Question

\(\Lambda\) saturated solution of \(\mathrm{II}_{2} \mathrm{~S}\) in \(0.1 \mathrm{M} \mathrm{IICl}\) at \(25^{\circ} \mathrm{C}\) contains a \(\mathrm{S}^{2-}\) ion concentration of \(10^{-23} \mathrm{~mol} \mathrm{~L}^{-1}\). The solubility products of some sulphidcs are: \(\mathrm{CuS}=10^{-44}\), \(\mathrm{FeS}=10^{-14} ; \mathrm{MnS}=10^{-15}\) and \(\mathrm{CdS}=10^{-25} .\) If \(0.01 \mathrm{M}\) solutions of these salts in \(1 \mathrm{M}\) IICl are saturated with \(\mathrm{H}_{2} \mathrm{~S}\), which of these will be precipitated? (1) All (2) All except MnS (3) All except MnS and FeS (4) Only CuS

Short answer

Only CuS precipitates.

Step-by-step solution

Determine Sulfide Ion Concentration

The given concentration of \(\text{S}^{2-}\) ion in \(0.1 \text{ M IICl}\) is \(10^{-23} \text{ mol L}^{-1}\).

Calculate Ion Product for Each Salt

For each salt, calculate the ion product using \( \text{IP} = [\text{M}^{2+}][\text{S}^{2-}] \). Here, \([\text{M}^{2+}] = 0.01 \text{ M} \) for all salts. \[ \text{IP}_{\text{CuS}} = 0.01 \times 10^{-23} = 10^{-25} \] \[ \text{IP}_{\text{FeS}} = 0.01 \times 10^{-23} = 10^{-25} \] \[ \text{IP}_{\text{MnS}} = 0.01 \times 10^{-23} = 10^{-25} \] \[ \text{IP}_{\text{CdS}} = 0.01 \times 10^{-23} = 10^{-25} \].

Compare Ion Product with Solubility Product

For precipitation to occur, \( \text{IP} \) must be greater than or equal to \( \text{K}_{\text{sp}} \). Compare the calculated \(\text{IP}\) for each salt with their respective \( \text{K}_{\text{sp}} \): \[ \text{CuS}: 10^{-25} = 10^{-44} \rightarrow \text{Precipitates} \] \[ \text{FeS}: 10^{-25} < 10^{-14} \rightarrow \text{Does not Precipitate} \] \[ \text{MnS}: 10^{-25} < 10^{-15} \rightarrow \text{Does not Precipitate} \] \[ \text{CdS}: 10^{-25} < 10^{-25} \rightarrow \text{Precipitates} \]

Determine Which Salts Precipitate

Based on the comparison, conclude which salts will precipitate. Both \( \text{CuS} \) and \( \text{CdS} \) precipitate.

Key concepts

sulfide ion concentration

In the given exercise, the concentration of sulfide ions \( \text{S}^{2-} \) in a saturated solution is critical. The sulfide ion concentration in this problem is given as \( 10^{-23} \text{ mol L}^{-1} \). This very low concentration impacts the solubility and precipitation of various sulfide salts. To understand why this matters, recall that sulfide ions react with metal ions to form insoluble sulfide compounds. The low concentration of \( \text{S}^{2-} \) means fewer ions are available to react, influencing whether a precipitate forms or not.

ion product calculation

Ion product (IP) is used to predict whether a precipitate will form in a solution. It is calculated by multiplying the concentrations of the ions involved. In this exercise, the ion product for each metal sulfide is calculated using the formula: \[ \text{IP} = [\text{M}^{2+}][\text{S}^{2-}] \] where \[ [\text{M}^{2+}] \] is the concentration of the metal ion, and \[ [\text{S}^{2-}] \] is the sulfide ion concentration. Given \[ [\text{M}^{2+}] = 0.01 \text{ M} \] for all salts, the ion product for them is: \[ \text{IP} = 0.01 \times 10^{-23} = 10^{-25} \]. This value will be compared to the solubility product constant to decide if precipitation occurs.

solubility product constant

The solubility product constant (Ksp) is a unique value for each compound that indicates how much of that compound can dissolve in water. It is temperature-dependent and specific for different salts. In this problem, the provided Ksp values for the sulfides are: \[ \text{CuS} = 10^{-44}, \text{FeS} = 10^{-14}, \text{MnS} = 10^{-15}, \text{CdS} = 10^{-25} \]. By comparing the calculated ion product (IP) with the Ksp, we can predict if precipitation occurs. For example, if \[ \text{IP} \] is greater than or equal to \[ \text{Ksp} \], the compound will precipitate.

precipitation criteria

To determine if precipitation occurs, we compare the ion product (IP) of a salt with its solubility product constant (Ksp). The criteria are: \[ \text{If IP _greater than Ksp_ -> Precipitation occurs} \] \[ \text{If IP _less than Ksp_ -> No precipitation} \]. For instance: \[ \text{CuS}: 10^{-25} = 10^{-44} \rightarrow \text{Precipitates} \] \[ \text{FeS}: 10^{-25} _less than 10^{-14} \rightarrow \text{Does not Precipitate} \] Using the criteria, we determine that \[ \text{CuS and CdS} \] will precipitate, while \[ \text{FeS and MnS} \] will not.

chemical equilibrium

In a saturated solution, a dynamic equilibrium exists between the solid and dissolved ions of a salt. This balance is described by the solubility product constant (Ksp). For any salt, at equilibrium: \[ \text{M}_{\text{solid}}^2+ + 2 \text{S}^{2-} \rightleftharpoons \text{M}^{2+} + 2 \text{S}^{2-} \] If a salt is in equilibrium, its ion product (IP) will be equal to \[ \text{Ksp} \]. Any disturbance, like adding more \[ \text{M}^{2+} \] or \[ \text{S}^{2-} \] ions, will shift the equilibrium. This principle guides us in predicting if a salt will precipitate.