Problem 90 - Chemistry An Atoms First Approach

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. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\)

Step-by-Step Solution

Answer

To separate the ions by selective precipitation, follow these steps for each group: a. Add NaCl to precipitate Ag+ as AgCl, then filter. Add NaOH to precipitate Cu²⁺ as Cu(OH)₂, then filter to get Mg²⁺. b. Add KI to precipitate Pb²⁺ as PbI₂, then filter. Add NaOH to precipitate Fe²⁺ as Fe(OH)₂, then filter to get Ca²⁺. c. Add NaOH to precipitate Bi³⁺ as Bi(OH)₃, then filter to get Pb²⁺.

Step 1: Precipitating \(\mathrm{Ag}^{+}\) ions

Add sodium chloride (NaCl) solution to the mixture. This will form insoluble silver chloride (AgCl) as per the following reaction: \(\mathrm{Ag}^{+} + \mathrm{Cl}^{-} \rightarrow \mathrm{AgCl}\). The AgCI precipitate can be filtered out to separate the silver ions from the other two ions.

Step 2: Precipitating \(\mathrm{Cu}^{2+}\) ions

To the remaining solution, add a sodium hydroxide (NaOH) solution. This will precipitate \(\mathrm{Cu}^{2+}\) ions as insoluble copper hydroxide (Cu(OH)₂) according to the reaction: \(\mathrm{Cu}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Cu(OH)}_{2}\). Filter the precipitate to separate copper ions from magnesium ions. b. \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\)

Step 1: Precipitating \(\mathrm{Pb}^{2+}\) ions

Add potassium iodide (KI) to the solution to form insoluble lead iodide (PbI₂): \(\mathrm{Pb}^{2+} + 2\mathrm{I}^{-} \rightarrow \mathrm{PbI}_{2}\). Filter out the precipitate to separate lead ions from the remaining ions.

Step 2: Precipitating \(\mathrm{Fe}^{2+}\) ions

To the remaining solution, add sodium hydroxide (NaOH) solution. This will precipitate \(\mathrm{Fe}^{2+}\) ions as insoluble iron hydroxide (Fe(OH)2): \(\mathrm{Fe}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Fe(OH)}_{2}\). Filter the precipitate to separate iron ions from calcium ions. c. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\)

Step 1: Precipitating \(\mathrm{Bi}^{3+}\) ions

Add sodium hydroxide (NaOH) solution to the mixture. Only \(\mathrm{Bi}^{3+}\) ions will precipitate as insoluble bismuth hydroxide (Bi(OH)₃): \(\mathrm{Bi}^{3+} + 3\mathrm{OH}^{-} \rightarrow \mathrm{Bi(OH)}_{3}\). Filter the precipitate to separate bismuth ions from lead ions.

Key Concepts

Precipitation Reactions

Understanding precipitation reactions is essential for mastering selective precipitation, a method used for separating ions based on their solubility differences. These reactions occur when solutions of two ionic compounds are mixed and an insoluble solid, known as a precipitate, forms. This solid can then be separated from the mixture through filtration.

For example, when a solution containing silver ions \( \mathrm{Ag}^{+} \) is mixed with sodium chloride, a white precipitate of silver chloride \( \mathrm{AgCl} \) forms. This is represented by the chemical equation: \( \mathrm{Ag}^{+} + \mathrm{Cl}^{-} \rightarrow \mathrm{AgCl} \). The precipitate can be removed by filtration, effectively separating the silver ions from the solution.

Solubility Rules

To predict and control precipitation reactions, knowledge of solubility rules is crucial. These rules are guidelines that help determine whether an ionic compound is likely to dissolve in water (soluble) or form a precipitate (insoluble). They state that certain ions form soluble compounds except with specific ions. For instance:

• Nitrates \( \mathrm{NO}_3^- \) are generally soluble.
• Silver \( \mathrm{Ag}^{+} \) is typically insoluble except with nitrates or perchlorates.
• Hydroxides \( \mathrm{OH}^{-} \) are usually insoluble, which is why adding sodium hydroxide to a solution containing copper ions \( \mathrm{Cu}^{2+} \) results in the formation of insoluble copper hydroxide \( \mathrm{Cu(OH)}_2 \).

When dealing with multiple ions, solubility rules make it easier to devise a strategy for separating them using selective precipitation—by adding the right reagent to precipitate specific ions while keeping others in solution.

Separation of Ions

Selective precipitation is a technique widely used in chemistry to separate a mixture of ions. By adding specific reagents that react with certain ions to form insoluble compounds, chemists can sequentially remove ions from a mixture.

Consider, for instance, a mixture containing lead \( \mathrm{Pb}^{2+} \) and bismuth \( \mathrm{Bi}^{3+} \) ions. To separate them, sodium hydroxide \( \mathrm{NaOH} \) is added. While both ions could potentially react with hydroxide, solubility rules indicate that bismuth hydroxide \( \mathrm{Bi(OH)}_3 \) is less soluble than lead hydroxide. Therefore, only bismuth hydroxide precipitates, and when it is filtered out, the remaining solution will predominantly contain lead ions. Through careful control of conditions and understanding ion reactivity and solubility, chemists can systematically isolate different ions from complex mixtures.