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Chapter 31: Problem 24

The ion that cannot be precipitated by both \(\mathrm{HCl}\) and \(\mathrm{H}_{2} \mathrm{~S}\) is (a) \(\mathrm{Pb}^{2+}\) (b) \(\mathrm{Cu}^{+}\) (c) \(\mathrm{Ag}^{+}\) (d) \(\mathrm{Sn}^{2+}\)

Short Answer

Expert verified
Ion \(\mathrm{Sn}^{2+}\) cannot be precipitated by both \(\mathrm{HCl}\) and \(\mathrm{H}_2\mathrm{S}\).

Step by step solution

01

Identify Ion Precipitated by HCl

Hydrochloric acid (HCl) can precipitate ions as chlorides if the metal chloride is insoluble in aqueous solution. From the given ions, \(\mathrm{Ag}^{+}\) is known to form an insoluble chloride, \(\mathrm{AgCl}\). Similarly, \(\mathrm{Pb}^{2+}\) forms \(\mathrm{PbCl}_2\) which is slightly soluble in cold water but precipitates out. \(\mathrm{Cu}^{+}\) and \(\mathrm{Sn}^{2+}\) ions do not generally form insoluble chlorides with HCl.
02

Identify Ion Precipitated by H2S in Acidic Solution

In acidic conditions, \(\mathrm{H}_2\mathrm{S}\) can precipitate metal sulfides which are insoluble. \(\mathrm{Pb}^{2+}\) forms \(\mathrm{PbS}\), \(\mathrm{Cu}^{+}\) forms \(\mathrm{Cu}_2\mathrm{S}\), \(\mathrm{Ag}^{+}\) forms \(\mathrm{Ag}_2\mathrm{S}\), and \(\mathrm{Sn}^{2+}\) forms \(\mathrm{SnS}\). All these are insoluble sulfides.
03

Conclusion Based on Precipitation

For an ion not to be precipitated by both \(\mathrm{HCl}\) and \(\mathrm{H}_2\mathrm{S}\), it must either form soluble chlorides or soluble sulfides. From steps 1 and 2, \(\mathrm{Sn}^{2+}\) does not form an insoluble chloride when treated with \(\mathrm{HCl}\) but forms an insoluble sulfide when treated with \(\mathrm{H}_2\mathrm{S}\). Therefore, \(\mathrm{Sn}^{2+}\) is the ion that cannot be precipitated by both \(\mathrm{HCl}\) and \(\mathrm{H}_2\mathrm{S}\).

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Key Concepts

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

Insoluble Chlorides
In the world of chemistry, precipitation reactions are vital for separating different ions in a solution. An important part of understanding these reactions is recognizing when a metal ion forms an insoluble chloride. When hydrochloric acid ( HCl ) is added to a solution, certain metal ions can become insoluble and precipitate out as chlorides. For example:
  • Silver ions ( Ag^{+} ) form silver chloride ( AgCl ), which is insoluble in water and precipitates as a solid.
  • Lead ions ( Pb^{2+} ) can form lead chloride ( PbCl_{2} ). Lead chloride is slightly soluble in cold water but becomes less soluble and precipitates upon cooling or in the presence of more chloride ions.
  • Copper ( Cu^{+} ) and tin ( Sn^{2+} ) do not typically form insoluble chlorides with HCl , allowing them to remain dissolved.
Understanding which ions form insoluble chlorides helps in identifying and separating them during chemical analysis.
Insoluble Sulfides
Just like chlorides, certain metal ions can form insoluble sulfides when treated with hydrogen sulfide ( H_2S ) in an acidic solution. This is another method to separate ions in a mixture. Let's go over a few examples:
  • Lead ions ( Pb^{2+} ) form lead sulfide ( PbS ), a characteristic black precipitate.
  • Copper ions ( Cu^{+} ) create copper(I) sulfide ( Cu_2S ), another insoluble black solid.
  • Silver ions ( Ag^{+} ) result in silver sulfide ( Ag_2S ), further illustrating the widespread occurrence of metal sulfides in chemical precipitation.
  • Tin ions ( Sn^{2+} ) become tin(II) sulfide ( SnS ), which is also insoluble, reinforcing its predictable precipitation behavior.
Knowing these reactions helps in the practical field of chemical ion identification, especially in processes requiring the removal or isolation of specific metal ions.
Chemical Ion Identification
Chemical ion identification is crucial in analytical chemistry, where the aim is to determine what ions are present in a given solution. This involves combining knowledge of insoluble chlorides and sulfides to predict reactions:
  • Reactions with HCl help identify ions that form insoluble chlorides by observing precipitates.
  • Using H_2S in acidic conditions allows us to identify metals that form insoluble sulfides.
  • Not knowing these typical reactions means missing out on vital data useful for separating ions.
For example, in identifying ions in a sample, knowing that Sn^{2+} doesn't form an insoluble chloride with HCl aids in recognizing its presence after sulfide testing. Being adept at identifying chemical ions through these reactions helps advance one's skills in chemical analysis and research.

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Most popular questions from this chapter

A solution of a metal ion when treated with KI gives a red precipitate which dissolves in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is (a) \(\mathrm{Pb}^{2+}\) (b) \(\mathrm{Hg}^{2+}\) (c) \(\mathrm{Cu}^{2+}\) (d) \(\mathrm{Co}^{2+}\)

A gas ' \(\mathrm{X}^{\prime}\) is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas ' \(\mathrm{Y}\) '. Identify ' \(\mathrm{X}^{\prime}\) and 'Y' (a) \(\mathrm{X}=\mathrm{CO}_{2}, \mathrm{Y}=\mathrm{Cl}_{2}\) (b) \(\mathrm{X}=\mathrm{Cl}_{2}, \mathrm{Y}=\mathrm{CO}_{2}\) (c) \(\mathrm{X}=\mathrm{Cl}_{2}, \mathrm{Y}=\mathrm{H}_{2}\) (d) \(\mathrm{X}=\mathrm{H}_{2}, \mathrm{Y}=\mathrm{Cl}_{2}\)

A solution which is \(10^{-3} \mathrm{M}\) each in \(\mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+}\) is treated with \(10^{-16} \mathrm{M}\) sulphide ion. If \(K_{s p}\) of \(\mathrm{MnS}, \mathrm{FeS}, \mathrm{ZnS}\) and \(\mathrm{HgS}\) are \(10^{-15}\), \(10^{-23}, 10^{-20}\) and \(10^{-54}\) respectively, which one will precipitate first? (a) \(\mathrm{FeS}\) (b) \(\mathrm{MgS}\) (c) \(\mathrm{HgS}\) (d) \(\mathrm{ZnS}\)

Sodium Carbonate cannot be used in place of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) for the identification of \(\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+}\) and \(\mathrm{Sr}^{2+}\) ions (in group V) during mixture analysis because : (a) \(\mathrm{Mg}^{2+}\) ions will also be precipitated. (b) Concentration of \(\mathrm{CO}_{3}^{2-}\) ions is very low. (c) Sodium ions will react with acid radicals. (d) \(\mathrm{Na}^{+}\)ions will interfere with the detection of \(\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+}, \mathrm{Sr}^{2+}\) ions.

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