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Chapter 17: Problem 63

In a group 1 analysis, a student obtained a precipitate containing both \(\mathrm{AgCl}\) and \(\mathrm{PbCl}_{2}\). Suggest one reagent that would enable her to separate \(\operatorname{AgCl}(s)\) from \(\mathrm{PbCl}_{2}(s)\)

Short Answer

Expert verified
Hot water can be used to separate \(\mathrm{AgCl}\) from \(\mathrm{PbCl}_{2}\).

Step by step solution

01

Understanding the Components

The precipitate contains both silver chloride (\(\mathrm{AgCl}\)) and lead(II) chloride (\(\mathrm{PbCl}_{2}\)). These are both insoluble chloride salts that can be separated based on their solubility differences in certain solvents.
02

Analyzing Solubility Properties

Silver chloride (\(\mathrm{AgCl}\)) is highly insoluble in water, while lead(II) chloride (\(\mathrm{PbCl}_{2}\)) is somewhat more soluble, especially in hot water. This means thermal manipulation can be explored for separation.
03

Identifying a Suitable Reagent

Since \(\mathrm{PbCl}_{2}\) is more soluble in hot water, heating the mixture and filtering allows \(\mathrm{PbCl}_{2}\) to dissolve in the hot water, leaving \(\operatorname{AgCl}(s)\) as a precipitate. Thus, hot water can be used as a reagent to separate the two.
04

Procedure for Separation

The mixture containing \(\mathrm{AgCl}\) and \(\mathrm{PbCl}_2\) is shaken with hot water. \(\mathrm{PbCl}_2\) dissolves and can be filtered off as the solution, leaving solid \(\operatorname{AgCl}(s)\) behind. Upon cooling, lead(II) chloride can eventually recrystallize.

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Key Concepts

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

Solubility
Solubility refers to the ability of a substance to dissolve in a solvent, forming a solution. It is an important property in chemical separation techniques. Different substances have different solubilities based on factors such as temperature, pressure, and the nature of the solvent.
  • For instance, \( \text{AgCl} \) is significantly less soluble in water compared to \( \text{PbCl}_2 \).
  • This difference in solubility is crucial for separating these compounds.
By understanding and exploiting these solubility differences, chemists can effectively separate and purify compounds. Solubility tends to increase with temperature; thus, heating can be a pivotal step in facilitating the separation process.
Group 1 Analysis
Group 1 analysis is an important aspect of qualitative inorganic analysis. It involves identifying specific ions present in a mixture and is often applied to ions that form insoluble chlorides, like silver and lead ions.
  • In this type of analysis, common reagents are used to selectively precipitate ions based on different properties.
  • For \( \text{AgCl} \) and \( \text{PbCl}_2 \), both are considered in Group 1 because they form precipitates with chloride ions that are typically insoluble.
Group 1 analysis simplifies the process of differentiating between ions by using reagents and conditions that target specific chemical characteristics, such as solubility. This methodical approach ensures accurate and efficient separation and identification of ions.
Precipitation
Precipitation is the process where a soluble substance forms an insoluble solid upon reaction, known as a precipitate. It's a key technique used in chemistry to separate different substances from a solution or mixture.
  • For example, when chloride ions are added to solutions of silver or lead, \( \text{AgCl} \) and \( \text{PbCl}_2 \) precipitate out.
  • In the context of this analysis, precipitation is essential for isolating these compounds from the rest of the solution.
The conditions under which precipitation occurs, including concentration, temperature, and the presence of other ions, can influence the efficiency and completeness of this separation process. By adjusting these conditions, chemists can control and optimize precipitation reactions.
Filtration
Filtration is a straightforward and essential technique used to separate solids from liquids, crucial in isolating precipitates. It involves passing a mixture through a filter that allows the liquid to pass while retaining the solid on the filter.
  • When \( \text{PbCl}_2 \) dissolves in hot water, filtration can be employed to separate \( \text{AgCl} \) from the solution containing dissolved lead chloride.
  • The solid \( \text{AgCl} \) remains as a residue, while the liquid filtrate contains dissolved substances.
After filtration, cooling the filtrate can lead to recrystallization and recovery of the dissolved component, \( \text{PbCl}_2 \), illustrating the utility of this technique for sequential separation and purification steps based on solubility properties.

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Most popular questions from this chapter

Silver chloride has a larger \(K_{\mathrm{sp}}\) than silver carbonate (see Table 17.2). Does this mean that the former also has a larger molar solubility than the latter?

Which of these ionic compounds will be more soluble in acid solution than in water: (a) \(\mathrm{BaSO}_{4},\) (b) \(\mathrm{PbCl}_{2}\), (c) \(\mathrm{Fe}(\mathrm{OH})_{3},\) (d) \(\mathrm{CaCO}_{3}\) ? Explain. (Hint: For each salt, determine any possible reaction between the anion and \(\mathrm{H}^{+}\) ions. \(.\)

A student carried out an acid-base titration by adding \(\mathrm{NaOH}\) solution from a buret to an Erlenmeyer flask containing \(\mathrm{HCl}\) solution and using phenolphthalein as indicator. At the equivalence point, he observed a faint reddish-pink color. However, after a few minutes, the solution gradually turned colorless. What do you suppose happened?

Write balanced equations and solubility product expressions for the solubility equilibria of these compounds: (a) \(\mathrm{CuBr}\) (b) \(\mathrm{ZnC}_{2} \mathrm{O}_{4}\) (c) \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) (d) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (e) \(\mathrm{AuCl}_{3}\) (f) \(\mathrm{Mn}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Calculate \(x\), the number of molecules of water in oxalic acid hydrate, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot x \mathrm{H}_{2} \mathrm{O},\) from the following data: \(5.00 \mathrm{~g}\) of the compound is made up to exactly \(250 \mathrm{~mL}\) solution and \(25.0 \mathrm{~mL}\) of this solution requires \(15.9 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) solution for neutralization.
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