Problem 259 - A Complete Resource Book in Chemistry for JEE Main

Question

Solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(\mathrm{S}\) mol litre \(^{-1} .\) The solubility product (Ksp) under the same condition is (a) \(4 \mathrm{~S}^{3}\) (b) \(3 \mathrm{~S}^{4}\) (c) \(4 \mathrm{~S}^{2}\) (d) \(\mathrm{S}^{3}\)

Step-by-Step Solution

Answer

The solubility product is (a) \(4S^3\).

Step 1: Write the Dissociation Equation

Calcium hydroxide, \(\text{Ca(OH)}_2\), dissociates in water according to the equation:\[ \text{Ca(OH)}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{OH}^- (aq) \] This indicates that one mole of \(\text{Ca(OH)}_2\) produces one mole of \(\text{Ca}^{2+}\) ions and two moles of \(\text{OH}^-\) ions.

Step 2: Express in Terms of Solubility

If the solubility of \(\text{Ca(OH)}_2\) is \(S\) mol L\(^{-1}\), then the concentration of \(\text{Ca}^{2+}\) ions is \(S\) and the concentration of \(\text{OH}^-\) ions is \(2S\), because each formula unit produces 2 \(\text{OH}^-\) ions.

Step 3: Write the Solubility Product Expression

The solubility product, \(K_{sp}\), of \(\text{Ca(OH)}_2\) is given by the following expression based on its equilibrium concentrations:\[ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \] Substituting the values in terms of \(S\), we get:\[ K_{sp} = S \cdot (2S)^2 \]

Step 4: Simplify the Expression

Simplify the expression obtained for \(K_{sp}\):\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \] Thus, \(K_{sp} = 4S^3\).

Key Concepts

Calcium Hydroxide Dissociation

Understanding the dissociation of calcium hydroxide, \(\text{Ca(OH)}_2\), is crucial to knowing how it dissolves in water and breaks down into its ions. When placed in water, calcium hydroxide dissociates according to the equation:\[\text{Ca(OH)}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{OH}^- (aq) \]This equation demonstrates that each mole of calcium hydroxide produces one mole of calcium ions \(\text{Ca}^{2+}\) and two moles of hydroxide ions \(\text{OH}^-\). This stoichiometry is fundamental, as it forms the basis for calculating the solubility product. By knowing the number of ions formed, we can understand how these ions contribute to the overall ionic concentration in the solution.
Learning this concept helps students predict how substances like calcium hydroxide behave in different solutions.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of ions in solution once equilibrium is reached. For \(\text{Ca(OH)}_2\), if its solubility is given as \(S\) mol/L, then the equilibrium concentration of \(\text{Ca}^{2+}\) ions is \(S\) mol/L. Since each unit of calcium hydroxide produces two \(\text{OH}^-\) ions, the concentration of \(\text{OH}^-\) ions is \(2S\) mol/L.

• Calcium Ions Concentration: Directly equals the solubility, \(S\).
• Hydroxide Ions Concentration: Doubled due to the two ions produced per formula unit, making it \(2S\).

Understanding this relationship is vital for solving problems involving ionic equilibrium and provides the groundwork for calculating the solubility product \(K_{sp}\). Translating solubility into equilibrium concentrations allows easy substitution into mathematical expressions.

Ksp Calculation

The solubility product constant, \(K_{sp}\), is a reflection of the solubility of a compound under conditions of dynamic equilibrium. For calcium hydroxide, it is calculated from the concentrations of dissociated ions. To find \(K_{sp}\), we use the expression:\[K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2\]Substituting the expressions based on cellular diffusion:\[K_{sp} = S \times (2S)^2\]This becomes:\[K_{sp} = S \times 4S^2 = 4S^3\]Thus, the solubility product \(K_{sp}\) equals \(4S^3\). Knowing \(K_{sp}\) is essential for understanding how soluble a compound is in a solution. It shows the limits of solute that can dissociate before both reactants and products settle into equilibrium. Such calculations are fundamental in chemistry for predicting the behavior of ionic compounds in aqueous solutions.