Answers to precipitation questions

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Answers to Questions.
Question One:

Reaction 1.

The ions before mixing are Pb²⁺(aq), NO₃^-(aq), Na⁺(aq) and Cl^-(aq).

The precipitate formed when the solutions are mixed is lead chloride, PbCl₂(s).

The precipitate forms because Pb²⁺(aq) react with Cl^-(aq) ions to form the insoluble (ionic) solid, lead chloride.
Question Two:

1. 
   (i) silver chloride AgCl(s) (ii) no precipitate (iii) calcium sulfate CaSO₄(s)
   
2. 
   Either Ag⁺(aq) + Cl^-(aq)  AgCl(s) OR Ca²⁺(aq) + SO₄^2-(aq)  CaSO₄(s)
   

Question Three:

Lead nitrate + potassium chloride  lead chloride + potassium nitrate OR

Lead nitrate + potassium chloride  potassium nitrate + lead chloride (order of the products doesn’t matter).
Question Four:

1. 
   A white precipitate would form (in a very pale green solution)
   
2. 
   It is a precipitation reaction.
   
3. 
   Barium nitrate will react with iron(II) sulfate to make barium sulfate and iron(II) nitrate. Barium sulfate is insoluble and will make a WHITE precipitate. A pale green solution of iron(II) nitrate will be formed. (Iron(II) compounds are usually a very pale green colour).
   

Ba²⁺(aq) + SO₄^2-(aq)  BaSO₄(s)
Question Five:

1. 
   (i) no precipitate (ii) lead chloride (iii) magnesium hydroxide
   
2. 
   Either Pb²⁺(aq) + 2Cl^-(aq)  PbCl₂(s) OR Mg²⁺(aq) + 2OH^-(aq)  Mg(OH)₂(s)
   

Question Six:

Barium chloride + magnesium sulfate  barium sulfate + magnesium chloride

Question Seven:
CaCl₂(aq) + K₂CO₃(aq)  2KCl(aq) + CaCO₃(s)
AgNO₃(aq) + NaCl(aq) NaNO₃(aq) + AgCl(s)
(These two because 2 solutions (aq) react to make a solution (aq) and a solid (s) – the solid is the precipitate).
Question Eight:

• 
  2AgNO₃(aq) + BaCl₂(aq)  2AgCl(s) + Ba(NO₃)₂(aq)
  

(These one only because 2 solutions (aq) react to make a solution (aq) and a solid (s) – the solid is the precipitate).

Question Nine:

1. 
   (i) calcium sulfate (ii) copper hydroxide (iii) no precipitate
   
2. 
   Ca²⁺(aq) + SO₄^2-(aq)  CaSO₄(s) OR Cu²⁺(aq) + 2OH^-(aq)  Cu(OH)₂(s)
   

Question Ten:

Magnesium chloride + sodium hydroxide  magnesium hydroxide + sodium chloride

Question Eleven:

1. 
   When pale green iron(II) nitrate solution is added to colourless sodium hydroxide solution, a green precipitate of iron(II) hydroxide forms and a colourless solution of sodium nitrate.
   
2. 
   Iron(II) hydroxide is an insoluble (ionic) solid
   
3. 
   Fe²⁺(aq) + 2OH^-(aq)  Fe(OH)₂(s)
   

Question Twelve:
CuCl₂(aq) + K₂CO₃(aq)  CuCO₃(s) + 2KCl(aq)


Question Thirteen:

1. 
   Calcium sulfate (ii) lead chloride (iii) zinc carbonate
   

Question Fourteen:

Sodium hydroxide + magnesium sulfate  sodium sulfate + magnesium hydroxide

Question Fifteen:

(a) CaCI₂(aq) + Na₂SO₄(aq) → CaSO₄(s) + 2NaCI (aq)

(b) Cu(NO₃)₂(aq) + 2NaOH(aq)  Cu(OH)₂(s) + 2NaNO₃(aq)
Question Sixteen: NOTE iron(III) hydroxide not examined from 2012 onwards

1. 
   When the colourless solution of sodium hydroxide is added to the orange solution of iron(III) chloride a brown precipitate of iron(III) hydroxide forms and a colourless solution of sodium chloride.
   
2. 
   Fe³⁺(aq) + 3OH^-(aq) → Fe(OH)₃(s)
   

Question Seventeen:

This is a precipitation reaction. Calcium ions, Ca²⁺, will react with hydroxide ions, OH^–, to form insoluble calcium hydroxide, Ca(OH)₂. This is a white precipitate. The other ions in the reaction, potassium, K⁺, and nitrate, NO₃^–, do not react with each other, so stay in solution. Since neither They are not written in the

The potassium nitrate is soluble in water and since neither ion (K⁺ and NO₃^-) is coloured, the solution will be colourless.

Ca²⁺ (aq) + 2OH^–(aq) → Ca(OH)₂(s)

2KOH + Ca(NO₃)₂ → 2KNO₃ + Ca(OH)₂
Question Eighteen:

When a green solution of iron(II) sulfate is added to a colourless solution of potassium carbonate solution, a green precipitate forms.

This precipitate is iron(II) carbonate. Iron(II) carbonate is insoluble in water so forms a

solid that will settle in the container. The potassium sulfate is soluble in water and since

neither ion (K⁺ and SO₄^2–) is coloured, the solution will be colourless.

FeSO₄(aq) + K₂CO₃(aq) → K₂SO₄(aq) + FeCO₃(s)

Fe²⁺(aq) + CO₃^2–(aq) → FeCO₃(s)

Question Nineteen:

This is a precipitation reaction. Calcium ions, Ca²⁺, will react with hydroxide ions, OH^–, to form insoluble calcium hydroxide, Ca(OH)₂. This is a white precipitate. The other ions in the reaction, sodium, Na⁺, and nitrate, NO₃^–, do not react with each other, so stay in solution.

The sodium nitrate is soluble in water and since neither ion (Na⁺ and NO₃^-) is coloured, the solution will be colourless.

Ca²⁺ (aq) + 2OH^–(aq) → Ca(OH)₂(s)

2NaOH + Ca(NO₃)₂ → 2NaNO₃ + Ca(OH)₂
Question Twenty:

A precipitate would form in beaker A. The ions in this mixture are Pb²⁺, NO₃^–, Na⁺ and Cl^–.

The lead ion and chloride ion do react to form the insoluble solid lead chloride. The sodium and nitrate ions are spectator ions and do not react.

A (white) precipitate of lead chloride will form in the bottom of the beaker.

Pb²⁺(aq) + 2Cl^–(aq) → PbCl₂(s)
(There would be no precipitate in Beaker B as iron(II) nitrate and magnesium sulfate are both soluble substances).