/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 The reagent(s) that can selectiv... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The reagent(s) that can selectively precipitate \(\mathrm{S}^{2-}\) from a mixture of \(\mathrm{S}^{2-}\) and \(\mathrm{SO}_{4}^{2-}\) in aqueous solution is (are) (a) \(\mathrm{CuCl}_{2}\) (b) \(\mathrm{BaCl}_{2}\) (c) \(\mathrm{Pb}\left(\mathrm{OOCCH}_{3}\right)_{2}\) (d) \(\mathrm{Na}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NO}\right]\)

Short Answer

Expert verified
The reagent is (d) \( \mathrm{Na}_2[\mathrm{Fe(CN)}_5 \mathrm{NO}] \).

Step by step solution

01

Identifying Selective Precipitation Concept

We need to determine which reagent can selectively precipitate sulfide ions \( \mathrm{S}^{2-} \) while leaving sulfate ions \( \mathrm{SO}_4^{2-} \) in solution. This involves finding a reagent that forms an insoluble compound with \( \mathrm{S}^{2-} \) and not with \( \mathrm{SO}_4^{2-} \).
02

Analyzing Each Option

(a) \( \mathrm{CuCl}_2 \): Copper(II) chloride reacts with sulfide ions to form copper(II) sulfide \( \mathrm{CuS} \), a very insoluble compound, precipitating \( \mathrm{S}^{2-} \). However, it doesn't react significantly with \( \mathrm{SO}_4^{2-} \).(b) \( \mathrm{BaCl}_2 \): Barium chloride reacts with both sulfide ions and sulfate ions. While it forms barium sulfide \( \mathrm{BaS} \), it also forms barium sulfate \( \mathrm{BaSO}_4 \), thus not being selective.(c) \( \mathrm{Pb(OOCCH}_3)_2 \): Lead(II) acetate can react to form lead(II) sulfide \( \mathrm{PbS} \) which precipitates \( \mathrm{S}^{2-} \). However, it also can form \( \mathrm{PbSO}_4 \), and thus is not entirely selective.(d) \( \mathrm{Na}_2[\mathrm{Fe(CN)}_5 \mathrm{NO}] \): Also known as sodium nitroprusside, this reagent reacts specifically with sulfide ions to form a black precipitate, not affecting \( \mathrm{SO}_4^{2-} \).
03

Evaluating the Selectivity of Reagents

Reagents that react to form a precipitate with \( \mathrm{S}^{2-} \) more favorably than with \( \mathrm{SO}_4^{2-} \) are selective for sulfide. Based on the analysis, sodium nitroprusside \( \mathrm{Na}_2[\mathrm{Fe(CN)}_5 \mathrm{NO}] \) only reacts with sulfide to form a precipitate, making it selective in aqueous solutions.
04

Conclusion

The reagent that selectively precipitates \( \mathrm{S}^{2-} \) from the mixture is sodium nitroprusside \( \mathrm{Na}_2[\mathrm{Fe(CN)}_5 \mathrm{NO}] \). It does not form a precipitate with \( \mathrm{SO}_4^{2-} \), hence it's the correct choice for selective precipitation.

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 91Ó°ÊÓ!

Key Concepts

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

Sulfide Ion Precipitation
In chemistry, selective precipitation is a key method of separating ions in a solution based on their different solubilities. When dealing with sulfide ions (\( \mathrm{S}^{2-} \)), certain reagents can cause them to form insoluble compounds and precipitate out of the solution. Let's examine the problem at hand: choosing a reagent that will selectively precipitate \( \mathrm{S}^{2-} \) from a mixture containing both sulfide and sulfate ions (\( \mathrm{SO}_{4}^{2-} \)).

The selective precipitation process hinges on finding a reagent that:
  • Forms an insoluble compound with sulfide ions.
  • Does not react, or reacts very minimally, with sulfate ions.
Among common reagents, \( \mathrm{CuCl}_{2} \) (Copper(II) chloride) stands out as a candidate because it forms copper(II) sulfide (\( \mathrm{CuS} \)), which is highly insoluble. However, copper does not form a similar insoluble product with sulfates. This property allows sulfides to be separated effectively in both qualitative and quantitative analysis.
Sodium Nitroprusside Reaction
Sodium nitroprusside, with the chemical formula \( \mathrm{Na}_{2} \left[ \mathrm{Fe(CN)}_{5} \mathrm{NO} \right] \), is known for its selective reaction with sulfide ions. This reagent specifically interacts with \( \mathrm{S}^{2-} \) ions and not with \( \mathrm{SO}_{4}^{2-} \) ions. In the reaction with sulfide ions, sodium nitroprusside forms a black-colored precipitate. This distinctive color change makes it a particularly useful qualitative test for the presence of sulfide ions.The reaction is primarily employed because:
  • The resulting precipitate's color is starkly different from the original solution, facilitating easy visualization.
  • Sodium nitroprusside does not interfere with other common anions like sulfates, making it extraordinarily selective.
This selectivity allows chemists to confidently detect and precipitate sulfide ions without the concern of cross-reactions with other ions in the solution.
Solubility of Sulfide and Sulfate
Understanding the solubility rules is critical when predicting the outcome of reactions. Sulfide ions (\( \mathrm{S}^{2-} \)) and sulfate ions (\( \mathrm{SO}_{4}^{2-} \)) exhibit different solubility behaviors, which can be leveraged in laboratory settings to achieve selective precipitation.

In general:
  • Sulfides tend to form highly insoluble compounds with metal cations. Examples include copper(II) sulfide (\( \mathrm{CuS} \)) and lead sulfide (\( \mathrm{PbS} \)).
  • Sulfates are usually more soluble, though there are exceptions such as barium sulfate (\( \mathrm{BaSO}_{4} \)), which is very insoluble.
This differential solubility is pivotal in methods that separate ions because it allows chemists to add specific reagents knowing which ions will stay dissolved and which will precipitate. In solutions where precise detection or measurement is required, this solubility contrast underpins effective separation techniques, thus playing a crucial role in fields like analytical chemistry.

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

Which one of the following is likely to give a precipitate with \(\mathrm{AgNO}_{3}\) solution ? \(\quad\) (a) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl}\) (b) \(\mathrm{CCl}_{4}\) (c) \(\mathrm{CHCl}_{3}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\)

While titrating dilute \(\mathrm{HCl}\) solution with aqueous \(\mathrm{NaOH}\), which of the following will not be required? (a) Burette and porcelain tile (b) Pipette and distilled water (c) Clamp and phenolphthalein (d) Bunsen burner and measuring cylinder

An organic compound is estimated through Dumas method and was found to evolve 6 moles of \(\mathrm{CO}_{2}, 4\) moles of \(\mathrm{H}_{2} \mathrm{O}\) and 1 mole of nitrogen gas. The formula of the compound is: (a) \(\mathrm{C}_{12} \mathrm{H}_{8} \mathrm{~N}\) (b) \(\mathrm{C}_{12} \mathrm{H}_{8} \mathrm{~N}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}\)

Which one among the following pairs of ions cannot be separated by \(\mathrm{H}_{2} \mathrm{~S}\) in dilute hydrochloric acid? (a) \(\mathrm{Bi}^{3+}, \mathrm{Sn}^{4+}\) (b) \(\mathrm{Al}^{3+}, \mathrm{Hg}^{2+}\) (c) \(\mathrm{Zn}^{2+}, \mathrm{Cu}^{2+}\) (d) \(\mathrm{Ni}^{2+}, \mathrm{Cu}^{2+}\)

When \(20.02 \mathrm{~g}\) of a white solid \(X\) is heated \(4.4 \mathrm{~g}\) of an acid gas \(A\) and \(1.8 \mathrm{~g}\) of a neutral gas \(B\) are evolved, leaving behind a solid residue \(Y\) of weight \(13.8\) g. \(A\) turns lime water milky and \(B\) condenses into a liquid which changes anhydrous copper sulphate blue. The aqueous solution of \(Y\) is alkaline to litmus and gives \(19.7 \mathrm{~g}\) of white precipitate \(Z\) with barium chloride solution. \(Z\) gives carbon dioxide with an acid. Identify \(A, B, X, Y\) and \(Z\)

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.