Problem 267 - A Complete Resource Book in Chemistry for JEE Main

Question

The molar solubility (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) of a sparingly soluble salt \(\mathrm{MX}_{4}\) is 's'. The corresponding solubility product is Ksp.s is given in terms of Ksp by the relation [2004] (a) \(\mathrm{s}=(\mathrm{Ksp} / 256)^{1 / 5}\) (b) \(\mathrm{s}=(128 \mathrm{Ksp})^{1 / 4}\) (c) \(\mathrm{s}=(\mathrm{Ksp} / 128)^{1 / 4}\) (d) \(\mathrm{s}=(256 \mathrm{Ksp})^{1 / 5}\)

Step-by-Step Solution

Answer

The correct relation is (a) \( \mathrm{s}=(\mathrm{Ksp} / 256)^{1 / 5}\).

Step 1: Identify the Salt's Dissociation Equation

The salt \( \text{MX}_4 \) dissociates in water to form one \( \text{M}^{4+} \) ion and four \( \text{X}^- \) ions. The dissolution reaction can be represented as: \[ \text{MX}_4(s) \rightleftharpoons \text{M}^{4+}(aq) + 4\text{X}^-(aq) \]

Step 2: Express Concentrations in Terms of 's'

Replace concentrations in terms of molar solubility \( s \). When \( s \) moles of \( \text{MX}_4 \) dissolve, the concentrations at equilibrium will be \( [\text{M}^{4+}] = s \) and \( [\text{X}^-] = 4s \).

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

Using the concentrations, we can write the expression for \( K_{sp} \) as: \[ K_{sp} = [\text{M}^{4+}][\text{X}^-]^4 = (s)(4s)^4 \] This simplifies to: \[ K_{sp} = s \cdot 256s^4 \]

Step 4: Simplify the Ksp Expression

Combine and simplify the terms from the expression: \[ K_{sp} = 256s^5 \]

Step 5: Solve for Molar Solubility 's' in Terms of Ksp

To find \( s \), solve the equation \( 256s^5 = K_{sp} \) for \( s \): \[ s^5 = \frac{K_{sp}}{256} \] Taking the fifth root on both sides, \[ s = \left(\frac{K_{sp}}{256}\right)^{1/5} \]

Step 6: Select the Correct Answer

Compare the result \( s = (\frac{K_{sp}}{256})^{1/5} \) with the given options. The correct choice is (a) \( \mathrm{s}=(\mathrm{Ksp} / 256)^{1 / 5}\).

Key Concepts

Molar Solubility

Molar solubility is an important concept in chemistry that tells us the number of moles of a solute that can dissolve in a liter of solution to reach saturation at a particular temperature. In the context of sparingly soluble salts, like \(\text{MX}_4\), molar solubility refers to the number of moles of \(\text{MX}_4\) that dissolve to form a saturated solution.

• "Saturated solution" means that no additional solute can dissolve; the solution contains the maximum amount of dissolved solute at equilibrium.
• Molar solubility is expressed in units of \(\text{mol L}^{-1}\).
• Consider that the solubility of compounds can vary greatly; sparingly soluble salts, like \(\text{MX}_4\), have low molar solubility in comparison to more soluble salts.

Understanding molar solubility helps predict how much salt can dissolve in a given amount of solvent, a key aspect in processes like precipitation and dissolution.

Dissociation Equation

One of the initial steps in determining the solubility product of any salt, is writing its dissociation equation. This equation represents how a solid salt dissociates into its component ions in an aqueous solution.
For the salt \( \text{MX}_4 \), the dissociation equation is:
\[ \text{MX}_4(s) \rightleftharpoons \text{M}^{4+}(aq) + 4\text{X}^-(aq) \]

• "(s)" indicates the salt is in solid form, while "(aq)" shows the ions are aqueous, meaning they are dissolved in water.
• The double arrow denotes a dynamic equilibrium; the reactions for dissociation and recombination happen at the same rate.
• The stoichiometry of the salt, which is 1:4 for \(\text{MX}_4\), guides the calculation of equilibrium concentrations. For every mole of \(\text{MX}_4\) that dissolves, one mole of \(\text{M}^{4+}\) and four moles of \(\text{X}^-\) ions are formed.

Crafting the dissociation equation provides the fundamental understanding needed to progress to equilibrium concentration calculations.

Equilibrium Concentration

Once the dissociation equation is established, the next step is identifying equilibrium concentrations, which describe the concentrations of ions when an equilibrium state is reached. Using the molar solubility "s" of the salt \( \text{MX}_4 \):

• If "s" moles of \(\text{MX}_4\) dissolve, \([\text{M}^{4+}] = s\) and \([\text{X}^-] = 4s\).
• Equilibrium concentration lets you see how the initial solute concentration changes after dissociation.
  

By substituting these concentrations into the solubility product (\(K_{sp}\)) expression, you can connect the molar solubility to the solubility product constant. This step is crucial for understanding the dynamics of dissolution in sparingly soluble salts and for solving related chemistry problems.

Chemical Equilibrium

The concept of chemical equilibrium is central in understanding reactions dynamics at a molecular level. Here, equilibrium represents a state where the rate of the forward reaction equals the rate of the backward reaction. In the case of sparingly soluble salts like \(\text{MX}_4\):

• The equilibrium lies slightly towards the undissolved salt, as it is sparingly soluble.
• We express equilibrium with the solubility product constant, \(K_{sp}\), which helps determine how a salt behaves in solution.
• \(K_{sp}\) is an equilibrium constant specific to dissolving solids, indicating the product of ion concentrations raised to the power of their stoichiometric coefficients.
  

Understanding chemical equilibrium allows students to predict the solubility of salts under different conditions, an essential skill in many practical and theoretical chemistry applications.