Lead(II) iodide is an insoluble salt. A student prepared it by mixing aqueous lead(II) nitrate with aqueous potassium iodide, as shown in Fig. 7.1. Fig. 7.1...

This question tests the precipitation route to an insoluble salt. When two soluble salts are mixed and the pair of ions that meet are insoluble together, they drop out of solution as a solid; everything else stays dissolved.

(a) Observations [2]. Lead(II) nitrate and potassium iodide are both colourless solutions, but on mixing the lead ions and iodide ions combine to form insoluble lead(II) iodide:

• On mixing: a bright yellow precipitate (of lead(II) iodide) forms [1].
• On standing: the yellow solid sinks / settles to the bottom, leaving a colourless solution above it [1].

(b) Obtaining a pure, dry sample [4]. Because the salt is insoluble it can be separated as a solid by filtration:

1. Filter the mixture; the lead(II) iodide is collected as the residue in the filter paper [1].
2. Wash the residue with distilled water (poured through the funnel) to rinse away soluble potassium nitrate clinging to the crystals [1].
3. Dry the solid, for example between sheets of filter paper or in a warm oven [1].
4. Pure, dry lead(II) iodide is left [1].

(c) Salt left in the filtrate [1]. The spectator ions are potassium and nitrate, so the soluble salt that remains dissolved in the filtrate is potassium nitrate [1].

Examination tip: the washing step is a common mark to lose. Without it, soluble impurity (here potassium nitrate) dries onto the crystals and the sample is not pure.