Question

For the equilibrium \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})\) \(K_{\mathrm{sp}}=1.3 \times 10^{-33}\) (a) What is the minimum \(p H\) at which \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) will precipitate from a solution that is \(0.075 \mathrm{M}\) in \(\mathrm{Al}^{3+} ?\) (b) A solution has \(\left[\mathrm{Al}^{3+}\right]=0.075 \mathrm{M}\) and \(\left[\mathrm{CH}_{3} \mathrm{COOH}\right]=1.00 \mathrm{M} .\) What is the maximum quantity of \(\mathrm{NaCH}_{3} \mathrm{COO}\) that can be added to \(250.0 \mathrm{mL}\) of this solution before precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) begins?

Short answer

The exact values will depend on the calculated results, but the solution should provide the minimum pH at which \(\mathrm{Al(OH)_{3}}\) precipitates and the maximum amount of sodium acetate that can be added before precipitation begins.

Step-by-step solution

Calculation of Concentration of OH- ions

For the reaction: \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\rightleftharpoons \mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})\), the dissociation can be written as: \(K_{\mathrm{sp}} = [\mathrm{Al^{3+}}][\mathrm{OH^-}]^{3}\). Given that the \(\mathrm{Al}^{3+}\) ion concentration is 0.075 M and that \(K_{\mathrm{sp}} = 1.3 \times 10^{-33}\), the hydroxide ion concentration can be determined by substituting these values into the equation to solve for \([\mathrm{OH^{-}}]\).

Calculation of pH

After getting the \(\mathrm{OH^{-}}\) concentration, convert this to \(\mathrm{pOH}\) using the formula: \(pOH = -log[\mathrm{OH^{-}}]\). After getting the \(\mathrm{pOH}\), you can find the pH by using the relationship: \(\mathrm{pH} = 14 - [\mathrm{pOH}]\). This will give the minimum pH at which \(\mathrm{Al(OH)_{3}}\) will precipitate from the solution.

Calculation of Maximum Quantity of Sodium Acetate (NaCH3COO)

First, calculate the concentration of \(\mathrm{OH^{-}}\) ions at equilibrium from Step 1. To find out the maximum quantity of sodium acetate (NaCH3COO) that can be added to the solution before precipitation of \(\mathrm{Al(OH)_{3}}\) begins, you need to find out the concentration of sodium acetate (which essentially provides \(\mathrm{OH^{-}}\) ions in the solution) that will initiate precipitation. Use M1V1 = M2V2 to calculate the new concentration of \(\mathrm{OH^{−}}\) due to NaCH3COO (where M1 and V1 are the initial molarity and volume, M2 is the final molarity, and V2 is the final volume). The number of moles of sodium acetate can then be calculated.

Conclusion of Results

Following these steps and using the appropriate equations will result in the solution of the two questions: the minimum pH at which \(\mathrm{Al(OH)_{3}}\) will precipitate and the maximum quantity of sodium acetate that can be added to the solution before precipitation of \(\mathrm{Al(OH)_{3}}\) begins.

Key concepts

Solubility Product (Ksp)

Understanding the Solubility Product, denoted as \( K_{\text{sp}} \), is crucial in predicting the precipitation of a compound from a solution. The \( K_{\text{sp}} \) value represents the equilibrium constant for the dissociation of a sparingly soluble compound. For example, in the equilibrium \( \mathrm{Al(OH)_3(s)} \rightleftharpoons \mathrm{Al^{3+}(aq)} + 3\mathrm{OH^{-}(aq)} \), the \( K_{\text{sp}} \) is expressed as \( [\mathrm{Al^{3+}}][\mathrm{OH^{-}}]^3 \). Given the \( K_{\text{sp}} \), one can calculate the concentration of ions in a solution when the system is at equilibrium.

Solubility products help determine how much of a compound can dissolve to form a particular concentration of ions. If the product of the concentration of ions exceeds the \( K_{\text{sp}} \), the salts will precipitate from the solution. Conversely, when the ion product is less than \( K_{\text{sp}} \), the solution remains unsaturated, and precipitation does not occur. Understanding \( K_{\text{sp}} \) is essential for predicting solubility and guiding experiments in chemistry.

Precipitation Reactions

A precipitation reaction occurs when two ionic substances in solution combine to form an insoluble salt, known as a precipitate. When dealing with equilibrium and solubility, understanding when and why a precipitate forms is key.

To determine if a reaction results in precipitation, you can compare the ion product \( Q \), which is calculated similarly to \( K_{\text{sp}} \), to the solubility product \( K_{\text{sp}} \).

• If \( Q > K_{\text{sp}} \), precipitation occurs.
• If \( Q = K_{\text{sp}} \), the system is at equilibrium, and the solution is saturated.
• If \( Q < K_{\text{sp}} \), no precipitation occurs as the solution is unsaturated.

The goal is to maintain the conditions of the solution just below the threshold where \( Q \) becomes greater than \( K_{\text{sp}} \), to prevent undesired precipitation.

pH Calculations

The concept of pH is important in reactions involving precipitation, especially when hydroxide ions are involved. The pH of a solution affects the solubility of compounds like \( \mathrm{Al(OH)_3} \).

To determine the pH at which precipitation occurs, start by calculating the equilibrium concentration of \( \mathrm{OH^{-}} \) ions using the \( K_{\text{sp}} \) expression. Once you have \([\mathrm{OH^{-}}]\), calculate the pOH using the formula: \[ \text{pOH} = -\log[\mathrm{OH^{-}}] \]. Then convert pOH to pH using: \[ \text{pH} = 14 - \text{pOH} \]. This method tells you the pH value below which precipitation of the insoluble compound may begin. For compounds like \( \mathrm{Al(OH)_3} \), precipitation starts at higher pH values where fewer \( \mathrm{OH^{-}} \) ions mean a higher acidity.

Molarity and Concentration

Understanding molarity and concentration is pivotal for stoichiometry and chemical equilibrium. Molarity is a measurement of the concentration of a solute in a solution expressed as moles per liter (M).

In reactions involving precipitates, molarity allows you to calculate the concentration of each ion once equilibrium is established.

• To find the molarity, use the formula: \[ M = \frac{n}{V} \] where \( n \) is the number of moles and \( V \) is the volume in liters.
• Molarity plays a key role in determining how much of a compound is present in the solution, as well as how much more can be added or needs to be removed before reaching a new equilibrium.
• Through stoichiometry, the initial molarity and volume of reactants in a solution can help calculate the resultant concentrations of products, guiding whether precipitation occurs.

This foundational understanding links directly to predicting and controlling chemical changes in experimental conditions.