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Chapter 7: Problem 76

Most sulfide compounds of the transition metals are insoluble in water. Many of these metal sulfides have striking and characteristic colors by which we can identify them. Therefore, in the analysis of mixtures of metal ions, it is very common to precipitate the metal ions by using dihydrogen sulfide (commonly called hydrogen sulfide), \(\mathrm{H}_{2} \mathrm{S}\). Suppose you had a mixture of \(\mathrm{Fe}^{2+}, \mathrm{Cr}^{3+},\) and \(\mathrm{Ni}^{2+} .\) Write net ionic equations for the precipitation of these metal ions by the use of \(\mathrm{H}_{2} \mathrm{S}\).

Short Answer

Expert verified
The net ionic equations for the precipitation of the given metal ions with hydrogen sulfide are: 1. For the precipitation of Fe虏鈦�: \[ Fe^{2+}(aq) + H_{2}S(g) \rightarrow FeS(s) + 2 H^{+}(aq) \] 2. For the precipitation of Cr鲁鈦�: \[ 2 Cr^{3+}(aq) + 3 H_{2}S(g) \rightarrow Cr_{2}S_{3}(s) + 6 H^{+}(aq) \] 3. For the precipitation of Ni虏鈦�: \[ Ni^{2+}(aq) + H_{2}S(g) \rightarrow NiS(s) + 2 H^{+}(aq) \]

Step by step solution

01

Precipitation Reaction of Fe虏鈦� with H鈧係

When Fe虏鈦� reacts with hydrogen sulfide (H鈧係), it results in the formation of precipitates of iron(II) sulfide (FeS). The balanced molecular equation for this reaction is: \[ Fe^{2+}(aq) + H_{2}S(g) \rightarrow FeS(s) + 2 H^+(aq) \] Now, we write the complete ionic equation and then cancel the common spectator ions to obtain the net ionic equation.
02

Net Ionic Equation for the Precipitation of Fe虏鈦� with H鈧係

The complete ionic equation for this reaction is: \[ Fe^{2+}(aq) + H_{2}S(g) \rightarrow FeS(s) + 2 H^{+}(aq) \] Since there are no spectator ions in this equation, the net ionic equation is the same as the complete ionic equation: \[ Fe^{2+}(aq) + H_{2}S(g) \rightarrow FeS(s) + 2 H^{+}(aq) \]
03

Precipitation Reaction of Cr鲁鈦� with H鈧係

When Cr鲁鈦� reacts with hydrogen sulfide (H鈧係), it results in the formation of precipitates of chromium(III) sulfide (Cr鈧係鈧�). The balanced molecular equation for this reaction is: \[ 2 Cr^{3+}(aq) + 3 H_{2}S(g) \rightarrow Cr_{2}S_{3}(s) + 6 H^{+}(aq) \] Now, we write the complete ionic equation and then cancel the common spectator ions to obtain the net ionic equation.
04

Net Ionic Equation for the Precipitation of Cr鲁鈦� with H鈧係

The complete ionic equation for this reaction is: \[ 2 Cr^{3+}(aq) + 3 H_{2}S(g) \rightarrow Cr_{2}S_{3}(s) + 6 H^{+}(aq) \] Since there are no spectator ions in this equation, the net ionic equation is the same as the complete ionic equation: \[ 2 Cr^{3+}(aq) + 3 H_{2}S(g) \rightarrow Cr_{2}S_{3}(s) + 6 H^{+}(aq) \]
05

Precipitation Reaction of Ni虏鈦� with H鈧係

When Ni虏鈦� reacts with hydrogen sulfide (H鈧係), it results in the formation of precipitates of nickel(II) sulfide (NiS). The balanced molecular equation for this reaction is: \[ Ni^{2+}(aq) + H_{2}S(g) \rightarrow NiS(s) + 2 H^{+}(aq) \] Now, we write the complete ionic equation and then cancel the common spectator ions to obtain the net ionic equation.
06

Net Ionic Equation for the Precipitation of Ni虏鈦� with H鈧係

The complete ionic equation for this reaction is: \[ Ni^{2+}(aq) + H_{2}S(g) \rightarrow NiS(s) + 2 H^{+}(aq) \] Since there are no spectator ions in this equation, the net ionic equation is the same as the complete ionic equation: \[ Ni^{2+}(aq) + H_{2}S(g) \rightarrow NiS(s) + 2 H^{+}(aq) \]

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Key Concepts

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

Transition Metal Sulfides
Transition metal sulfides are formed when transition metals react with sulfide ions. These compounds are typically insoluble in water and often appear in striking and distinctive colors. This makes them useful for identifying specific metal ions in chemical analysis. For example:
  • Iron(II) sulfide (FeS): formed when iron ions react with sulfide ions, usually appearing as a black precipitate.
  • Chromium(III) sulfide (Cr鈧係鈧�): created from chromium ions and sulfide, known for its dark green color.
  • Nickel(II) sulfide (NiS): composed of nickel ions and sulfide, typically forms a black or brown-black presence.
The precipitation of these sulfides occurs because the individual ions combine to create a solid that separates from the solution. The colorful nature of these precipitates can help in the visual detection and analysis of metal ions.
Precipitation Reactions
Precipitation reactions are a type of chemical reaction in which two soluble ions in solution react to form an insoluble solid called a precipitate. In the context of sulfide chemistry, this occurs when metal ions such as \[ Fe^{2+}, Cr^{3+}, \text{and} Ni^{2+} \] react with hydrogen sulfide (\[ H_{2}S \]).The process often follows these steps:
  • Mix solutions containing metal ions with \[ H_{2}S \] gas.
  • Sulfide ions (\[ S^{2-} \]) from \[ H_{2}S \] react with metal ions.
  • A solid precipitate forms, removing these ions from the solution.
  • The net ionic equations for these reactions do not include spectator ions, showing only the ions directly involved in forming the precipitate.
Using precipitation reactions, chemists can separate and identify specific ions based on their unique precipitates. This makes them especially valuable in analytical chemistry.
Hydrogen Sulfide Chemistry
Hydrogen sulfide (H鈧係) is a colorless gas with a distinct rotten egg smell. In aqueous solutions, it behaves as a weak acid and dissociates to release sulfide ions (\[ S^{2-} \]), which are critical in forming metal sulfide precipitates.Here are some key aspects:
  • Weak Acid Behavior: Hydrogen sulfide only partially dissociates in water, providing sulfide ions needed for precipitation reactions.
  • Role in Reactions: Acts as a reagent to supply \[ S^{2-} \] ions for reactions with transition metal ions.
  • Environmental Presence: Found naturally in environments such as volcanic gases and in the breakdown of organic matter, often posing a hazard due to its toxicity and flammability.
Despite its simplicity, hydrogen sulfide is crucial in analytical and industrial processes that require the formation of metal sulfides.

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Most popular questions from this chapter

What is a synthesis or combination reaction? Give an example. Can such reactions also be classified in other ways? Give an example of a synthesis reaction that is also a combustion reaction. Give an example of a synthesis reaction that is also an oxidation-reduction reaction, but which does not involve combustion.

On the basis of the general solubility rules given in Table \(7.1,\) predict the identity of the precipitate that forms when aqueous solutions of the following substances are mixed. If no precipitate is likely, indicate which rules apply. a. sodium sulfate, \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), and calcium chloride, \(\mathrm{CaCl}_{2}\) b. ammonium iodide, \(\mathrm{NH}_{4} \mathrm{I}\), and silver nitrate, \(\mathrm{AgNO}_{3}\) c. potassium phosphate, \(\mathrm{K}_{3} \mathrm{PO}_{4}\), and lead(II) nitrate, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) d. sodium hydroxide, \(\mathrm{NaOH}\), and iron(III) chloride, \(\mathrm{FeCl}_{3}\) e. potassium sulfate, \(\mathrm{K}_{2} \mathrm{SO}_{4}\), and sodium nitrate, \(\mathrm{NaNO}_{3}\) f. sodium carbonate, \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) and barium nitrate, \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\)

Balance each of the following oxidation-reduction reactions. In each, indicate which substance is being oxidized and which is being reduced. a. \(\mathrm{Na}(s)+\mathrm{S}(s) \rightarrow \mathrm{Na}_{2} \mathrm{S}(s)\) b. \(\operatorname{Mg}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MgO}(s)\) c. \(\mathrm{Ca}(s)+\mathrm{F}_{2}(g) \rightarrow \mathrm{CaF}_{2}(s)\) d. \(\operatorname{Fe}(s)+\mathrm{Cl}_{2}(g) \rightarrow \operatorname{Fe} \mathrm{Cl}_{3}(s)\)

Balance each of the following equations that describe synthesis reactions. a. \(\operatorname{Co}(s)+S(s) \rightarrow C o_{2} S_{3}(s)\) b. \(\mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}_{2}(g)\) c. \(\mathrm{FeO}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{FeCO}_{3}(s)\) d. \(\mathrm{Al}(s)+\mathrm{F}_{2}(g) \rightarrow \mathrm{AlF}_{3}(s)\) e. \(\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s)\)

For the reaction \(\mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{MgCl}_{2}(s),\) illustrate how electrons are gained and lost during the reaction.
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