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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.

Short Answer

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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 by step solution

01

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.
02

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.
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The concentrations of various cations in seawater, in moles per liter, are $$ \begin{array}{llllll} \hline \text { Ion } & \mathrm{Na}^{+} & \mathrm{Mg}^{2+} & \mathrm{Ca}^{2+} & \mathrm{A}^{3+} & \mathrm{Fe}^{3+} \\ \text { Molarity }(M) & 0.46 & 0.056 & 0.01 & 4 \times 10^{-7} & 2 \times 10^{-7} \\ \hline \end{array} $$ (a) At what \(\left[\mathrm{OH}^{-}\right]\) does \(\mathrm{Mg}(\mathrm{OH})_{2}\) start to precipitate? (b) At this concentration, will any of the other ions precipitate? (c) If enough \(\mathrm{OH}^{-}\) is added to precipitate \(50 \%\) of the \(\mathrm{Mg}^{2+}\), what percentage of each of the other ions will precipitate? (d) Under the conditions in (c), what mass of precipitate will be ob- tained from one liter of seawater?

Shown below is a representation of the ionic solid \(\mathrm{MX}\), where \(\mathrm{M}\) cations are represented by squares and \(X\) anions are represented by circles. Fill in the box after the arrow to represent what happens to the solid after it has been completely dissolved in water. For simplicity, do not represent the water molecules.

Given \(K_{s p}\) and the equilibrium concentration of one ion, calculate the equilibrium concentration of the other ion. (a) cadmium(II) hydroxide: \(K_{\text {sp }}=2.5 \times 10^{-14} ;\left[\mathrm{Cd}^{2+}\right]=1.5 \times 10^{-6} M\) (b) copper(II) arsenate \(\left(\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\right): K_{\mathrm{sp}}=7.6 \times 10^{-36} ;\left[\mathrm{AsO}_{4}^{3-}\right]=\) \(2.4 \times 10^{-4} M\) (c) zinc oxalate: \(K_{s p}=2.7 \times 10^{-8} ;\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=8.8 \times 10^{-3} M\)

Lead azide, \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}\), is used as a detonator in car airbags. The impact of a collision causes \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}\) to be converted into an enormous amount of gas that fills the airbag. At \(25^{\circ} \mathrm{C}\), a saturated solution of lead azide is prepared by dissolving \(25 \mathrm{mg}\) in water to make \(100.0 \mathrm{~mL}\) of solution. What is \(K_{\mathrm{sp}}\) for lead azide?

Calculate the molar solubility of gold(I) chloride \(\left(K_{s p}=2.0 \times 10^{-13}\right)\) in \(0.10 \mathrm{M} \mathrm{NaCN}\). The complex ion formed is \(\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\) with \(K_{\mathrm{f}}=2 \times 10^{38}\). Ignore any other competing equilibrium systems.

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