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

Expert verified
To separate the ions in each group by selective precipitation, we can follow these steps: a. For \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\): 1. Precipitate \(\mathrm{Ag}^{+}\) by adding a chloride source, such as \(\mathrm{NaCl}\), forming insoluble \(\mathrm{AgCl}\). 2. Precipitate \(\mathrm{Cu}^{2+}\) by adding a hydroxide source, such as \(\mathrm{NaOH}\), forming insoluble \(\mathrm{Cu(OH)_2}\). b. For \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\): 1. Precipitate \(\mathrm{Pb}^{2+}\) by adding a sulfate source, like \(\mathrm{Na_2SO_4}\), forming insoluble \(\mathrm{PbSO_4}\). 2. Precipitate \(\mathrm{Fe}^{2+}\) by adding a hydroxide source, such as \(\mathrm{NaOH}\), forming insoluble \(\mathrm{Fe(OH)_2}\). c. For \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\): 1. Precipitate \(\mathrm{Bi}^{3+}\) by adding a hydroxide source, like \(\mathrm{NaOH}\), forming insoluble \(\mathrm{Bi(OH)_3}\).

Step by step solution

01

1. Precipitation of \(\mathrm{Ag}^{+}\) ions

To precipitate the \(\mathrm{Ag}^{+}\) ions, we can add a source of chloride ion (\(\mathrm{Cl}^-\)), such as \(\mathrm{NaCl}\), since silver chloride \(\mathrm{(AgCl)}\) is insoluble in water (Ksp_{\(\mathrm{AgCl}\)} \(= 1.8 \times 10^{-10}\)). The resulting equation is: \[\mathrm{Ag}^{+}(aq) + \mathrm{Cl}^-(aq) \rightarrow \mathrm{AgCl}(s)\]
02

2. Precipitation of \(\mathrm{Cu}^{2+}\) ions

Next, to precipitate the \(\mathrm{Cu}^{2+}\) ions remaining in solution, we can add a source of hydroxide ions (\(\mathrm{OH}^-\)), such as \(\mathrm{NaOH}\), since copper(II) hydroxide \(\mathrm{(Cu(OH)_2)}\) is insoluble in water (Ksp_{\(\mathrm{Cu(OH)_2}\)} \(= 2.2 \times 10^{-20}\)). The \(\mathrm{Mg}^{2+}\) ions will remain in solution because magnesium hydroxide \(\mathrm{(Mg(OH)_2)}\) has a higher solubility (Ksp_{\(\mathrm{Mg(OH)_2}\)} \(= 5.6 \times 10^{-12}\)). The resulting equation is: \[\mathrm{Cu}^{2+}(aq) + 2\mathrm{OH}^-(aq) \rightarrow \mathrm{Cu(OH)_2}(s)\] b. Separating \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\) ions:
03

1. Precipitation of \(\mathrm{Pb}^{2+}\) ions

To precipitate the \(\mathrm{Pb}^{2+}\) ions, we can add a source of sulfate ion (\(\mathrm{SO_4}^{2-}\)), such as \(\mathrm{Na_2SO_4}\), since lead(II) sulfate \(\mathrm{(PbSO_4)}\) is insoluble in water (Ksp_{\(\mathrm{PbSO_4}\)} \(= 2.0 \times 10^{-8}\)). The \(\mathrm{Ca}^{2+}\) and \(\mathrm{Fe}^{2+}\) ions will remain in solution due to the higher solubility of their sulfates. The resulting equation is: \[\mathrm{Pb}^{2+}(aq) + \mathrm{SO_4}^{2-}(aq) \rightarrow \mathrm{PbSO_4}(s)\]
04

2. Precipitation of \(\mathrm{Fe}^{2+}\) ions

Next, to precipitate the \(\mathrm{Fe}^{2+}\) ions remaining in solution, we can add a source of hydroxide ions (\(\mathrm{OH}^-\)), such as \(\mathrm{NaOH}\). The \(\mathrm{Ca}^{2+}\) ions will remain in solution because calcium hydroxide \(\mathrm{(Ca(OH)_2)}\) has a higher solubility (Ksp_{\(\mathrm{Ca(OH)_2}\)} \(= 6.5 \times 10^{-6}\)). The resulting equation is: \[\mathrm{Fe}^{2+}(aq) + 2\mathrm{OH}^-(aq) \rightarrow \mathrm{Fe(OH)_2}(s)\] c. Separating \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\) ions:
05

1. Precipitation of \(\mathrm{Bi}^{3+}\) ions

To precipitate the \(\mathrm{Bi}^{3+}\) ions, we can add a source of hydroxide ions (\(\mathrm{OH}^-\)), such as \(\mathrm{NaOH}\), since bismuth(III) hydroxide \(\mathrm{(Bi(OH)_3)}\) is insoluble in water (Ksp_{\(\mathrm{Bi(OH)_3}\)} \(= 5.0 \times 10^{-25}\)). The \(\mathrm{Pb}^{2+}\) ions will remain in the solution as lead(II) hydroxide \(\mathrm{(Pb(OH)_2)}\) has a higher solubility (Ksp_{\(\mathrm{Pb(OH)_2}\)} \(= 4.0 \times 10^{-15}\)). The resulting equation is: \[\mathrm{Bi}^{3+}(aq) + 3\mathrm{OH}^-(aq) \rightarrow \mathrm{Bi(OH)_3}(s)\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant (Ksp)
The solubility product constant, commonly denoted as Ksp, is a useful quantitative measure of the solubility of slightly soluble ionic compounds. It is defined as the product of the concentrations of the ions involved in the solid, each raised to the power of its coefficient in the balanced equation, when the compound is in equilibrium with its ions in solution.

Understanding Ksp is crucial when dealing with precipitation reactions, as it allows a chemist to predict whether a solid will precipitate under certain conditions. The lower the Ksp value, the less soluble the compound is. For instance, in the precipitation of silver chloride (AgCl), the Ksp is exceedingly low (\(1.8 \times 10^{-10}\)), indicating that upon mixing with a chloride source, Ag+ ions are likely to form a precipitate.

In the case of copper(II) hydroxide, the Ksp (\(2.2 \times 10^{-20}\)) is even lower, making it very insoluble and therefore a prime candidate for precipitation to separate Cu2+ from other ions in the mixture.
Precipitation Reactions
Precipitation reactions involve the formation of an insoluble solid called a precipitate from the reaction of two soluble compounds in solution. This process is employed in the technique of selective precipitation, which is based on the differences in the solubility of ionic compounds.

By adding a specific reagent that will react with one or more of the ions but not with others, one can systematically separate ions in a mixture. For instance, when Pb2+ ions are mixed with sulfate ions, lead(II) sulfate (PbSO4), which has a low solubility product constant, precipitates out of the solution. This selective formation of precipitates is widely used in various applications, including water treatment, mineral extraction, and analytical chemistry.
Ionic Separation
Ionic separation is a critical process in which different ions in a solution are separated based on their individual solubility. Selective precipitation, aided by the understanding of Ksp values, is one of the primary methods for achieving this separation.

By carefully choosing the right precipitating agent and controlling the concentration of ions, it is possible to selectively remove certain ions while leaving others in solution. For example, by adjusting the pH level in the solution through the addition of hydroxide ions (OH-), we can selectively precipitate out ions such as Cu2+ or Bi3+, since their respective hydroxides have very low solubility. This practical approach helps in purifying compounds, removing impurities, and preparing substances for further analysis or application.

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Most popular questions from this chapter

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(7.2 \times 10^{-2}-M \mathrm{KIO}_{3}\) solution is \(6.0 \times 10^{-9} \mathrm{~mol} / \mathrm{L}\). Calculate the \(K_{\text {sp }}\) value for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\).

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}\) b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of EDTA \(^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is \(\begin{aligned} \mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) & \\ K &=1.1 \times 10^{18} \end{aligned}\) Consider a solution with \(0.010\) mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 L of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing \(0.050 M\) NasEDTA. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

The active ingredient of Pepto-Bismol is the compound bismuth subsalicylate, which undergoes the following dissociation when added to water: \(\begin{aligned} \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}(a q) \\\ &+\mathrm{Bi}^{3+}(a q)+\mathrm{OH}^{-}(a q) \quad K=? \end{aligned}\) If the maximum amount of bismuth subsalicylate that reacts by this reaction is \(3.2 \times 10^{-19} \mathrm{~mol} / \mathrm{L}\), calculate the equilibrium constant for the preceding reaction.

A solution contains \(0.018\) mole each of \(\mathrm{I}^{-}, \mathrm{Br}^{-}\), and \(\mathrm{Cl}^{-}\). When the solution is mixed with 200. \(\mathrm{mL}\) of \(0.24 \mathrm{M} \mathrm{AgNO}_{3}\) what mass of \(\mathrm{AgCl}(s)\) precipitates out, and what is \(\left[\mathrm{Ag}^{+}\right] ?\) Assume no volume change. $$ \begin{aligned} \text { AgI: } K_{\text {sp }} &=1.5 \times 10^{-16} \\ \mathrm{AgBr:} K_{\text {sp }} &=5.0 \times 10^{-13} \\ \mathrm{AgCl}: K_{\text {sp }} &=1.6 \times 10^{-10} \end{aligned} $$

Will a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) form if \(1.0 \mathrm{~mL}\) of \(1.0 \mathrm{M}\) \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(1.0 \mathrm{~L}\) of \(5.0 \mathrm{M} \mathrm{NH}_{3}\) ? \(\mathrm{Cd}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q)\) \(K=1.0 \times 10^{7}\) \(\mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\) \(K_{\mathrm{sp}}=5.9 \times 10^{-15}\)

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