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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, follow these methods: (a) Add a small amount of Na鈧係 to separate Zn虏鈦 and Cd虏鈦 ions, and filter the mixture to isolate the CdS precipitate. (b) Add a strong acid like HCl to dissolve Fe(OH)鈧 and filter the Cr(OH)鈧. (c) Add ammonium oxalate to form an insoluble precipitate of MgC鈧侽鈧 and filter to separate K鈦 ions. (d) Add a chloride source, such as NaCl, to separate Ag鈦 and Mn虏鈦 ions, and filter the mixture to isolate the AgCl precipitate.

Step by step solution

01

Separating Zn虏鈦 and Cd虏鈦 ions

To separate Zn虏鈦 and Cd虏鈦 ions, you can take advantage of their different solubilities with different anions. One common way to do this is by using sodium sulfide (Na鈧係), which forms insoluble sulfides with both ions but with different solubility products: Zn虏鈦 + S虏鈦 鈫 ZnS (s) Cd虏鈦 + S虏鈦 鈫 CdS (s) The solubility products of the two sulfides are different: Ksp (ZnS) = \(6.3\times10^{-18}\) Ksp (CdS) = \(5.2\times10^{-27}\) To separate the two ions, first, add a small amount of Na鈧係 to the mixture. Since CdS has a much lower solubility product, Cd虏鈦 ions will precipitate as CdS, while most Zn虏鈦 ions will remain in the solution. Filter the mixture to isolate the CdS precipitate.
02

Separating Cr(OH)鈧 and Fe(OH)鈧 compounds

The mixture of Cr(OH)鈧 and Fe(OH)鈧 can be separated by taking advantage of their different solubilities in acidic or basic solutions. When a strong acid is added to this mixture, Fe(OH)鈧 will readily dissolve, while Cr(OH)鈧 will remain insoluble: Fe(OH)鈧 + 3H鈦 鈫 Fe鲁鈦 + 3H鈧侽 Add a strong acid, such as HCl, to the mixture, and then filter the solution to isolate the insoluble Cr(OH)鈧. The Fe鲁鈦 ions will remain in the acidic solution.
03

Separating Mg虏鈦 and K鈦 ions

To separate Mg虏鈦 and K鈦 ions, you can take advantage of the fact that Mg虏鈦 forms an insoluble precipitate with ammonium oxalate ((NH鈧)鈧侰鈧侽鈧), while K鈦 does not: Mg虏鈦 + C鈧侽鈧劼测伝 鈫 MgC鈧侽鈧 (s) Add ammonium oxalate to the solution containing Mg虏鈦 and K鈦 ions. The Mg虏鈦 ions will form an insoluble precipitate of MgC鈧侽鈧, while K鈦 ions will remain in the solution. Filter the mixture to separate the MgC鈧侽鈧 precipitate from the solution containing K鈦 ions.
04

Separating Ag鈦 and Mn虏鈦 ions

To separate Ag鈦 and Mn虏鈦 ions, you can take advantage of the fact that Ag鈦 forms an insoluble precipitate with chloride ions (Cl鈦), while Mn虏鈦 ions do not: Ag鈦 + Cl鈦 鈫 AgCl (s) Add a source of chloride ions, such as NaCl, to the solution containing Ag鈦 and Mn虏鈦 ions. The Ag鈦 ions will form an insoluble precipitate of AgCl, while Mn虏鈦 ions will remain in the solution. Filter the mixture to separate the AgCl precipitate from the solution containing Mn虏鈦 ions.

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Key Concepts

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

Separation of Ions
Understanding the separation of ions plays a pivotal role in qualitative analysis chemistry. It lays the groundwork for identifying the composition of unknown samples in a mixture. Specifically, the ability to separate ions depends on exploiting their differential reactions with reagents to produce distinct substances that can be isolated through physical means such as filtration.

For example, in a mixture of Zn2+ and Cd2+ ions, the selective precipitation of one ion over the other is performed. By introducing sodium sulfide to the solution, Cd2+ precipitates as CdS due to its significantly lower solubility product compared to ZnS. This difference in solubility allows for separation through filtration. The fundamental principle here is that each ion has unique chemical properties that when understood, can be manipulated to allow for separation from its counterparts in a mixture.
Solubility Products
The solubility product constant, or Ksp, is an essential parameter in the field of chemistry, especially when predicting the formation of a precipitate. Defined as the product of the ionic concentrations, each raised to the power of its stoichiometric coefficient, the Ksp provides insights into the solubility of compounds under equilibrium conditions. The lower the Ksp, the less soluble the compound is, and vice versa.

Interpreting Solubility Products

In the context of separating ions, consider ZnS and CdS. The Ksp value of ZnS is higher, implying that it is more soluble than CdS. When sodium sulfide is added to a solution containing both Zn2+ and Cd2+, CdS precipitates first because its lower Ksp indicates a lower solubility in water. This distinct solubility allows chemists to predict and control the separation of ions in a mixture. To determine the amount of ion that can be dissolved in a solution, the Ksp serves as a key guide. By calculating the ionic product and comparing it with the Ksp, one can predict whether a precipitate will form.
Precipitation Reactions
Precipitation reactions are central to the separation of ions and are based on the principle that certain ionic compounds have limited solubility in water and will form a solid, known as a precipitate, when the product of the ionic concentrations exceeds the solubility product. By introducing an appropriate reagent that forms an insoluble compound with a specific ion, a chemist can selectively separate that ion in the form of a precipitate.

Let's illustrate this with the separation of Ag+ and Mn2+ ions. Upon addition of a chloride source like NaCl, Ag+ quickly reacts to form AgCl, a precipitate, due to the low solubility of silver chloride. This reaction does not affect Mn2+ ions as they remain in solution. Utilizing this technique, a chemist can target specific ions to be removed from a solution, thus achieving the desired separation. Precipitation reactions not only serve to separate ions but also help in quantifying their concentration through gravimetric analysis.

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Most popular questions from this chapter

You have to prepare a \(\mathrm{pH} 4.80\) buffer, and you have the following \(0.10 \mathrm{M}\) solutions available: formic acid, sodium formate, propionic acid, sodium propionate, phosphoric acid, and sodium dihydrogen phosphate. Which solutions would you use? How many milliliters of each solution would you use to make approximately a liter of the buffer?

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

The solubility-product constant for barium permanganate, \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\), is \(2.5 \times 10^{-10}\). Assume that solid \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\) is in equilibrium with a solution of \(\mathrm{KMnO}_{4} .\) What concentration of \(\mathrm{KMnO}_{4}\) is required to establish a concentration of \(2.0 \times 10^{-8} M\) for the \(\mathrm{Ba}^{2+}\) ion in solution?

Show that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to \(\mathrm{pK}_{a}\) for the acid.

Predict whether the equivalence point of each of the following titrations is below, above, or at \(\mathrm{pH}\) ? (a) \(\mathrm{NaHCO}_{3}\) titrated with \(\mathrm{NaOH}\), (b) \(\mathrm{NH}_{3}\) titrated with \(\mathrm{HCl}\), (c) KOH titrated with \(\mathrm{HBr}\).

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