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Chapter 16: Problem 6

What happens to the \(K_{\mathrm{xg}}\) value of a solid as the temperature of the solution changes? Consider both increasing and decreasing temperatures, and explain your answer.

Short Answer

Expert verified
In summary, the effect of temperature on the solubility product constant (\(K_{\mathrm{xg}}\)) of a solid depends on the nature of the dissolution process. If the reaction is endothermic, \(K_{\mathrm{xg}}\) will increase as the temperature of the solution increases and vice versa. If the reaction is exothermic, \(K_{\mathrm{xg}}\) will decrease as the temperature of the solution increases and vice versa. This behavior can be explained by Le Chatelier's Principle, which states that a system at equilibrium will adjust itself to counteract any changes in conditions, such as temperature, to restore a new equilibrium.

Step by step solution

01

Understanding the Solubility Product Constant

The solubility product constant (\(K_{\mathrm{xg}}\)) is an equilibrium constant that represents the solubility of a sparingly soluble ionic compound in a solution. It is determined by the concentrations of the ions that make up the compound at saturation. Higher values of \(K_{\mathrm{xg}}\) indicate greater solubility, while lower values indicate lower solubility.
02

Le Chatelier's Principle

To answer the given question, it is necessary to consider Le Chatelier's Principle, which states that when a system at equilibrium is subjected to a change in a condition (e.g., temperature, pressure, or concentration), the system will adjust itself to counteract the effect of the change and restore a new equilibrium. In the context of solubility, Le Chatelier's Principle helps us predict how the solubility of a compound will be affected by changes in temperature.
03

Solubility and Temperature: Endothermic Reactions

In general, the solubility of a solid is affected by temperature due to the heat absorbed or released during the dissolution process. When the dissolution of a solid is endothermic (heat is absorbed), solubility increases with increasing temperature. This is because an increase in temperature provides the necessary energy to overcome the lattice energy of the ionic solid, promoting dissolution and increasing the solubility product constant (\(K_{\mathrm{xg}}\)).
04

Solubility and Temperature: Exothermic Reactions

Conversely, when the dissolution of a solid is exothermic (heat is released), solubility decreases with increasing temperature. In this case, the increase in temperature shifts the equilibrium towards the solid phase, as the system tries to counteract the increase in temperature by favoring the reaction that absorbs heat. As a result, the solubility product constant (\(K_{\mathrm{xg}}\)) decreases, indicating lower solubility.
05

Conclusion

In summary, the effect of temperature on the solubility product constant (\(K_{\mathrm{xg}}\)) of a solid depends on the nature of the dissolution process. If the reaction is endothermic, \(K_{\mathrm{xg}}\) will increase as the temperature of the solution increases and vice versa. If the reaction is exothermic, \(K_{\mathrm{xg}}\) will decrease as the temperature of the solution increases and vice versa. This behavior can be explained by Le Chatelier's Principle, which states that a system at equilibrium will adjust itself to counteract any changes in conditions, such as temperature, to restore a new equilibrium.

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Most popular questions from this chapter

Solutions of sodium thiosulfate are used to dissolve unexposed \(\mathrm{AgBr}\left(K_{\mathrm{sp}}=5.0 \times 10^{-13}\right)\) in the developing process for blackand-white film. What mass of AgBr can dissolve in \(1.00 \mathrm{~L}\) of \(0.500 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} ? \mathrm{Ag}^{+}\) reacts with \(\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\) to form a complex ion:

The copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times 10^{-6}\). Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}\) : $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 \mathrm{M} \mathrm{NaCl}\).

Calculate the final concentrations of \(\mathrm{K}^{+}(a q), \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)\), \(\mathrm{Ba}^{2+}(a q)\), and \(\mathrm{Br}^{-}(a q)\) in a solution prepared by adding \(0.100 \mathrm{~L}\) of \(0.200 \mathrm{M} \mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to \(0.150 \mathrm{~L}\) of \(0.250 \mathrm{M} \mathrm{BaBr}_{2}\). (For \(\mathrm{BaC}_{2} \mathrm{O}_{4}\) \(\left.K_{\mathrm{sp}}=2.3 \times 10^{-\mathrm{s}} .\right)\)

What mass of \(\mathrm{ZnS}\left(K_{\text {? }}=2.5 \times 10^{-22}\right.\) ) will dissolve in \(300.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} ?\) Ignore the basic properties of \(\mathrm{S}^{2-}\) .

A solution saturated with a salt of the type \(\mathrm{M}_{3} \mathrm{X}_{2}\) has an osmotic pressure of \(2.64 \times 10^{-2}\) atm at \(25^{\circ} \mathrm{C}\). Calculate the \(K_{s p}\) value for the salt, assuming ideal behavior.
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