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Consider the following observation \(\mathrm{M}^{\mathrm{n}+}+\mathrm{HCl} \longrightarrow \mathrm{white}\) precipitate \(\stackrel{\Delta}{\longrightarrow}\) Water soluble The metal ion \(\mathrm{M}^{\mathrm{n}+}\) will be (a) \(\mathrm{Pb}^{2+}\) (b) \(\mathrm{Hg}^{2+}\) (c) \(\mathrm{Sn}^{2+}\) (d) \(\mathrm{Ag}^{+}\)

Short Answer

Expert verified
The metal ion Mn+ is Pb2+.

Step by step solution

01

Identify Reaction Conditions

We observe that the compound forms a white precipitate when combined with HCl. This suggests a chloride complex or a chloride precipitate formation. Furthermore, the addition of heat (D) dissolves this precipitate in water, indicating a specific reaction characteristic.
02

Analyze Reactivity with HCl

Among the given ions, lead (Pb^{2+}), mercury (Hg^{2+}), tin (Sn^{2+}), and silver (Ag^{+}) can form chlorides that precipitate with HCl. Therefore, these elements might be candidates based on the first reaction step with HCl.
03

Assess Solubility with Heat

After assessing the formation of a white precipitate, it is important to consider how it reacts when heated. PbCl_2, which forms from Pb^{2+}, is known to be sparingly soluble in water at higher temperatures, unlike  the chlorides of  other options that are either insoluble even with heat or soluble without the need for heating.
04

Choose the Metal Ion

Considering the formation of a white precipitate in reaction with HCl and partial solubility in hot water, the most appropriate choice is Pb^{2+}. Lead(II) chloride is the only one that fits both conditions: it precipitates as a white solid and becomes soluble in hot water.

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Key Concepts

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

Chloride Complex Formation
When certain metal ions react with hydrochloric acid (HCl), they can form what is known as a chloride complex. This involves the metal ions combining with chloride ions. A complex makes an insoluble compound, appearing as a white precipitate. This reaction with chloride ions is crucial in identifying metal ions in solution.

Key metal ions that form chloride complexes include lead (Pb\(^{2+}\)), mercury (Hg\(^{2+}\)), tin (Sn\(^{2+}\)), and silver (Ag\(^{+}\)). Each of these ions forms different types of chlorides, though all may produce a white precipitate.

Chlorine complexes highlight the unique nature of coordination chemistry, where ligands (like chloride ions) bond to a central metal ion, forming a larger complex. This plays a critical role in various chemical reactions.
Solubility Rules
Solubility rules are guidelines that help predict the solubility of different compounds in water. They are fundamental to understanding how metal chlorides behave when exposed to heat and solvents. Most simple metal chlorides are either soluble or insoluble in water without external factors affecting them. However, lead(II) chloride (PbCl\(_2\)) is special.

Here are some solubility rules to remember:
  • Most chloride salts are soluble, but those of silver, lead, and mercury are exceptions.
  • PbCl\(_2\) is not very soluble in cold water but becomes more soluble at higher temperatures.
  • Mercury, silver, and tin chlorides generally remain insoluble even when heated.
Knowing these rules helps determine which metal ion might be present based on whether its chloride form will dissolve after heating.
Reaction with HCl
When metal ions interact with hydrochloric acid, they undergo a reaction that often results in the formation of a precipitate. The type and behavior of this precipitate can indicate the presence of specific metal ions. Here is how the reaction typically unfolds:
  • Metal ions in solution meet HCl, resulting in an insoluble metal chloride.
  • This metal chloride appears as a white precipitate in the solution.
  • Application of heat can change the solubility of this precipitate, revealing more about its identity.
The process of adding HCl to metal ions and observing precipitation is a classic test in chemistry, critical for identifying unknown metal ions.
Metal Ion Identification
Metal ion identification through chemical reactions is a crucial analytical process in chemistry. Here, we focus on using reactions with HCl and solubility behavior to pinpoint the specific metal ion:

Given a reaction that produces a white precipitate with HCl, and observing how it responds to heating gives clues about the ion's identity. If the precipitate dissolves upon heating, it suggests the presence of lead(II) ion, owing to the unique solubility properties of lead(II) chloride.

Let's recap the role of solubility in identification:
  • Silver and mercury chlorides remain insoluble, even when heated, eliminating them as candidates.
  • Tin chloride may mislead initially, but its reactivity differs from that of lead.
  • Lead chloride proves unique due to its increased solubility in hot water, making it distinct.
Each metal ion's specific characteristics become apparent through these steps, confirming their identity with confidence.

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Most popular questions from this chapter

An organic compound ' \(\mathrm{X}\) ' with a molecular formula \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\) when reacts with excess of methyl magnesium chloride following by acidification gives an alcohol ' \(\mathrm{Y}\) ' only, when this alcohol ' \(\mathrm{Y}\) ' is oxidized with \(\mathrm{NaOCl}\) followed by acidification, acidic acid is formed. Here the compound ' \(\mathrm{X}\) ' is (a) Propyl methanoate (b) Isopropyl methanoate (c) Ethyl ethanoate (d) Both (a) and (b)

A light greenish coloured salt was soluble in water. On passing \(\mathrm{H}_{2} \mathrm{~S}\) into the solution, a black ppt. was obtained which dissolved readily in \(\mathrm{HCl}\). The metal ion present is (a) \(\mathrm{Fe}^{2+}\) (b) \(\mathrm{Mn}^{2+}\) (c) \(\mathrm{Ni}^{2+}\) (d) \(\mathrm{Co}^{2+}\)

In the chemical reactions, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}+\mathrm{CHCl}_{3}+3 \mathrm{KOH} \longrightarrow\) (a) \(+(\mathrm{b})+3 \mathrm{H}_{2} \mathrm{O}\) the compounds (a) and (b) are respectively: (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CONH}_{2}\) and \(3 \mathrm{KCl}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NC}\) and \(\mathrm{K}_{2} \mathrm{CO}_{3}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NC}\) and \(3 \mathrm{KCl}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CN}\) and \(3 \mathrm{KCl}\)

Three separate samples of a solution of single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. \(\mathrm{NaCl}\) and one formed a black precipitate with \(\mathrm{H}_{2} \mathrm{~S} .\) The salt could be (a) \(\mathrm{AgNO}_{3}\) (b) \(\mathrm{MnSO}_{4}\) (c) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) (d) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\)

Which of the following sulphates are soluble in water? (a) \(\mathrm{PbSO}_{4}\) (b) \(\mathrm{CuSO}_{4}\) (c) \(\mathrm{BaSO}_{4}\) (D) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\)

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