Problem 51 - A Complete Resource Book in Chemistry for JEE Main

Question

Which among the following is the least soluble? (a) \(\mathrm{MnS}\left(\mathrm{K}_{\mathrm{sp}}=7 \times 10^{-16}\right)\) (b) \(\mathrm{FeS}\left(\mathrm{K}_{\mathrm{sp}}=4 \times 10^{-15}\right)\) (c) \(\mathrm{PtS}\left(\mathrm{K}_{\mathrm{sp}}^{4}=8 \times 10^{-73}\right)\) (d) \(\mathrm{NiS}\left(\mathrm{K}_{\mathrm{sp}}=3 \times 10^{-12}\right)\)

Step-by-Step Solution

Answer

PtS is the least soluble with a K_{sp} of approximately 5.7 imes 10^{-19}.

Step 1: Understanding Solubility Product Constant

The solubility product constant (K_{sp}) indicates the solubility of a compound in water. A smaller K_{sp} value means the compound is less soluble. We will compare the given K_{sp} values to determine which compound is the least soluble.

Step 2: Identify Given K_{sp} Values

Review the K_{sp} values for each compound: - MnS: K_{sp} = 7 imes 10^{-16} - FeS: K_{sp} = 4 imes 10^{-15} - PtS: K_{sp}^{4} = 8 imes 10^{-73} - NiS: K_{sp} = 3 imes 10^{-12}.

Step 3: Dealing with the Fourth Power in PtS

The K_{sp}^{4} for PtS is given as 8 imes 10^{-73}. In comparing directly, we must first take the fourth root to find the actual K_{sp} for PtS: K_{sp}( ext{PtS}) = (8 imes 10^{-73})^{1/4} = 8^{1/4} imes 10^{-73/4} = 1.68 imes 10^{-18.25}.

Step 4: Compare K_{sp} Values

Look at the adjusted K_{sp} values to determine which is smallest: - MnS: K_{sp} = 7 imes 10^{-16} - FeS: K_{sp} = 4 imes 10^{-15} - PtS: K_{sp} = 5.7 imes 10^{-19} - NiS: K_{sp} = 3 imes 10^{-12} Based on these calculations, PtS has the smallest K_{sp} value, indicating that it is the least soluble.

Key Concepts

Solubility comparison

In the context of solubility, the solubility product constant, denoted as \( K_{sp} \), serves as a key indicator of how soluble a compound is in water. The general rule of thumb is: the smaller the \( K_{sp} \), the less soluble the compound. This is because a smaller \( K_{sp} \) value indicates that fewer ions are released into the solution, demonstrating limited solubility.

When tasked with comparing solubility, as outlined in the exercise, it's crucial to look at the \( K_{sp} \) values provided for each compound. The exercise provides four compounds, each with a specific \( K_{sp} \). A deeper understanding of these values allows us to rank the compounds based on their solubility. For instance, \( MnS \) with a \( K_{sp} = 7 \times 10^{-16} \) is more soluble than \( PtS \), once you've adjusted \( PtS \)'s given \( K_{sp} \) from \( 8 \times 10^{-73} \) to \( 5.7 \times 10^{-19} \), which turns out to be the smallest and thus least soluble.

Inorganic chemistry

Inorganic chemistry primarily deals with substances that aren't based on carbon-hydrogen bonds. A significant part of this field includes studying salts and minerals, which often participate in equilibrium reactions such as dissolution and precipitation. Understanding solubility through \( K_{sp} \) involves applying knowledge from equilibrium concepts.

Compounds like sulfides (MnS, FeS, PtS, NiS) are a classic part of inorganic chemistry. Their solubility is fundamental when predicting the extent to which they dissolve in aqueous solutions. This entails consideration of different factors such as temperature, ion interactions, and the presence of other compounds. In exam questions, you often compare solubility by using the solubility product constant, an application of Le Chatelier's principle and equilibrium constants, which are cornerstones of inorganic chemistry.

Ksp calculations

Calculating solubility product constants requires an understanding of equilibrium principles. The \( K_{sp} \) value represents the product of the concentrations of the ions each raised to the power of their coefficients in the balanced equation of dissociation.

For example, if you have a compound \( AB_2 \) which dissociates as \( AB_2 \rightarrow A^{2+} + 2B^{-} \), the \( K_{sp} \) expression is \([A^{2+}][B^{-}]^2\). In solving the exercise given, the complex calculation for \( PtS \), with its initial expression of \( K_{sp}^{4} = 8 \times 10^{-73} \), requires taking the fourth root to find the feasible \( K_{sp} \). This modification is essential for accurate comparison with \( MnS \), \( FeS \), and \( NiS \). By computing the fourth root for \( PtS \), students can accurately compare solubility limits using the refined \( K_{sp} \) value, thus pinpointing \( PtS \) as the compound with the least solubility.