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The metal ion which is precipitated when \(\mathrm{H}_{2} \mathrm{~S}\) is passed with HCl (a) \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{Mn}^{2+}\) (c) \(\mathrm{Ni}^{2+}\) (d) \(\mathrm{Zn}^{2+}\)

Short Answer

Expert verified
(a) \(\mathrm{Cd}^{2+}\) precipitates as \(\mathrm{CdS}\).

Step by step solution

01

Understanding the context

Hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)) is a gas that reacts with metal ions to form metal sulfides, which are typically insoluble. To identify which metal ion will precipitate, consider the solubility of its sulfide in the presence of hydrochloric acid (\(\mathrm{HCl}\)).
02

Analyzing each option

Evaluate the given metal ions: - \(\mathrm{Cd}^{2+}\): Cadmium sulfide (\(\mathrm{CdS}\)) is insoluble in water and will precipitate even in the presence of HCl.- \(\mathrm{Mn}^{2+}\): Manganese sulfide (\(\mathrm{MnS}\)) is slightly soluble in acidic solutions such as HCl, so it does not precipitate.- \(\mathrm{Ni}^{2+}\): Nickel sulfide (\(\mathrm{NiS}\)) requires more specific conditions to precipitate and does not typically precipitate in the presence of HCl.- \(\mathrm{Zn}^{2+}\): Zinc sulfide (\(\mathrm{ZnS}\)) might precipitate, but not readily in acidic solutions, because it is more soluble in such conditions than cadmium sulfide.
03

Concluding which precipitates

Given the solubility conditions, \(\mathrm{Cd}^{2+}\) will readily precipitate as \(\mathrm{CdS}\) when hydrogen sulfide is passed in the presence of hydrochloric acid. This is because \(\mathrm{CdS}\) is not soluble even in acidic solutions.

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Key Concepts

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

Metal Sulfides
Metal sulfides are formed when metal ions react with hydrogen sulfide ( H_2S ). These compounds are generally insoluble in water. This property makes metal sulfides very useful in separating different metal ions from a solution in a process called precipitation. This insolubility is crucial for chemists who conduct experiments that include the detection or isolation of specific ions. In these processes:
  • Metal ions are combined with H_2S .
  • Upon reaction, metal sulfides precipitate out of the solution.
  • This helps in separating the metal ions based on their sulfide solubility.
Knowing how metal sulfides behave in various conditions, such as when acids are present, is key to successfully using this reaction in analysis.
Solubility
Solubility refers to the extent to which a substance can dissolve in a solvent. In chemistry, understanding the solubility of metal sulfides is essential. If you can predict which metals will precipitate from a solution, it can aid in identifying and studying them. This is particularly true for precipitation reactions involving hydrogen sulfide:
  • Metal ions like Cd^{2+} form sulfides with very low solubility, leading them to precipitate easily.
  • Other metal ions may form slightly soluble sulfides. In such cases, they might not precipitate as readily.
Cadmium sulfide ( CdS ), for instance, is known for its very low solubility, even in the presence of acidic solutions. This is why it easily precipitates when exposed to H_2S in the presence of HCl .
Qualitative Analysis
Qualitative analysis is a method used in chemistry to identify the chemical constituents of a sample. This process often relies on precipitation reactions to separate and identify ions. Here's how it typically works:
  • React your sample with different reagents to precipitate certain ions.
  • Observe which elements form solid precipitates.
  • Use these observations to deduce their identity.
For instance, when dealing with metal ions, adding H_2S in a hydrochloric acid medium can determine which metal ions are in the mix. Using such methods, one can conclude that cadmium ions would be indicated by an insoluble CdS precipitate, while manganese might not precipitate as MnS due to its higher solubility.
Hydrochloric Acid
Hydrochloric acid ( HCl ) is a strong acid often used in chemical reactions to influence the solubility of compounds. In the context of precipitation reactions, it has a special role:
  • It decreases the solubility of certain metal sulfides.
  • By controlling the pH, HCl helps in selectively precipitating specific ions from a solution.
In qualitative analysis, HCl can alter the solubility of metal sulfides. For example, cadmium sulfide remains insoluble even in acidic conditions created by HCl and, therefore, precipitates out. Meanwhile, other metal sulfides like those of zinc or manganese might remain dissolved and not precipitate, allowing chemists to target certain ions effectively.

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Most popular questions from this chapter

How do we differentiate between \(\mathrm{Fe}^{3+}\) and \(\mathrm{Cr}^{3+}\) in group III? (a) by increasing \(\mathrm{NH}_{4}{ }^{+}\)ion concentration (b) by decreasing \(\mathrm{OH}^{-}\)ion concentration (c) by adding excess of \(\mathrm{NH}_{4} \mathrm{OH}\) solution (d) both (a) and (b)

Three separate samples of a solution of single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. \(\mathrm{NaCl}\) and one formed a black precipitate with \(\mathrm{H}_{2} \mathrm{~S} .\) The salt could be (a) \(\mathrm{AgNO}_{3}\) (b) \(\mathrm{MnSO}_{4}\) (c) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) (d) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\)

Identify the products \(\mathrm{A}\) and \(\mathrm{B}\) here. (1) \(2 \mathrm{Ag}^{+}\)(excess) \(+\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow \mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) \(\stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{A}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (2) \(2 \mathrm{NO}_{2}^{-}+2 \mathrm{I}^{-} \stackrel{\text { acid medium }}{\longrightarrow} \mathrm{B}+\mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (a) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{O}, \mathrm{B}=\mathrm{N}_{2}\) (b) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{~S}, \mathrm{~B}=\mathrm{N}_{2} \mathrm{O}\) (c) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{O}, \mathrm{B}=2 \mathrm{NO}_{2}\) (d) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{~S}, \mathrm{~B}=2 \mathrm{NO}\)

Which is not correct regarding the titration of ferrous ammonium sulphate and \(\mathrm{KMnO}_{4} ?\) (a) Here \(\mathrm{KMnO}_{4}\) acts like an oxidant and a self indicator (b) Here high temperature is maintained during the titration (c) Here ferrous sulphate is oxidized into ferric sulphate. (d) Both (b) and (c)

For the preparation of lyophobic colloids the methods used are normally (a) By dissolution into water (b) Condensation method (c) Dispersion method (d) Both (b) and (c)

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