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 [2002] (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 (Ksp) is \(4\mathrm{~S}^{3}\).

Step 1: Understand the Dissolution Reaction

The dissolution of calcium hydroxide \(\mathrm{Ca(OH)}_2\) in water can be represented as: \[ \mathrm{Ca(OH)}_2 (s) \rightleftharpoons \mathrm{Ca}^{2+} (aq) + 2\mathrm{OH}^- (aq) \] This equation shows that one mole of \( \mathrm{Ca(OH)}_2 \) dissolves to produce one mole of \( \mathrm{Ca}^{2+} \) ions and two moles of \( \mathrm{OH}^- \) ions in solution.

Step 2: Express Ion Concentrations in Terms of Solubility

If the molar solubility of \( \mathrm{Ca(OH)}_2 \) is \( \mathrm{S} \) mol/L, then the concentration of \( \mathrm{Ca}^{2+} \) ions in solution is also \( \mathrm{S} \) because one mole of \( \mathrm{Ca(OH)}_2 \) provides one mole of \( \mathrm{Ca}^{2+} \). The concentration of \( \mathrm{OH}^- \) ions is \( 2\mathrm{S} \) as dissolution produces two moles of \( \mathrm{OH}^- \) for each mole of \( \mathrm{Ca(OH)}_2 \) that dissolves.

Step 3: Write the Expression for Solubility Product (Ksp)

The solubility product \( K_{sp} \) is expressed as the product of the concentrations of the ions each raised to the power of their coefficients in the balanced equation. For \( \mathrm{Ca(OH)}_2 \), \[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^-]^2 \] Substitute ion concentration expressions in terms of solubility \( \mathrm{S} \): \[ K_{sp} = (\mathrm{S})(2\mathrm{S})^2 \]

Step 4: Simplify the Ksp Expression

Simplify the equation \[ K_{sp} = (\mathrm{S})(2\mathrm{S})^2 = (\mathrm{S})(4\mathrm{S}^2) = 4\mathrm{S}^3 \] Thus, the solubility product \( K_{sp} \) is equal to \( 4\mathrm{S}^3 \).

Key Concepts

Solubility of Calcium Hydroxide

Understanding the solubility of calcium hydroxide is essential when studying chemical equilibria. Calcium hydroxide, often known as slaked lime, is slightly soluble in water. Its ability to dissolve is quantified as 'molar solubility', which represents the number of moles of calcium hydroxide that can be dissolved in a litre of water. When calcium hydroxide dissolves, it splits into calcium ions and hydroxide ions. This is described by the dissolution equation: \[ \mathrm{Ca(OH)}_2 (s) \rightleftharpoons \mathrm{Ca}^{2+} (aq) + 2\mathrm{OH}^- (aq) \]Here, the calcium ions and hydroxide ions increase until equilibrium is achieved. The molar solubility "S" is the concentration of calcium ions, and it directly reflects how much \(\mathrm{Ca(OH)}_2\) can be dissolved under specific conditions.

Dissolution Reaction

A dissolution reaction shows how a substance dissolves in water, representing the process with a balanced chemical equation. For calcium hydroxide, its dissolution can be written as follows: \[ \mathrm{Ca(OH)}_2 \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^- \]This balance highlights two main points:- One mole of calcium hydroxide produces one mole of calcium ions.- It generates two moles of hydroxide ions for every mole that dissolves.Knowing this helps us establish how ions behave in solution. Plus, the balanced equation provides insight into the ratio and concentration of the ions, essential for calculating solubility and reactions in aqueous solutions.

Ion Concentration Expression

The ion concentration expression is crucial for deducing solubility products. It shows how the concentrations of ions are related to the solubility of a compound. When calcium hydroxide dissolves, the concentration of ions can be written in terms of its solubility \(\mathrm{S}\):- \(\mathrm{Ca}^{2+}\) ion concentration is \(\mathrm{S}\) mol/L.- \(\mathrm{OH}^-\) ion concentration is \(2\mathrm{S}\) mol/L, since each unit of calcium hydroxide results in two hydroxide ions.We use these expressions to find the solubility product \(K_{sp}\), representing how many ions can exist in a saturated solution before the compound starts settling out of the solution. This is calculated by the expression: \[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^-]^2 = (\mathrm{S})(2\mathrm{S})^2 = 4\mathrm{S}^3 \]This solubility product tells us the extent to which a compound can be dissolved, playing a critical role in determining reaction directions and solubility under varying conditions.