Problem 64 - Chemistry: Principles and Reactions

Question

Which of the following statements are true? (a) For an insoluble metallic salt, \(K_{\text {sp }}\) is always less than 1 . (b) More \(\mathrm{PbCl}_{2}\) can be dissolved at \(100^{\circ} \mathrm{C}\) than at \(25^{\circ} \mathrm{C}\). One can conclude that dissolving \(\mathrm{PbCl}_{2}\) is an exothermic process. (c) When strips of copper metal are added to a saturated solution of \(\mathrm{Cu}(\mathrm{OH})_{2}\), a precipitate of \(\mathrm{Cu}(\mathrm{OH})_{2}\) can be expected to form because of the common ion effect.

Step-by-Step Solution

Answer

Question: Evaluate the following statements as true or false based on the given analysis: a) For an insoluble metallic salt, Ksp is always less than 1. b) More PbCl2 can be dissolved at 100°C than at 25°C, so this must be an exothermic process. c) When strips of copper metal are added to a saturated solution of Cu(OH)2, a precipitate of Cu(OH)2 can be expected to form because of the common ion effect. Answer: a) True, b) False, c) False

Step 1: Statement (a) evaluation

For an insoluble metallic salt, Ksp is always less than 1. Let's break down the concept of Ksp, or solubility product constant. Ksp is an equilibrium constant for the dissolution process of slightly soluble salts in water. The smaller the Ksp, the less soluble the salt is in water. With this information, we can conclude that the statement is true. For an insoluble metallic salt, Ksp is always less than 1 because the salt is less soluble in water, and its solubility equilibria would lie predominantly to the left, which leads to a smaller value of Ksp.

Step 2: Statement (b) evaluation

More \(PbCl_2\) can be dissolved at \(100^{\circ}C\) than at \(25^{\circ}C\). One can conclude that dissolving \(PbCl_2\) is an exothermic process. To analyze this statement, we need to consider how solubility is affected by temperature. Generally, for most salts, solubility increases with an increase in temperature in the case of endothermic processes. However, the solubility decreases with temperature increase for exothermic processes. Since the dissolution of \(PbCl_2\) is more at higher temperature (\(100^{\circ}C\)) than at lower temperature (\(25^{\circ}C\)), this indicates that its dissolution process is endothermic, not exothermic. Thus, the statement is false.

Step 3: Statement (c) evaluation

When strips of copper metal are added to a saturated solution of \(Cu(OH)_2\), a precipitate of \(Cu(OH)_2\) can be expected to form because of the common ion effect. The common ion effect states that adding an ion that is common to the ions in an existing equilibrium will cause a shift in the equilibrium position to accommodate the additional ion and minimize its effect. In this case, the copper strips will not affect the equilibrium, as there is no common ion between the copper metal and the ions in the solution of \(Cu(OH)_2\). Therefore, the statement is false because the copper strips will not cause any shift in the equilibrium, and no additional precipitate will form due to the common ion effect. In summary: - Statement (a) is true. - Statement (b) is false. - Statement (c) is false.

Key Concepts

Ksp evaluation

The Solubility Product Constant, often represented as Ksp, is a crucial parameter in chemistry that helps us understand how a salt dissolves in water.
It is specifically used for slightly soluble ionic compounds.

• Ksp is an equilibrium constant that signifies the product of the ion concentrations each raised to the power of their stoichiometric coefficients in a saturated solution.
• If a compound is very soluble in water, its Ksp will be high.
• Conversely, low or small Ksp values suggest limited solubility, as seen in insoluble or sparingly soluble salts.

For example, if we consider an insoluble salt like PbCl2, its Ksp will typically be less than 1 because it dissolves minimally in water.
This tendency for a small Ksp value stems from the equilibrium of the dissolution process leaning predominantly toward the undissolved solid.
Understanding Ksp is vital for predicting whether a precipitate will form when solutions are mixed.

Common Ion Effect

The common ion effect can significantly influence the solubility of compounds in a solution.
It arises when a compound shares an ion with that already in the solution.

• The presence of a common ion suppresses the solubility of the ionic compound by shifting the dissolution equilibrium.
• This shift occurs to decrease the concentration of the additional common ion.
• The common ion effect is a direct application of Le Châtelier's principle, where the equilibrium shifts to counterbalance any disturbance.

For example, if you have a saturated solution of Cu(OH)2 and you add a compound containing Cu2+, the solubility of Cu(OH)2 decreases due to the presence of this common ion.
This results in the formation of a precipitate as the solution can't hold the excess ions.

Endothermic and Exothermic Processes

Endothermic and exothermic processes describe the heat exchange that occurs during chemical reactions or dissolving processes.
Understanding these concepts helps predict how a substance will behave with temperature changes.

• In endothermic processes, heat is absorbed from the surroundings, and the reaction requires energy input.
• As the temperature increases, solubility typically increases in endothermic reactions because heat aids in overcoming intermolecular forces.
• Conversely, exothermic processes release heat into the surroundings and often show a decrease in solubility with increasing temperature.

For instance, if PbCl2 dissolves more readily at higher temperatures, it suggests the dissolution is endothermic.
This is because the process needs heat to dissolve more of the salt.
Grasping these concepts can predict the behavior of different salts and reactions under various temperature conditions.