Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is formed when \(\mathrm{H}_{2} \mathrm{~S}\) is passed through acidified solution of \(\mathrm{CuSO}_{4}\) ? (a) \(\mathrm{CuSO}_{4}\) (b) \(\mathrm{CuS}\) (c) \(\mathrm{CuHSO}_{4}\) (d) \(\mathrm{Cu}_{2} \mathrm{~S}\)

Short Answer

Expert verified
The product formed is \(\mathrm{CuS}\), the correct answer is (b).

Step by step solution

01

Understanding the Reaction

The problem involves passing hydrogen sulfide (\(\mathrm{H}_2 \mathrm{~S}\)) through an acidified copper(II) sulfate (\(\mathrm{CuSO}_4\)) solution. We need to determine the product of this chemical reaction.
02

Chemical Equation Setup

When \(\mathrm{H}_2 \mathrm{~S}\) is added to \(\mathrm{CuSO}_4\), a double displacement reaction occurs. Copper(II) ions (\(\mathrm{Cu^{2+}}\)) react with \(\mathrm{H}_2 \mathrm{~S}\) to form copper(II) sulfide (\(\mathrm{CuS}\)) and hydrogen sulfate ions (\(\mathrm{H}^{+}\)). The reaction is: \[ \mathrm{CuSO}_4 + \mathrm{H}_2\mathrm{S} \rightarrow \mathrm{CuS} + \mathrm{H}_2\mathrm{SO}_4 \]
03

Identifying the Product

The copper(II) sulfide (\(\mathrm{CuS}\)) formed is an insoluble black precipitate, indicating a typical reaction in this context, where copper ions react with sulfide ions to form a precipitate.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Double Displacement Reaction
In chemistry, a double displacement reaction, also known as a metathesis reaction, occurs when parts of two ionic compounds are exchanged to form two new compounds. These reactions are characterized by the exchange of ions between the reacting substances. In essence, it involves the anions and cations of two different molecules swapping places, leading to the formation of two entirely new compounds.
In the case of copper(II) sulfate (\(\mathrm{CuSO}_4\)) and hydrogen sulfide (\(\mathrm{H}_2 \mathrm{~S}\)), the copper ions (\(\mathrm{Cu}^{2+}\)) and the sulfate ions (\(\mathrm{SO}_4^{2-}\)) from the copper sulfate react with the hydrogen and sulfide ions from the hydrogen sulfide. The copper ions form a new compound with the sulfide ions. Similarly, the hydrogen ions combine with the sulfate ions to form sulfuric acid (\(\mathrm{H}_2 \mathrm{SO}_4\)). This type of reaction is predictable and often results in the formation of at least one insoluble product, called a precipitate. Understanding this concept can help in predicting products of such reactions easily.
Precipitation Reaction
A precipitation reaction occurs when two soluble salts are mixed together, resulting in an insoluble solid known as a precipitate. This happens when the product of the reaction does not dissolve in water, causing it to separate from the liquid solution into a solid form.
In the scenario described, when hydrogen sulfide (\(\mathrm{H}_2 \mathrm{~S}\)) reacts with copper(II) sulfate (\(\mathrm{CuSO}_4\)), copper(II) sulfide (\(\mathrm{CuS}\)) forms as a precipitate. This black solid is insoluble in water, and its formation is evidence of a successful precipitation reaction.
Precipitation reactions are useful in a variety of contexts, including chemical analysis and synthesis. They allow chemists to isolate certain compounds and study their properties. Recognizing these reactions is crucial, as the formation of a precipitate often indicates that a specific ionic interaction has taken place.
Copper Sulfide Formation
Copper sulfide (\(\mathrm{CuS}\)) is a chemical compound that results from the reaction between copper ions (\(\mathrm{Cu}^{2+}\)) and sulfide ions (\(\mathrm{S}^{2-}\)). This formation is a crucial part of many chemical reactions involving sulfide ions. When hydrogen sulfide gas is bubbled through an acidified copper sulfate solution, copper(II) ions from the solution react with sulfide ions from the gas.
This process essentially swaps the ions, forming copper sulfide and liberating sulfuric acid in the process. The resulting copper sulfide appears as a black precipitate, indicating the reaction's completion. The solid nature of copper sulfide highlights that it is not soluble in the aqueous medium in which the reaction occurs.
This reaction and the production of copper sulfide are important in various industrial applications and laboratories, where the formation of this compound can be utilized for separations and purifications based on insolubility.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Anhydrous zinc chloride is prepared by heating of (a) \(\mathrm{ZnCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Zn}+\mathrm{HgCl}_{2}\) (c) \(\mathrm{Zn}+\mathrm{Cl}\) (d) both \((\mathrm{b})\) and \((\mathrm{c})\)

\(\mathrm{Fe}^{3+}\) can be detected by the reaction (1) \(\mathrm{Fe}^{3+}+\left[\mathrm{Fe}^{\mathrm{II}}(\mathrm{CN})_{6}\right]^{4} \longrightarrow \mathrm{Fe}^{\mathrm{III}}\left[\mathrm{Fe}^{\mathrm{ll}}(\mathrm{CN})_{6}\right]\) Deep blue (2) \(\mathrm{Fe}^{3+}+\left[\mathrm{Fe}^{\mathrm{ll}}(\mathrm{CN})_{6}\right]^{3} \longrightarrow \mathrm{Fe}^{\mathrm{Il}}\left[\mathrm{Fe}^{\mathrm{III}}(\mathrm{CN})_{6}\right]\) (3) \(\mathrm{Fe}^{3+}+\mathrm{SCN}^{-} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\mathrm{C}} \begin{gathered}\text { Brown } \\\ {\left[\mathrm{Fe}(\mathrm{SCN})\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\right]^{2^{+}}}\end{gathered}\) Blood red (a) 1 and 2 only (b) 2 and 3 only (c) 1,2 and 3 (d) 3 only

Galvanization is applying a coating of? [2016] (a) \(\mathrm{Cu}\) (b) \(\mathrm{Zn}\) (c) \(\overline{\mathrm{Cr}}\) (d) \(\mathrm{Pb}\)

Mixture of \((\mathrm{X})=0.02\) moles of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\) and \(0.02\) mole of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}\) was prepared in 2 litre of solution 1 litre of mixture \((\mathrm{X})+\) excess \(\mathrm{AgNO}_{3} \longrightarrow(\mathrm{Y})\) 1 litre of mixture \((\mathrm{X})+\) excess \(\mathrm{BaCl}_{2} \longrightarrow(\mathrm{Z})\) (a) \(0.01,0.01\) (b) \(0.02,0.01\) (c) \(0.01,0.02\) (d) \(0.02,0.02\)

Which of the following statement is correct with reference to the ferrous and ferric ions? 1\. \(\mathrm{Fe}^{3+}\) gives brown colour with potassium ferricyanide 2\. \(\mathrm{Fe}^{2+}\) gives blue precipitate with potassium ferricyanide 3\. \(\mathrm{Fe}^{3+}\) gives red colour with potassium thiocyanate 4\. \(\mathrm{Fe}^{2+}\) gives brown colour with ammonium thiocyanate. (a) 1,4 (b) 1,2 (c) 2,3 (d) all of these

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free