Problem 15 - Chemistry in 30 days

Question

If the solubility of \(\mathrm{PbBr}_{2}\) is ' \(S^{\prime}\) gram mol per litre, considering \(80 \%\) ionization, its ionic product is (a) \(2 S^{3}\) (b) \(4 s^{2}\) (c) \(45^{3}\) (d) \(2.5^{-1}\)

Step-by-Step Solution

Answer

The ionic product is approximately \(2S^3\), so the correct answer is (a).

Step 1: Write the Dissolution and Ionization Equation

Consider the dissociation of \(\mathrm{PbBr}_2\) in water: \(\mathrm{PbBr}_2 (s) \rightleftharpoons \mathrm{Pb^{2+}} (aq) + 2 \mathrm{Br^-} (aq)\). Given that it's \(80\%\) ionized, the effective concentration of ions formed is reduced by \(20\%\).

Step 2: Define Solubility in Terms of 'S'

The molar solubility \(S^{\prime}\) represents the concentration of \(\mathrm{Pb}^{2+}\). Since \(80\%\) is ionized, the concentration of \(\mathrm{Pb}^{2+}\) in solution is \(0.8S\) and \(\mathrm{Br^-}\) will be \(2 \times 0.8S = 1.6S\).

Step 3: Write Expression for Ionic Product

The ionic product \((K_{sp})\) for \(\mathrm{PbBr}_2\) is \([\mathrm{Pb^{2+}}][\mathrm{Br^-}]^2\). Substituting the effective concentrations we have, \(K_{sp} = (0.8S)(1.6S)^2\).

Step 4: Calculate the Ionic Product

Calculate the ionic product using \(K_{sp} = (0.8S)(1.6S)^2 = (0.8S)(2.56 S^2) = 2.048 S^3\). Simplifying further gives us approximately \(2S^3\).

Step 5: Find the Correct Option

Compare the calculated expression \(2S^3\) with the options given. (a) matches the expression.

Key Concepts

Solubility

Solubility refers to the ability of a substance to dissolve in a solvent to form a homogeneous solution. When we talk about the solubility of a compound, it usually means the amount of the compound that can dissolve in a solvent, like water, at a specified temperature and pressure.
In our example,

• we consider the solubility of lead(II) bromide,
• which is denoted as gram moles per litre.
• Since chemical equilibria are involved, the concept of molar solubility is particularly relevant.

It allows us to measure how much of the compound has dissolved to reach saturation in the solution at equilibrium.

Ionization

Ionization is the process by which atoms or molecules gain a positive or negative charge by gaining or losing electrons, often in solution.
For salts like PbBr₂, upon being added to water, they split into their constituent ions:

• The formula can be represented as PbBr₂ = Pb²⁺ + 2 Br^-.
• This equation describes the full ionization process.
• Since only 80% of PbBr₂ gets ionized, the effective concentrations are less compared to full ionization.
• It highlights the fraction of ions in the solution, 0.8S for Pb²⁺ and 1.6S for Br^-.

Ionization is crucial because it directly influences the ionic strength and conductivity of solutions.

Dissolution Equation

A dissolution equation provides a symbolic representation of the dissolution process, where a solid compound separates into ions in a solvent. For PbBr₂, the equation is:

• PbBr₂ (s) ⇌ Pb²⁺ (aq) + 2Br^- (aq).

This equation encompasses both the dissociation of the solid into its ions and their propagation into the solution. It shows the stoichiometry of the ion formation.
Understanding this allows us to set up expressions for ionic product calculations.
In the context of equilibrium, these equations play an important role by showing the whole picture of how a substance interacts and starts to dissolve in a solution.

Molar Solubility

Molar solubility is defined as the number of moles of a solute that can dissolve per litre of solution before the solution becomes saturated. It is directly related to the Ksp (solubility product constant) of a compound.
For PbBr₂,

• the molar solubility is denoted by S', which represents the concentration of Pb²⁺ ions in the solution.
• Knowing this allows us to calculate the ionic product using these effective ion concentrations.
• Since the degree of ionization affects effective concentrations, understanding molar solubility helps predict the behavior of dissolved species in the solution.

The relationship to the ionic product shows the interaction between a compound's dissolving property and its equilibrium position within the solution.