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The equilibrium constant for the equation $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \leftrightharpoons\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}(a q) $$ is \(K_{f}=40\) at \(25^{\circ} \mathrm{C}\). Calculate the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in a solution buffered at \(\mathrm{pH}=12.00\) at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The solubility of \(\mathrm{Al(OH)_3}\) is 0.4 M in the buffered solution.

Step by step solution

01

Identify Given Values and Formulas

We are given the equilibrium constant \(K_{f} = 40\) for the reaction \(\mathrm{Al(OH)_3(s)} + \mathrm{OH}^- (aq) \leftrightharpoons [\mathrm{Al(OH)_4}]^-(aq)\) at 25掳C. We need to find the solubility of \(\mathrm{Al(OH)_3}\) in a solution with \(\text{pH} = 12.00\). The relevant formula is \(\text{pH} + \text{pOH} = 14\), which allows us to calculate \(\text{pOH}\) and thus \([\mathrm{OH}^-]\).
02

Calculate [OH鈦籡 from pH

Given \(\text{pH} = 12.00\), we find \(\text{pOH} = 14 - 12 = 2\). Using the relation \([\mathrm{OH}^-] = 10^{-\text{pOH}}\), we calculate \([\mathrm{OH}^-] = 10^{-2} = 0.01\, \text{M}\).
03

Write the Equilibrium Expression

The expression for the equilibrium constant \(K_{f}\) is: \[ K_{f} = \frac{[\mathrm{Al(OH)_4}^-]}{[\mathrm{OH}^-]} \]
04

Replace Concentrations in the Equilibrium Expression

Since the initial amount of \([\mathrm{Al(OH)_4}^-]\) is 0, let its concentration at equilibrium be \(x\). Therefore, according to the reaction, the change in \([\mathrm{Al(OH)_4}^-]\) is also \(x\), which is equal to solubility, while \([\mathrm{OH}^-]\) remains approximately 0.01 M. Substitute into the expression: \[ 40 = \frac{x}{0.01} \]
05

Solve for x

Solving the equation \(40 = \frac{x}{0.01}\), we find: \[ x = 40 \times 0.01 = 0.4 \]\,\text{M}. This provides the solubility of \(\mathrm{Al(OH)_3}\) in the buffered solution.

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Key Concepts

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

Solubility
Solubility refers to the ability of a substance, often a solid, to dissolve in a solvent, forming a solution at a specific temperature and pressure. In the context of the given exercise, we are looking at the solubility of aluminum hydroxide, \( \text{Al(OH)}_3 \). Solubility is typically expressed in terms of concentration, like molarity (mol/L), which indicates how much of the substance can be dissolved.

Solubility depends on a variety of factors:
  • Temperature: Generally, an increase in temperature can increase the solubility of solids in liquids.
  • pH: For compounds that produce ions in solutions, pH can significantly affect solubility. A high pH means the solution is more basic, influencing the solubility of \( \text{Al(OH)}_3 \) since hydroxide ions \( \text{OH}^- \) are involved.
  • The presence of other substances: Common ions can reduce solubility, while others can complex with ions, increasing solubility.
In the exercise, we're specifically determining how much \( \text{Al(OH)}_3 \) can dissolve in a solution with a \( \text{pH} = 12 \). This involves calculating the \( \text{OH}^- \) concentration at this pH and using it to determine how much solid dissolves before reaching equilibrium.
Buffer Solutions
A buffer solution is a special type of solution that can resist pH changes upon the addition of an acid or base. It is composed of weak acids and their corresponding weak bases or vice versa. Buffer solutions maintain a fairly constant pH level, which is very useful for certain calculations and reactions.

In this exercise, a basic buffer environment is created because the solution is maintained at \( \text{pH} = 12 \). When working in such a high pH solution, the concentration of hydroxide ions is dominant, specifically \( [\text{OH}^-] = 0.01 \, \text{M} \). This stable \( \text{pH} \) is crucial because:
  • Consistency: It ensures that calculations, like those for \( \text{OH}^- \) concentration, are consistent and reliable.
  • Predictability: It allows us to predict the behavior of \( \text{Al(OH)}_3 \), including its solubility in the solution, based on its equilibrium constant \( K_f \).
  • Experimental conditions: In laboratory settings, buffer solutions mimic biological or environmental conditions, supporting accurate biochemical reactions or corrosion studies.
So, the buffered solution context is a key element that stabilizes the conditions of the solubility experiment allowing for precise calculations.
Chemical Equilibrium
Chemical equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction in a chemical process. At this point, the concentrations of reactants and products remain constant.

In the problem provided, we are dealing with an equilibrium constant, \( K_f = 40 \), for the dissolution of aluminum hydroxide in an aqueous hydroxide solution.

Understanding chemical equilibrium involves:
  • Equilibrium Constant \( K \): It defines the ratio of concentrations of products to reactants at equilibrium. For the dissolution of \( \text{Al(OH)}_3 \), it shapes our understanding of how much product forms from the solid and hydroxide ions.
  • Le Chatelier's Principle: If a system at equilibrium experiences a change, it will adjust to counter the change. Here, changes in \( \text{OH}^- \) availability influence the solubility of \( \text{Al(OH)}_3 \).
  • Dynamic Nature: Even though concentrations stay static at equilibrium, molecules constantly interact. This means some \( \text{Al(OH)}_3 \) dissolves while some \( [\text{Al(OH)}_4]^- \) reforms into solid \( \text{Al(OH)}_3 \).
By using these concepts, we can predict and calculate how \( \text{Al(OH)}_3 \) reacts and what solute concentrations will exist in a buffered solution.

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Most popular questions from this chapter

Potassium perchlorate, \(\mathrm{KClO}_{4}(s)\), is soluble in water to the extent of \(0.70\) grams per \(100.0 \mathrm{~mL}\) at \(0^{\circ} \mathrm{C}\). Calculate the value of \(K_{\text {sp }}\) of \(\mathrm{KClO}_{4}(s)\) at \(0{ }^{\circ} \mathrm{C}\).

Indicate for which of the following compounds the solubility increases as the \(\mathrm{pH}\) of the solution is lowered: (a) \(\mathrm{PbCrO}_{4}(s)\) (b) \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\) (c) \(\mathrm{Hg}_{2} \mathrm{I}_{2}(s)\) (d) \(\mathrm{Ag}_{2} \mathrm{O}(s)\)

It is observed that a precipitate forms when a \(2.0 \mathrm{M} \mathrm{KOH}(a q)\) solution is added dropwise to a \(0.20 \mathrm{M} \mathrm{Zn}\left(\mathrm{ClO}_{4}\right)_{2}(a q)\) solution and that, similarly, a precipitate forms when a \(2.0 \mathrm{M} \mathrm{KOH}(a q)\) solution is added dropwise to a \(0.20 \mathrm{M} \mathrm{Mg}\left(\mathrm{ClO}_{4}\right)_{2}(a q)\) solution. However, on further addition of \(\mathrm{KOH}(a q)\), the precipitate from the \(\mathrm{Zn}\left(\mathrm{ClO}_{4}\right)_{2}(a q)\) solution dissolves, whereas the precipitate from the \(\mathrm{Mg}\left(\mathrm{ClO}_{4}\right)_{2}(a q)\) solution does not. Explain these observations using balanced chemical equations. What can we conclude about the properties of these respective metal ions?

When aqueous ammonia is added to a solution containing copper(II) nitrate, a solid precipitate forms. However, when additional aqueous ammonia is added, the precipitate redissolves, forming a soluble \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(a q)\) complex ion. Write the balanced chemical equation for each of these reactions.

Magnesium oxalate, \(\mathrm{MgC}_{2} \mathrm{O}_{4}(s)\), is sparingly soluble in water. Predict the effect on its solubility when (a) the solution is made more acidic (b) the solution is made more basic (c) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(s)\) is added to the solution

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