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Chapter 16: 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 difference between sulfide precipitates insoluble in acidic solutions and those insoluble in basic solutions is due to the varying concentration of H+ ions in the environment. In acidic solutions, high H+ ion concentration causes sulfide ions to react and form hydrogen sulfide gas, leading to a higher insolubility of sulfide precipitates. Conversely, in basic solutions, low H+ ion concentration prevents this reaction, resulting in a lower insolubility of sulfide precipitates.

Step by step solution

01

Understand Solubility

Solubility is the property of a solid, liquid or gaseous chemical substance called solute to dissolve in a solid, liquid or gaseous solvent. In this case, we are concerned with the solubility of sulfide precipitates in acidic and basic solutions.
02

Sulfides in acidic solutions

In acidic solutions, the concentration of H+ ions is high. When a sulfide precipitate is exposed to an acidic environment, the following reaction occurs: \[ S^{2-} + 2H^+ \rightarrow H_2S \] As H+ ions react with the sulfide ions (S2-), hydrogen sulfide gas (H2S) is formed. This reaction reduces the concentration of sulfide ions in the solution, making it more likely for the sulfide precipitate to dissolve. Therefore, sulfide precipitates tend to be more insoluble in acidic solutions.
03

Sulfides in basic solutions

In basic solutions, the concentration of H+ ions is low. When a sulfide precipitate is exposed to a basic environment, there are not enough H+ ions to cause the same reaction as in acidic solutions. As a result, sulfide ions will not form hydrogen sulfide gas (H2S) and will remain in the solution. This leads to an increase in the concentration of sulfide ions, making it less likely for the sulfide precipitate to dissolve. Therefore, sulfide precipitates tend to be more insoluble in basic solutions.
04

Conclusion

The difference between the two groups of sulfide precipitates lies in the concentration of H+ ions in acidic and basic solutions and the behavior of sulfide ions in these environments. In acidic solutions, high H+ ions concentration leads to the formation of hydrogen sulfide gas, making sulfides more insoluble. In basic solutions, low H+ ions concentration inhibits the formation of hydrogen sulfide gas, making sulfides less insoluble.

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Most popular questions from this chapter

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{~mol} / \mathrm{L}\). b. The solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\).

a. Calculate the molar solubility of \(\mathrm{Sr} \mathrm{F}_{2}\) in water, ignoring the basic properties of \(\mathrm{F}^{-} .\) (For \(\mathrm{SrF}_{2}, K_{\mathrm{se}}=7.9 \times 10^{-10} .\) ) b. Would the measured molar solubility of \(\mathrm{Sr} \mathrm{F}_{2}\) be greater than or less than the value calculated in part a? Explain. c. Calculate the molar solubility of \(\mathrm{SrF}_{2}\) in a solution buffered at \(\mathrm{pH}=2.00 .\left(K_{\mathrm{a}}\right.\) for \(\mathrm{HF}\) is \(7.2 \times 10^{-4} .\) )

The solubility of the ionic compound \(\mathrm{M}_{2} \mathrm{X}_{3}\), having a molar mass of \(288 \mathrm{~g} / \mathrm{mol}\), is \(3.60 \times 10^{-7} \mathrm{~g} / \mathrm{L}\). Calculate the \(K_{\mathrm{sp}}\) of the compound.

Calculate the solubility of \(\mathrm{AgCN}(s)\left(K_{\mathrm{sp}}=2.2 \times 10^{-12}\right)\) in a solution containing \(1.0 M \mathrm{H}^{+} .\left(K_{\mathrm{a}}\right.\) for \(\mathrm{HCN}\) is \(6.2 \times 10^{-10}\).)

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) is \(4.8 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\). b. The solubility of \(\mathrm{BiI}_{3}\) is \(1.32 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\).
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