Hydrogen sulfide is bubbled into a solution that is 0.10M in both Pb^2+ and Fe^2+ and 0.30 M in

step 1 Starting with the question, considering the first reaction, we have H2S plus 2H2O, which gives 2

Hydrogen sulfide is bubbled into a solution that is 0.10M in...

Hydrogen sulfide is bubbled into a solution that is 0.10$M$ in both $\mathrm{Pb}^{2+}$ and $\mathrm{Fe}^{2+}$ and 0.30 $\mathrm{M}$ in HCl. After the solution has come to equilibrium it is saturated with $\mathrm{H}_{2} \mathrm{S}\left(\left[\mathrm{H}_{2} \mathrm{S}\right]=0.10 M\right) .$ What concentrations of $\mathrm{Pb}^{2+}$ and $\mathrm{Fe}^{2+}$ remain in the solution? For a saturated solution of $\mathrm{H}_{2} \mathrm{S}$ we can use the equilibrium: $$ \mathrm{H}_{2} \mathrm{S}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{S}^{2-}(a q) \quad K=1.0 \times 10^{-26} $$ (Hint: The $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$ changes as metal sulfides precipitate. $)$

Hydrogen sulfide is bubbled into a solution that is 0.10$M$ in both $\mathrm{Pb}^{2+}$ and $\mathrm{Fe}^{2+}$ and 0.30 $\mathrm{M}$ in HCl. After the solution has come to equilibrium it is saturated with $\mathrm{H}_{2} \mathrm{S}\left(\left[\mathrm{H}_{2} \mathrm{S}\right]=0.10 M\right) .$ What concentrations of $\mathrm{Pb}^{2+}$ and $\mathrm{Fe}^{2+}$ remain in the solution? For a saturated solution of $\mathrm{H}_{2} \mathrm{S}$ we can use the equilibrium:
$$
\mathrm{H}_{2} \mathrm{S}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{S}^{2-}(a q) \quad K=1.0 \times 10^{-26}
$$
(Hint: The $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$ changes as metal sulfides precipitate. $)$

Key Concepts

Acid-Dependent Equilibria

This refers to the equilibria involving weak acids such as hydrogen sulfide (H?S) in solution. H?S does not completely dissociate but instead establishes an equilibrium where its degree of ionization depends on the pH of the solution. In a highly acidic medium (with a significant concentration of H?O? from HCl), the formation of S²? from H?S is heavily suppressed, which in turn influences the extent to which metal sulfides can precipitate.

Common Ion Effect

The common ion effect describes the shift in equilibrium that occurs when an ion common to the equilibrium is added to the system. Here, the presence of high concentrations of H?O? from HCl shifts the H?S ionization equilibrium toward the undissociated acid, greatly reducing the concentration of free sulfide ions. This suppression of S²? concentration limits the precipitation of metal sulfides despite the introduction of H?S into the solution.

Precipitation Reactions

Precipitation reactions occur when two soluble ions in solution combine to form an insoluble compound. In this context, metal ions such as Pb²? and Fe²? can react with sulfide ions (S²?) to form low-solubility metal sulfides, which precipitate out of solution, thereby decreasing the concentration of the free metal ions in the mixture.

Solubility Product (Ksp)

The solubility product constant, or Ksp, is used to describe the equilibrium between a sparingly soluble solid and its ions in solution. For metal sulfides, it represents the maximum product of the ion concentrations that can exist in solution without precipitation occurring. In this problem, while Ksp values for the metal sulfides are not provided explicitly, the effective concentration of S²?, controlled by another equilibrium, plays a similar role in determining how much precipitate forms.

Transcript

Please give Ace some feedback