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A solution, when diluted with water and boiled, gives a white precipitate. On addition of excess \(\mathrm{NH}_{4} \mathrm{Cl} /\) \(\mathrm{NH}_{4} \mathrm{OH}\), the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in \(\mathrm{NH}_{4} \mathrm{OH} / \mathrm{NH}_{4} \mathrm{Cl}\) (a) \(\mathrm{Zn}(\mathrm{OH})_{2}\) (b) \(\mathrm{Al}(\mathrm{OH})_{3}\) (c) \(\mathrm{Mg}(\mathrm{OH})\) (d) \(\mathrm{Ca}(\mathrm{OH})\)

Short Answer

Expert verified
The precipitate that dissolves is (a) \(\mathrm{Zn(OH)_2}\).

Step by step solution

01

Understand the Reaction Process

When a solution is diluted with water and then boiled, a white precipitate forms. This indicates the presence of a metal hydroxide that is not fully soluble in water.
02

Consider the Reaction with NH4Cl/NH4OH

The precipitate decreases in volume, which indicates that some form of displacement reaction is occurring. Excess \( \mathrm{NH}_{4}\mathrm{Cl} / \mathrm{NH}_{4}\mathrm{OH} \) will dissolve specific metal hydroxides.
03

Evaluate the Solubility Behavior

After adding excess \( \mathrm{NH}_{4}\mathrm{Cl} / \mathrm{NH}_{4}\mathrm{OH} \), the volume of the precipitate reduces, leaving behind a white gelatinous precipitate. This suggests a distinction in solubility between compounds.
04

Identify Solubility Properties of Each Compound

- \( \mathrm{Zn} ( \mathrm{OH})_2 \) can dissolve in ammonium solutions to form complex ions.- \( \mathrm{Al} ( \mathrm{OH})_3 \) is gelatinous and doesn't dissolve in ammonium.- \( \mathrm{Mg} ( \mathrm{OH})_2 \) and \( \mathrm{Ca} ( \mathrm{OH})_2 \) do not dissolve as much in \( \mathrm{NH}_{4}\mathrm{OH} / \mathrm{NH}_{4}\mathrm{Cl} \).
05

Conclusion

The precipitate which dissolves in excess \( \mathrm{NH}_{4}\mathrm{OH} / \mathrm{NH}_{4}\mathrm{Cl} \) and reduces volume while leaving a gelatinous residue is \( \mathrm{Zn} ( \mathrm{OH})_2 \), due to the formation of a soluble complex.

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Key Concepts

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

Precipitation Reactions
In inorganic chemistry, precipitation reactions occur when two soluble substances react to form an insoluble solid, known as a precipitate. This solid emerges from the solution because it has a lower solubility in the given solvent. Precipitation reactions are straightforward to identify.
  • They often result in a visible solid forming in a previously clear solution.
  • The precipitate typically has a different color or texture compared to the solution.
When a solution is diluted with water and boiled, certain metal hydroxides can precipitate because they become less soluble under these conditions. In the exercise, the initial white precipitate suggests the formation of a metal hydroxide, like zinc or aluminum hydroxide, both of which exhibit this behavior. Understanding the specific solubility rules for different ions helps to predict and identify the products of precipitation reactions effectively. These concepts are crucial for solving the given problem as they lay the foundation for predicting which substances will form solid precipitates in solution.
Complex Ion Formation
Complex ion formation occurs when a metal ion binds to one or more molecules or anions, forming a complex ion that can alter the solubility of the metal ion. In this context, adding excess \( \mathrm{NH}_4 \mathrm{OH} / \mathrm{NH}_4 \mathrm{Cl} \) affects how the precipitate behaves.
  • Zinc hydroxide \( \mathrm{Zn}(\mathrm{OH})_2 \) is known to dissolve in ammonium solutions because of the formation of complex ions like \([\mathrm{Zn}(\mathrm{NH}_3)_4]^{2+}\).
  • This complex formation reduces the solid precipitate, making it soluble in the solution.
In the exercise, when \( \mathrm{Zn}(\mathrm{OH})_2 \) dissolves in the ammonium solution, a complex ion is formed, leading to a decrease in the observable precipitate. This behavior distinguishes zinc hydroxide from aluminous hydroxides. For students, recognizing complex ion formation is key to understanding why certain precipitates dissolve in specific conditions. This helps in correctly identifying the compound behaviors in these reactions.
Amphoteric Hydroxides
Amphoteric hydroxides are intriguing because they can react with both acids and bases. Zinc hydroxide \( \mathrm{Zn}(\mathrm{OH})_2 \) is a classic example of an amphoteric hydroxide.
  • These hydroxides dissolve in basic solutions like \( \mathrm{NH}_4 \mathrm{OH} \) by forming complex ions.
  • They also react with acids, illustrating their dual ability to interact with different pH environments.
This type of behavior is crucial for solving the exercise. When the precipitate in the solution is treated with \( \mathrm{NH}_4 \mathrm{Cl} / \mathrm{NH}_4 \mathrm{OH} \), only amphoteric hydroxides will dissolve. This hints that the behavior is typical of zinc hydroxide, which decreases in mass as it converts to a soluble form. Understanding amphoteric behavior is essential for predicting the interactions of hydroxides in both acidic and basic mediums, illustrating why zinc hydroxide specifically reacts in the given problem.

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Most popular questions from this chapter

Anhydrous zinc chloride is prepared by heating of (a) \(\mathrm{ZnCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Zn}+\mathrm{HgCl}_{2}\) (c) \(\mathrm{Zn}+\mathrm{Cl}\) (d) both \((\mathrm{b})\) and \((\mathrm{c})\)

Amongst the following, the lowest degree of paramagnetism per mole of the compound at \(298 \mathrm{~K}\) will be shown by (a) \(\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CuSO}_{4} .5 \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{FeSO}_{4} .6 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NiSO}_{4} .6 \mathrm{H}_{2} \mathrm{O}\)

In the extraction of silver, \(\mathrm{Ag}_{2} \mathrm{~S}\) is dissolved in (a) \(\mathrm{HNO}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (c) \(\mathrm{HCl}\) (d) \(\mathrm{KCN}\)

In the process of extraction of gold roasted gold ore \(+\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{O}_{2} \rightarrow}[\mathrm{X}]+\mathrm{OH}^{-}\) \([\mathrm{X}]+\mathrm{Zn} \longrightarrow[\mathrm{Y}]+\mathrm{Au}\) \([\mathrm{X}]\) and \([\mathrm{Y}]\) are (a) \([\mathrm{X}]=\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-},[\mathrm{Y}]=\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{-2}\) (b) \([\mathrm{X}]=\left[\mathrm{Au}(\mathrm{CN})_{4}\right]^{-3},[\mathrm{Y}]=\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{-2}\) (c) \([\mathrm{X}]=\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-},[\mathrm{Y}]=\left[\mathrm{Zn}(\mathrm{CN})_{6}\right]^{-4}\) (d) \([\mathrm{X}]=\left[\mathrm{Au}(\mathrm{CN})_{4}\right]^{-},[\mathrm{Y}]=\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{-2}\)

Which of the following alloys contain \(\mathrm{Cu}\) and \(\mathrm{Zn} ?\) 1\. bronze 2\. brass 3\. gun metal 4\. type metal (a) \(2,3,4\) (b) \(1,2,3\) (c) 12 (d) all

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