What is in crude salt, and the plan
Salt made by drying sea water or from rock salt is called crude salt. Besides NaCl it holds Ca²⁺, Mg²⁺ and SO₄²⁻, and some sand and mud.
The plan is called chemical precipitation: add a reagent whose ion joins the unwanted ion to make a solid that does not dissolve (a precipitate). Then filter it away.
- First dissolve the crude salt in water and filter off the sand and mud.
- Then remove each unwanted ion with a reagent, in the right order.
- Filter, neutralise, and evaporate.
Removing sulfate, magnesium and calcium
| Add | It removes | Solid made (white) |
|---|---|---|
| BaCl₂ (a little more than needed) | SO₄²⁻ | BaSO₄ |
| NaOH | Mg²⁺ | Mg(OH)₂ |
| Na₂CO₃ | Ca²⁺ and the extra Ba²⁺ | CaCO₃ and BaCO₃ |
Ionic equations: Ba²⁺ + SO₄²⁻ → BaSO₄, Mg²⁺ + 2OH⁻ → Mg(OH)₂, Ca²⁺ + CO₃²⁻ → CaCO₃.
Why this order? Na₂CO₃ must come after BaCl₂. If Ba²⁺ is added last, the extra Ba²⁺ has nothing to remove it and stays in the salt. Barium ions are poisonous, so this matters.
Filter, neutralise and crystallise
Filter once, when all the solids have formed. The filtrate (clear liquid) still has a little extra OH⁻ and CO₃²⁻ from our reagents. Add dilute HCl drop by drop until the liquid is just neutral: OH⁻ + H⁺ → H₂O and CO₃²⁻ + 2H⁺ → H₂O + CO₂ (you see small bubbles). HCl adds only Cl⁻ and H⁺, which do no harm.
Last, heat the liquid in an evaporating dish until crystals begin to appear, then let it cool. Do not dry it out completely, or it spits. The white crystals are pure NaCl.
Check: dissolve a little product and add BaCl₂ and Na₂CO₃. No white cloudiness means the ions have gone.
Part two: sulfur changing its oxidation number
The oxidation number tells how many electrons an atom has lost (plus) or gained (minus) in a compound. Sulfur has a wide range:
| Substance | Oxidation number of S |
|---|---|
| H₂S, hydrogen sulfide | −2 |
| S, sulfur | 0 |
| SO₂, sulfur dioxide | +4 |
| H₂SO₄, sulfuric acid | +6 |
Going up = oxidation (loss of electrons), going down = reduction.
Sulfur conversions you can run
- 0 → +4: S + O₂ → SO₂ (sulfur burns with a blue flame). Oxygen oxidises sulfur.
- +4 → +6: 2SO₂ + O₂ → 2SO₃ (with a V₂O₅ catalyst in factories); SO₃ + H₂O → H₂SO₄.
- +6 → +4: Cu + 2H₂SO₄ (hot, concentrated) → CuSO₄ + SO₂ + 2H₂O. Here the acid is reduced.
- −2 and +4 → 0: 2H₂S + SO₂ → 3S + 2H₂O. Yellow solid sulfur appears from a gas mixture.
- Test for SO₂: it turns purple acidified KMnO₄ colourless, because SO₂ is a reducing agent.
Safety: SO₂ and H₂S gases are harmful. Do these in a fume cupboard, as a teacher demonstration or in micro-scale.
Try it: predict, then check
Before step 6 predict: if I add Na₂CO₃ first and BaCl₂ last, which ion is left? Then try it. At home: stir 2 spoons of coarse salt in a glass of water and let it settle. Pour off the clear part through a clean cloth. Heat a little on a steel lid (with a grown-up) and compare the crystals with table salt. Count how much mud stayed behind.
Key formulas and definitions
- Ba²⁺ + SO₄²⁻ → BaSO₄ (white solid)
- Mg²⁺ + 2OH⁻ → Mg(OH)₂ (white solid)
- Ca²⁺ + CO₃²⁻ → CaCO₃ and Ba²⁺ + CO₃²⁻ → BaCO₃
- OH⁻ + H⁺ → H₂O and CO₃²⁻ + 2H⁺ → H₂O + CO₂
- Oxidation numbers of S: H₂S −2, S 0, SO₂ +4, H₂SO₄ +6; up = oxidation, down = reduction
Worked examples
1. A sample of crude salt solution holds 0.96 g of sulfate ions (SO₄²⁻, 96 g/mol). What mass of BaSO₄ (233 g/mol) forms with excess BaCl₂?
Moles of sulfate = 0.96 ÷ 96 = 0.01 mol. Ba²⁺ + SO₄²⁻ → BaSO₄ is 1 : 1, so 0.01 mol BaSO₄. Mass = 0.01 × 233 = 2.33 g.
2. Which reagent removes Mg²⁺, and what is the solid? Write the ionic equation.
NaOH removes Mg²⁺. The solid is Mg(OH)₂. Mg²⁺ + 2OH⁻ → Mg(OH)₂.
3. A student adds Na₂CO₃ before BaCl₂. Which impurity is left in the salt, and why?
Extra Ba²⁺ is left. The carbonate was already used up on Ca²⁺ before barium was added, so nothing is left to remove the excess Ba²⁺.
4. Find the oxidation number of sulfur in SO₂ and in H₂SO₄.
SO₂: S + 2(−2) = 0, so S = +4. H₂SO₄: 2(+1) + S + 4(−2) = 0, so S = +6.
Common mistakes
- Adding Na₂CO₃ before BaCl₂, which leaves extra barium ions in the salt.
- Filtering after each reagent and again, instead of once at the end. One filtration after all solids form is enough.
- Skipping the dilute HCl, so the salt keeps extra OH⁻ and CO₃²⁻ and tastes bitter.
- Mixing up oxidation and reduction: going up the oxidation number ladder (−2 to +6) is oxidation.