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Which of the following is not precipitated as sulphide by passing \(\mathrm{H}_{2} \mathrm{~S}\) in presence of conc. HCl (a) arsenic (b) lead (c) copper (d) cadmium

Short Answer

Expert verified
Arsenic does not precipitate as a sulphide under these conditions.

Step by step solution

01

Understanding Sulphide Precipitation

When passing \(H_{2}S\) through a solution, metal ions can form insoluble sulphides that precipitate out. However, the presence of conc. HCl suppresses the sulphide precipitation of some metal ions. Specifically, under these conditions, the metal sulphides of lead, copper, and cadmium are more likely to precipitate, whereas arsenic sulphide does not precipitate.
02

Analyzing Each Option

1. **Arsenic** - Typically forms soluble sulphides in acidic environments, so it does not precipitate as a sulphide in the presence of conc. HCl.2. **Lead** - Lead sulphide (PbS) is poorly soluble and will precipitate in the presence of \(H_{2}S\) regardless of conc. HCl.3. **Copper** - Copper sulphide (CuS) forms readily in acidic solutions and will precipitate.4. **Cadmium** - Cadmium sulphide (CdS) is also insoluble and will precipitate.From this analysis, arsenic is the element whose sulphide does not normally precipitate in these conditions.
03

Conclusion

Based on the solubility of respective sulphides, arsenic does not form a precipitate as its sulphide in the presence of \(H_{2}S\) and conc. HCl. The others (lead, copper, cadmium) form insoluble sulphides that precipitate.

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Key Concepts

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

Insoluble Sulphides
Insoluble sulphides are compounds formed when metal ions react with sulphide ions and create a solid that doesn't dissolve in water. This process is crucial in chemistry for separating or identifying certain metals in a mixture.
When hydrogen sulphide (H鈧係) is introduced into a solution containing metal ions, these ions can combine with the sulphide ions from H鈧係 to form insoluble sulphides. This makes them precipitate out of the solution as a solid.
These metal sulphides are often characterized by their low solubility, meaning they don鈥檛 dissolve easily in water or even in acidic conditions.
  • **Lead Sulphide (PbS)** - A common example of an insoluble sulphide that forms as a black precipitate when H鈧係 is bubbled through a lead solution.
  • **Copper Sulphide (CuS)** - Also forms a black precipitate and is poorly soluble in water.
  • **Cadmium Sulphide (CdS)** - Recognized by its yellow precipitate, CdS is another example of an insoluble sulphide.
This behavior is used in qualitative inorganic analysis to determine the presence of certain metal ions in a solution.
Metal Ions
Metal ions are atoms of metal that carry a positive charge due to the loss of one or more electrons. In an aqueous solution, these ions are typically surrounded by water molecules and can participate in various chemical reactions.
They play a crucial role in forming sulphide precipitates. Different metal ions react differently with sulphide ions based on their charge and size, affecting which sulphides are insoluble and thus precipitate out.
**Impact on Precipitation:**
  • Metal ions like copper (Cu虏鈦), lead (Pb虏鈦), and cadmium (Cd虏鈦) readily form insoluble sulphides in sulphide precipitation processes.
  • Some metal ions, such as arsenic (as As鲁鈦 or As鈦碘伜), form soluble sulphides in acidic environments, meaning they stay dissolved and do not precipitate out in the presence of concentrated HCl.
Understanding the behavior of different metal ions helps chemists select the right conditions for precipitating specific compounds.
Hydrogen Sulphide
Hydrogen sulphide ( H鈧係 ) is a colorless gas that smells like rotten eggs. In chemistry, it's used as a reagent because it can donate sulphide ions, which are essential for sulphide precipitation.
By introducing H鈧係 into metal ion solutions, chemists can cause certain metal ions to react with sulfide ions, forming a precipitate of the metal sulphide. This is an essential step in separating and identifying different metals.
**Key Characteristics:**
  • H鈧係 behaves as a weak acid in water and releases sulphide ions (S虏鈦), which are crucial for forming metal sulphides.
  • It is particularly effective in forming precipitates in less acidic environments, as strong acids can limit its ability to release sufficient sulphide ions into the solution.
This property is why it's often used in analytical chemistry to induce precipitation of metal ions as sulphides.
Element Precipitation Conditions
Element precipitation conditions refer to the environment that allows certain compounds to precipitate out of solution. In the case of sulphide precipitation, this involves adjusting factors like pH and the presence of other chemicals, such as concentrated HCl.
**Influence of pH and Concentrated HCl:**
  • In more acidic conditions, especially in the presence of concentrated HCl, the ability of H鈧係 to form insoluble precipitates changes.
  • Some metal ions, like lead, copper, and cadmium, form insoluble sulphides even in acidic solutions because their sulphides are stable and poorly soluble. These conditions discourage arsenic sulphide from forming a precipitate because it remains soluble.
By understanding these conditions, one can control which elements precipitate as sulphides, aiding in the separation and analysis of complex mixtures in solutions.

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Most popular questions from this chapter

Which one among the following pairs of ions can not be separated by \(\mathrm{H}_{2} \mathrm{~S}\) in dilute hydrochloric acid? (a) \(\mathrm{Bi}^{3+}, \mathrm{Sn}^{4+}\) (b) \(\mathrm{Al}^{3+}, \mathrm{Hg}^{2+}\) (c) \(\mathrm{Zn}^{2+}, \mathrm{Cu}^{2+}\) (d) \(\mathrm{Ni}^{2+}, \mathrm{Cu}^{2+}\)

Borax \(\left[\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} .10 \mathrm{H}_{2} \mathrm{O}\right]\) when heated on platinum loop it gives a dark transparent glass like bead. The hot bead is dipped in the salt till it reacts with transition metal oxide. It produces characteristic bead of meta borate. $$ \begin{array}{ll} \text { Colour of the bead } & \text { Ion } \\ \text { Blue green or light blue } & \mathrm{Cu}^{2+} \\ \text { Yellow } & \mathrm{Fe}^{2+} \text { or } \mathrm{Fe}^{3+} \\ \text { Green } & \mathrm{Cr}^{3+} \\ \text { Violet } & \mathrm{Mn}^{2+} \\ \text { Dark blue } & \mathrm{Co}^{2+} \\ \text { Brown } & \mathrm{Ni}^{2+} \end{array} $$ The hybridisation of \(\mathrm{B}\) in borax is (a) sp (b) \(\mathrm{sp}^{2}\) (c) \(\mathrm{sp}^{3}\) (d) Both (B) and (C)

Mark the correct statement. (a) V group basic radicals precipitate as carbonates (b) IV group basic radicals precipitate as sulphides (c) I group basic radicals precipitate as chlorides (d) all the above statements are correct

A solution, when diluted with \(\mathrm{H}_{2} \mathrm{O}\) 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) \(\mathrm{Mg}(\mathrm{OH})_{2}\) (d) \(\mathrm{Ca}(\mathrm{OH})_{2}\)

A salt which on heating with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) gives violet vapours is (a) bromide (b) nitrate (c) iodide (d) sulphate

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