Question

A solution containing \(\mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+}\) with a molar concentration of \(10^{-3} \mathrm{M}\) each is treated with \(10^{-16} \mathrm{M}\) sulphide ion solution. Which ions will precipitate first if \(K_{s p}\) of \(\mathrm{MnS}, \mathrm{FeS}, \mathrm{ZnS}\) and \(\mathrm{HgS}\) are \(10^{-15}, 10^{-23}, 10^{-20}\) and \(10^{-54}\) respectively? (a) FeS (b) \(\mathrm{MnS}\) (c) \(\mathrm{HgS}\) (d) \(\mathrm{ZnS}\)

Short answer

FeS will precipitate first because it has the highest Q/Ksp ratio, meaning its ion product exceeds the Ksp by the largest margin compared to MnS, ZnS, and HgS.

Step-by-step solution

Understand Precipitation and Solubility Product

To determine which ions will precipitate first, we need to compare the ion product (Q) with the solubility product constant (Ksp) for each compound. Precipitation begins when the ion product (concentration of cation times concentration of anion) exceeds the solubility product (Ksp). The compound with the lowest Ksp in relation to its ion product will precipitate first.

Calculate Ion Product (Q) for Each Compound

The ion product (Q) for a compound like M^(2+)S is given by \( Q = [\mathrm{M}^{2+}][\mathrm{S}^{2-}] \).Since we are given molar concentration of sulphide ions and the M^(2+), Q can be calculated for each metal sulphide.

Determine Which Compound Precipitates First

Compare Q with Ksp for each compound. The one with the Q value that exceeds its Ksp the most is the one that precipitates first. Here, we look for the highest Q/Ksp ratio to determine which compound precipitates first.

Calculating Q/Ksp Ratio

Calculate the ratio Q/Ksp for MnS, FeS, ZnS, and HgS using the provided molar concentrations and given Ksp values.

Key concepts

Ion Product

The ion product is a term used in the context of solution chemistry, particularly when discussing the solubility of sparingly soluble salts. It reflects the theoretical product of the concentrations of the ions that would form if a salt were to dissolve in solution. For example, if we consider a salt represented by the formula AB, where A is a cation and B is an anion, the ion product (\(Q\)) would be written as:
\[ Q = [A^+][B^-] \]
This equation shows the product of the molar concentration of the cation \[A^+\] and the molar concentration of the anion \[B^-\]. In the context of precipitation reactions, the ion product is used to predict whether a precipitate will form when two solutions are mixed together. If the ion product exceeds the solubility product constant (\(K_{sp}\)) of the salt, a precipitate will begin to form. This comparison helps chemists understand the conditions under which a compound will precipitate out of solution.

Solubility Product Constant

The solubility product constant (\(K_{sp}\)) is a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. It is the product of the concentrations of the ions at equilibrium, each raised to the power of their stoichiometric coefficient in the balanced equation for the dissolution of the salt. For instance, for a general salt AB that dissociates into \[ A^+ \] and \[ B^- \], the \(K_{sp}\) is expressed as:
\[ K_{sp} = [A^+]^m[B^-]^n \]
Where \(m\) and \(n\) represent the coefficients of the ions in the balanced dissolution equation. The \(K_{sp}\) value is constant for a given salt at a particular temperature and provides a measure of the salt's solubility in water. If the ion product is less than the \(K_{sp}\), the solution is unsaturated and more salt can dissolve; If it's equal, the solution is at equilibrium; and if it exceeds the \(K_{sp}\), the solution is supersaturated and a precipitate is expected to form. By comparing the ion product and the \(K_{sp}\), chemists can predict which compounds will precipitate under given conditions.

Ksp Calculations

Ksp calculations are an essential aspect of understanding precipitation chemistry. They allow us to quantitatively determine the solubility of a compound in a solution and predict whether a precipitate will form. When given the concentrations of ions in a solution, one can calculate the ion product and compare it to the known \(K_{sp}\) values to assess the likelihood of precipitation. The calculation steps typically involve:

1. Determining the ions involved in the equilibrium and their concentrations.
2. Applying the relationship \( Q = [Cation][Anion] \), where Q is the ion product.
3. Comparing the computed Q value with the corresponding \(K_{sp}\) of the salt.

In the provided example, we calculate Q for each metal sulphide using the given ion concentrations and use the individual \(K_{sp}\) values to determine which salt precipitates first. The calculations are an exercise in understanding the dynamic nature of equilibria in solution and the factors that influence the solubility of ionic compounds.

Selective Precipitation

Selective precipitation is a procedure used in chemistry to separate ions from a solution, based on their differing solubilities. By introducing a precipitant that selectively forms a solid with one or more types of ions, while leaving others in solution, it is possible to separate a mixture into its components. This is a practical application of the principles of precipitation chemistry, and it often relies on careful Ksp calculations to determine the proper conditions under which a specific ion will precipitate.
In a mixture of ions like \(\mathrm{Mn}^{2+}\), \(\mathrm{Fe}^{2+}\), \(\mathrm{Zn}^{2+}\), and \(\mathrm{Hg}^{2+}\), selecting the right precipitant and manipulating the concentrations can lead to the precipitation of one ion over the others. The process involves controlling the conditions so that the ion product for one ion reaches its \(K_{sp}\) before the others do. This highly controlled approach is invaluable in analytical chemistry, metal refining, and waste treatment, allowing for the efficient recovery or removal of specific components of a mixture.