Question

For sparingly soluble salt \(\mathrm{A}_{\mathrm{p}} \mathrm{B}_{q}\), the relationship of its solubility product \(\left(\mathrm{L}_{\mathrm{s}}\right)\) with its solubility (s) is: (a) \(L_{\mathrm{s}}=\mathrm{s}^{p+q} \mathrm{p}^{\mathrm{p}} q^{\mathrm{q}}\) (b) \(\mathrm{L}_{\mathrm{s}}=\mathrm{s}^{\mathrm{p}+q} \mathrm{p}^{q} \mathrm{q}^{\mathrm{p}}\) (c) \(\mathrm{L}_{\mathrm{s}}=\mathrm{s}^{\mathrm{pq}} \mathrm{p}^{\mathrm{p}} \mathrm{q}^{q}\) (d) \(L_{\mathrm{s}}=\mathrm{s}^{\mathrm{pq}}(\mathrm{pq})^{\mathrm{p}+q}\)

Short answer

Option (a): \( L_s = s^{p+q} p^p q^q \) is correct.

Step-by-step solution

Understanding the Dissolution Equation

The dissociation of a sparingly soluble salt \( \mathrm{A}_{\mathrm{p}} \mathrm{B}_{q} \) in water can be represented as: \[ \mathrm{A}_{\mathrm{p}} \mathrm{B}_{q} \leftrightarrow \mathrm{pA}^{\mathrm{q}+} + \mathrm{qB}^{\mathrm{p}-} \] Here, \( \mathrm{A} \) and \( \mathrm{B} \) are the ions formed in the solution, and \( \mathrm{p} \) and \( \mathrm{q} \) are their respective stoichiometric coefficients.

Writing the Expression for Solubility Product, \( L_s \)

The solubility product \( L_s \) for the reaction is given by:\[ L_s = [\mathrm{A}^{q+}]^p [\mathrm{B}^{p-}]^q \] Assuming the solubility of the salt is \( s \), the concentration of ions can be written as \( \mathrm{p}s \) for \( \mathrm{A}^{q+} \) and \( \mathrm{q}s \) for \( \mathrm{B}^{p-} \).

Substituting Ion Concentrations

Substitute the concentrations into the solubility product expression:\[ L_s = (\mathrm{p}s)^p (\mathrm{q}s)^q \] Expand the expression:\[ L_s = \mathrm{p}^p s^p \times \mathrm{q}^q s^q \] This can be simplified to:\[ L_s = s^{p+q} \times \mathrm{p}^p \times \mathrm{q}^q \]

Compare and Match the Expression

The derived expression is:\[ L_s = s^{p+q} \times p^p \times q^q \]Compare with the provided options:- (a) \( L_{\mathrm{s}}=s^{p+q} p^{p} q^{q} \) - (b) \( L_{\mathrm{s}}=s^{p+q} p^{q} q^{p} \) - (c) \( L_{\mathrm{s}}=s^{pq} p^{p} q^{q} \) - (d) \( L_s=s^{pq}(pq)^{p+q} \)Clearly, option (a) matches our derived expression.

Key concepts

Sparingly Soluble Salt

The term "sparingly soluble salt" refers to a type of salt that does not dissolve thoroughly in water. In other words, only a small amount of it is able to go into the solution. This happens because the salt has a low solubility product, denoted as \( L_s \). A classic example of a sparingly soluble salt is \( \mathrm{A}_{\mathrm{p}} \mathrm{B}_{q} \), where the salt breaks down into its constituent ions when dissolved in water. The solubility of such salts is often measured using the "solubility product constant."

• The solubility product helps determine the extent to which a salt will dissolve.
• It is particularly useful for predicting precipitation reactions.

Understanding sparingly soluble salts is essential in areas like chemical separation methods and environmental chemistry, where controlling the concentration of ions is crucial.

Stoichiometric Coefficients

Stoichiometric coefficients are numbers that appear in front of compounds in a chemical equation. They indicate the number of molecules or moles of a substance involved in the reaction. In the dissolution equation for sparingly soluble salts, \( \mathrm{A}_{\mathrm{p}} \mathrm{B}_{q} \), the coefficients \( \mathrm{p} \) and \( \mathrm{q} \) show how many ions of \( \mathrm{A}^{\mathrm{q+}} \) and \( \mathrm{B}^{\mathrm{p-}} \) are produced, respectively.

• These coefficients are crucial for writing balanced chemical equations.
• They influence the solubility product \( L_s \) by affecting the concentration terms in the equations.

Therefore, correctly identifying stoichiometric coefficients is essential for predicting the concentrations of ions formed during the dissolution process.

Ion Concentrations

Ion concentrations refer to the number of particles of ions in a solution, usually expressed in molarity (moles per liter). In the context of sparingly soluble salts, these concentrations are crucial in calculating the solubility product, \( L_s \).When a salt like \( \mathrm{A}_{\mathrm{p}} \mathrm{B}_{q} \) dissolves, it produces \( \mathrm{pA}^{\mathrm{q+}} \) and \( \mathrm{qB}^{\mathrm{p-}} \). If \( s \) is the solubility of the salt, then the ion concentrations can be described as:

• \( [\mathrm{A}^{\mathrm{q+}}] = \mathrm{p}s \)
• \( [\mathrm{B}^{\mathrm{p-}}] = \mathrm{q}s \)

These expressions for ion concentrations are used in forming the solubility product equation. A proper understanding is necessary for computations involving ionic equilibria in various chemical systems.

Dissolution Equation

A dissolution equation represents how a compound breaks down into its ions when it dissolves in a solvent, typically water. For sparingly soluble salts, understanding this equation is key to determining how well the salt will dissolve.For instance, the dissolution of \( \mathrm{A}_{\mathrm{p}} \mathrm{B}_{q} \) is expressed as:\[\mathrm{A}_{\mathrm{p}} \mathrm{B}_{q} \leftrightarrow \mathrm{pA}^{\mathrm{q+}} + \mathrm{qB}^{\mathrm{p-}}\]This equation highlights the stoichiometric relationship between the salt and its resulting ions. Each molecule of the salt produces \( \mathrm{p} \) cations and \( \mathrm{q} \) anions.

• It sets the basis for writing the solubility product equation.
• It helps in determining the ratio and concentration of ions formed.

The dissolution equation is essential in fields like analytical chemistry and chemical engineering, where dissolution patterns influence practical applications.