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Chapter 17: Problem 79

In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\), (b) \(\mathrm{Cr}(\mathrm{OH})_{2}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{4}\), (d) \(\mathrm{Ag}^{*}\) and \(\mathrm{Mn}^{2+}\), Suggest how each mixture might be separated.

Short Answer

Expert verified
To separate the given mixtures, use the following methods: a) Add excess H鈧係 to the acidic solution containing Zn虏鈦� and Cd虏鈦� ions, which will precipitate CdS. Filter the mixture to obtain separated ions. b) Add an acidic solution to the mixture of Cr(OH)鈧� and Fe(OH)鈧�. Cr(OH)鈧� will dissolve, forming Cr鲁鈦� ions, while Fe(OH)鈧� remains undissolved. Filter the mixture to obtain separated components. c) Add sodium oxalate (Na鈧侰鈧侽鈧�) to the solution containing Mg虏鈦� and K鈦� ions. The insoluble magnesium oxalate (MgC鈧侽鈧�) will precipitate, while potassium oxalate remains soluble. Filter the mixture to separate the ions. d) Add a solution of NaCl to the mixture of Ag鈦� and Mn虏鈦� ions. The Ag鈦� ions will form a white precipitate of AgCl, while Mn虏鈦� ions remain in solution as soluble MnCl鈧�. Filter the mixture to obtain separated ions.

Step by step solution

01

Mixture (a): Zn2+ and Cd2+ ions

To separate Zn^2+ and Cd^2+ ions, we will use the difference in their solubilities in H鈧係 solutions. When excess H鈧係 is added to the acidic solution containing both ions, Zn ions do not precipitate as ZnS, but Cd^2+ ions will precipitate as CdS thanks to its lower solubility. Thus, we can separate the two ions by filtration.
02

Mixture (b): Cr(OH)2 and Fe(OH)3

To separate Cr(OH)鈧� and Fe(OH)鈧�, we can use the difference in their solubilities in acidic solutions. When an acidic solution is added to the mixture, Cr(OH)鈧� will dissolve in acid, forming Cr鲁鈦� ions, while Fe(OH)鈧� remains undissolved due to its lower solubility in acidic solutions. We can then filter the mixture to separate insoluble Fe(OH)3 from the Cr鲁鈦� ions in the solution.
03

Mixture (c): Mg2+ and K4 ions

To separate Mg虏鈦� and K鈦� ions in the mixture, we can take advantage of the difference in solubilities of their respective salts. If we add sodium oxalate (Na鈧侰鈧侽鈧�) to the solution, the virtually insoluble magnesium oxalate (MgC鈧侽鈧�) will form a white precipitate, while the potassium oxalate remains soluble. We can separate the precipitated MgC鈧侽鈧� from the potassium ions in the solution by filtration.
04

Mixture (d): Ag+ and Mn2+ ions

To separate Ag鈦� and Mn虏鈦� ions, we can utilize the difference in solubilities of their chlorides. When a solution of NaCl is added to the mixture, the Ag鈦� ions will form a white precipitate of AgCl since it is insoluble in water, while Mn虏鈦� ions will remain in solution as soluble MnCl鈧�. We can separate the precipitated AgCl from 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 of Ions
The separation of ions in a mixture is a fundamental process in qualitative analytical chemistry. It's used to identify and quantify the components within a sample. To achieve separation, chemists often exploit the unique characteristics of each ion, such as their charge, size, or their specific chemical reactivity.

For instance, when faced with a mixture of zinc and cadmium ions, a selective precipitation method can be employed. By adding hydrogen sulfide to an acidic solution containing these ions, only cadmium ions react to form an insoluble compound, cadmium sulfide (CdS), while zinc ions remain in solution. After this, a simple filtration can separate the solid precipitate from the liquid phase, effectively isolating one ion from the other.
Solubility Differences
Ions and compounds have differing solubilities in various solvents, which is a key principle used in separation techniques. Solubility differences are particularly useful in processes such as dissolving, precipitating, and selectively crystallizing substances.

Depending on whether a substance is more soluble in an acidic or basic environment, chemists can adjust the pH of a solution to encourage the dissolution of one component while another remains unaltered. For instance, chromium hydroxide dissolves in acidic solutions, transforming into soluble chromium ions, while iron hydroxide does not, allowing the two to be easily separated by filtration.
Precipitation Reactions
Precipitation reactions are at the heart of many separation techniques. These reactions occur when two soluble salts react in a solution to form an insoluble solid, known as a precipitate. The precipitate can then be removed by filtration.

By adding a reagent that specifically reacts with one ion in a solution, chemists can form a solid precipitate. This is seen with magnesium and potassium ions 鈥� when sodium oxalate is added to their mixture, only magnesium forms an insoluble oxalate, which precipitates out and can be filtered, while potassium remains in solution.
Filtration Technique
Filtration is a mechanical or physical operation used to separate solids from fluids (liquids or gases) by interposing a medium through which only the fluid can pass. The solid particles are trapped by the filter medium, also known as the filter.

In our previous examples, filtration is the final step after a precipitate has been formed. For instance, after forming a precipitate of silver chloride in the presence of sodium chloride, the insoluble AgCl can be separated from the remaining solution containing soluble MnCl鈧� through the use of filtration. This simple technique is instrumental in a wide range of chemical analyses and synthesis applications.

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Most popular questions from this chapter

How many microliters of \(1.000 \mathrm{M} \mathrm{NaOH}\) solution must be added to \(25.00 \mathrm{~mL}\) of a \(0.1000 \mathrm{M}\) solution of lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to produce a buffer with \(\mathrm{pH}=3.75 ?\)

An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered off, the \(\mathrm{pH}\) is adjusted to about 1 and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled in; a precipitate again forms. After filtering off this precipitate, the pH is adjusted to 8 and \(\mathrm{H}_{2} \mathrm{~S}\) is again added; no precipitate forms. No precipitate forms upon addition of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). (See Figure 7.23.) The remaining solution shows a yellow color in a flame test (see Figure 7.22). Based on these observations, which of the following compounds might be present, which are definitely present, and which are definitely absent: CdS, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{HgO}_{3}, \mathrm{ZnSO}_{4}, \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\), and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) ?

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the pH of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) is adjusted to \(8.0 ?\) (b) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when \(100 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{AgNO}{ }_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(5.0 \times 10^{-2} \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution?

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (e) \(\mathrm{CuBr}\).

Furoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of (a) a solution formed by adding \(25.0 \mathrm{~g}\) of furoic acid and \(30.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.250 \mathrm{~L}\) of solution, (b) a solution formed by mixing \(30.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and \(20.0 \mathrm{~mL}\). of \(0.22 \mathrm{M} \mathrm{NaC} \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\) and diluting the total volume to \(125 \mathrm{~mL}\). (c) a solution prepared by adding \(50.0 \mathrm{~mL}\) of \(1.65 \mathrm{M} \mathrm{NaOH}\) solution to \(0.500 \mathrm{~L}\) of \(0.0850 \mathrm{M} \mathrm{HC} \mathrm{H}_{3} \mathrm{O}_{3}\).
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