) red, (b) orange, (c) yellow.
  4. The pink solution turns colourless at the end point (from pink to colourless as the alkali is neutralised).
  5. The rough titre (26.20) is ignored. 25.40 and 25.50 are within 0.10 cm³ of each other: they are concordant. 25.70 differs from 25.50 by 0.20 cm³, so it is not concordant and is left out. Mean = (25.40 + 25.50) ÷ 2 = 25.45 cm³.
  6. To find the approximate volume at the end point, so the accurate titrations can be run quickly then slowly near the end point; and to learn what the colour change looks like.