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At \(25^{\circ} \mathrm{C}\), the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in water is \(2.86 \times 10^{-9} \mathrm{M}\). What are the equilibrium concentrations of the cation and the anion in a saturated solution?

Short Answer

Expert verified
In a saturated solution of Al(OH)3 at 25掳C with solubility \(2.86 \times 10^{-9} \mathrm{M}\), the equilibrium concentrations of Al3+ and OH- are \(2.86 \times 10^{-9} \mathrm{M}\) and \(8.58 \times 10^{-9} \mathrm{M}\), respectively.

Step by step solution

01

Write the balanced chemical equation for Al(OH)3's dissolution

Al(OH)3 (s) 鈫 Al3+ (aq) + 3 OH- (aq)
02

Set up the expression for Ksp

For a balanced chemical equation 'aA(s) 鈫 bB(aq) + cC(aq)', the solubility product constant (Ksp) can be defined as: Ksp = [B]^b 脳 [C]^c Using the balanced chemical equation for the dissolution of Al(OH)3, Ksp = [Al3+] 脳 [OH-]^3
03

Calculate the equilibrium concentrations of cation and anion

Since the solubility of Al(OH)3 is given as \(2.86 \times 10^{-9}\) M, we have the same solubility for Al3+ as it is in a 1:1 stoichiometry with Al(OH)3. Therefore, [Al3+] = \(2.86 \times 10^{-9}\) M. Now, the balanced chemical equation shows that there is a 1:3 stoichiometry between Al3+ and OH-. Hence, for each Al3+ ion, there will be three OH- ions in the solution. Thus, to find [OH-], we multiply the given molar concentration of Al3+ by 3: [OH-] = 3 脳 [Al3+] = 3 脳 \(2.86 \times 10^{-9}\) M = \(8.58 \times 10^{-9}\) M Therefore, the equilibrium concentrations of Al3+ and OH- are \(2.86 \times 10^{-9}\) M and \(8.58 \times 10^{-9}\) M, respectively.

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Key Concepts

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

Equilibrium Concentrations
When we talk about equilibrium concentrations, we're referring to the concentrations of ions or molecules in a solution at equilibrium, where the rates of the forward and backward reactions are equal. In the case of solubility, this involves a solid dissolving in water until no more can dissolve, leading to a saturated solution. At this point, the concentration of dissolved ions reaches a stable state, which we call the equilibrium concentration.

For Al(OH)鈧, its dissolution reaches equilibrium when it turns into Al鲁鈦 ions and OH鈦 ions in water. The given solubility of Al(OH)鈧 tells us how many moles of Al(OH)鈧 can dissolve per liter of water, reaching this stable state. Using the solubility value, we can directly find the equilibrium concentration of Al鲁鈦, because the solid dissociates to produce one Al鲁鈦 ion for each formula unit that dissolves.

Calculating for OH鈦 involves considering the stoichiometry of the reaction, as each Al(OH)鈧 produces three OH鈦 ions. Knowing how to determine these concentrations is key to understanding the solubility product, Ksp.
Stoichiometry
Stoichiometry helps us understand the quantitative relationships in chemical reactions. It is based on the balanced chemical equation and allows us to convert between different chemical species. In the dissolution reaction of Al(OH)鈧, stoichiometry plays a crucial role in figuring out how much of each ion is produced.

The balanced equation for the dissolution is: Al(OH)鈧 (s) 鈫 Al鲁鈦 (aq) + 3 OH鈦 (aq).

This tells us that one mole of Al(OH)鈧 produces one mole of Al鲁鈦 and three moles of OH鈦 ions. Thus, if the solubility of Al鲁鈦 is given, as in our example, we multiply this value by three to find the OH鈦 concentration, because there are three hydroxide ions for every one aluminum ion formed. This stoichiometric relationship is central to interconverting the concentrations of solute particles.

Understanding stoichiometry helps clarify how much of each ion will be in the solution once equilibrium is reached.
Dissolution Reaction
A dissolution reaction involves a solid substance dissolving in a solvent to form a solution, and in the context of salts like Al(OH)鈧, it's a reversible process. This means equilibrium can be achieved, as both the dissolution and precipitation processes may occur simultaneously until a balance is reached.

In the example of Al(OH)鈧, when it dissolves in water, it dissociates into its constituent ions: one Al鲁鈦 ion and three OH鈦 ions. At the molecular level, you can envision solid Al(OH)鈧 particles dissociating into ions that move into the solution. It's at dissolution that solubility characteristics of a compound are defined, and a balanced equation is key to understanding how many ions of each type are produced.

This balanced process defines how a material dissolves and allows us to express the equilibrium state through the solubility product constant, or Ksp, which quantifies the saturated concentration of ions in the solution at equilibrium.

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Most popular questions from this chapter

If \(k_{\mathrm{f}}>k_{\mathrm{r}}\), will \(K_{\mathrm{eq}}\) be less than 1 or greater than 1? Explain your answer.

How would the value of the equilibrium constant for a one-step reaction calculated as \(k_{\mathrm{f}} / k_{\mathrm{r}}\) compare with the value calculated from the concentrations of all substances present at equilibrium?

Ethyl acetate, a solvent used as nail polish remover, is produced by the reaction $$ \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftarrows \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \Delta E=+180.5 \mathrm{~kJ} $$ Ethyl acetate (a) Rewrite this reaction with the word heat in it. (b) Does the amount of ethyl acetate in the equilibrium mixture increase or decrease when the temperature is raised?

Diamond and graphite are two forms of elemental carbon. Under the appropriate conditions they will be in equilibrium with each other: \(C_{\text {diamond }} \rightleftarrows C_{\text {graphite }}\) If graphite is subjected to very high pressure and temperature, it will convert into the diamond form. (a) Is the above equilibrium reaction exothermic or endothermic? Explain how you know. (b) Which form, graphite or diamond, has the higher density? (Hint: Think about what increasing the pressure of a gas does to its density. It works the same for the solid and liquid phases as well.)

What is "dynamic" about the equilibrium that is established when a sparingly soluble salt is added to water?

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