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Chapter 15: Problem 14

Sulfide precipitates are generally grouped as sulfides insoluble in acidic solution and sulfides insoluble in basic solution. Explain why there is a difference between the two groups of sulfide precipitates.

Short Answer

Expert verified
The primary difference between the two groups of sulfide precipitates lies in the types of solutions they are insoluble in and the chemical processes governing their solubility. Sulfides insoluble in acidic solutions have higher affinity for H+ ions and form insoluble precipitates due to the reduction of sulfide ion concentration. On the other hand, sulfides insoluble in basic solutions have higher affinity for OH- ions and form insoluble metal hydroxides, leading to a similar reduction of metal ion concentration. Understanding these differences helps predict the solubility behavior of metal sulfide compounds, essential for various industrial processes and environmental applications.

Step by step solution

01

Understanding Solubility of Sulfide Precipitates

The solubility of a compound is determined by its ability to dissociate into its constituent ions in a given type of solution. In the case of sulfide precipitates, we have metal sulfide compounds (MxSy) that can dissociate into metal cations (Mx+) and sulfide anions (S^(2-)). Solubility trends of these compounds often depend on the nature of the metal cation and the sulfide anion as well as the properties of the acidic or basic solution present.
02

Sulfides Insoluble in Acidic Solution

The solubility of sulfide compounds in acidic solutions is determined by the presence of hydrogen ions (H+). H+ ions react with sulfide ions (S^(2-)) to form hydrogen sulfide (H2S) gas. This reaction reduces the concentration of sulfide ions in the solution, leading to the precipitation of the insoluble metal sulfide compound. Therefore, metal sulfide compounds with higher affinity for H+ ions tend to be insoluble in acidic solutions. Examples include silver sulfide (Ag2S), copper sulfide (CuS), and lead sulfide (PbS).
03

Sulfides Insoluble in Basic Solution

In basic solutions, the solubility of metal sulfide compounds depends on the presence of hydroxide ions (OH-). OH- ions react with metal cations (Mx+) to form metal hydroxide compounds (Mx(OH)y), reducing the concentration of metal ions in the solution. The remaining dissolved sulfide ions (S^(2-)) can then react with water molecules to form HS^- ions, which are soluble. Therefore, metal sulfide compounds with higher affinity for OH- ions tend to be insoluble in basic solutions. Examples include zinc sulfide (ZnS), nickel sulfide (NiS), and cadmium sulfide (CdS).
04

The Difference Between the Two Groups of Sulfide Precipitates

The primary difference between the two groups of sulfide precipitates lies in the types of solutions they are insoluble in and the chemical processes governing their solubility. Sulfides insoluble in acidic solutions have higher affinity for H+ ions and form insoluble precipitates due to the reduction of sulfide ion concentration. On the other hand, sulfides insoluble in basic solutions have higher affinity for OH- ions and form insoluble metal hydroxides, leading to a similar reduction of metal ion concentration. Understanding the differences between these two groups of sulfide precipitates helps in predicting the solubility behavior of metal sulfide compounds in different types of solutions, which is essential in many industrial processes and environmental applications.

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Most popular questions from this chapter

Assuming that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is \(1.6 \times 10^{-7}\) \(\operatorname{mol} / \mathrm{L}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{sp}}\) for this salt. Ignore any potential reactions of the ions with water.

The concentration of \(\mathrm{Pb}^{2+}\) in a solution saturated with \(\mathrm{PbBr}_{2}(s)\) is \(2.14 \times 10^{-2} \mathrm{M} .\) Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{PbBr}_{2}\).

The \(\mathrm{Hg}^{2+}\) ion forms complex ions with \(\mathrm{I}^{-}\) as follows: $$\begin{aligned} \mathrm{Hg}^{2+}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}^{+}(a q) & & K_{1}=1.0 \times 10^{8} \\ \mathrm{HgI}^{+}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{2}(a q) & & K_{2}=1.0 \times 10^{5} \\ \mathrm{HgI}_{2}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{3}^{-}(a q) & & K_{3}=1.0 \times 10^{9} \\ \mathrm{HgI}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{4}^{2-}(a q) & & K_{4}=1.0 \times 10^{8} \end{aligned}$$ A solution is prepared by dissolving 0.088 mole of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) and 5.00 moles of NaI in enough water to make 1.0 L of solution. a. Calculate the equilibrium concentration of \(\left[\mathrm{HgI}_{4}^{2-}\right] .\) b. Calculate the equilibrium concentration of \(\left[\mathrm{I}^{-}\right] .\) c. Calculate the equilibrium concentration of \(\left[\mathrm{Hg}^{2+}\right]\).

A friend tells you: "The constant \(K_{\mathrm{sp}}\) of a salt is called the solubility product constant and is calculated from the concentrations of ions in the solution. Thus, if salt A dissolves to a greater extent than salt \(\mathbf{B}\), salt \(\mathbf{A}\) must have a higher \(K_{\mathrm{sp}}\) than salt \(\mathbf{B}\)." Do you agree with your friend? Explain.

Devise as many ways as you can to experimentally determine the \(K_{\mathrm{sp}}\) value of a solid. Explain why each of these would work.
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