Ionic equations and precipitation

Question 1

What do ionic substances form?

Answer 1

• Ionic substances form ionic latittices. The ions are fixed in position within the solid lattice.
• The lattice of ionic substances break apart in water and the separate ions can move.
• In precipitation reactions, the reactant ions can move in solution but must collide to react.
• In a precipitation reaction involving mixing two different ionic solutions, there is an extremely high collision frequency between the ions of the two solutions, so the reaction is extremely fast. The insoluble precipitate is formed so fast it appears to be instant.

Question 2

When barium chloride is added to sodium sulfate this is the word equation:

Answer 2

barium chloride + sodium sulfate ⇒ barium sulfate + sodium chloride

Question 3

Testing for ions.

Answer 3

• Precipitation reactions of barium chloride with sulfate ions is used as a test for sulfate ions.
• Precipitation reactions of lead nitrate with halide ions is used as a test for iodide and chloride ions.

Question 4

Preparing a clean and dry sample of an insoluble salt

Answer 4

• Stage 1: Mix: the two solutions to make barium sulfate and sodium chloride.
• Stage 2: Filter: the precipitate of barium sulfate, which stays in the filter paper.
• Stage 3: Wash: with distilled water to remove traces of the sodium chloride solution.
• Stage 4: Dry: by leaving the precipitate in a warm place for the water to evaporate.

Question 5

Spectator ions

Answer 5

When lead nitrate and sodium iodide react, only lead ions react with iodide ions as these are precipitated out of the solution. The other ions do not directly take part in the reaction. They are called spectator ions.

Question 6

Word equation of lead nitrate and sodium iodide.

Answer 6

lead nitrate_(aq) + sodium iodide_(aq) ⇒ lead iodide_(s) + sodium nitrate_(aq)

Question 7

Symbol equation of lead nitrate and sodium iodide.

Answer 7

Pb(NO₃)_2(aq) + 2NaI_(aq) ⇒ PbI_2(s) + 2NaNO_3(aq)

Question 8

Ionic equation of lead and iodide.

Answer 8

Pb²⁺_(aq) + 2I^-_(aq) ⇒ PbI_2(aq)