Problem 111 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility of \(\mathrm{AgCl}\) in moles per litre when its solubility product is \(1.56 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\) is (a) \(0.576 \times 10^{-8} \mathrm{~mol} /\) litre (b) \(1.056 \times 10^{-4} \mathrm{~mol} /\) litre (c) \(1.249 \times 10^{-5} \mathrm{~mol} /\) litre (d) \(1.478 \times 10^{-6} \mathrm{~mol} /\) litre

Step-by-Step Solution

Answer

The solubility of AgCl is \(1.249 \times 10^{-5}\) mol/L (option c).

Step 1: Understanding Solubility Product

The solubility product constant (Ksp) for a salt like AgCl is a measure of the extent to which the salt can dissociate in water to form its constituent ions under equilibrium conditions. For AgCl, this can be represented by the equation: \( \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \). Here, the Ksp is given as \(1.56 \times 10^{-10}\).

Step 2: Express Solubility in Terms of Ksp

Let the molar solubility of AgCl be \(s\) mol/L. At equilibrium, the concentration of Ag\(^+\) ions and Cl\(^-\) ions will each be \(s\) mol/L as well, because each formula unit of AgCl produces one Ag\(^+\) ion and one Cl\(^-\) ion. Thus, \( K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s \times s = s^2 \).

Step 3: Solve for Molar Solubility

Given \( K_{sp} = 1.56 \times 10^{-10} \), we have: \( s^2 = 1.56 \times 10^{-10} \). To find \(s\), take the square root of both sides: \( s = \sqrt{1.56 \times 10^{-10}} \).

Step 4: Perform the Calculation

Calculate \(s\) by taking the square root: \( s = \sqrt{1.56 \times 10^{-10}} \approx 1.249 \times 10^{-5} \text{ mol/L} \).

Step 5: Conclusion

The molar solubility of \(\text{AgCl}\) in water at \(25^{\circ} \mathrm{C}\) given the Ksp of \(1.56 \times 10^{-10}\) is \(1.249 \times 10^{-5} \text{ mol/L} \). This matches option (c).

Key Concepts

Molar Solubility

Understanding molar solubility is essential when predicting how much of a salt can dissolve in a solvent like water to form a saturated solution. Molar solubility specifically refers to the number of moles of a solute that can be dissolved per liter of solution to reach saturation. This concept is particularly crucial when dealing with sparingly soluble salts, like silver chloride (AgCl).
To find molar solubility, we often use the solubility product constant, or Ksp, which provides a direct relationship between solubility and equilibrium concentrations of the ions formed in solution. For example, if the Ksp of AgCl is given as a small value like 1.56 \( \times \) 10^{-10}, it indicates that only a tiny amount of the salt can dissolve in water at equilibrium.
In exercises, you'll set up an equation based on the dissociation of the salt (like AgCl dissociating into Ag\(^+\) and Cl\(^-\)) and equate it to Ksp to solve for the molar solubility \(s\). Thus, for AgCl: \([\mathrm{Ag}^+] = [\mathrm{Cl}^-] = s\), leading to \(K_{sp} = s^2\). Solving this gives you the molar solubility.

Equilibrium

Equilibrium in the context of solubility refers to the state where the rate at which the salt dissolves into its ions equals the rate at which the ions recombine to form the solid salt. This dynamic balance means no net change occurs in the concentration of ions, though exchanges continually happen.
For a saturated solution of a salt like AgCl, which slowly dissolves, equilibrium is achieved when the solution can dissolve no more solute under given conditions. This steady state is described mathematically by the Ksp value. Ksp is derived under equilibrium conditions and is temperature-dependent, indicating at 25°C, this particular salt achieves equilibrium solubility represented by its Ksp value.
The concept of equilibrium helps us predict and control the conditions under which more solute might dissolve or precipitate from a solution, making it invaluable in chemical processes and reactions.

Dissociation

Dissociation refers to the process by which a compound breaks down into its constituent ions in a solution. For salts like AgCl, dissociation occurs when the salt is added to water, resulting in silver ions (Ag\(^+\)) and chloride ions (Cl\(^-\)) separating and dispersing throughout the solvent.
This process is vital as it determines the ionic concentration in solution, directly affecting the overall solubility and the solubility product constant (Ksp). Understanding dissociation is also key to calculating molar solubility since each dissociation of AgCl produces equal amounts of Ag\(^+\) and Cl\(^-\), leading to the formulation \([\mathrm{Ag}^+] = [\mathrm{Cl}^-] = s\), where \(s\) is the molar solubility.
Dissociation is influenced by factors such as temperature and presence of other ions in the solution, which may shift the equilibrium position, enhancing or reducing the degree of solubility, as proposed by Le Chatelier's principle.