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How might you separate the following pairs of ions by the addition of a single reagent? Include formulas for the major products of the reactions. (a) \(\mathrm{Fe}^{3+}\) and \(\mathrm{Na}^{+}\) (b) \(\mathrm{Cr}^{3+}\) and \(\mathrm{Fe}^{3+}\) (c) \(\mathrm{Fe}^{3+}\) and \(\mathrm{Cu}^{2+}\)

Short Answer

Expert verified
(a) Precipitate \(\mathrm{Fe}^{3+}\) with \(\mathrm{OH}^-\); (b) Precipitate \(\mathrm{Cr}^{3+}\) with \(\mathrm{NH_3}\); (c) Precipitate \(\mathrm{Cu}^{2+}\) with \(\mathrm{H_2S}\).

Step by step solution

01

Identifying the Reagent for Pair (a)

To separate \(\mathrm{Fe}^{3+}\) from \(\mathrm{Na}^{+}\), we need a reagent that reacts with \(\mathrm{Fe}^{3+}\) but not with \(\mathrm{Na}^{+}\). Hydroxide ions (\(\mathrm{OH}^-\)) can precipitate \(\mathrm{Fe}^{3+}\) from solution as \(\mathrm{Fe(OH)_3}\), while \(\mathrm{Na}^+\) ions remain in solution as \(\mathrm{NaOH}\) is soluble. Major products: \(\mathrm{Fe(OH)_3}\).
02

Choosing the Precipitant for Pair (b)

To separate \(\mathrm{Cr}^{3+}\) from \(\mathrm{Fe}^{3+}\), use a reagent that selectively precipitates one ion. Adding \(\mathrm{NH_3}\) will cause \(\mathrm{Cr}^{3+}\) to form \(\mathrm{Cr(OH)_3}\) precipitate, while \(\mathrm{Fe}^{3+}\) can form complexes that remain in solution. Major products: \(\mathrm{Cr(OH)_3}\).
03

Separate the Ions in Pair (c)

For \(\mathrm{Fe}^{3+}\) and \(\mathrm{Cu}^{2+}\), adding a sulfide ion source, such as \(\mathrm{H_2S}\), will precipitate \(\mathrm{Cu}^{2+}\) as \(\mathrm{CuS}\), while \(\mathrm{Fe}^{3+}\) remains in solution under acidic conditions. Major products: \(\mathrm{CuS}\).

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Key Concepts

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

Chemical Precipitation
Chemical precipitation is a critical technique in separating ions from a solution. It involves adding a reagent that reacts with dissolved ions to form an insoluble solid, known as a precipitate. This is a useful method for purifying mixtures or for isolating specific components.
For example, when separating ions like \(\mathrm{Fe}^{3+}\) from \(\mathrm{Na}^{+}\), a suitable reagent such as hydroxide ions can be added. These ions interact with \(\mathrm{Fe}^{3+}\) to form an insoluble compound, \(\mathrm{Fe(OH)_3}\), which can be easily removed.
  • Effective for removing unwanted ions from solutions.
  • Forms a solid precipitate while leaving other ions in solution.
  • Commonly used in water treatment and mineral processing.
The choice of reagent is essential, as it determines which ions will precipitate and which will remain soluble. Chemical precipitation is not universal; it requires a thorough knowledge of the chemical properties of the involved ions.
Hydroxide Ion
Hydroxide ions (\(\mathrm{OH}^-\)) play a significant role in chemical processes, especially in precipitation reactions. They are negatively charged and can readily combine with metal ions to form insoluble compounds.
When \(\mathrm{OH}^-\) ions are added to a solution containing \(\mathrm{Fe}^{3+}\), they form \(\mathrm{Fe(OH)_3}\), an orange-brown solid. This precipitate can be separated from the solution, leaving \(\mathrm{Na}^+\) ions, which do not react with \(\mathrm{OH}^-\), in the solution.
  • Used to precipitate many metal ions as hydroxides.
  • Commonly available in solutions of bases like sodium hydroxide (\(\mathrm{NaOH}\)).
  • Key component in water purification and treatment processes.
The use of hydroxide ions is a powerful approach in separating ions due to its ability to selectively precipitate specific metals.
Ammonia Complexation
Ammonia complexation involves the addition of ammonia (\(\mathrm{NH}_3\)) to a solution to form complexes with metal ions. This technique is valuable for differentiating ions based on their ability to form soluble complexes.
In a solution containing \(\mathrm{Cr}^{3+}\) and \(\mathrm{Fe}^{3+}\), adding ammonia causes \(\mathrm{Cr}^{3+}\) to precipitate as \(\mathrm{Cr(OH)_3}\), while \(\mathrm{Fe}^{3+}\) forms a complex that remains dissolved.
  • Allows selective precipitation when multiple metallic ions are present.
  • Forms soluble complexes with ions that can be separated by their solubility.
  • Used in qualitative analysis and metal recovery processes.
This selective process is crucial for isolating specific ions and indicates the importance of understanding ion-complex interactions.
Sulfide Precipitation
Sulfide precipitation is an effective technique for separating metal ions based on their ability to form insoluble metal sulfides. This is often achieved by adding sulfide ions (from sources like hydrogen sulfide, \(\mathrm{H_2S}\), to the solution.
In the case of separating \(\mathrm{Fe}^{3+}\) from \(\mathrm{Cu}^{2+}\), sulfide ions can react with \(\mathrm{Cu}^{2+}\) to form copper sulfide (\(\mathrm{CuS}\)), a black precipitate. \(\mathrm{Fe}^{3+}\), on the other hand, remains in solution under acidic conditions.
  • Efficient at removing heavy metals from solutions.
  • Forms distinct, often colored, precipitates useful for identification.
  • Applied in environmental and industrial processes for decontamination.
Choosing the right conditions, such as pH, is critical in ensuring that only the desired metals are precipitated.

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Most popular questions from this chapter

Spinach contains a lot of iron but is not a good source of dietary iron because nearly all the iron is tied up in the oxalate complex \(\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\) (a) The formation constant \(K_{\mathrm{f}}\) for \(\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\) is \(3.3 \times 10^{20}\). Calculate the equilibrium concentration of free \(\mathrm{Fe}^{3+}\) in a \(0.100 \mathrm{M}\) solution of \(\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\). (Ignore any acid-base reactions.) (b) Under the acidic conditions in the stomach, the \(\mathrm{Fe}^{3+}\) concentration should be greater because of the reaction $$ \begin{aligned} \left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}(a q) &+6 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \\ \mathrm{Fe}^{3+}(a q)+3 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Show, however, that this reaction is nonspontaneous under standard-state conditions. (For \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}, K_{\mathrm{a} 1}=5.9 \times 10^{-2}\) and \(\left.K_{\mathrm{a} 2}=6.4 \times 10^{-5} .\right)\) (c) Draw a crystal field energy-level diagram for \(\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\), and predict the number of unpaired electrons. \(\left(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right.\) is a weak-field bidentate ligand.) (d) Draw the structure of \(\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\). Is the complex chiral or achiral?

What is the formula, including the charge, for each of the following complexes? (a) An iron(III) complex with six water ligands (b) A nickel(II) complex with two ethylenediamine and two bromide ligands (c) A platinum(II) complex with two chloride and two ammonia ligands

Which of the following compounds is amphoteric? (a) \(\mathrm{Cr}(\mathrm{OH})_{2}\) (b) \(\mathrm{Fe}(\mathrm{OH})_{2}\) (c) \(\mathrm{Cr}(\mathrm{OH})_{3}\) (d) \(\mathrm{Fe}(\mathrm{OH})_{3}\)

Which of the following metals have only one oxidation state? (a) \(\underline{V}\) (b) \(\mathrm{Al}\) (c) Co (d) Sc

Describe the bonding in \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\), using both crystal field theory and valence bond theory. Include the appropriate crystal field \(d\) -orbital energy-level diagram and the valence bond orbital diagram. Which model allows you to predict the number of unpaired electrons? How many do you expect?

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