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In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{+}\), (d) \(\mathrm{Ag}^{+}\) and \(\mathrm{Mn}^{2+} .\) Suggest how each mixture might be separated.

Short Answer

Expert verified
To separate the given mixtures, the following techniques can be used: (a) Add hydrogen sulfide (H鈧係) to the solution containing Zn虏鈦 and Cd虏鈦 ions. This causes ZnS to precipitate while CdS stays in solution. (b) Dissolve both Cr(OH)鈧 and Fe(OH)鈧 in an acidic solution, then add excess NaOH to the mixture. Cr(OH)鈧 will re-precipitate and can be separated by filtration. (c) Add NH鈧凜l to the mixture containing Mg虏鈦 and K鈦 ions. Mg虏鈦 ions will form a complex with NH鈧凜l and precipitate as magnesium chloride (MgCl鈧.6H鈧侽), while K鈦 ions will stay in the solution. (d) Add an excess of chloride ions (Cl鈦) to the solution containing Ag鈦 and Mn虏鈦 ions. Ag鈦 ions will precipitate as silver chloride (AgCl), while Mn虏鈦 ions remain in the solution.

Step by step solution

01

a) Separation of Zn虏鈦 and Cd虏鈦 ions

To separate these two ions, we could add a sulfide source, such as hydrogen sulfide (H鈧係) gas, to the solution containing Zn虏鈦 and Cd虏鈦. Due to the difference in solubility of their sulfides, ZnS will precipitate out while CdS will stay in solution: \(6 \,\mathrm{Zn}^{2+}(aq) + 3 \,\mathrm{H}_{2}\mathrm{S}(g) \rightarrow 2 \,\mathrm{Zn}_{3}\mathrm{S}_{2}(s) + 6 \mathrm{H}^{+}\, (aq)\) After precipitating ZnS, filtration can be used to separate the solid from the remaining Cd虏鈦 ions in the solution.
02

b) Separation of Cr(OH)鈧 and Fe(OH)鈧

Both Cr(OH)鈧 and Fe(OH)鈧 are insoluble in water and have a similar appearance. However, a difference in their chemical properties can be utilized for their separation. We can start by adding an acidic solution to dissolve both hydroxides, forming Cr鲁鈦 and Fe鲁鈦 ions. Next, we can add an excess of a strong, alkaline solution such as NaOH; only Cr(OH)鈧 will re-precipitate due to its lower solubility product than Fe(OH)鈧: \(\mathrm{Cr}^{3+}(aq) + 3 \text{OH}^-(aq) \rightarrow \mathrm{Cr(OH)}_{3}(s)\) Now, we can separate the Cr(OH)鈧 precipitate from the solution, which still contains the Fe鲁鈦 ions, by filtration.
03

c) Separation of Mg虏鈦 and K鈦 ions

Mg虏鈦 and K鈦 ions can be separated based on the difference in their solubility in ammonium chloride (NH鈧凜l) solution. We can add NH鈧凜l to the mixture containing Mg虏鈦 and K鈦 ions. The Mg虏鈦 ions will form a complex with NH鈧凜l and precipitate as magnesium chloride (MgCl鈧.6H鈧侽), while K鈦 ions will stay in the solution: \(\mathrm{Mg}^{2+}(aq) + 2\, \mathrm{Cl}^{-}(aq) + 6\, \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{MgCl}_{2}\cdot 6\, \mathrm{H}_{2}\mathrm{O}(s)\) We can then use filtration to separate the solid magnesium chloride from the K鈦 ions that remain in the solution.
04

d) Separation of Ag鈦 and Mn虏鈦 ions

To separate the Ag鈦 and Mn虏鈦 ions, we can use a selective precipitation method by adding an excess of chloride ions (Cl鈦). Due to the very low solubility product of AgCl, only Ag鈦 ions will precipitate, forming silver chloride (AgCl), while Mn虏鈦 ions remain in the solution: \(\mathrm{Ag}^+ (aq) + \mathrm{Cl}^- (aq) \rightarrow \mathrm{AgCl}(s)\) We can then separate the solid AgCl from the remaining Mn虏鈦 ions in the solution using filtration.

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Key Concepts

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

Separation Techniques
Qualitative analysis often involves separating ions from a mixture, which is where separation techniques come into play. These methods are crucial in determining the composition of a mixture by selectively isolating its components. Common techniques include filtration, which is used to separate a precipitated solid from a solution, and precipitation, where we induce a solid鈥檚 formation by adding a reagent that causes certain ions to become insoluble. Separation strategies often rely on differences in solubility, chemical reactivity, or physical properties of the components. For instance, in our exercise, using hydrogen sulfide gas allows us to precipitate zinc sulfide while cadmium sulfide remains in solution, showcasing a technique dependent on differential solubilities.
Precipitation Reactions
Precipitation reactions are a cornerstone of qualitative analysis, designed to create an insoluble product from a solution that drives the separation process. The general principle is to add a reagent that reacts with specific ions to form a solid, allowing it to be filtered out. In our examples, hydrogen sulfide gas is used to react with zinc ions to form zinc sulfide, a solid, while cadmium ions remain in solution. This method is sensitive to the concentrations of the reactants and the nature of the precipitating agent, playing a critical role in isolating particular ions from mixtures.
Ionic Solubility
Ionic solubility refers to the extent to which ions dissolve in a solvent, usually water. It is an essential concept when deciding if a precipitation reaction will occur. In practical scenarios, like our exercise, solubility differences make it possible to selectively precipitate ions. For instance, when adding chloride ions to a mixture of silver and manganese ions, only silver ions form an insoluble compound, silver chloride, due to its low solubility compared to manganese compounds. Understanding solubility guides the choice of reagents in qualitative analysis, enabling effective separation of ions based on solubility discrepancies.
Solubility Product
The solubility product, denoted as K ext{sp}, is a numerical value representing a compound's solubility in water. It is determined by the product of the concentrations of its constituent ions at saturation. In separation processes, the K ext{sp} helps predict whether a salt will precipitate under given conditions. For example, knowing that silver chloride has a very low K ext{sp}, we can anticipate its precipitation in the presence of chloride ions. Manipulating conditions like ion concentration can shift the equilibrium, influencing precipitation and aiding in the separation of closely related ions, as seen when distinguishing between chromium and iron hydroxides based on their differing K ext{sp} values.
Chemical Properties of Ions
The chemical properties of ions, including their reactivity and preferred compounds, substantially influence separation techniques. For instance, chromium and iron hydroxides both dissolve in acidic solutions, forming Cr鲁鈦 and Fe鲁鈦 ions. However, when an alkaline solution is added, only chromium hydroxide re-precipitates due to its chemical properties, specifically its solubility in basic conditions compared to iron hydroxide. This selectivity is pivotal, allowing chemists to determine suitable conditions for selective precipitation. Understanding the unique chemical behaviors of ions ensures effective design of qualitative analysis protocols and accurate identification of components within a complex mixture.

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Most popular questions from this chapter

A solution contains \(1.0 \times 10^{-4} \mathrm{Ca}^{2+}(a q)\) and \(1.0 \times 10^{-4}\) \(\mathrm{La}^{3+}(a q) .\) If \(\mathrm{NaF}\) is added, will \(\mathrm{CaF}_{2}\left(K_{s p}=3.9 \times 10^{-11}\right)\) or \(\mathrm{LaF}_{3}\left(K_{s p}=2 \times 10^{-19}\right)\) precipitate first? Specify the concentration of \(\mathrm{F}^{-}(a q)\) needed to begin precipitation.

Which of the following salts will be substantially more soluble in an \(\mathrm{HNO}_{3}\) solution than in pure water: (a) \(\mathrm{BaSO}_{4}\), (b) \(\mathrm{CuS},\) (c) \(\mathrm{Cd}(\mathrm{OH})_{2}\) (d) \(\mathrm{PbF}_{2}\), (e) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ?\)

(a) Will \(\mathrm{Co}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.020 \mathrm{M}\) solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) is adjusted to \(8.5 ?(\mathbf{b})\) Will \(\mathrm{AgIO}_{3}\) precipitate when \(20 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{AgIO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{NaIO}_{3} ?\left(K_{s p}\right.\) of \(\mathrm{AgIO}_{3}\) is \(3.1 \times 10^{-8} .\) )

Excess \(\mathrm{Ca}(\mathrm{OH})_{2}\) is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 -mL sample titrated with \(\mathrm{HCl}\) requires \(11.23 \mathrm{~mL}\) of \(0.0983 \mathrm{MHCl}\) to reach the end point. Calculate \(K_{s p}\) for \(\mathrm{Ca}(\mathrm{OH})_{2} .\) Compare your result with that in Appendix D. Suggest a reason for any differences you find between your value and the one in Appendix D.

What is the \(\mathrm{pH}\) of a solution made by mixing \(0.40 \mathrm{~mol}\) \(\mathrm{NaOH}, 0.25 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{HPO}_{4}\), and \(0.30 \mathrm{~mol} \mathrm{H}_{3} \mathrm{PO}_{4}\) with water and diluting to \(2.00 \mathrm{~L} ?\)

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