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(a) Precipitation of the group 4 cations (Figure \(17.22)\) requires a basic medium. Why is this so? (b) What is the most significant difference between the sulfides precipitated in group 2 and those precipitated in group \(3 ?\) (c) Suggest a procedure that would serve to redissolve the group 3 cations following their precipitation.

Short Answer

Expert verified
(a) The precipitation of group 4 cations requires a basic medium because it provides hydroxide ions (OH^-) which react with the cations, forming insoluble hydroxides and precipitating them from the solution. (b) The most significant difference between group 2 and group 3 sulfides is their solubility in acidic solutions. Group 2 sulfides are insoluble in dilute acids, while group 3 sulfides are soluble and generate hydrogen sulfide (H2S) when reacted with dilute acids. (c) To redissolve group 3 cations following their precipitation, add a dilute acid like hydrochloric acid (HCl) or nitric acid (HNO3) to the mixture of their respective sulfides. The acid will react with the sulfides and convert them into dissolved soluble salts and release hydrogen sulfide gas, effectively redissolving the group 3 cations back into the solution.

Step by step solution

01

Explain the need for basic medium in precipitation of group 4 cations

Group 4 cations include Pb^2+, Zn^2+, and Ni^2+. These cations form insoluble salts when they react with hydroxide ions (OH^-). The basic medium generates OH^-, which reacts with the cations and precipitates them as hydroxides. The reaction is given as: \(M^{2+} + 2OH^- \rightarrow M(OH)_2\) Where M represents the group 4 cation. Without a basic medium, hydroxide ions would not be present, and the cations would remain dissolved in solution.
02

Determine the most significant difference between group 2 and group 3 sulfides

Group 2 sulfides include HgS, CuS, and PbS, while group 3 sulfides include FeS, CoS, and NiS. The most significant difference between these two groups is their solubility in acidic solutions. Group 2 sulfides are insoluble in dilute acids, while group 3 sulfides are soluble when reacted with dilute acids such as hydrochloric acid (HCl) or nitric acid (HNO3), which generate hydrogen sulfide (H2S). The reaction is given as: \(MS_{(group 3)} + 2HCl \rightarrow MCl_2 + H_2S\)
03

Propose a procedure to redissolve group 3 cations after precipitation

As mentioned earlier, group 3 sulfides are soluble in dilute acids. A suitable procedure to redissolve group 3 cations after precipitation would involve adding a dilute acid like hydrochloric acid (HCl) or nitric acid (HNO3) to the mixture of their respective sulfides. This acid will react with the sulfides and convert them into dissolved soluble salts and release hydrogen sulfide gas, effectively redissolving the group 3 cations back into the solution.

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Key Concepts

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

Basic Medium in Precipitation
Understanding the role of a basic medium in precipitation is crucial in inorganic chemistry, especially for Group 4 cations like Pb2+, Zn2+, and Ni2+. These cations readily combine with hydroxide ions (OH-) to form insoluble hydroxides. But why is a basic medium needed?

A basic medium provides the necessary OH- ions for the reaction to occur. When these cations encounter hydroxide ions, they undergo a precipitation reaction, forming solid hydroxides which fall out of the solution. The general reaction for a group 4 cation (M) is shown as:
\[M^{2+} + 2OH^- \rightarrow M(OH)_2\]
Without sufficient OH- ions, which are supplied by the basic medium, these cations would not precipitate and remain in solution. It's the availability of hydroxide ions that drives the formation of solid hydroxides, facilitating the separation of cations from a mixture.

Comprehending this concept aids students in grasping how various cation groups can be separated by manipulating the pH of the solution. In laboratory practice, careful control of the solution's pH is essential for successful precipitation and isolation of specific metals.
Solubility of Sulfides
Sulfides of different metals have varying solubility properties, which is particularly evident when comparing Group 2 and Group 3 sulfides. Sulfides from Group 2 such as HgS, CuS, and PbS, are characterized by their insolubility in dilute acids, while Group 3 sulfides like FeS, CoS, and NiS display solubility under these conditions.

The distinct behavior of these sulfides is due to their different chemical structures and bond characteristics. Solubility in acid is a critical property, as it allows for the selective separation of metals. For Group 3 sulfides, when exposed to dilute acids like hydrochloric acid (HCl), they react and dissolve, forming soluble chlorides and releasing hydrogen sulfide gas:\[MS_{(group 3)} + 2HCl \rightarrow MCl_2 + H_2S\]This property is exploited in qualitative analysis for the separation of metal ions. By understanding the solubility rules, students can predict the outcome of reactions and identify substances based on their chemical behavior when treated with acids. It's also a factor to consider in environmental and industrial processes, where the handling of sulfide-containing materials can give rise to sulfur compounds, influencing both procedure and environmental safety.
Acidic Solution Reactions
Acidic solution reactions play an important role in the context of precipitating and redissolving metal sulfides. In these reactions, acids serve to solubilize certain metal sulfides, which can be an integral step in chemical analysis or recovery processes.Group 3 metal sulfides, which we already know to be soluble in dilute acids, can be redissolved after precipitation. By adding an acid such as hydrochloric acid (HCl) or nitric acid (HNO3), the sulfide precipitate reacts and converts back into dissolved ions and a gaseous byproduct. For example:\[CoS_{(s)} + 2HCl_{(aq)} \rightarrow CoCl_2_{(aq)} + H_2S_{(g)}\]By manipulating the reaction conditions, one can selectively redissolve certain metals, helping in their identification or separation from a mixture. This demonstrates how acidity can reverse the precipitation process, offering a method to recover metals from their sulfide forms. Acidic solution reactions are a testament to the dynamic nature of chemistry, where the same substance can undergo different transformations under varying conditions, underscoring the importance of understanding reactivity and solubility principles.

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Most popular questions from this chapter

A sample of \(0.2140 \mathrm{~g}\) of an unlenown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0950 \mathrm{M} \mathrm{NaOH}\). The acid required \(27.4 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be \(6.50 .\) What is the \(K_{a}\) for the unknown acid?

Suppose that a 10-mL sample of a solution is to be tested for \(\mathrm{Cl}^{-}\) ion by addition of 1 drop \((0.2 \mathrm{~mL})\) of \(0.10 \mathrm{M}\) \(\mathrm{AgNO}_{3}\). What is the minimum number of grams of \(\mathrm{Cl}^{-}\) that must be present for \(\mathrm{AgCl}(\mathrm{s})\) to form?

Seawatercontains \(0.13 \%\) magnesium by mass, and has a density of \(1.025 \mathrm{~g} / \mathrm{mL}\). What fraction of the magnesium can be removed by adding a stoichiometric quantity of \(\mathrm{CaO}\) (that is, one mole of \(\mathrm{CaO}\) for each mole of \(\mathrm{Mg}^{2+}\) )?

How many milliliters of \(0.105 \mathrm{M} \mathrm{HCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH}\), (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3}\), (c) \(125.0 \mathrm{~mL}\) of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

A solution containing an unknown number of metal ions is treated with dilute \(\mathrm{HCl}\); no precipitate forms. The \(\mathrm{pH}\) is adjusted to about 1 , and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. Again, no precipitate forms. The pH of the solution is then adjusted to about 8 . Again, \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). No precipitate forms.

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