Identification of metallic and non-metallic ions III
Francine Taylor-Campbell, Contributor

Last week, we looked at the reactions of some anions and cations. This week we hope to use the knowledge gained to answer questions which require the identification of these ions. Let us examine the questions given below.

Question 1

Identify each lettered substance below by giving its name and formula (with the aid of equations).

Q is a sodium salt. When a solution of Q was added to aqueous silver nitrate, a white precipitate, R, was formed. This precipitate did not dissolve in dilute nitric acid.

S is a sodium salt. When a solution of S was added to aqueous barium nitrate, a white precipitate, T, was formed. When diluted nitric acid was added to T, it dissolved and a gas, U, which turned lime water milky.

When aqueous sodium hydroxide was added to aqueous zinc chloride, a white precipitate, V, was formed. This precipitate dissolved in excess of sodium hydroxide to form a colourless solution in which zinc was present as a salt, W.

ANSWERS

(a) The ion present is Cl^-, which means that R is AgCl (silver chloride) and Q is NaCl (sodium chloride). Ag⁺(aq) + Cl^-(aq) = AgCl (white precipitate).

(b) The gas produced, U, is CO₂, which means that the ion is CO₃^2-. T is BaCO₃ (barium carbonate), hence the sodium salt S is Na2CO3 (sodium carbonate).

Ba²+(aq) + CO₃^2- (aq) == BaCO₃ (s) white

BaCO₃(s) + HNO₃ (aq) == Ba(NO₃)₂(aq) + H₂O (l) + CO₂(g)

(c) V is zinc hydroxide (Zn(OH)₂) and in excess NaOH, W, which is sodium zincate (Na₂ZnO₂), is formed. Zn²⁺ (aq) + 2OH^- (aq) == Zn(OH)₂ (s)

Zn(OH)₂(s) + 2NaOH (aq) == Na₂ZnO2 (aq) + 2H₂O (l)

Question 2

Describe what is observed (with the aid of equations) in each of the following reactions.

(i) Aqueous NaOH is added to aq iron(III) sulphate

(ii) Dilute HCl is added to solid sodium carbonate

(iii) Aqueous barium chloride is added to dilute sulphuric acid

(iv) Aqueous silver nitrate is added to aq sodium chloride.

(8 marks)

ANSWERS

(a) A rusty-brown precipitate is formed, which is insoluble in excess NaOH.

Fe³⁺(aq) + 3OH-(aq) == Fe(OH)₃ (s)

(b) Sodium carbonate would dissolve and a gas would be given off

Na₂CO₃(s) + 2HCl (aq) == 2NaCl (aq) + H₂O(l) + CO₂ (g)

(c) A white precipitate will be seen, which will remain insoluble.

Ba²⁺(aq) + SO₄^2-(aq) == BaSO₄(s)

(d) The silver nitrate would react with the chloride ions to form a white precipitate of silver chloride

Ag⁺(aq) + Cl^-(aq) == AgCl (s)

Lieutenant Osmar Fiddler receives a plaque for being the top student to be trained as military pilots at the Jamaica Military Aviation School, from Minister of National Security Colonel Trevor MacMillan, recently. In the background are Lieutenants Melecia Sinclair and Victor Dawkins, who, along with Fiddler, became the first persons to be trained as military pilots at the school. - Contributed

Francine Taylor-Campbell is an independent contributor.