Question

Determine whether \(\mathrm{Cd}^{2+}\) can be separated from \(\mathrm{Zn}^{2+}\) by bubbling \(\mathrm{H}_{2} \mathrm{~S}\) through a \(0.3 \mathrm{M} \mathrm{HCl}\) solution that contains \(0.005 \mathrm{M} \mathrm{Cd}^{2+}\) and \(0.005 \mathrm{M} \mathrm{Zn}^{2+} .\left(K_{\mathrm{spa}}\right.\) for CdS is \(8 \times 10^{-7} .\) )

Short answer

CdS can be separated if the \\( K_{sp} \\) for ZnS is sufficiently small, confirming selective precipitation of CdS.

Step-by-step solution

Calculate H+ Concentration

In a 0.3 M HCl solution, \([H^+] = 0.3 \, ext{M}\) because HCl is a strong acid and dissociates completely.

Write Solubility Product Expressions

The solubility product expression for \( CdS \) is \( K_{sp} = [Cd^{2+}][S^{2-}] \). For \( CdS \), we have \( K_{sp} = 8 \times 10^{-7} \). For \( ZnS \), the expression will be similar: \( K_{sp} = [Zn^{2+}][S^{2-}] \), although we don't need its value immediately.

Calculate Sulfide Ion Concentration for CdS Precipitation

To find \( [S^{2-}] \) needed for \( CdS \) precipitation: \[ [S^{2-}]_{CdS} = \frac{K_{sp, CdS}}{[Cd^{2+}]} = \frac{8 \times 10^{-7}}{0.005} = 1.6 \times 10^{-4} \]

Determine if H2S Can Separate Cd2+ and Zn2+

To separate \( Cd^{2+} \) from \( Zn^{2+} \), the condition \( [S^{2-}] < [S^{2-}]_{ZnS} \) must be true. Knowing that \( ZnS \) generally has a smaller \( K_{sp} \), if \( K_{sp, \, ZnS} \) is smaller than \( 1.6 \times 10^{-4} \, ext{M} \), then \( ZnS \) won't precipitate.

Conclusion

If \( [S^{2-}] \) in the solution reaches \( 1.6 \times 10^{-4} \) M, and the \( K_{sp} \) value for ZnS indeed suggests it is lesser than this sulfide threshold, then \( Cd^{2+} \) can be selectively precipitated without precipitating \( Zn^{2+} \).

Key concepts

Solubility Product Constant

The solubility product constant, often abbreviated as \( K_{sp} \), is a crucial concept in understanding how substances dissolve, or don't dissolve, in a solution. It describes the point at which solutes in an ionic solid will reach equilibrium in a saturated solution. This means no more solute can dissolve at a given temperature, such as when you mix salt in water and it won't dissolve further.

In the case of precipitating cadmium sulfide \( (CdS) \), its \( K_{sp} \) value, given as \( 8 \times 10^{-7} \), dictates when the cadmium and sulfide ions combine to form a solid. The lower the \( K_{sp} \), the less soluble the compound is in water.

This constant is determined by multiplying the molar concentrations of the dissociated ions at expansion to the power of their stoichiometric coefficients. For \( CdS \), this is given by:

• \( [Cd^{2+}] \times [S^{2-}] = 8 \times 10^{-7} \)

Recognizing this value helps us predict at what point cadmium ions will begin to precipitate out of solution, leaving other ions behind if managed correctly.

Chemical Separation Techniques

Chemical separation techniques allow us to separate components of a mixture based on their chemical properties. One such technique involves differences in solubility due to \( K_{sp} \) values. For this exercise, we're interested in separating cadmium ions \((Cd^{2+})\) from zinc ions \((Zn^{2+})\). By understanding the differing \( K_{sp} \) values of compounds like \( CdS \) and \( ZnS \), we can selectively precipitate one ion while the other remains dissolved.

• Adding \( H_2S \) to an acidic solution allows sulfide ions to react with metal ions to form precipitates.
• By carefully managing sulfide ion concentration, we can precipitate \( Cd^{2+} \) into \( CdS \) without affecting \( Zn^{2+} \).
• This depends on the condition \( [S^{2-}] \) needed to start precipitation of the second ion.

Below a certain \([S^{2-}]\), only \( CdS \) precipitates, preventing \( Zn^{2+} \) from forming \( ZnS \). Such precise control allows us to achieve thorough chemical separation, focusing on solubility differences between the two metal sulfides.

Aqueous Solutions Chemistry

In aqueous solution chemistry, compounds exist in a dissolved state, making them available for reactions. The interaction between water molecules and solutes forms the basis of these solutions. Water works as a solvent, breaking apart ionic compounds into ions. This environment allows ions to react with one another, as seen in the separation of \( Cd^{2+} \) and \( Zn^{2+} \) ions by selectively precipitating \( CdS \).

Adding \( H_2S \) introduces sulfide ions \((S^{2-})\) into the mixture, interacting with metal ions. The concentration of \( H^+ \) in the acidic \( HCl \) solution plays a key role here, as it affects the sulfide ion level, setting the stage for precipitation. A strong acid like \( HCl \) dissociates completely, dictating the \( S^{2-} \) concentration needed. As the reaction proceeds, the formation of a solid precipitate is a sign of exceeded \( K_{sp} \), visible as a turbidity in the solution.

Through this understanding of aqueous solutions, we manipulate reactions, allowing selective precipitation —a practical application in chemical separation.