الكيمياء (بالإنجليزية) الصف الثالث الثانوي، الأسبوع 8: Aluminium, iron(II), iron(III) and calcium

Cations, and the idea of a titration

الأسبوع 8 · الباب الثاني (٣): الكشف عن الكاتيونات (المجموعات التحليلية الست والنحاس (II) والألومنيوم والحديد (II) والحديد (III) والكالسيوم) وبداية التحليل الكمي: المعايرة والمحلول القياسي ونقطة النهاية والأدلة · 3 دروس

Aluminium, iron(II), iron(III) and calcium

Chemistry Sec 3 Term 1, Chapter 2: pp. 39–43

GoalRecognise Al³⁺, Fe²⁺, Fe³⁺ and Ca²⁺ by their precipitates with ammonium hydroxide, sodium hydroxide and ammonium carbonate.

New words

gelatinous precipitate · a soft, jelly-like precipitate, such as white Al(OH)₃hydroxide · a compound of a metal with OH⁻, such as Fe(OH)₃flame test · a colour test in which a salt in a flame gives a colour: brick red for calcium (paper only here)

Explanation

Group 3 cations form hydroxides when ammonium hydroxide is added. Aluminium gives white, gelatinous Al(OH)₃: Al₂(SO₄)₃ + 6NH₄OH → 2Al(OH)₃ + 3(NH₄)₂SO₄. The book shows that Al(OH)₃ dissolves in extra sodium hydroxide, Al(OH)₃ + NaOH → NaAlO₂ + 2H₂O, a short way of writing the aluminate ion. Iron(II) gives a white precipitate that turns green-white in air: FeSO₄ + 2NaOH → Fe(OH)₂ + Na₂SO₄. Iron(III) gives a reddish-brown precipitate: FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl, and the same kind of precipitates form with NH₄OH. In group 5, calcium gives white calcium carbonate with ammonium carbonate: CaCl₂ + (NH₄)₂CO₃ → CaCO₃ + 2NH₄Cl. It dissolves in dilute HCl with a gas, CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, and in water with CO₂, CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂. It is confirmed by dilute sulphuric acid, which gives white CaSO₄ (only sparingly soluble), and by a brick-red flame, which is paper only here. Ammonia solution, sodium hydroxide and flames are never used at home.

Cation	Reagent	Result
Al³⁺	NH₄OH	white gelatinous Al(OH)₃; dissolves in excess NaOH
Fe²⁺	NaOH	white Fe(OH)₂, green-white in air
Fe³⁺	NaOH	reddish-brown Fe(OH)₃
Ca²⁺	(NH₄)₂CO₃	white CaCO₃; dissolves in dilute HCl

Worked examples

	Left	Right
Al	2	2
S	3	3
O	12 + 6 = 18	6 + 12 = 18
N	6	6
H	30	6 + 24 = 30

Count the atoms. Al₂(SO₄)₃ + 6NH₄OH → 2Al(OH)₃ + 3(NH₄)₂SO₄. Left: Al 2, S 3, O 12 + 6 = 18, N 6, H 6 × 5 = 30. Right: Al 2, O 6 + 12 = 18, H 6 + 24 = 30, S 3, N 6. The equation is balanced, and two aluminium ions need six hydroxide groups.

Cation	With NH₄OH	Excess NaOH
Al³⁺	white	dissolves
Fe²⁺	white	stays
Fe³⁺	reddish-brown	stays

Which test is ambiguous? Aluminium and iron(II) both give a white precipitate with ammonium hydroxide, so that result alone cannot tell them apart. Iron(III) is the least ambiguous, because its reddish-brown colour is unlike the others. To separate Al³⁺ from Fe²⁺, add excess NaOH: Al(OH)₃ dissolves, and an iron(II) hydroxide precipitate does not (an extra fact) and changes colour in air.

Calcium in steps. (1) CaCl₂ + (NH₄)₂CO₃ → CaCO₃ + 2NH₄Cl gives a white precipitate. (2) Dilute HCl dissolves it with fizzing: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, the same reaction as a carbonate in last week’s test. (3) CaCl₂ + H₂SO₄ → CaSO₄ + 2HCl confirms calcium with a white solid that only dissolves a little. Count Ca 1, Cl 2, S 1, O 4, H 2 on both sides of step 3.

Class activity

Air changes colours, with a grown-up. A grown-up cuts an apple in half (or a banana). Put one half cut-side up on a plate and wrap the other half tightly in plastic wrap. After half an hour, compare the two cut faces. The apple changes colour because of the oxygen in air, like iron(II) hydroxide, which turns green-white in air. Do not try to make any of the book’s precipitates, because ammonia solution and sodium hydroxide are lab chemicals. Then write a three-row table: cation, reagent and colour seen, from the book facts above.

Practice

Complete and balance: AlCl₃ + NH₄OH → ?

Complete and balance: FeCl₂ + NaOH → ? What happens to the colour of the precipitate in air?

Iron(III) sulphate reacts with sodium hydroxide: Fe₂(SO₄)₃ + 6NaOH → 2Fe(OH)₃ + 3Na₂SO₄. What colour is the precipitate, and how many moles of Fe(OH)₃ form from 0.10 mol of Fe₂(SO₄)₃?

0.050 mol of FeCl₃ reacts with excess NaOH: FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl. Find the mass of Fe(OH)₃ (Fe = 56, O = 16, H = 1).