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) \(1.249 \times 10^{-5} \mathrm{~mol} / \mathrm{litre}\) (d) \(1.478 \times 10^{-6} \mathrm{~mol} /\) litre

Short answer

The solubility of \( \mathrm{AgCl} \) is approximately \( 1.249 \times 10^{-5} \text{ mol/L} \), which corresponds to option (c).

Step-by-step solution

Understand the Solubility Product

The solubility product, denoted as \( K_{sp} \), is a constant at a given temperature which indicates the level at which a solute can dissolve in solution. For \( \text{AgCl} \), \( K_{sp} = [\text{Ag}^+][\text{Cl}^-] = 1.56 \times 10^{-10} \) at 25°C. This means the product of the concentrations of the resulting ions must equal this value at equilibrium.

Set Up the Expression for Solubility

When \( \text{AgCl} \) dissolves, it dissociates into its constituent ions: \( \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \). Let the solubility of \( \text{AgCl} \) be \( s \) moles per liter, then \([\text{Ag}^+] = s\) and \([\text{Cl}^-] = s\). Substitute these into the solubility product expression: \( s \cdot s = s^2 \), to give \( s^2 = 1.56 \times 10^{-10} \).

Solve the Equation for Solubility

To find \( s \), take the square root of both sides of the equation: \( s = \sqrt{1.56 \times 10^{-10}} \). Calculating this gives \( s \approx 1.25 \times 10^{-5} \) moles per liter.

Identify the Correct Answer

Based on the calculation in Step 3, compare the calculated solubility value to the provided options. The correct answer, corresponding to \( 1.25 \times 10^{-5} \text{ mol/L} \), is option (c) \( 1.249 \times 10^{-5} \text{ mol/L} \).

Key concepts

Equilibrium

In the world of chemistry, equilibrium is a crucial concept, especially when discussing solutions and solubility. Equilibrium refers to the state when the dissolution and precipitation processes in a solution occur at the same rate. At this point, the concentration of ions in the solution remains constant over time.
For a salt like silver chloride (\(\text{AgCl}\)), equilibrium is established when it is dissolved in water and dissociates into its ions (\(\text{Ag}^+\) and \(\text{Cl}^-\)). The system reaches a balance where the rate of \(\text{AgCl}\) dissolving is equal to the rate of \(\text{Ag}^+\) and \(\text{Cl}^-\) ions recombining to form the solid.

• In an equilibrium state, identifiable by the constant product of ion concentrations, the term solubility product (\(K_{sp}\)) quantifies this balance.


• This constant is crucial in determining how much of the salt can dissolve in the solution.

Understanding equilibrium helps predict how changes in conditions, like temperature or concentration, will affect the solubility of compounds.

Dissolution Equation

The dissolution of silver chloride in water can be expressed as a chemical equation. This dissolution equation, \(\text{AgCl}\; \rightleftharpoons\; \text{Ag}^+ + \text{Cl}^-\), showcases the dynamic process of dissolving and precipitating.

• The double arrow (\(\rightleftharpoons\)) signifies a reversible reaction, indicating the dissolution and the reformation of \(\text{AgCl}\) are happening at a steady state.


• Each silver chloride molecule dissociates into one \(\text{Ag}^+\) ion and one \(\text{Cl}^-\) ion, reflecting a 1:1 stoichiometry of the dissolution.

The solubility product constant \(K_{sp}\) represents the mathematical expression of this equilibrium. It is derived from the concentrations of the dissociated ions:\[K_{sp} = [\text{Ag}^+][\text{Cl}^-]\]This equation highlights the relationship between the concentrations of ions in the dissolved state, establishing a framework to calculate solubility and predict the behavior of the salt in various conditions.

Concentration Calculation

Calculating the concentration of ions in a solution at equilibrium is pivotal for understanding the solubility of a compound. In the case of \(\text{AgCl}\), we start by letting the solubility be \(s\) moles/liter. Here is how you calculate it:

• The dissolution equation shows that each molecule of \(\text{AgCl}\) generates one \(\text{Ag}^+\) ion and one \(\text{Cl}^-\) ion, so \([\text{Ag}^+] = s\) and \([\text{Cl}^-] = s\).


• Inserting these expressions into the solubility product equation, you get:\[s^2 = K_{sp} = 1.56 \times 10^{-10}\]
• To find the solubility \(s\), take the square root:\[s = \sqrt{1.56 \times 10^{-10}}\]\[\approx 1.25 \times 10^{-5}\, \text{mol/L}\]

This calculation shows that the concentration of dissolved \(\text{AgCl}\) is approximately \(1.25 \times 10^{-5}\) moles per liter. Understanding this process helps in predicting the solubility under different scenarios and is vital for applications in chemistry and industry.