Question

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

Short answer

To separate the ions in each group by selective precipitation, follow these steps: Group a: \(\mathrm{Ag}^+, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\) 1. Add NaCl to selectively precipitate \(\mathrm{AgCl}\). Filter to remove the precipitate, leaving \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cu}^{2+}\) in the solution. 2. Add Na2S to selectively precipitate \(\mathrm{CuS}\). Filter to remove the precipitate, leaving \(\mathrm{Mg}^{2+}\) in the solution. Group b: \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\) 1. Add Na2SO4 to precipitate \(\mathrm{PbSO}_4\) and \(\mathrm{Bi}_2(\mathrm{SO}_4)_3\). Filter to remove precipitates. 2. Dissolve precipitates in HCl, and add NaOH to selectively precipitate \(\mathrm{Bi(OH)}_3\). Filter to remove the precipitate, leaving \(\mathrm{Pb}^{2+}\) in the solution. Group c: \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\) 1. Add Na2SO4 to selectively precipitate \(\mathrm{PbSO}_4\). Filter to remove the precipitate, leaving \(\mathrm{Ca}^{2+}\) and \(\mathrm{Fe}^{2+}\) in the solution. 2. Add NaOH to selectively precipitate \(\mathrm{Fe(OH)}_2\). Filter to remove the precipitate, leaving \(\mathrm{Ca}^{2+}\) in the solution.

Step-by-step solution

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

To selectively precipitate \(\mathrm{Ag}^{+}\) ions, add a chloride salt, such as sodium chloride (NaCl). The \(\mathrm{Ag}^{+}\) ions will react with the chloride ions to form silver chloride (AgCl), an insoluble compound. The reaction can be represented as: \[\mathrm{Ag}^{+} + \mathrm{Cl}^{-} \rightarrow \mathrm{AgCl(s)}\] Filter the solution to remove the AgCl precipitate, and the remaining solution will contain \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cu}^{2+}\) ions. 2. \(\mathrm{Mg}^{2+} \) and \(\mathrm{Cu}^{2+} \)

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

To selectively precipitate \(\mathrm{Cu}^{2+}\) ions, add a sulfide salt, such as sodium sulfide (Na2S). The \(\mathrm{Cu}^{2+}\) ions will react with the sulfide ions to form copper sulfide (CuS), an insoluble compound. The reaction can be represented as: \[\mathrm{Cu}^{2+} + \mathrm{S}^{2-} \rightarrow \mathrm{CuS(s)}\] Filter the solution to remove the CuS precipitate, and the remaining solution will contain \(\mathrm{Mg}^{2+}\) ions. #Group b: Pb2+, Bi3+#

Precipitate \(\mathrm{Pb}^{2+}\) and \(\mathrm{Bi}^{3+}\) ions

In this case, both ions can be precipitated together by adding a sulfate salt, such as sodium sulfate (Na2SO4). The \(\mathrm{Pb}^{2+}\) and \(\mathrm{Bi}^{3+}\) ions will react with the sulfate ions to form lead sulfate (PbSO4) and bismuth sulfate (Bi2(SO4)3), both insoluble compounds. The reaction can be represented as: \[\mathrm{Pb}^{2+} + \mathrm{SO}_4^{2-} \rightarrow \mathrm{PbSO}_4(s)\] \[\mathrm{2Bi}^{3+} + \mathrm{3SO}_4^{2-} \rightarrow \mathrm{Bi}_2(\mathrm{SO}_4)_3(s)\] Filter the solution to remove the PbSO4 and Bi2(SO4)3 precipitates. Then, dissolve the mixed precipitate in a dilute solution of hydrochloric acid (HCl) to obtain a solution containing \(\mathrm{Pb}^{2+}\) and \(\mathrm{Bi}^{3+}\) ions.

Separate \(\mathrm{Pb}^{2+}\) and \(\mathrm{Bi}^{3+}\) ions

Now, add a small amount of sodium hydroxide (NaOH) to the solution, which will selectively precipitate the \(\mathrm{Bi}^{3+}\) ions as bismuth hydroxide (Bi(OH)3). The reaction can be represented as: \[\mathrm{Bi}^{3+} + \mathrm{3OH}^{-} \rightarrow \mathrm{Bi(OH)}_3(s)\] Filter the solution to remove the Bi(OH)3 precipitate, and the remaining solution will contain \(\mathrm{Pb}^{2+}\) ions. #Group c: Pb2+, Ca2+, Fe2+#

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

To selectively precipitate \(\mathrm{Pb}^{2+}\) ions, add a sulfate salt, such as sodium sulfate (Na2SO4). The \(\mathrm{Pb}^{2+}\) ions will react with the sulfate ions to form lead sulfate (PbSO4), an insoluble compound. The reaction can be represented as: \[\mathrm{Pb}^{2+} + \mathrm{SO}_4^{2-} \rightarrow \mathrm{PbSO}_4(s)\] Filter the solution to remove the PbSO4 precipitate, and the remaining solution will contain \(\mathrm{Ca}^{2+}\) and \(\mathrm{Fe}^{2+}\) ions.

Precipitate \(\mathrm{Fe}^{2+}\) ions

To selectively precipitate \(\mathrm{Fe}^{2+}\) ions, add sodium hydroxide (NaOH) solution to the mixture. The \(\mathrm{Fe}^{2+}\) ions will react with the hydroxide ions to form iron(II) hydroxide (Fe(OH)2), an insoluble compound. The reaction can be represented as: \[\mathrm{Fe}^{2+} + \mathrm{2OH}^{-} \rightarrow \mathrm{Fe(OH)}_2(s)\] Filter the solution to remove the Fe(OH)2 precipitate, and the remaining solution will contain \(\mathrm{Ca}^{2+}\) ions.