Solutions and Titration

Lessons

Senior 5 · Chemistry

Solutions and Titration

5 lessons

Lesson 2 of 5

Acid–Base Titrations and Back Titration

25 min

Measuring what you cannot dissolve

Farmers spread agricultural lime, mostly calcium carbonate, on acidic soils. Its value depends on how much calcium carbonate it really contains. Titrating it directly is awkward: the powder does not dissolve in water, and it reacts with acid slowly, fizzing as it goes, so the end point drifts. The answer is an indirect method, the back titration. First, though, you need a sound direct titration technique.

The direct titration

Figure 11.2: Titration apparatus.

Rinse the pipette with the solution to be pipetted, and transfer an aliquot (e.g. 25.00 cm³) to a conical flask. Add 2–3 drops of indicator.
Rinse the burette with the titrant, fill it, run some out to fill the tip, and record the initial reading at eye level.
Do a rough titration to find the approximate end point, then accurate titrations, adding dropwise near the end point while swirling.
Repeat until two or more titres are concordant (within 0.10 cm³) and average only those.

Titration type	pH at equivalence	Suitable indicator
strong acid + strong base (HCl + NaOH)	7	methyl orange or phenolphthalein
strong acid + weak base (HCl + $\mathrm{Na_2CO_3}$ or $\mathrm{NH_3}$ )	below 7	methyl orange (red ↔ yellow, pH 3.1–4.4)
weak acid + strong base ( $\mathrm{CH_3COOH}$ + NaOH)	above 7	phenolphthalein (colourless ↔ pink, pH 8.2–10.0)
weak acid + weak base	near 7, no sharp change	none suitable

The calculation, in four steps

(1) moles of the standard, $n = c V$ ; (2) the mole ratio from the balanced equation; (3) moles of the unknown; (4) its concentration, $c = \frac{n}{V}$ , allowing for any dilution.

A conical flask at the end point of an acid–base titration with phenolphthalein. The alkali just added has turned the top layer deep pink, while the mixed liquid below is only faintly pink: swirl after each addition. Photo: 384, CC BY-SA 4.0, via Wikimedia Commons

Worked example

Problem. 25.00 cm³ of 0.1000 mol dm⁻³ sodium carbonate needs a mean titre of 23.45 cm³ of hydrochloric acid, with methyl orange. Calculate the concentration of the acid.

Solution. $\mathrm{Na_2CO_3 + 2HCl \rightarrow 2NaCl + CO_2 + H_2O}$

$n(\mathrm{Na_2CO_3}) = 0.1000 \times 0.02500 = 2.500 \times 10^{-3}\,\mathrm{mol} \qquad n(\mathrm{HCl}) = 2 \times 2.500 \times 10^{-3} = 5.000 \times 10^{-3}\,\mathrm{mol}$

$c(\mathrm{HCl}) = \frac{5.000 \times 10^{-3}}{0.02345} = 0.2132\,\mathrm{mol\,dm^{-3}}$

Answer: 0.2132 mol dm⁻³. The acid is now a secondary standard.

Worked example

Problem. 25.00 cm³ of vinegar is diluted to 250.0 cm³. 25.00 cm³ portions of the diluted vinegar need 21.60 cm³ of 0.1000 mol dm⁻³ NaOH with phenolphthalein. Find the concentration of ethanoic acid in the vinegar in mol dm⁻³ and g dm⁻³ (M = 60.05 g mol⁻¹).

Solution. $\mathrm{CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O}$ (1 : 1)

$n(\mathrm{NaOH}) = 0.1000 \times 0.02160 = 2.160 \times 10^{-3}\,\mathrm{mol} \qquad c_{\text{diluted}} = \frac{2.160 \times 10^{-3}}{0.02500} = 0.08640\,\mathrm{mol\,dm^{-3}}$

The vinegar was diluted tenfold: $c = 0.8640\,\mathrm{mol\,dm^{-3}}$ , or $0.8640 \times 60.05 = 51.9\,\mathrm{g\,dm^{-3}}$ .

Answer: 0.864 mol dm⁻³, 51.9 g dm⁻³ (about 5.2 g per 100 cm³). Phenolphthalein is used because this is a weak acid–strong base titration.

Back titration

A back titration is used when the substance (a) is insoluble in water, like calcium carbonate or magnesium hydroxide in antacid tablets; (b) is volatile, like ammonia; (c) reacts too slowly with the titrant for a sharp end point.

Figure 11.3: The logic of a back titration.

Worked example

Problem. 1.250 g of agricultural lime is added to 50.00 cm³ of 1.000 mol dm⁻³ HCl (an excess). When the fizzing stops, the solution is made up to 250.0 cm³. 25.00 cm³ portions need 26.40 cm³ of 0.1000 mol dm⁻³ NaOH. Calculate the percentage by mass of $\mathrm{CaCO_3}$ in the lime (M = 100.09 g mol⁻¹), assuming nothing else in it reacts with acid.

Solution.

Total acid: $n = 1.000 \times 0.05000 = 0.05000\,\mathrm{mol}$ .

Excess acid in 25.00 cm³: $0.1000 \times 0.02640 = 2.640 \times 10^{-3}\,\mathrm{mol}$ ; in 250.0 cm³: $0.02640\,\mathrm{mol}$ .

Acid that reacted: $0.05000 - 0.02640 = 0.02360\,\mathrm{mol}$ .

$\mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O}$ , so $n(\mathrm{CaCO_3}) = \frac{0.02360}{2} = 0.01180\,\mathrm{mol}$ ; mass $= 0.01180 \times 100.09 = 1.181\,\mathrm{g}$ .

$\%\,\mathrm{CaCO_3} = \frac{1.181}{1.250} \times 100 = 94.5\,\%$

Answer: 94.5 %.

Warning

The two errors that ruin back titrations: forgetting to scale the titrated excess back up to the whole solution (here, × 10), and using the wrong mole ratio for the first reaction. Write both equations before you start.

Safety: 1 mol dm⁻³ HCl and NaOH irritate skin and are dangerous in the eyes; wear goggles. First aid (CDC/NIOSH): flush eyes or skin with plenty of water immediately and for at least 15 minutes; get medical help for any eye exposure.

Tip

Draw the "money box" in Figure 11.3 for every back titration: total paid in (known), change returned (titrated), so amount spent = total − change. The amount spent is what reacted with your sample.

Paul Morgan, A Level Chemistry: How to Solve Back Titration Calculations. Follow the percentage of calcium carbonate in limestone step by step, and pause to try each step before he does.

✦

Practice

Name a suitable indicator for titrating (a) aqueous ammonia with HCl; (b) ethanoic acid with KOH; and give the colour change in each.
Titres of 24.70 (rough), 24.30, 24.55, 24.35 and 24.30 cm³ are recorded. Which titres should be averaged, and what is the mean?
20.00 cm³ of NaOH needs 18.75 cm³ of 0.1050 mol dm⁻³ $\mathrm{H_2SO_4}$ . Calculate the concentration of the NaOH.
2.00 g of an impure sample of NaOH is dissolved and made up to 250.0 cm³. 25.00 cm³ needs 23.80 cm³ of 0.200 mol dm⁻³ HCl. Calculate the percentage purity of the NaOH.
0.500 g of an ammonium sulfate fertiliser is boiled with 50.00 cm³ of 0.2000 mol dm⁻³ NaOH until no more ammonia is given off. The remaining NaOH needs 27.30 cm³ of 0.1000 mol dm⁻³ HCl. Calculate the percentage of nitrogen in the fertiliser.
Explain why the ammonium content of a fertiliser is found by a back titration and not by titrating it directly with NaOH.
A student uses phenolphthalein for titrating sodium carbonate with HCl and records a titre about half the expected one. Explain.

Show answersHide answers

(a) methyl orange, yellow to red (orange at the end point); (b) phenolphthalein, colourless to pink.
24.30, 24.35 and 24.30 (within 0.10 cm³); mean = 24.32 cm³.
$n(\mathrm{H_2SO_4}) = 0.1050 \times 0.01875 = 1.969 \times 10^{-3}$ ; $n(\mathrm{NaOH}) = 3.938 \times 10^{-3}$ ; $c = 0.1969\,\mathrm{mol\,dm^{-3}}$ .
$n(\mathrm{HCl}) = 4.760 \times 10^{-3}$ per 25 cm³, so $0.04760\,\mathrm{mol}$ NaOH in total; mass = 1.904 g; purity = 95.2 %.
NaOH added $= 0.01000\,\mathrm{mol}$ ; excess $= 2.730 \times 10^{-3}$ ; reacted with $\mathrm{NH_4^{+}}$ $= 7.27 \times 10^{-3}\,\mathrm{mol}$ (1 : 1), which contains $7.27 \times 10^{-3}$ mol N = 0.1018 g; % N = 20.4 %.
$\mathrm{NH_4^{+}}$ is a very weak acid, so a direct titration has no sharp end point, and the ammonia formed would escape. Boiling with excess NaOH drives off all the $\mathrm{NH_3}$ , and the excess strong base is easy to titrate.
Phenolphthalein changes colour (pH about 8–9) when the carbonate has been converted only to hydrogencarbonate, $\mathrm{CO_3^{2-} + H^{+} \rightarrow HCO_3^{-}}$ , which uses half the acid. Methyl orange changes only after the second step, $\mathrm{HCO_3^{-} + H^{+} \rightarrow CO_2 + H_2O}$ .

Sources and credits

Indicator ranges: CRC Handbook of Chemistry and Physics (acid–base indicators).
First aid: NIOSH Pocket Guide to Chemical Hazards, Hydrogen chloride and Sodium hydroxide, US Centers for Disease Control and Prevention.
Image: Phenolphthalein in Flask.jpg by 384, via Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: How to Solve Back Titration Calculations, Paul Morgan, embedded from YouTube.
The text, Figures 11.2 and 11.3, the table, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Measuring what you cannot dissolve

2The direct titration

3The calculation, in four steps

4Back titration

Measuring what you cannot dissolve

The direct titration

The calculation, in four steps

Back titration