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A white substance, \(X\) described as highly poisonous, dissolves in water. Its aqueous solution produces a scarlet precipitate with KI, soluble in excess KI solution giving a clear and colourless solution, Y. The aqueous solution of \(X\) gives a yellow precipitate with \(\mathrm{NaOH}\). What may be observed on adding \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{NaOH}\) to \(\mathrm{Y} ?\) (a) Brown precipitate (b) Black precipitate (c) Grey precipitate (d) Bright red precipitate

Short Answer

Expert verified
A brown precipitate forms.

Step by step solution

01

Identify X

The description of substance X suggests it is lead(II) nitrate, \( \text{Pb(NO}_3)_2 \), because it dissolves in water and forms a scarlet precipitate with KI, indicating the formation of lead(II) iodide, \( \text{PbI}_2 \). This precipitate is known to dissolve in excess KI to form a colorless solution due to the formation of a lead(II) iodide complex.
02

Identify Y

The scarlet precipitate \( \text{PbI}_2 \) dissolves in excess \( \text{KI} \), forming a soluble complex \( \text{K}_2[\text{PbI}_4] \). This clear and colorless solution is Y.
03

Understand the Reaction with NH4Cl and NaOH

When \( \text{NH}_4\text{Cl} \) and \( \text{NaOH} \) are added to solution Y (\( \text{K}_2[\text{PbI}_4] \)), they produce ammonium iodide and precipitate white \( \text{Pb(OH)}_2 \). On heating, the color may change.
04

Determine the Precipitate Color

The reaction of \( \text{PbI}_2 \) in solution leading to \( \text{Pb(OH)}_2 \) and subsequent heating or oxygenation typically leads to a brown color, consistent with the formation of additional lead oxide species in solution, suggesting a brown precipitate.

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Key Concepts

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

Lead(II) Nitrate
Lead(II) Nitrate, chemically represented as \( \text{Pb(NO}_3)_2 \), is a commonly used lead compound in various chemical reactions. It is known for its high solubility in water, which makes it a useful source of lead ions in aqueous solutions. When dissolved, it dissociates to form lead ions \( \text{Pb}^{2+} \) and nitrate ions \( \text{NO}_3^- \).
These lead ions are highly reactive and play a crucial role in precipitation reactions, especially when combined with other ions to form insoluble salts. Due to its toxicity, handle lead(II) nitrate with care in well-ventilated areas, and always follow safety protocols. Regularly replacing lead ions with less toxic alternatives in educational settings is a good practice.
Lead(II) Iodide
Lead(II) Iodide, \( \text{PbI}_2 \), is an interesting compound due to its characteristic color and insoluble nature. In pure form, it appears as a yellow crystalline solid. Upon reaction with potassium iodide (KI) in solution, it initially forms as a scarlet precipitate.
This vivid scarlet coloration helps in visually identifying the formation of lead(II) iodide in educational experiments. Moreover, the peculiar solubility properties of \( \text{PbI}_2 \) make it dissolve in excess iodide ions, forming a soluble complex. This complex nature not only fascinates students but also demonstrates the dynamic equilibrium of chemical systems.
Chemical Precipitation
Chemical precipitation is a foundational concept in chemistry, especially in understanding solution chemistry and reactions. It occurs when ions in solution combine to form an insoluble compound, known as a precipitate.
In the exercise, the precipitation of lead(II) iodide from the reaction of lead(II) nitrate and potassium iodide is a classic example. As the lead ions \( \text{Pb}^{2+} \) encounter iodide ions \( \text{I}^- \), they form \( \text{PbI}_2 \), which, due to its poor solubility in water, precipitates out of the solution.
  • The formation of a solid precipitate clearly indicates a chemical change.
  • Observing the color change provides a visible cue for reaction progress.
  • The reaction's reversibility illustrates chemical equilibrium concepts.
Understanding these reactions helps students connect theoretical knowledge with practical laboratory skills.
Complex Ions
Complex ions are specially structured ions that consist of a central metal ion bonded to one or more ligands. In the given scenario, when excess potassium iodide is added to the scarlet precipitate formed by lead(II) iodide, a complex ion is formed.
This ion is \( \text{K}_2[\text{PbI}_4] \), resulting in a clear and colorless solution. The process of forming complex ions highlights the solubility principles in chemistry.
Key aspects of complex ions include:
  • They enhance solubility of otherwise insoluble compounds.
  • Formation often shifts reaction equilibrium, leading to interesting solubility changes.
  • Visual observation, such as the transition from colored to colorless solutions, makes understanding complex ion formation intuitive.
Introducing students to complex ions enriches their comprehension of the intricate balance and versatility in chemical systems.

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Most popular questions from this chapter

How do we differentiate between \(\mathrm{Fe}^{3+}\) and \(\mathrm{Cr}^{3+}\) in group III? (a) by increasing \(\mathrm{NH}_{4}{ }^{+}\)ion concentration (b) by decreasing \(\mathrm{OH}^{-}\)ion concentration (c) by adding excess of \(\mathrm{NH}_{4} \mathrm{OH}\) solution (d) both (a) and (b)

Identify the correct order of solubility of \(\mathrm{Na}_{2} \mathrm{~S}, \mathrm{CuS}\) and \(\mathrm{ZnS}\) in aqueous medium. (a) \(\mathrm{Na}_{2} \mathrm{~S}>\mathrm{CuS}>\mathrm{ZnS}\) (b) \(\mathrm{Na}_{2} \mathrm{~S}>\mathrm{ZnS}>\mathrm{CuS}\) (c) \(\mathrm{CuS}>\mathrm{ZnS}>\mathrm{Na}_{2} \mathrm{~S}\) (d) \(\mathrm{ZnS}>\mathrm{Na}_{2} \mathrm{~S}>\mathrm{CuS}\)

Which of the following salts will form colourless beads in borax bead test? (a) \(\mathrm{CoCl}_{2}\) (b) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (d) \(\mathrm{CoC}_{2} \mathrm{O}_{4}\)

When \(\mathrm{H}_{2} \mathrm{~S}\) gas is passed through the HCl containing aqueous solution of \(\mathrm{CuCl}_{2}, \mathrm{HgCl}_{2}, \mathrm{BiCl}_{3}\) and \(\mathrm{CoCl}_{2}\), it does not precipitate out (a) \(\operatorname{CoS}\) (b) \(\mathrm{HgS}\) (c) \(\mathrm{CuS}\) (d) \(\mathrm{Bi}_{2} \mathrm{~S}_{3}\)

A green crystalline compound shows the following tests: (i) Its aqueous solution gives a dense white precipitate with Barium chloride solution (ii) Its aqueous solution gives a rose-red precipitate with dimethyl glyoxime and \(\mathrm{NH}_{4} \mathrm{OH}\). The compound is (a) \(\mathrm{NiSO}_{4}\) (b) \(\mathrm{FeSO}_{4}\) (c) \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (d) \(\mathrm{FeCl}_{3}\)

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