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 (Ksp) is \( 4S^3 \), option (a).

Step 1: Write the Dissociation Equation

Calcium hydroxide dissociates in water according to the following equation:\[ \text{Ca(OH)}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq) \]Each formula unit of \( \text{Ca(OH)}_2 \) produces one \( \text{Ca}^{2+} \) ion and two \( \text{OH}^- \) ions.

Step 2: Define the Solubility (S)

Let \( S \) be the molar solubility of \( \text{Ca(OH)}_2 \), meaning the concentration of \( \text{Ca}^{2+} \) ions in the solution is \( S \) mol/L, and the concentration of \( \text{OH}^- \) ions is \( 2S \) mol/L because for each formula unit that dissolves two hydroxide ions are produced.

Step 3: Write the Expression for Ksp

The solubility product expression is given by:\[ \text{Ksp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \]Substituting the concentrations from Step 2 gives:\[ \text{Ksp} = (S)(2S)^2 \]

Step 4: Simplify the Expression

Simplify the expression for \( \text{Ksp} \):\[ \text{Ksp} = S \times (4S^2) = 4S^3 \]

Step 5: Determine the Correct Option

Compare the simplified expression for \( \text{Ksp} \) with the given options. The correct expression for \( \text{Ksp} \) is \( 4S^3 \), which corresponds to option (a).

Key Concepts

Solubility of Calcium Hydroxide

Calcium hydroxide is a compound that is slightly soluble in water. Its solubility is determined by its ability to dissolve to form a saturated solution.
The solubility, often denoted as \( S \), is the concentration of a solute that can dissolve in water to form a saturated solution. For calcium hydroxide, this can be expressed as \( S \) mol/L.
This means in a saturated solution of calcium hydroxide, the concentration of calcium ions \( \text{Ca}^{2+} \) is equal to \( S \) mol/L.

• The solubility depends on factors such as temperature, pressure, and the presence of other ions.
• Increasing the temperature often increases solubility, allowing more calcium hydroxide to dissolve.

It is this equilibrium state that is crucial for calculating the solubility product (Ksp), which describes the product of the concentrations of the ions in a saturated solution.

Dissociation Equation

The dissociation equation describes how a compound separates into its ions when it dissolves in water.
For calcium hydroxide, the dissociation can be expressed with the following reaction:
\[ \text{Ca(OH)}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq) \]
Here, each mole of calcium hydroxide produces one mole of \( \text{Ca}^{2+} \) ions and two moles of \( \text{OH}^- \) ions.

• This balanced equation is fundamental for understanding the relationship between the dissolved ions and the original compound.
• It helps in determining the concentrations of the ions, which is crucial for calculating the Ksp.

Knowing the dissociation equation allows us to derive the concentrations of the ions in terms of the solubility \( S \). This is key when relating the solubility to the solubility product.

Ksp Calculation

The solubility product, or Ksp, is a constant that quantifies the product of the ion concentrations in a saturated solution at equilibrium.
For calcium hydroxide, the expression is derived from:
\[ \text{Ksp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \]
Given that the concentration of \( \text{Ca}^{2+} \) is \( S \) and that of \( \text{OH}^- \) is \( 2S \), you can substitute these values into the Ksp expression:
\[ \text{Ksp} = (S)(2S)^2 = S \times 4S^2 = 4S^3 \]

• The calculation is straightforward once the ions' concentrations are known from the dissociation equation.
• The value of Ksp is unique to the compound and specific conditions such as temperature.

Ksp is essential in predicting whether a precipitate will form when two solutions are mixed and helps understand the dissolution of salts in different environments.