Problem 17 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A solution which is \(10^{-3} \mathrm{M}\) each in \(\mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+}\) is treated with \(10^{-16} \mathrm{M}\) sulphide ion. 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, which one will precipitate first? (a) \(\mathrm{FeS}\) (b) \(\mathrm{MgS}\) (c) \(\mathrm{HgS}\) (d) \(\mathrm{ZnS}\)

Step-by-Step Solution

Answer

HgS will precipitate first due to its higher ion product compared to its solubility product.

Step 1: Identify the Ion Product

First, calculate the ion product (MxnS) for each metal sulfide by multiplying the concentration of the metal ion by the concentration of sulfide ion: \(IP_{MnS} = [Mn^{2+}][S^{2-}] = 10^{-3} \times 10^{-16}\), \(IP_{FeS} = [Fe^{2+}][S^{2-}] = 10^{-3} \times 10^{-16}\), \(IP_{ZnS} = [Zn^{2+}][S^{2-}] = 10^{-3} \times 10^{-16}\), \(IP_{HgS} = [Hg^{2+}][S^{2-}] = 10^{-3} \times 10^{-16}\).

Step 2: Compare Ion Product with Solubility Product

Next, compare each ion product with its respective solubility product (K_{sp}). If the ion product exceeds the solubility product, precipitation occurs. \(10^{-19} \) for MnS \( (IP_{MnS} < 10^{-15}), \) \(10^{-19}\) for FeS \( (IP_{FeS} < 10^{-23}), \) \(10^{-19}\) for ZnS \( (IP_{ZnS} < 10^{-20}), \) \(10^{-19}\) for HgS \( (IP_{HgS} > 10^{-54}).\)

Step 3: Determine the First Precipitate

From the comparison, since \(IP_{HgS}\) is greater than \(K_{spHgS}\) while others have \(IP < K_{sp}\), HgS will precipitate first.

Key Concepts

Precipitation Reaction

A precipitation reaction occurs when ions in a solution combine to form a solid, known as a precipitate. This solid separates from the liquid because it is insoluble or only slightly soluble in the solution. For a precipitation reaction to occur, the concentration of ions in the solution must exceed the solubility product constant, often denoted as \(K_{sp}\). If the ions’ concentration exceeds this threshold, they start forming solid particles, leading to precipitation.
In the context of the given exercise, when a solution containing metal ions \((Mn^{2+}, Fe^{2+}, Zn^{2+}, Hg^{2+})\) is treated with sulfide ions \((S^{2-})\), these metal ions can form metal sulfides \((MnS, FeS, ZnS, HgS)\). The metal sulfide with the ion product greater than its solubility product is the one that will precipitate out first. Therefore, understanding when a precipitation reaction is initiated depends significantly on comparing ion products with their respective solubility products.

Ion Product

The ion product is a calculation used to predict whether a precipitate will form in a solution. It is calculated by multiplying the molar concentrations of the ions that form the compound. An ion product can be thought of as a snapshot of the ionic conditions in a solution at a given time.
In the exercise example, the ion product for each metal sulfide is determined by the product of the concentration of metal ions and sulfide ions: \([M^{2+}][S^{2-}]\). For instance, for MnS, it is calculated as \(10^{-3} \times 10^{-16}\).
After calculating the ion product, it is compared to the compound's solubility product (\(K_{sp}\)). If the ion product is greater than \(K_{sp}\), the compound will precipitate; otherwise, it remains dissolved. This comparison is essential in predicting whether precipitation will occur under specific conditions.

Metal Sulfide

Metal sulfide compounds are formed when metal ions combine with sulfide ions \((S^{2-})\). These compounds are usually insoluble in water, making them prone to precipitate from solutions. Each metal sulfide has a characteristic solubility product \((K_{sp})\) that indicates how much of the compound can dissolve in a solution before precipitation occurs.
In the exercise, metal sulfide compounds like \(MnS, FeS, ZnS,\) and \(HgS\) have different \(K_{sp}\) values. The smaller the \(K_{sp}\), the less soluble the compound is, and the more likely it will precipitate out of solution first. Specifically, \(HgS\) has the smallest \(K_{sp}\), which is why it precipitates first when the ion product exceeds this value, as demonstrated in the solution.

Chemical Equilibrium

Chemical equilibrium in the context of solubility occurs when the rate of dissolution of the compound into ions equals the rate of precipitation back into the solid form. This state is characterized by the solubility product \(K_{sp}\), which defines the maximum concentration of ions in a saturated solution where no more solid can dissolve.
In equilibrium, the concentrations of the reacting ions forming a precipitate do not change. When these concentrations (represented by the ion product) exceed the solubility product, it disrupts the equilibrium, leading to the formation of a solid precipitate. This concept of equilibrium is crucial for understanding how and when different compounds will precipitate under given conditions, as seen in determining which metal sulfide precipitates first in the exercise.