Problem 38
Question
The metal ion which is precipitated when \(\mathrm{H}_{2} \mathrm{~S}\) is passed with \(\mathrm{HCl}\) (a) \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{Mn}^{2+}\) (c) \(\mathrm{Ni}^{2+}\) (d) \(\mathrm{Zn}^{2+}\)
Step-by-Step Solution
Verified Answer
The metal ion that precipitates is \( \text{Cd}^{2+} \).
1Step 1: Understand the Reaction
When hydrogen sulfide gas (\( H_2S \)) is passed into a solution, it can lead to the precipitation of metal sulfides. The acidity (like the presence of \( HCl \)) affects which metal ions will precipitate.
2Step 2: Consider the Effect of \( HCl \)
In the presence of \( HCl \), the concentration of \( S^{2-} \) ions is reduced, affecting precipitation reactions. Only sulfides of metals with very low solubility will precipitate under these conditions.
3Step 3: Examine Solubility Products
\( CdS \), \( MnS \), \( NiS \), and \( ZnS \) have different solubility products (\( K_{sp} \) values). Among them, \( CdS \) has a very low \( K_{sp} \), meaning it is less soluble and more prone to precipitate even at low \( S^{2-} \) concentrations.
4Step 4: Compare Metal Ions
Cadmium sulfide (\( CdS \)) will precipitate in the presence of \( HCl \) and \( H_2S \) because it has a considerably lower solubility product compared to the others. \( MnS \), \( NiS \), and \( ZnS \) require more basic conditions to precipitate.
5Step 5: Conclusion
Since \( Cd^{2+} \) forms a very poorly soluble sulfide, \( CdS \), it is the ion that precipitates when \( H_2S \) is passed in the presence of \( HCl \).
Key Concepts
Solubility ProductMetal IonsHydrogen SulfideSolubility
Solubility Product
The solubility product, denoted as \(K_{sp}\), is a key concept in understanding how substances dissolve in a solution. It is a constant for a specific substance at a given temperature and represents the product of the concentrations of the ions in a saturated solution, each raised to the power of its coefficient in the balanced chemical equation. For a slightly soluble salt like \(CdS\), the expression for the solubility product is given by:
- \(K_{sp} = [Cd^{2+}][S^{2-}]\)
Metal Ions
Metal ions are simply atoms of metals that have lost electrons and carry a positive charge. In precipitation reactions, these ions usually react with anions to form insoluble compounds called precipitates.
The behavior of metal ions in solution is determined by the solubility of their possible compounds.
Cadmium ions (\(Cd^{2+}\)), manganese ions (\(Mn^{2+}\)), nickel ions (\(Ni^{2+}\)), and zinc ions (\(Zn^{2+}\)) react differently with sulfide ions depending on the acidity of the solution. A lower pH decreases the concentration of free \(S^{2-}\) ions, influencing which metal ions will precipitate.
The behavior of metal ions in solution is determined by the solubility of their possible compounds.
Cadmium ions (\(Cd^{2+}\)), manganese ions (\(Mn^{2+}\)), nickel ions (\(Ni^{2+}\)), and zinc ions (\(Zn^{2+}\)) react differently with sulfide ions depending on the acidity of the solution. A lower pH decreases the concentration of free \(S^{2-}\) ions, influencing which metal ions will precipitate.
- **Cadmium (\(Cd^{2+}\))** - Forms \(CdS\), which precipitates readily even under acidic conditions.
- **Manganese (\(Mn^{2+}\))**, **Nickel (\(Ni^{2+}\))**, **Zinc (\(Zn^{2+}\))** - These ions are more soluble in acidic solutions and require basic conditions to form their respective insoluble sulfides.
Hydrogen Sulfide
Hydrogen sulfide, with the chemical formula \(H_2S\), is a colorless gas known for its distinct smell of rotten eggs. When dissolved in water, it provides sulfide ions (\(S^{2-}\)), which can lead to precipitation of metal ions as metal sulfides.
The presence of additional acids like \(HCl\) influences this process. The added hydrogen ions from \(HCl\) reduce the concentration of \(S^{2-}\) ions by forming \(HS^{-}\) ions, so only compounds with very low solubility products will form precipitates in such an acidic environment.
This is particularly evident when \(H_2S\) is passed through a solution containing various metal ions; only those ions capable of forming compounds with very low solubility, such as \(CdS\), will precipitate.
The presence of additional acids like \(HCl\) influences this process. The added hydrogen ions from \(HCl\) reduce the concentration of \(S^{2-}\) ions by forming \(HS^{-}\) ions, so only compounds with very low solubility products will form precipitates in such an acidic environment.
This is particularly evident when \(H_2S\) is passed through a solution containing various metal ions; only those ions capable of forming compounds with very low solubility, such as \(CdS\), will precipitate.
Solubility
Solubility refers to the ability of a substance to dissolve in a solvent, forming a homogeneous solution at a specific condition of temperature and pressure. It is a crucial factor in determining whether a solid will form when ions in a solution interact.
Low solubility equates to a high tendency to precipitate as it indicates that very little of the substance can remain dissolved. For metal sulfides like \(CdS\), low aqueous solubility speaks to the formation of solid precipitates.
In the context of our example, cadmium sulfide (\(CdS\)) has low solubility compared to other metal sulfides, which results in its precipitation when \(H_2S\) is introduced in the presence of \(HCl\).
This process occurs because \(CdS\)'s solubility product is exceeded, causing it to fall out of solution as a solid salt.
Low solubility equates to a high tendency to precipitate as it indicates that very little of the substance can remain dissolved. For metal sulfides like \(CdS\), low aqueous solubility speaks to the formation of solid precipitates.
In the context of our example, cadmium sulfide (\(CdS\)) has low solubility compared to other metal sulfides, which results in its precipitation when \(H_2S\) is introduced in the presence of \(HCl\).
This process occurs because \(CdS\)'s solubility product is exceeded, causing it to fall out of solution as a solid salt.
Other exercises in this chapter
Problem 36
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The gas that turns lime water milky is (a) \(\mathrm{SO}_{2}\) (b) \(\mathrm{CO}_{2}\) (c) both of these (d) none of these
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Which of the following is not a preliminary test used to detect ions (a) brown ring test (b) borax bead test (c) flame test (d) permanganate test
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