Question

Can \(\mathrm{Fe}^{2+}\) be separated from \(\mathrm{Sn}^{2+}\) by bubbling \(\mathrm{H}_{2} \mathrm{~S}\) through a \(0.3 \mathrm{M} \mathrm{HCl}\) solution that contains \(0.01 \mathrm{M} \mathrm{Fe}^{2+}\) and \(0.01 \mathrm{M} \mathrm{Sn}^{2+} ?\) A saturated solution of \(\mathrm{H}_{2} \mathrm{~S}\) has \(\left[\mathrm{H}_{2} \mathrm{~S}\right] \approx 0.10 \mathrm{M}\). Values of \(K_{\text {spa }}\) are \(6 \times 10^{2}\) for \(\mathrm{FeS}\) and \(1 \times 10^{-5}\) for SnS.

Short answer

Yes, \(\mathrm{Fe}^{2+}\) remains in solution while \(\mathrm{Sn}^{2+}\) precipitates as \(\mathrm{SnS}\).

Step-by-step solution

Understand the Chemical Reaction

When bubbling \( \mathrm{H}_2 \mathrm{S} \) through the solution, \( \mathrm{Fe}^{2+} \) and \( \mathrm{Sn}^{2+} \) can react with \( \mathrm{S}^{2-} \). The following reactions may occur:\[ \mathrm{Fe}^{2+} + \mathrm{S}^{2-} \rightleftharpoons \mathrm{FeS} \]\[ \mathrm{Sn}^{2+} + \mathrm{S}^{2-} \rightleftharpoons \mathrm{SnS} \]

Calculate Ion Product for Precipitation

The ion product \( Q_{sp} \) for precipitation is given by the equation:\[ Q_{sp} = [\text{M}^{2+}][\text{S}^{2-}] \]Assuming complete dissociation of \( \mathrm{H}_2 \mathrm{S} \) in water, the concentration of \( \mathrm{S}^{2-} \) is the same as \([\mathrm{H}_2 \mathrm{S}] \approx 0.10 \mathrm{M} \).

Compare Ion Product with Solubility Product

Determine if \( Q_{sp} \) is greater than \( K_{sp} \) for each sulfide to decide if precipitation occurs.For \( \mathrm{FeS} \):\[ Q_{sp, Fe} = [0.01][0.10] = 0.001 \]\[ Q_{sp, Fe} = 0.001 \,\text{is much less than}\, K_{sp, Fe} = 6 \times 10^{2} \]For \( \mathrm{SnS} \):\[ Q_{sp, Sn} = [0.01][0.10] = 0.001 \]\[ Q_{sp, Sn} = 0.001 \,\text{is greater than}\, K_{sp, Sn} = 1 \times 10^{-5} \]

Conclusion on Separation

Since \( Q_{sp, Sn} \) exceeds \( K_{sp, Sn} \), \( \mathrm{Sn}^{2+} \) will precipitate as \( \mathrm{SnS} \). However, since \( Q_{sp, Fe} \) is less than \( K_{sp, Fe} \), \( \mathrm{Fe}^{2+} \) will remain in solution. Thus, \( \mathrm{Fe}^{2+} \) and \( \mathrm{Sn}^{2+} \) can be separated.

Key concepts

Solubility Product

Understanding the solubility product is crucial for predicting the formation of precipitates in solutions. The solubility product constant, denoted as \(K_{sp}\), represents the extent to which a compound can dissolve in water. It is the product of the molar concentrations of the ions in a saturated solution of a sparingly soluble salt, each raised to the power of its stoichiometric coefficient.
For instance, for a generic salt \(AB\) that dissolves into \(A^+\) and \(B^-\), the expression for \(K_{sp}\) is:

• \(K_{sp} = [A^+][B^-]\)

In the context of our specific problem, this constant helps us determine whether compounds like \(\text{FeS}\) or \(\text{SnS}\) will remain dissolved or form a precipitate when \(H_2S\) is bubbled through the solution.

Ion Product

The ion product, \(Q_{sp}\), is similar in concept to the solubility product. However, while the solubility product refers to equilibrium conditions, the ion product deals with current concentrations in a solution not necessarily at equilibrium.
To calculate \(Q_{sp}\), you use the actual concentrations of the ions. For a compound \(MX\) dissolving into \(M^+\) and \(X^-\), the ion product is:

• \(Q_{sp} = [M^+][X^-]\)

If \(Q_{sp} > K_{sp}\), precipitation will occur, meaning more solid will form. If \(Q_{sp} < K_{sp}\), the solution remains unsaturated and no precipitate forms. This concept allows us to predict the behavior of our solution with \(\text{Fe}^{2+}\) and \(\text{Sn}^{2+}\) ions.

Precipitation Reaction

Precipitation reactions occur when ions in a solution form an insoluble compound, precipitating out of the solution. This is guided by comparing \(Q_{sp}\) to \(K_{sp}\).
Here's a quick guide to predict precipitation:

• If \(Q_{sp} > K_{sp}\), the solution is supersaturated, and precipitation will occur.
• If \(Q_{sp} = K_{sp}\), the solution is at equilibrium with the solid and saturated, which typically leads to no change.
• If \(Q_{sp} < K_{sp}\), the solution is unsaturated, and no precipitate will form.

In our problem, \(\text{SnS}\) precipitates because \(Q_{sp, Sn} > K_{sp, Sn}\), while \(\text{Fe}^{2+}\) remains dissolved as \(Q_{sp, Fe} < K_{sp, Fe}\).

FeS and SnS

Iron(II) sulfide \((\text{FeS})\) and tin(II) sulfide \((\text{SnS})\) are both products of possible precipitation when \(\text{Fe}^{2+}\) and \(\text{Sn}^{2+}\) react with \(\text{S}^{2-}\) ions, respectively.
The \(K_{sp}\) values are markedly different for these two compounds:

• \(K_{sp, FeS} = 6 \times 10^2\)
• \(K_{sp, SnS} = 1 \times 10^{-5}\)

This large disparity means that \(\text{Sn}^{2+}\) readily forms a precipitate because even low concentrations of \(\text{Sn}^{2+}\) will exceed its \(K_{sp}\) when \(\text{S}^{2-}\) is present. Meanwhile, \(\text{FeS}\) remains dissolved due to its much higher \(K_{sp}\), enabling the separation of these metal ions in solution.

Chemical Equilibrium

Chemical equilibrium describes the state in which the concentrations of reactants and products remain constant over time because the forward and reverse reactions occur at the same rate. It is a crucial principle in understanding how and why precipitation reactions might occur or stop.
For equilibrium involving \(\text{Fe}^{2+}\) and \(\text{Sn}^{2+}\) ions in our solution:

• When \(\text{SnS}\) precipitates, it alters the equilibrium as the concentration of \(\text{Sn}^{2+}\) ions decreases in solution.
• This shift allows \(\text{Fe}^{2+}\) to remain in the dissolved state since the equilibrium conditions favor no precipitation at these concentrations.

Understanding chemical equilibrium helps explain the behavior of ions in solution and the conditions under which separation through precipitation is possible.