Question

\(K_{\mathrm{sp}}\) of \(\mathrm{AgCl}\) at \(18^{\circ} \mathrm{C}\) is \(1.8 \times 10^{10}\). If \(\mathrm{Ag}\) of solution is \(4 \times 10^{3}\) mol/litre, the Cl that must exceed before \(\mathrm{AgCl}\) is precipitated would be (1) \(4.5 \times 10^{-8} \mathrm{~mol} /\) litre (2) \(7.2 \times 10^{-13} \mathrm{~mol} /\) litrc (3) \(4.0 \times 10^{-3} \mathrm{~mol} /\) litre (4) \(4.5 \times 10^{-7} \mathrm{~mol} /\) itre

Short answer

The \(Cl^- concentration must exceed \)4.5 \times 10^{-8}\( mol/L for \)AgCl$ to precipitate. The correct answer is option (1).

Step-by-step solution

Understanding the Problem

Given the solubility product constant (\(K_{\text{sp}}\)) of \(AgCl\) and the concentration of \(Ag^+\) ions in the solution, determine the concentration of \(Cl^-\) ions required to start precipitating \(AgCl\).

Writing the Solubility Product Expression

The solubility product expression for \(AgCl\) is given by \(K_{\text{sp}} = [Ag^+] \times [Cl^-]\).

Substitute Given Values into Equation

Given \(K_{\text{sp}} = 1.8 \times 10^{-10}\) and \([Ag^+] = 4 \times 10^{-3}\) mol/L, substitute these values into the solubility product expression: \(1.8 \times 10^{-10} = (4 \times 10^{-3}) \times [Cl^-]\).

Solve for \([Cl^-]\)

Rearrange the equation to solve for \([Cl^-]\): \([Cl^-] = \frac{1.8 \times 10^{-10}}{4 \times 10^{-3}}\).

Perform the Calculation

Perform the division: \([Cl^-] = \frac{1.8 \times 10^{-10}}{4 \times 10^{-3}} = 4.5 \times 10^{-8}\) mol/L.

Identify the Correct Answer

Among the given options, \(4.5 \times 10^{-8}\) mol/L matches with option (1).

Key concepts

Ksp (Solubility Product Constant)

In chemistry, the solubility product constant, represented as Ksp, is an equilibrium constant that applies to the solubility of ionic compounds. When an ionic compound dissolves in water, it dissociates into its constituent ions. Ksp expresses the product of the concentrations of these ions, each raised to the power of its coefficient in the balanced equation. For example, if we have a compound AB that dissolves into A⁺ and B⁻ ions, the Ksp expression is: \( K_{\text{sp}} = [A^+] \times [B^-] \). This constant is specific to a particular ionic compound at a particular temperature. A low Ksp value indicates that the compound is not very soluble in water. In the original exercise, the Ksp of AgCl (Silver Chloride) at 18°C is provided as \( 1.8 \times 10^{-10} \).

Precipitation of Ionic Compounds

Precipitation occurs when the product of the ion concentrations in a solution exceeds the Ksp of the ionic compound. In simple terms, if the concentration of dissolved ions in a solution is too high, the excess ions will combine to form a solid precipitate, falling out of the solution. This process helps us understand at what point a certain amount of ions will start forming a solid compound again. In our provided exercise, we want to find the concentration of chloride ions (Cl⁻) that will cause AgCl to start precipitating. We use the given Ksp and the concentration of Ag⁺ ions. By rearranging and solving the Ksp expression, we can find the critical concentration of Cl⁻ that initiates precipitation.

Equilibrium Concentrations

Equilibrium concentrations refer to the steady-state concentrations of reactants and products in a chemical reaction that has reached equilibrium. In the context of solubility equilibria, such as with the dissolution of an ionic compound, equilibrium is reached when the rate of dissolution equals the rate of precipitation. For example, in the dissolution of AgCl, we write the equilibrium as follows: \( K_{\text{sp}} = [Ag^+] \times [Cl^-] \). Given that the system is at equilibrium, the concentrations of Ag⁺ and Cl⁻ ions will remain constant as long as no additional ions are added or removed. Solving the provided exercise involves calculating the equilibrium concentration of Cl⁻ given the Ksp and the concentration of Ag⁺, which helps to determine when precipitation will occur.