الأسبوع 7 · الباب الثاني (٢): بقية الأنيونات: الثيوكبريتات والنيتريت ومجموعة حمض الكبريتيك المركز (الكلوريد والبروميد واليوديد والنيترات واختبار الحلقة البنية) ومجموعة كلوريد الباريوم (الفوسفات والكبريتات) وتدريب من إعداد Darslab على تصنيف الأنيونات · 3 دروس
3
The barium chloride group, and a decision chart for anions
Chemistry Sec 3 Term 1, Chapter 2: pp. 35–39
GoalIdentify phosphate and sulphate with barium chloride, and use a three-step decision chart for the twelve anions (Darslab’s own).
New words
phosphate · the ion PO₄³⁻; it gives a white precipitate with BaCl₂ that dissolves in dilute HClsulphate · the ion SO₄²⁻; it gives a white precipitate with BaCl₂ that does not dissolve in dilute HCldecision chart · a plan of tests in a fixed order, with each result sending you to the next step
Explanation
The third group does not react with dilute HCl or with concentrated H₂SO₄, so it is shown by barium chloride solution. Phosphate gives white barium phosphate, 2Na₃PO₄ + 3BaCl₂ → Ba₃(PO₄)₂ + 6NaCl, which dissolves in dilute HCl. Silver nitrate gives yellow Ag₃PO₄, Na₃PO₄ + 3AgNO₃ → Ag₃PO₄ + 3NaNO₃, soluble in ammonia and in nitric acid. Sulphate gives white barium sulphate, Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl, which does not dissolve in dilute HCl; with lead(II) acetate it gives white PbSO₄. Be careful: the sulphate precipitate does not dissolve in dilute HCl, the phosphate one does. The book has no summary of the three groups, so here is a decision chart of our own: first add dilute HCl and look for a gas; if there is none, test the solid with concentrated H₂SO₄; if there is still nothing, add BaCl₂ solution. Then use the second test of each anion to decide. This is a thinking tool and all reactions stay on paper.
Sulphate or phosphate? Both give a white precipitate with BaCl₂. Now add dilute HCl to each precipitate. The one that dissolves is Ba₃(PO₄)₂, so that salt is the phosphate. The one that stays is BaSO₄, so that salt is the sulphate.
Use the chart. A solid salt gives no gas with dilute HCl. With concentrated H₂SO₄ it gives a colourless gas that makes white fumes with ammonia. The chart sends it to step 2, and the white fumes point to HCl, so the anion is chloride. Silver nitrate would give white AgCl that dissolves in concentrated ammonia.
Which test separates them? Sodium chloride from sodium iodide: concentrated H₂SO₄ gives violet vapour only with the iodide, or silver nitrate gives white AgCl against yellow AgI. Sodium sulphite from sodium sulphate: dilute HCl frees SO₂ only from the sulphite, while BaCl₂ followed by dilute HCl leaves a precipitate only for the sulphate.
Class activity
Build your own chart, paper only. Write the twelve anions on small cards. Put them in three piles by the group test (dilute HCl, concentrated H₂SO₄, BaCl₂). Inside each pile, put the second test on the back of each card (a paper colour, a gas name or a precipitate colour). Give your pile to a friend, who picks a card from the front and names the test from the back without turning it. No chemical is used: it is a memory and reasoning game.
Practice
1
What do you expect when BaCl₂ solution is added to potassium phosphate, K₃PO₄? Write the equation.
A white precipitate: 2K₃PO₄ + 3BaCl₂ → Ba₃(PO₄)₂ + 6KCl (K 6, P 2, O 8, Ba 3, Cl 6 on both sides).
2
A white precipitate formed with BaCl₂ stays unchanged when dilute HCl is added. Which anion is present?
Sulphate, because BaSO₄ does not dissolve in dilute HCl.
3
Put each anion in group 1, 2 or 3 of the chart: nitrite, bromide, sulphate, nitrate.
nitrite: group 1; bromide: group 2; sulphate: group 3; nitrate: group 2.
4
0.020 mol of sodium sulphate reacts with excess BaCl₂. What mass of BaSO₄ forms (Ba = 137, S = 32, O = 16)?
Ratio 1 : 1, so n(BaSO₄) = 0.020 mol; M = 233 g/mol; m = 0.020 × 233 = 4.66 g.
