Question

What will be the solubility of \(\mathrm{AgCl}\) in \(0.05 \mathrm{M} \mathrm{NaCl}\) aqueous solution if solubility product of \(\mathrm{AgCl}\) is \(1.5 \times 10^{-10} ?\) (a) \(3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(0.05 \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(1.5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\) (d) \(3 \times 10^{9} \mathrm{~mol} \mathrm{~L}^{-1}\)

Short answer

The solubility of AgCl in 0.05 M NaCl aqueous solution is \(3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\).

Step-by-step solution

Understand the concept of solubility product (Ksp)

The solubility product, Ksp, represents the maximum amount of a solid that can dissolve in a saturated solution. It is the product of the concentrations of the ions, each raised to the power of their respective coefficients in the dissociation equation. For AgCl, which dissociates into Ag+ and Cl-, the Ksp expression is Ksp = [Ag+][Cl-].

Write the dissolution equation of AgCl

Silver chloride (AgCl) dissociates into silver ions (Ag+) and chloride ions (Cl-) in solution according to the following equation:AgCl(s) ⇌ Ag+(aq) + Cl-(aq).

Account for common ion effect

In the presence of a common ion (NaCl providing additional Cl- ions), the solubility of AgCl will decrease. This is because the equilibrium will shift to the left according to Le Chatelier's principle to counter the increase in Cl- concentration.

Set up the expression using Ksp and the common ion concentration

We need to set up the equation for the solubility product (Ksp) of AgCl taking into account the presence of Cl- from NaCl:Ksp = [Ag+][Cl-] where [Cl-] is 0.05 M since it comes from the NaCl. The concentration of Ag+ will be equal to the solubility of AgCl, which we'll call 's'.

Solve for the solubility 's' of AgCl

The concentration of Ag+ (s) will be equal to the solubility of AgCl. Putting values into the equation from Ksp we get1.5 x 10^{-10} = s(0.05).Solving for s, we divide both sides by 0.05 to gets = (1.5 x 10^{-10}) / 0.05 = 3 x 10^{-9} M.

Key concepts

Understanding the Common Ion Effect

The common ion effect refers to the decrease in solubility of an ionic compound when a solution already contains one of the ions present in the compound. This occurs because the addition of a common ion shifts the equilibrium of the dissolution process. It's linked to Le Chatelier's principle, which predicts how an equilibrium will respond to changes in concentration, temperature, or pressure. In the case of AgCl in NaCl solution, NaCl provides additional Cl- ions, which lead to a shift in the equilibrium towards the left, favoring the formation of solid AgCl, thus reducing its solubility.

Let's explore how this works mathematically. The presence of a common ion in solution means that the concentration of that ion is higher than it would be in pure water. When you add NaCl to a solution, you increase the concentration of Cl- ions. According to the solubility product (Ksp), for the dissolving process of AgCl, represented as AgCl(s) ⇌ Ag+(aq) + Cl-(aq), the product of the molarities [Ag+] and [Cl-] should remain constant at equilibrium. Thus, when [Cl-] increases because of the added NaCl, the equilibrium shifts to reduce the number of Ag+ ions that can dissolve, leading to a lower solubility of AgCl.

Le Chatelier's Principle in Action

Le Chatelier's principle provides insight into the behavior of a system at equilibrium when it experiences external stress. In the context of the solubility of compounds, the 'stress' can be the addition of a common ion, a change in temperature, or a change in pressure. The principle states that the equilibrium will shift to minimize that stress. For our AgCl example, when Cl- ions are introduced into the solution through Naïve, the system faces an increased concentration of Cl- ions. To minimize this disturbance, the system responds by shifting the equilibrium towards the reactants (AgCl), thus reducing the ionization and consequently the solubility of AgCl.

It's important to remember that Le Chatelier's principle is a qualitative tool; it tells us the direction of the shift but not the magnitude. To quantitatively determine the effect on solubility, we return to the solubility product expressions and carry out calculations that take into account the increased concentration of the common ion.

Dissolution Equation Essentials

A dissolution equation expresses the dissolving of a solid substance to form an ionic solution. In the case of AgCl, the dissolution equation is represented by AgCl(s) ⇌ Ag+(aq) + Cl-(aq). This equation illustrates the equilibrium between the undissolved solid and the ions in solution. When we say that a compound like AgCl has a given Ksp, we're saying that at equilibrium in a saturated solution, the product of the concentrations of the ions produced from the salt is constant.

This constant is crucial because it helps us calculate the solubility of the substance in various conditions, such as in the presence of a common ion. By understanding and using the dissolution equation, students can set up solubility problems effectively. You'll often need to refer to the dissolution equation to understand how solubility changes with the presence of additional ions in the solution, like in the context of the common ion effect.