Coaching Centre
Gravimetric analysis and precipitation titrations
Quantitative analysis by mass and by precipitation titration: finding sulfate in a fertiliser by gravimetric analysis, finding chloride by the Mohr method, and the errors that affect each.
The bag says "21% nitrogen, 24% sulfur". A farmer spreading 200 kg of it per hectare across 500 hectares is paying for those numbers. Fertiliser makers and regulators check the label with methods that need nothing more than a precise balance, a burette and careful hands. They are some of the oldest methods in chemistry, and in skilled hands they are still among the most accurate.
From "what" to "how much"
Pages 8B and 8C were qualitative. Quantitative analysis measures how much of a substance is present. Two classic methods use the same precipitation reactions:
- Gravimetric analysis: precipitate the ion completely, then weigh the dry precipitate.
- Precipitation titration: add a solution of known concentration until the ion has just been used up, and measure the volume.
Both depend on a precipitate that is very insoluble (so essentially all of the ion is removed) and on a known formula (so moles of precipitate convert to moles of the ion).
Gravimetric analysis
To find the sulfate in a fertiliser:
- Weigh the sample accurately and dissolve it in deionised water. Add a little dilute HCl so that carbonate or phosphate can't precipitate with barium.
- Heat the solution and add barium chloride solution in excess, slowly and with stirring. Hot, slow precipitation gives larger crystals that are easier to filter.
- Test for completeness: let the solid settle and add a drop more BaCl₂ to the clear liquid. No new cloudiness means all the sulfate has gone.
- Filter through pre-weighed fine (ashless) filter paper or a sintered glass crucible, and wash the solid with warm deionised water to remove soluble ions.
- Dry in an oven at about 110 °C, cool in a desiccator, weigh, and repeat until two weighings agree: drying to constant mass.
A 1.250 g sample of fertiliser gave 0.874 g of dry BaSO₄. Calculate the percentage by mass of sulfate in the fertiliser.
Errors in gravimetric analysis
| Error | Effect on the mass of precipitate | Effect on the result |
|---|---|---|
| Precipitate not dried to constant mass (still wet) | too high | too high |
| Not washed: soluble salts dry onto the solid | too high | too high |
| Fine particles pass through the filter paper | too low | too low |
| Not enough BaCl₂, so some sulfate stays in solution | too low | too low |
| Other ions co-precipitate (e.g. BaCO₃ if not acidified) | too high | too high |
These are systematic errors: repeating the experiment the same way won't remove them, so they reduce accuracy (and validity), not just precision.
- Sulfate of ammonia, (NH₄)₂SO₄, sample 1.25 g, 100 mL of 0.100 M BaCl₂: the mass of BaSO₄ is 2.208 g and the sulfate is 72.7%. Check this by calculation.
- Lower the BaCl₂ to 90 mL. The result falls to 69.2%. Use the moles shown to explain why.
- Switch to lawn food and find the smallest volume of BaCl₂ that still gives the true value (30.5%).
- Turn on each error in turn and predict whether the result will go up or down before you look.
Precipitation titrations
A precipitation titration measures the volume of a standard solution needed to precipitate all of an ion. The most common is the Mohr method for chloride, used for seawater, bore water, food and blood. Silver nitrate in the burette is added to the chloride sample, which contains a few drops of potassium chromate indicator:
$\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$ (white)
Silver chromate is more soluble than silver chloride, so it can't form while there is still a significant amount of chloride left. Just after the equivalence point, the first excess silver ions react with chromate to give a permanent red-brown tinge:
$2\text{Ag}^+(aq) + \text{CrO}_4^{2-}(aq) \rightarrow \text{Ag}_2\text{CrO}_4(s)$ (red-brown)
The solution must be close to neutral (pH 6.5–10). In acid, chromate turns into dichromate and the end point is late; in strongly basic solution, brown silver oxide precipitates. Two other methods are worth knowing about: the Volhard method (add excess Ag⁺, then back-titrate the leftover Ag⁺ with thiocyanate, using Fe³⁺ as the indicator, which works in acid), and the Fajans method (an adsorption indicator such as dichlorofluorescein that turns pink on the surface of the precipitate).
A 25.00 mL sample of estuary water was titrated with 0.1000 mol/L AgNO₃ using chromate indicator. The average titre was 18.45 mL. Calculate the chloride concentration in g/L.
Test yourself
In a gravimetric analysis, a precipitate is heated and weighed repeatedly until its mass stays the same. This is done to
A constant mass shows no more water is being lost.
A student forgets to wash the BaSO₄ precipitate before drying it. The calculated percentage of sulfate will be
Excess BaCl₂ and other dissolved salts are left behind when the water evaporates, adding mass.
20.00 mL of a chloride solution needs 16.00 mL of 0.0500 mol/L AgNO₃ in a Mohr titration. The chloride concentration is
n(Ag⁺) = 0.0500 × 0.01600 = 8.00 × 10⁻⁴ mol = n(Cl⁻); c = 8.00 × 10⁻⁴ ÷ 0.02000 = 0.0400 mol/L.
Why does red-brown silver chromate only appear at the end point of a Mohr titration?
Ag₂CrO₄ only forms when there is excess Ag⁺, after the chloride has precipitated.
Name each step of the gravimetric analysis.
How would each error affect the calculated percentage of sulfate?
In gravimetric analysis the precipitate is dried to constant . Barium chloride is added in so that all the sulfate precipitates. In the Mohr method the indicator is potassium , and the end point is the first permanent colour.
To check the label on a bag of sulfate of ammonia, (NH₄)₂SO₄, a 1.502 g sample was dissolved and treated with excess barium chloride. The dry precipitate weighed 2.587 g. (a) Calculate the percentage of sulfate in the sample. (b) The theoretical value for pure (NH₄)₂SO₄ is 72.7%. Explain one possible reason for the difference.
(a) $n(\text{BaSO}_4) = \dfrac{2.587}{233.4} = 0.01108$ mol $= n(\text{SO}_4^{2-})$. $m(\text{SO}_4^{2-}) = 0.01108 \times 96.07 = 1.065$ g. Percentage $= \dfrac{1.065}{1.502} \times 100 = 70.9\%$.
(b) The result is 1.8 percentage points below the theoretical value. One possible reason is that some fine BaSO₄ passed through the filter paper, so the mass of precipitate (and the calculated sulfate) was too low. A fine filter paper or sintered crucible, and digesting the precipitate while warm to grow larger crystals, would reduce this. Alternatively, the fertiliser itself may not be pure (NH₄)₂SO₄: it may contain moisture or an anti-caking agent, so less than 72.7% of its mass is sulfate.