Question

A solution is \(0.05 \mathrm{M}\) in \(\mathrm{Cu}^{2+},\) in \(\mathrm{Hg}^{2+},\) and in \(\mathrm{Mn}^{2+}\). Which sulfides will precipitate if the solution is made to be \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{S}(\mathrm{aq})\) and \(0.010 \mathrm{M} \mathrm{HCl}(\mathrm{aq}) ?\) For \(\mathrm{CuS}\), \(K_{\mathrm{spa}}=6 \times 10^{-16} ;\) for \(\mathrm{HgS}, K_{\mathrm{spa}}=2 \times 10^{-32} ;\) for \(\mathrm{MnS}\), \(K_{\mathrm{spa}}=3 \times 10^{7}\).

Short answer

HgS will precipitate first as it has the smallest Ksp value. Then, CuS will precipitate as it has the second smallest Ksp value. However, MnS will not precipitate as its Qsp is not higher than its Ksp value.

Step-by-step solution

Understand the fundamentals

The condition for precipitation to occur is when the ion product exceeds the solubility product (Ksp). In other words, a precipitate will form when [M2+][S2-] > Ksp. Precipitation will start with the metal whose sulfide has the lowest solubility product. To find the ion product, it’s important to know the concentration of the sulfide ion (S2-).

Calculate the concentration of the sulfide ion

\(\mathrm{H}_{2} \mathrm{S}\) will ionize in two steps: \(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{H}^{+}+\mathrm{HS}^{-}\) and \(\mathrm{HS}^{-} \rightarrow \mathrm{H}^{+}+\mathrm{S}^{2-}\). For the first reaction: \(K_{a1}=[H+][HS^{-}]/[H2S]\). For the second reaction: \(K_{a2}=[H+][S^{2-}]/[HS^{-}]\). Because HCl is a strong acid, we can reasonably assume that [H+] = 0.010 M. Hence, \([S^{2-}] = K_{a2}/[H+]\). Since we are not given Ka1 and Ka2, we will use the fact that for weak acids, Ka1 >> Ka2, thus conclude that the contribution to [S2-] will be primarily from the second ionization.

Compare the ion product with solubility product constants

Calculate the Qsp, for each metal sulfide. Qsp = ([M2+][S2-]). Where [M2+] = 0.05 M and [S2-] was calculated in Step2. Now, compare Qsp with the Ksp values for each metal sulfide. The sulfides will precipitate in order of increasing Ksp value, starting with the ion whose Qsp is more than the Ksp.

Key concepts

Chemical Equilibrium

Chemical equilibrium plays a central role in various chemical processes and is particularly important when discussing solubility products and precipitation reactions. It is the point at which the rate of the forward reaction equals the rate of the reverse reaction, meaning the concentrations of reactants and products remain constant over time. In the context of solubility, chemical equilibrium refers to the balance between the solid phase of a substance and its dissolved ions in solution. For a substance like a metal sulfide, the equilibrium involves its dissociation into metal ions and sulfide ions.

Solubility product constant, represented by the symbol Ksp, is a specific type of equilibrium constant that measures the solubility of a compound. The lower the Ksp value, the less soluble the compound is. When the solution contains ions that can form a compound with a Ksp, if the product of their concentrations (ion product) exceeds that constant, the compound will begin to precipitate, shifting the equilibrium towards the solid phase.

In the exercise given, the solubility products of various metal sulfides guide us to determine which of them will precipitate. By comparing the ion product with the Ksp, we know if the system is at equilibrium (when Qsp = Ksp) or if precipitation will occur (when Qsp > Ksp). This understanding is the backbone of predicting precipitation reactions in a solution.

Precipitation Reactions

Precipitation reactions are a subset of chemical reactions that result in the formation of an insoluble solid, known as a precipitate, from the reaction of soluble substances in a solution. They occur when the ion product of the reactants exceeds the solubility product (Ksp) of the potential precipitate.

In a practical sense, this means that the product of the concentrations of the reactive ions in solution must exceed a certain threshold for a precipitate to form. The steps provided in the exercise explain how to determine the point at which this occurs by calculating the ion product, Qsp, and comparing it to the given Ksp values for various metal sulfides.

Understanding precipitation is fundamental for numerous scientific and industrial processes, such as water purification, mineral extraction, and pharmaceutical formulation. Students are taught to calculate the concentrations of ions in solution, use the Ksp to predict the formation of precipitates, and comprehend how manipulating the concentrations, as shown in the exercise, affects the outcome.

Ion Product

The ion product, often represented as Qsp, is the product of the molar concentrations of the ions involved in a potential precipitate at any given moment, not just at equilibrium. It's crucial for predicting whether a precipitate will form in a solution. If the ion product is less than the solubility product (Ksp), no precipitate will form since the solution is unsaturated regarding the possible precipitate. Conversely, if the ion product is greater, the solution is supersaturated, and precipitation is likely.

In the solved exercise, calculating the Qsp was a critical step that required understanding the ionization of hydrogen sulfide in solution and the impact of the hydrogen ion concentration from hydrochloric acid on the availability of sulfide ions. Once the ion product for each metal sulfide could be determined, it was possible to systematically identify which sulfides will precipitate by comparing these values to the corresponding Ksp.

This concept not only helps in solving textbook problems but also in designing experiments and industrial processes where controlling the concentrations of ions is necessary to achieve a desired chemical outcome, such as minimizing scale buildup in boilers or separating desirable compounds in ore processing.