Problem 79 - A Complete Resource Book in Chemistry for JEE Main

Question

Solubility of \(\mathrm{MX}_{2}\) type electrolyte is \(0.5 \times 10^{-4} \mathrm{~mol}\) \(\mathrm{L}^{-1}\). Then find out Ksp of electrolytes. (a) \(5 \times 10^{-12}\) (b) \(25 \times 10^{-10}\) (c) \(1 \times 10^{-13}\) (d) \(5 \times 10^{-13}\)

Step-by-Step Solution

Answer

(d) \(5 \times 10^{-13}\)

Step 1: Understanding the Dissolution Equation

The dissolution of the electrolyte \(\mathrm{MX}_2\) can be represented by the equation:\[\mathrm{MX}_2 (s) \rightleftharpoons \mathrm{M}^{2+} (aq) + 2\mathrm{X}^- (aq)\] This implies that 1 mole of \(\mathrm{MX}_2\) will produce 1 mole of \(\mathrm{M}^{2+}\) ions and 2 moles of \(\mathrm{X}^-\) ions.

Step 2: Determining Ion Concentrations

Given the solubility of \(\mathrm{MX}_2\) is \(0.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\), the concentration of \(\mathrm{M}^{2+}\) will be the same as the solubility, which is \(0.5 \times 10^{-4}\). For \(\mathrm{X}^-\), the concentration will be twice the solubility: \(2 \times 0.5 \times 10^{-4} = 1 \times 10^{-4}\).

Key Concepts

MX2 type electrolyte

When dealing with an electrolyte of the form \(\mathrm{MX}_2\), it's important to understand its composition and behavior in a solution. The formula \(\mathrm{MX}_2\) indicates a compound made of one metal ion \((\mathrm{M}^{2+})\) and two halide or other anionic species \((\mathrm{X}^- )\). This means that when \(\mathrm{MX}_2\) dissolves in water, it separates into one cationic species and two anionic species. This is a typical behavior for salts like calcium fluoride \((\mathrm{CaF}_2)\), where one unit of the electrolyte results in three separate ions in the solution. Such knowledge is vital when predicting the behavior of substances in a chemical solution as it directly affects the ion concentration calculations and the solubility product constant \((K_{sp})\). Understanding the structure of \(\mathrm{MX}_2\) compounds lays a strong foundation for studying chemical equilibria and solubility phenomena.

Ion Concentrations

Once we understand the dissolution equation of \(\mathrm{MX}_2\), the next step is determining ion concentrations in the solution. Given that the solubility of \(\mathrm{MX}_2\) is \(0.5 \times 10^{-4}\; \mathrm{mol}\; \mathrm{L}^{-1}\), we can derive the concentration of the ions produced. The important points here are:

• The concentration of \(\mathrm{M}^{2+}\) ions will be equal to the solubility, which is \(0.5 \times 10^{-4}\).
• Since two anions are released per formula unit, the concentration of \(\mathrm{X}^{-}\) ions will be twice that, calculating as:\[2 \times 0.5 \times 10^{-4} = 1 \times 10^{-4}\; \mathrm{mol}\; \mathrm{L}^{-1}\]

These concentrations are crucial for calculating the solubility product \(K_{sp}\). Accurate calculation of these values allows chemists to better understand the balance between salts and dissolved ions in solutions.

Dissolution Equation

The dissolution equation captures the essence of how \(\mathrm{MX}_2\) behaves when it dissolves in a solution. It is represented as:\[\mathrm{MX}_2 (s) \rightleftharpoons \mathrm{M}^{2+} (aq) + 2\mathrm{X}^- (aq)\]This predicts that one mole of solid \(\mathrm{MX}_2\) splits into one mole of \(\mathrm{M}^{2+}\) ions and two moles of \(\mathrm{X}^-\) ions in the solution.The concept of reversibility, as indicated by the arrow pointing both directions, means the reaction can either proceed forward (dissolving the solid into ions) or go backward (forming the solid from ions). Knowing this equilibrium expression allows us to write the equilibrium constant, also called the solubility product \((K_{sp})\), for this process as:\[K_{sp} = [\mathrm{M}^{2+}][\mathrm{X}^-]^2\]By inserting the calculated ion concentrations, one can solve for \(K_{sp}\), thus predicting the solubility behavior in different conditions. Understanding this dissolution helps guide processes like purification, recrystallization, and predicting how substances behave in natural or industrial settings.