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

Short Answer

Expert verified
Use hot water to dissolve PbCl₂, separating it from AgCl.

Step by step solution

01

Understand Solubility Differences

AgCl and PbCl₂ have different solubility characteristics in hot water. While AgCl is poorly soluble in both hot and cold water, PbCl₂ is more soluble in hot water.
02

Select the Reagent Based on Solubility

To separate AgCl and PbCl₂, we can use hot water to dissolve PbCl₂. This separation takes advantage of the higher solubility of PbCl₂ in hot water compared to AgCl.
03

Apply the Reagent

Add hot water to the mixture of AgCl and PbCl₂. Stir the mixture to ensure that PbCl₂ dissolves into the hot water, while AgCl remains as a solid.
04

Perform Separation

Filter the solution. The hot water will contain dissolved PbCl₂, while AgCl will remain as a solid on the filter paper. This process effectively separates AgCl from PbCl₂.

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Key Concepts

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

Solubility Differences
Solubility differences arise when different substances exhibit varying abilities to dissolve in particular solvents like water. These differences are crucial in many separation techniques because they determine how substances can be selectively dissolved out of a mixture. For instance, silver chloride (\(\mathrm{AgCl}\)) has low solubility in water whether the water is hot or cold. Conversely, lead(II) chloride (\(\mathrm{PbCl_2}\)) becomes significantly more soluble in hot water, even though it's also relatively insoluble in cold water.
This property allows chemists to use temperature changes as a method of separating components of a mixture based on their solubility characteristics. By carefully selecting the correct conditions, one can selectively dissolve one component while leaving the other behind. This principle underlies many practical applications in laboratory settings and industrial processes.
Group 1 Analysis
Group 1 analysis refers to the systematic approach used in qualitative chemical analysis to identify and separate cations within the first group of metals. This group primarily includes ions that form insoluble chlorides, such as silver ion (\(\mathrm{Ag^+}\)), lead ion (\(\mathrm{Pb^{2+}}\)), and mercury(I) ion (\(\mathrm{Hg_2^{2+}}\)).
The process involves adding reagents that cause specific precipitation reactions. Due to the nature of group 1 ions, they can be initially precipitated out together as chlorides by adding hydrochloric acid. Subsequently, further treatment and analysis rely on exploiting different chemical properties such as variable solubility in selected solvents to achieve separation and identification.
  • This method is a fundamental part of classical qualitative analysis schemes in chemistry.
  • It's particularly useful in educational settings for teaching practical laboratory separation techniques.
  • This type of analysis forms the basis for more advanced comprehension of analytical chemistry concepts.
Precipitate Separation
Precipitate separation is a critical laboratory technique involving the separation of solid particles (precipitates) from a liquid mixture. It often relies on differences in physical or chemical properties among the components of a mixture. In our scenario involving \(\mathrm{AgCl}\) and \(\mathrm{PbCl_2}\), these compounds are initially present as a mixed precipitate.
To separate \(\mathrm{AgCl}\) from \(\mathrm{PbCl_2}\), understanding each compound's solubility differences becomes key. By heating the mixture with hot water, \(\mathrm{PbCl_2}\) dissolves, whereas \(\mathrm{AgCl}\) does not. This forms the basis for dividing the components into distinct phases using simple filtration.
  • The process often involves steps like washing, decanting, and using solvents selectively.
  • Using a filter paper can aid in capturing the remaining solid particles after one of the components is dissolved.
  • This is a fundamental exercise in separation based on solubility differences, often taught in basic chemistry labs.
Hot Water Solubility
Hot water solubility refers to how a substance's ability to dissolve in water changes with an increase in temperature. Generally, increased temperature enhances the solubility of solids in liquids, which is not always the case for all solutes but holds true for many ionic compounds.
In the context of \(\mathrm{PbCl_2}\) and \(\mathrm{AgCl}\), hot water becomes a selective solvent that can be used to dissolve \(\mathrm{PbCl_2}\) while leaving \(\mathrm{AgCl}\) undissolved. This is due to the nature of the compounds' crystal lattice energies and solvation dynamics with the water molecules. Such selective solubility is particularly useful in practical separation techniques where the aim is to isolate one component from a mixture while maintaining others in solid form.
  • This technique highlights the relationship between temperature and solubility, offering insights into thermodynamic principles.
  • The method is straightforward, requiring only heat application and subsequent filtration of the soluble component.
  • Understanding this principle is crucial for efficient laboratory manipulations in both educational and industrial chemistries.

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Most popular questions from this chapter

In principle, amphoteric oxides, such as \(\mathrm{Al}_{2} \mathrm{O}_{3}\) and \(\mathrm{BeO}\) can be used to prepare buffer solutions because they possess both acidic and basic properties (see Section 16.11). Explain why these compounds are of little practical use as buffer components.

The maximum allowable concentration of \(\mathrm{Pb}^{2+}\) ions in drinking water is 0.05 ppm (i.e., \(0.05 \mathrm{~g}\) of \(\mathrm{Pb}^{2+}\) in 1 million grams of water). Is this guideline exceeded if an underground water supply is at equilibrium with the mineral anglesite \(\left(\mathrm{PbSO}_{4}\right)\left(K_{\mathrm{sp}}=1.6 \times 10^{-8}\right) ?\)

Acid-base reactions usually go to completion. Confirm this statement by calculating the equilibrium constant for each of the following cases: (a) A strong acid reacting with a strong base. (b) A strong acid reacting with a weak base \(\left(\mathrm{NH}_{3}\right) .(\mathrm{c})\) A weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) reacting with a strong base. (d) A weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) reacting with a weak base \(\left(\mathrm{NH}_{3}\right)\). (Hint: Strong acids exist as \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions and strong bases exist as \(\mathrm{OH}^{-}\) ions in solution. You need to look up \(K_{\mathrm{a}}, K_{\mathrm{b}}\), and \(K_{\mathrm{w}}\).)

The \(K_{\mathrm{a}}\) of a certain indicator is \(2.0 \times 10^{-6} .\) The color of HIn is green and that of \(\mathrm{In}^{-}\) is red. A few drops of the indicator are added to an \(\mathrm{HCl}\) solution, which is then titrated against an \(\mathrm{NaOH}\) solution. At what \(\mathrm{pH}\) will the indicator change color?

Sketch the titration curve of a weak acid with a strong base like the one shown in Figure 17.4 . On your graph, indicate the volume of base used at the equivalence point and also at the half-equivalence point, that is, the point at which half of the acid has been neutralized. Explain how the measured \(\mathrm{pH}\) at the half-equivalence point can be used to determine \(K_{\mathrm{a}}\) of the acid.

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