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Chapter 15: Problem 1

Which of the following will affect the total amount of solute that can dissolve in a given amount of solvent? a. The solution is stirred. b. The solute is ground to fine particles before dissolving. c. The temperature changes.

Short Answer

Expert verified
Changing the temperature (Option C) will affect the total amount of solute that can dissolve in a given amount of solvent, as it influences the solubility of substances in different ways. For most solid solutes, increased temperature leads to increased solubility, while decreased temperature leads to decreased solubility. However, the solubility of gases in liquids generally decreases as temperature increases.

Step by step solution

01

Option A: The solution is stirred

Stirring a solution helps to distribute the solute particles evenly throughout the solvent, thus speeding up the dissolution process. However, stirring does not change the total amount of solute that can dissolve in a given amount of solvent. It only affects the rate of dissolution, so this option does not affect solubility.
02

Option B: The solute is ground to fine particles before dissolving

Grinding the solute into finer particles increases the surface area of the solute exposed to the solvent, which may increase the rate of dissolution. However, like stirring, this action does not affect the total amount of solute that can dissolve in a given amount of solvent. The solubility remains the same, only the speed at which the solute dissolves into the solvent increases.
03

Option C: The temperature changes

Changing the temperature can affect the total amount of solute that can dissolve in a given amount of solvent. Temperature affects the solubility of substances in different ways. For most solid solutes, an increase in temperature leads to an increase in solubility, while a decrease in temperature leads to a decrease in solubility. This is due to the fact that dissolution is often an endothermic process, meaning it requires heat. In contrast, the solubility of gases in liquids generally decreases as temperature increases. Based on the analysis of each option, we can conclude that: Option C (Temperature changes) will affect the total amount of solute that can dissolve in a given amount of solvent.

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Most popular questions from this chapter

Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s),\) but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-} .\) In terms of solubility, \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the \(\mathrm{pH}\) dependence of the solubility of Al(OH) \(_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3},\) as a function of \(\left[\mathrm{H}^{+}\right],\) obeys the equation $$ S=\left[\mathbf{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right] $$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q) $$ c. The value of \(K\) is 40.0 and \(K_{\mathrm{sp}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\) Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the pH range \(4-12\).

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