Checking a Fertiliser Label
A student analyses a 1.32 g sample of ammonium sulphate fertiliser, labelled 95% pure. The sample is warmed with excess sodium hydroxide and the ammonia produced is passed into 50.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid. The solution remaining is then titrated with 0.500 mol dm⁻³ sodium hydroxide; 13.0 cm³ is required to neutralise the excess acid, using methyl orange indicator. The student reports that the fertiliser is only about 30% pure and concludes that the manufacturer has cheated. A teacher points out that the method may have lost ammonia before it reached the acid, and asks the class to check the calculation rather than trust the headline figure. The apparatus used was a simple test-tube delivery tube, not a closed distillation set. (M: (NH₄)₂SO₄ = 132 g mol⁻¹; NH₃ = 17; HCl = 36.5; NaOH = 40.)
1. Calculate the percentage purity of the fertiliser from the data, showing each step, and decide whether the 95% label is supported.
Read the worked answer
Initial acid:
n(HCl)total = 0.0500 dm³ × 0.500 mol dm⁻³ = 0.0250 mol
Excess acid titrated with NaOH:
n(NaOH) = 0.0130 dm³ × 0.500 mol dm⁻³ = 0.00650 mol
NaOH + HCl → NaCl + H₂O, so the excess HCl is 0.00650 mol.
Acid that reacted with ammonia:
n(HCl reacted) = 0.0250 − 0.00650 = 0.0185 mol
NH₃ + HCl → NH₄Cl, so n(NH₃) = 0.0185 mol.
Ammonium sulphate:
(NH₄)₂SO₄ + 2NaOH → Na₂SO₄ + 2NH₃ + 2H₂O
One mole of (NH₄)₂SO₄ gives 2 mol NH₃:
n((NH₄)₂SO₄) = 0.0185 ÷ 2 = 0.00925 mol
mass = 0.00925 mol × 132 g mol⁻¹ = 1.221 g
percentage purity = (1.221 g ÷ 1.32 g) × 100 = 92.5%
This is close to the 95% label, not 30%. The student probably used the 0.00650 mol of NaOH as if it were the ammonia, which would give about 32.5% purity. The label is broadly supported within a few percent.
2. The teacher suggests some ammonia escaped before it reached the acid. Explain how this would affect the calculated purity, and describe one improvement to the apparatus.
Read the worked answer
If ammonia escapes, less NH₃ reaches the acid. Less HCl is therefore neutralised by ammonia, so more HCl remains for the NaOH titration. The NaOH titre becomes larger, the calculated n(NH₃) becomes smaller, and the calculated mass and percentage purity become lower than the true value. This is consistent with 92.5% being slightly below the 95% label.
A suitable improvement is to use a closed distillation apparatus: heat the fertiliser with NaOH in a flask, connect it by a tightly fitted delivery tube to a receiver containing the acid, and keep all joints sealed. The acid should be in a cooled flask so that the ammonia is absorbed efficiently. This reduces loss of ammonia before it reacts.
3. State two assumptions made in the calculation and explain whether each would tend to make the result too high or too low.
Read the worked answer
All the ammonia comes from ammonium sulphate. If the sample also contains another ammonium compound, the measured ammonia would be greater than that from (NH₄)₂SO₄ alone. The calculated mass of ammonium sulphate would then be too high, so the purity result would be too high. If the impurity is inert, this assumption is harmless.
The reaction with NaOH is complete and no ammonia is lost. If warming is too gentle or time is too short, some ammonium sulphate may not react, or some ammonia may remain dissolved. Less ammonia would be measured, so the result would be too low.
A third assumption is that the HCl is exactly 0.500 mol dm⁻³ and that it does not react with anything else; any error here would directly affect the titre and hence the calculated purity.