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

The addition of \(\mathrm{HCl}(\mathrm{aq})\) to a solution containing several different cations produces a white precipitate. The filtrate is removed and treated with \(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq})\) in 0.3 M HCl. No precipitate forms. Which of the following conclusions is (are) valid? Explain. (a) \(\mathrm{Ag}^{+}\) or \(\mathrm{Hg}_{2}^{2+}\) (or both) is probably present. (b) \(\mathrm{Mg}^{2+}\) is probably not present. (c) \(\mathrm{Pb}^{2+}\) is probably not present. (d) \(\mathrm{Fe}^{2+}\) is probably not present.

Short Answer

Expert verified
Option (a), (c), and (d) are potentially correct. \(\mathrm{Ag}^{+}\) or \(\mathrm{Hg}_{2}^{2+}\) (or both) might be present because they form a white precipitate with HCl. \(\mathrm{Pb}^{2+}\) and \(\mathrm{Fe}^{2+}\) are likely not present, as they wouldn't form precipitate with H2S under the conditions. \(\mathrm{Mg}^{2+}\) could be present without affecting the results.

Step by step solution

01

Analyze reaction with HCl

The addition of HCl to the solution results in a white precipitate, indicating the presence of ions that react with chloride ions to form a white insoluble substance. The most probable ions are the silver ion (\(Ag^{+}\)) and the mercury(I) ion (\(Hg_{2}^{2+}\)), as silver chloride (\(AgCl\)) and mercury(I) chloride (\(Hg_{2}Cl_{2}\)) are both white precipitates. So, option (a) is potentially correct.
02

Analyze reaction with H2S in HCl

The addition of H2S to the filtrate doesn't result in any precipitate. This indicates that the ions that react with H2S to form a precipitate in acidic medium are not present. These ions could include \(\mathrm{Pb}^{2+}\), which produces black lead(II) sulfide (\(PbS\)), and \(\mathrm{Fe}^{2+}\), which produces green ferrous sulfide (\(FeS\)). So, options (c) and (d) are potentially correct.
03

Non-reactivity with either reagent

\(\mathrm{Mg}^{2+}\) does not react with either HCl or H2S to form a precipitate, so does not influence the results reported from the tests. So, whether or not \(\mathrm{Mg}^{2+}\) is present in the solution, it can't be determined from the facts provided. Thus, option (b) makes an unsupported assumption and is probably not correct.

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.

Precipitation Reactions
Precipitation reactions are a type of chemical reaction that occurs in solution when two soluble reactants combine to form an insoluble product, known as a precipitate. These reactions typically involve ionic compounds where cations and anions come together to form a solid.

In the given exercise, adding hydrochloric acid ( HCl ) to the solution initiates a precipitation reaction, forming a white precipitate. This occurs because chloride ions ( Cl^- ) from HCl react with certain cations in the solution to form insoluble chlorides.
  • Silver chloride ( AgCl ) and mercury(I) chloride ( Hg_2Cl_2 ) are common white precipitates, as they are not soluble in water.
  • The presence of these precipitates suggests the initial presence of Ag^+ and Hg_2^{2+} ions.
Understanding how specific ions form precipitates is vital for determining the composition of a solution.
Cation Identification
Cation identification involves determining which positive ions, or cations, are present in a solution. In qualitative analysis, this is typically done by observing the behavior of ions in various chemical tests. The process can identify unknown cations based on their reactivity to different reagents.

For example, in our exercise, the appearance of a white precipitate upon adding HCl suggests the presence of certain cations like Ag^+ or Hg_2^{2+} .
Observing the lack of reaction after adding H_2S in acidic conditions further aids in the cation identification:
  • Pb^{2+} typically forms a precipitate with H_2S , leading to the inference it is not present.
  • Similarly, since Fe^{2+} also does not result in a precipitate here, it is likely absent.
Identifying cations effectively helps in understanding the chemical nature of solutions.
Chemical Tests in Solutions
Chemical tests in solutions are critical for analyzing the composition of different ions. These tests often involve adding specific reagents that will react in characteristic ways with certain ions to provide valuable information. The exercise offers excellent examples of such tests:

The test using HCl demonstrates how the presence of chloride-reactive cations forms visible precipitates. The absence of a precipitate when using H_2S in acidic HCl further refines our understanding of the solution's composition.
  • Some ions like Mg^{2+} do not react with these reagents and hence, require different tests for identification.
  • Chemical tests are integral to laboratory analyses, providing data to ensure accurate identification of components.
Mastering these tests enhances one's ability to conduct accurate qualitative analyses.

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

\(\mathrm{PbCl}_{2}(\mathrm{s})\) is considerably more soluble in \(\mathrm{HCl}(\mathrm{aq})\) than in pure water, but its solubility in \(\mathrm{HNO}_{3}(\mathrm{aq})\) is not much different from what it is in water. Explain this difference in behavior.

Both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cu}^{2+}\) are present in the same aqueous solution. Which of the following reagents would work best in separating these ions, precipitating one and leaving the other in solution: \(\mathrm{NaOH}(\mathrm{aq}), \mathrm{HCl}(\mathrm{aq})\), \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq}),\) or \(\mathrm{NH}_{3}(\mathrm{aq}) ?\) Explain your choice.

Appendix E describes a useful study aid known as concept mapping. Using the methods presented in Appendix \(\mathrm{E},\) construct a concept map that links the various factors affecting the solubility of slightly soluble solutes.

Excess \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) is shaken with water to produce a saturated solution. A 50.00 mL sample of the clear saturated solution is withdrawn and requires \(10.7 \mathrm{mL}\) of \(0.1032 \mathrm{M} \mathrm{HCl}\) for its titration. What is \(K_{\mathrm{sp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2} ?\)

Calculate \(\left[\mathrm{Cu}^{2+}\right]\) in a \(0.10 \mathrm{M} \mathrm{CuSO}_{4}(\) aq) solution that is also \(6.0 \mathrm{M}\) in free \(\mathrm{NH}_{3}\). \(\mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(\mathrm{aq})\) \(K_{\mathrm{f}}=1.1 \times 10^{13}\)

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