Question

Describe how you could separate the ions in each of the following groups by selective precipitation. a. \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\) c. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\)

Short answer

To separate the ions in each group by selective precipitation: a. For \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\): 1. Add dilute HCl, which forms \(\mathrm{AgCl}\) precipitate. Filter to separate \(\mathrm{AgCl}\). 2. Add NaOH, which forms \(\mathrm{Cu(OH)_2}\) precipitate. Filter to separate \(\mathrm{Cu(OH)_2}\) and obtain the remaining \(\mathrm{Mg}^{2+}\). b. For \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\): 1. Add NaOH, forming \(\mathrm{Pb(OH)_2}\) and \(\mathrm{Fe(OH)_2}\) precipitates. Filter and save them. 2. Dissolve the saved precipitates in dilute HCl, then add K2CrO4, forming \(\mathrm{PbCrO_4}\) precipitate. Filter to separate \(\mathrm{PbCrO_4}\) and obtain the remaining \(\mathrm{Fe}^{2+}\). c. For \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\): 1. Add dilute HCl, forming \(\mathrm{PbCl_2}\) precipitate. Filter to separate \(\mathrm{PbCl_2}\) and obtain the remaining \(\mathrm{Bi}^{3+}\).

Step-by-step solution

Precipitate \(\mathrm{Ag}^{+}\)

Add a few drops of dilute hydrochloric acid (HCl) to the solution containing the ions. \(\mathrm{Ag}^{+}\) will form a precipitate with chloride ions as \(\mathrm{AgCl}\) while \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cu}^{2+}\) will not: \[\mathrm{Ag}^{+} + \mathrm{Cl}^{-} \rightarrow \mathrm{AgCl (s)}\] Then, filter the precipitate to separate the \(\mathrm{AgCl}\) from the other ions.

Precipitate \(\mathrm{Cu}^{2+}\)

Add a few drops of a sodium hydroxide (NaOH) solution to the remaining filtrate. \(\mathrm{Cu}^{2+}\) will form a blue precipitate with hydroxide ions as \(\mathrm{Cu(OH)_2}\) while \(\mathrm{Mg}^{2+}\) will not: \[\mathrm{Cu}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Cu(OH)_2 (s)}\] Filter the precipitate, and \(\mathrm{Mg}^{2+}\) will remain in the solution. b. Separate \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\)

Precipitate \(\mathrm{Pb}^{2+}\) and \(\mathrm{Fe}^{2+}\)

Add a few drops of sodium hydroxide (NaOH) solution to the solution containing the ions. Both \(\mathrm{Pb}^{2+}\) and \(\mathrm{Fe}^{2+}\) will form precipitates with hydroxide ions while \(\mathrm{Ca}^{2+}\) will not: \[\mathrm{Pb}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Pb(OH)_2 (s)}\] \[\mathrm{Fe}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Fe(OH)_2 (s)}\] Filter the precipitates and save them for the next step.

Separate \(\mathrm{Pb}^{2+}\) from \(\mathrm{Fe}^{2+}\)

Dissolve the mixture of \(\mathrm{Pb(OH)_2}\) and \(\mathrm{Fe(OH)_2}\) in dilute hydrochloric acid (HCl). \(\mathrm{Pb(OH)_2}\) will form \(\mathrm{Pb}^{2+}\) and \(\mathrm{Fe(OH)_2}\) will form \(\mathrm{Fe}^{2+}\). Add a few drops of potassium chromate (K2CrO4) to this solution. \(\mathrm{Pb}^{2+}\) will form a yellow \(\mathrm{PbCrO_4}\) precipitate while \(\mathrm{Fe}^{2+}\) will not: \[\mathrm{Pb}^{2+} + \mathrm{CrO_4}^{2-} \rightarrow \mathrm{PbCrO_4 (s)}\] Filter this precipitate to separate it from the \(\mathrm{Fe}^{2+}\). c. Separate \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\)

Precipitate \(\mathrm{Pb}^{2+}\)

Add a few drops of dilute hydrochloric acid (HCl) to the solution containing the ions. \(\mathrm{Pb}^{2+}\) will form \(\mathrm{PbCl_2}\) precipitate while \(\mathrm{Bi}^{3+}\) will not: \[\mathrm{Pb}^{2+} + 2\mathrm{Cl}^{-} \rightarrow \mathrm{PbCl_2 (s)}\] Filter the precipitate to separate the \(\mathrm{PbCl_2}\) from \(\mathrm{Bi}^{3+}\).