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Chapter 22: Problem 48

Iron(II) sulfide, \(\mathrm{FeS}(s)\), is used as the pigment in black paint. A sample of \(\mathrm{FeS}(s)\) is suspected of containing lead(II) sulfide, \(\operatorname{PbS}(s)\), which can cause lead poisoning if ingested. Suggest a scheme based on \(\mathrm{pH}\) for separating \(\mathrm{FeS}(s)\) from \(\mathrm{PbS}(s)\).

Short Answer

Expert verified
Separate FeS from PbS by dissolving FeS in acid, then filter out insoluble PbS.

Step by step solution

01

Understand the Compounds

Identify the key components in the sample: Iron(II) sulfide (\(\mathrm{FeS}\)) and lead(II) sulfide (\(\mathrm{PbS}\)). Both are sulfide minerals but have different chemical properties.
02

Compare Solubility in Acidic Conditions

\(\mathrm{FeS}\) is more soluble in acidic conditions compared to \(\mathrm{PbS}\). When placed in an acidic solution, \(\mathrm{FeS}\) will dissolve more readily.
03

Adjusting pH for Separation

Add a dilute acid to the mixture to selectively dissolve \(\mathrm{FeS}\) while minimizing the dissolution of \(\mathrm{PbS}\). \(\mathrm{HCl}\) or \(\mathrm{H_2SO_4}\) might be appropriate acids for this step.
04

Filtration

After the dissolution of \(\mathrm{FeS}\), filter the solution to separate the undissolved \(\mathrm{PbS}\). The \(\mathrm{PbS}\) will remain solid and can be removed by filtration.
05

Recovery of \(\mathrm{FeS}\)

Neutralize the acidic solution to precipitate \(\mathrm{FeS}\) once again if needed. This is done by gradually adding a base until neutralization.

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Key Concepts

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

Iron(II) sulfide
Iron(II) sulfide, represented as \( \mathrm{FeS} \), is a compound used in various applications, including as a pigment in black paint. This compound results from combining iron and sulfur in a solid form and is characterized by its black color. There are essential aspects concerning its chemical nature:
  • Composition: It's a binary compound consisting of iron and sulfur in a 1:1 ratio.
  • Formation: \( \mathrm{FeS} \) forms through a direct reaction between iron and sulfur.
  • Properties: It is known for low solubility in water but reacts more promptly under acidic conditions.
Understanding these properties is key for processes like separation, as its solubility variations under different pH levels are vital analytical angles to consider when attempting to isolate this from other substances.
Lead(II) sulfide
Lead(II) sulfide, or \( \mathrm{PbS} \), is another sulfide compound often found in ore deposits. Recognized for its role as the main mineral in galena, \( \mathrm{PbS} \) is primarily noted for its high lead content and distinct properties:
  • Composition: Like \( \mathrm{FeS} \), this is also composed of sulfur and lead in a simple 1:1 ratio.
  • Toxicity: Known to cause health issues if ingested, making containment and proper handling necessary.
  • Solubility: Displaying poor solubility, \( \mathrm{PbS} \) remains intact unless under harsh conditions, differentiating it from \( \mathrm{FeS} \) in acidic settings.
Due to its lower solubility under acidic conditions, \( \mathrm{PbS} \) can be effectively separated from \( \mathrm{FeS} \) by employing methods that exploit these variations in chemical behavior.
pH and Solubility
The concept of pH plays a critical role in the chemistry of solubility, particularly for compounds such as \( \mathrm{FeS} \) and \( \mathrm{PbS} \). pH is a numeric scale used to specify the acidity or basicity of an aqueous solution:
  • A lower pH level indicates a more acidic environment, which often leads to increased solubility for many metal sulfides like \( \mathrm{FeS} \).
  • By contrast, substances like \( \mathrm{PbS} \) have markedly lower solubility in acidic conditions, which is key to differentiating it during separation processes.
  • Adjusting the pH can dramatically shift the solubility profiles of compounds and is an essential tool in chemical separation techniques.
Here, understanding and manipulating the pH enables the selective dissolution of \( \mathrm{FeS} \), leaving \( \mathrm{PbS} \) largely unaffected and facilitating their separation.
Separation Techniques
Separation techniques are crucial for differentiating and isolating chemical components from mixtures, such as \( \mathrm{FeS} \) and \( \mathrm{PbS} \) in the given problem. Here, a common approach involves exploiting differences in solubility and physical states:
  • pH Adjustment: Adding a dilute acid like \( \mathrm{HCl} \) selectively dissolves \( \mathrm{FeS} \) due to its higher solubility in an acidic environment.
  • Filtration: Following dissolution, filtering the solution helps physically separate the undissolved \( \mathrm{PbS} \) as a solid. This method effectively removes \( \mathrm{PbS} \) from the dissolved iron ions.
  • Recovery: By neutralizing the filtered acidic solution, \( \mathrm{FeS} \) can be precipitated back if required, completing the separation process.
Each step in these techniques is structured to ensure maximum separation efficiency, with a controlled approach towards handling potentially harmful compounds.
Acid-Base Reactions
Acid-base reactions are foundational in chemistry and draw upon the interaction of acids and bases to induce changes in compounds. In the context of separating \( \mathrm{FeS} \) from \( \mathrm{PbS} \), understanding this reaction type is key:
  • Acid Dissolution: Adding an acid encourages the dissolution of \( \mathrm{FeS} \), a process driven by the formation of soluble iron ions and hydrogen sulfide gas.
  • Base Neutralization: To precipitate dissolved \( \mathrm{FeS} \), a base is added, neutralizing the solution and reversing the dissolution.
  • Equilibrium: These reactions reach a balance point or equilibrium, indicating the completion of certain processes like precipitation or dissolution.
Grasping these reactions enables one to plan and execute chemical processes efficiently, facilitating the accurate separation of compounds through well-designed experiments.

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Most popular questions from this chapter

Copper(I) ions in aqueous solution react with \(\mathrm{NH}_{3}(a q)\) at \(25^{\circ} \mathrm{C}\) according to $$ \begin{array}{r} \mathrm{Cu}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \leftrightharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q) \\ K_{\mathrm{f}}=6.3 \times 10^{10} \mathrm{M}^{-2} \end{array} $$ Calculate the solubility of \(\operatorname{CuBr}(s)\) in a solution in which the equilibrium concentration of \(\mathrm{NH}_{3}(a q)\) is \(0.15 \mathrm{M}\) at \(25^{\circ} \mathrm{C}\).

Copper(I) ions in aqueous solution react with \(\mathrm{NH}_{3}(a q)\) according to $$ \mathrm{Cu}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \leftrightharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q) $$ with \(K_{\mathrm{f}}=6.3 \times 10^{10} \mathrm{M}^{-2}\). Calculate the solubility in grams per liter of \(\operatorname{CuBr}(s)\) in \(0.50 \mathrm{M} \mathrm{NH}_{3}(a q)\) at \(25^{\circ} \mathrm{C}\).

Calculate the solubility in grams per liter of \(\operatorname{AgBr}(s)\) in \(0.35 \mathrm{M} \mathrm{NH}_{3}(a q) .\) Take \(K_{i}=2.0 \times 10^{7} \mathrm{M}^{-2}\) for the complexation reaction of \(\mathrm{Ag}^{+}(a q)\) to produce \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q)\). Assume a temperature of \(25^{\circ} \mathrm{C}\).

Why is it possible to separate a mixture of \(\mathrm{Pb}^{2+}(a q)\) and \(\mathrm{Hg}_{2}^{2+}(a q)\) by selectively precipitating out the mercury as \(\mathrm{Hg}_{2} \mathrm{I}_{2}(s) ;\) but not possible to separate the same mixture by selectively precipitating out the lead as \(\mathrm{PbI}_{2}(s) ?\)

In each of the following cases, the two solutions indicated are mixed. In cases for which a precipi. tate forms on mixing, write the complete chemical equation and the net ionic equation. If no precipi. tate forms, then write "no reaction." Use the solubil. ity rules (Chapter 10 ) and assume that all solutions before mixing are \(0.20 \mathrm{M}\) and that equal volumes of the two solutions are mixed. (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}\left(\mathrm{ClO}_{4}\right)_{2}(a q) \rightarrow\) (b) \(\mathrm{AgNO}_{3}(a q)+\mathrm{NaClO}_{4}(a q) \rightarrow\) (c) \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{COO}(a q) \rightarrow\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{AgF}(a q) \rightarrow\)
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