Egypt Secondary 3 Chemistry, Week 8: Cations: analytical groups, copper(II)

Cations, and the idea of a titration

Week 8 · Chapter 2 (C): detecting cations (six analytical groups: copper(II), aluminium, iron(II), iron(III), calcium) and the start of quantitative analysis: titration, standard solutions, end point and indicators · 3 lessons

Cations: analytical groups, copper(II)

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

GoalExplain why cations are separated into analytical groups, and identify copper(II) in group 2.

New words

cation (basic radical) · the positive ion of a salt, such as Cu²⁺, Al³⁺ or Ca²⁺analytical group · cations whose salts have similar solubility, precipitated together by one group reagentsparingly soluble · dissolving only a very little in water, so it forms a precipitate

Explanation

Basic radicals, the cations, are harder to identify than anions for three reasons. There are very many of them, they disturb each other’s tests, and one metal can have more than one oxidation state, such as iron(II) and iron(III). So cations are sorted into six analytical groups by the solubility of their salts, and each group has its own group reagent. Think of a set of sieves: each reagent removes one family from the solution, and the next reagent works on what is left. In group 1, the chlorides of silver, mercury(I) and lead(II) are sparingly soluble, so dilute HCl precipitates them. In group 2, the cations give insoluble sulphides when hydrogen sulphide is passed through an acid solution; copper(II) gives black copper sulphide: CuSO₄ + H₂S → CuS + H₂SO₄. Hydrogen sulphide is a poison gas, so this equation is for paper only. Group 3 and group 5 appear next lesson; groups 4 and 6 are not needed in this course.

Group	Reagent	What forms	Example
1	dilute HCl	chlorides	Ag⁺, Pb²⁺
2	H₂S in acid solution	sulphides	Cu²⁺
3	NH₄OH	hydroxides	Al³⁺, Fe²⁺, Fe³⁺
4	—	not needed here	—
5	(NH₄)₂CO₃	carbonates	Ca²⁺
6	—	not needed here	—

Worked examples

Which group? A white precipitate of AgCl forms when dilute HCl is added to a clear solution: silver is in group 1, because its chloride is sparingly soluble. A black precipitate forms when H₂S is passed into an acid solution of a copper salt: copper is in group 2, as black CuS.

ƒIonic equation

Cu²⁺ + S²⁻ → CuS

Atoms and charge. CuSO₄ + H₂S → CuS + H₂SO₄: Cu 1, S 2, H 2, O 4 on both sides. In ion form, Cu²⁺ + S²⁻ → CuS: the charge is +2 − 2 = 0 on the left and 0 on the right. This is why a salt of Cu²⁺ and a sulphide always form one-to-one.

How much precipitate? 0.050 mol of CuSO₄ in solution reacts completely with H₂S (paper only). The ratio is 1 : 1, so n(CuS) = 0.050 mol. With Cu = 63.5 and S = 32, M(CuS) = 95.5 g/mol, so m = 0.050 × 95.5 = 4.775 g, about 4.78 g.

Class activity

Sieve game, in the kitchen with a grown-up. Mix a spoon each of dry rice, lentils and flour in a bowl (food only, do not taste the mixture). Use a colander to take out the biggest grains first, then a tea strainer for the next size, and see what is left. The order matters, like group reagents: each tool removes one family and the next works on the rest. Write the three families as groups A, B and C, and make a table with the tool and the family it removes. Wash your hands and put the food to the bin afterwards.

Practice

Which analytical group is each of these in: AgCl, PbCl₂, CuS?

Complete and balance: CuCl₂ + H₂S → ?

0.10 mol of CuCl₂ reacts completely with H₂S. What mass of CuS forms (Cu = 63.5, S = 32)?

Give three reasons why cations are harder to analyse than anions.