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Because of the insolubility of so many lead(II) compounds, the usual source of lead(II) ions in solution is lead(II) nitrate solution – and that will be assumed in all the following examples.

If a little sodium hydroxide solution is added to colourless lead(II) nitrate solution, a white precipitate of lead(II) hydroxide is produced.

\text{Pb}^{2+}_{(aq)} + 2OH^-_{(aq)} \longrightarrow \text{Pb}(OH)_{2(s)}

If more sodium hydroxide solution is added, the precipitate redissolves to give a colourless solution which might be called sodium plumbate(II) solution – but could be called by a lot of alternative names depending on exactly how the formula is written!

\text{Pb}(OH)_{2(s)} + 2OH^-_{(aq)} \longrightarrow {\text{Pb}O_2}^{2-}_{(aq)} + 2H_2O_{(l)}

Note: These equations are simplifications. You will get complexes formed involving hydroxide ions, but the formulae of these aren't very clear-cut. I am using these particular versions of the equations to keep them in line with the corresponding reaction between lead(II) oxide and sodium hydroxide solution on the oxides of Group 4 page – also a simplification!

Lead(II) chloride can be made as a white precipitate by adding a solution containing chloride ions to lead(II) nitrate solution. You could use things like sodium chloride solution to provide the chloride ions, but it is usually easier just to add some dilute hydrochloric acid.

\text{Pb}^{2+}_{(aq)} + 2\text{Cl}^-_{(aq)} \longrightarrow \text{Pb}\text{Cl}_{2(s)}

Note: If you add concentrated hydrochloric acid to excess, the lead(II) chloride precipitate will dissolve again. Complex ions like PbCl42- are produced, and these are soluble in water.

If you add colourless potassium iodide solution (or any other source of iodide ions in solution) to a solution of lead(II) nitrate, a bright yellow precipitate of lead(II) iodide is produced.

\text{Pb}^{2+}_{(aq)} + 2I^-_{(aq)} \longrightarrow \text{Pb}I_{2(s)}