Question

Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.4 \times 10^{-5}\).

Short answer

The pH of the 0.050 M Al(NO₃)₃ solution is approximately 3.28.

Step-by-step solution

Write the hydrolysis reaction

The hydrolysis reaction for Al(H₂O)₆³⁺ is as follows: \[ Al(H_2O)_6^{3+} + H_2O \rightleftharpoons H_3O^+ + Al(H_2O)_5(OH)^{2+} \]

Construct the ICE table

We will use an ICE (Initial, Change, Equilibrium) table to keep track of the changes in concentrations during the hydrolysis reaction: \[ \begin{array}{c|c|c|c} & [Al(H_2O)_6^{3+}] & [H_3O^+] & [Al(H_2O)_5(OH)^{2+}] \\ \hline I & 0.050 & 0 & 0 \\ C & -x & x & x \\ E & 0.050 - x & x & x \end{array} \]

Write the equilibrium expression and solve for x

Use the Ka value to solve for x. since Ka = 1.4 × 10⁻⁵. Therefore: \[ K_a = \frac{[H_3O^+][Al(H_2O)_5(OH)^{2+}]}{[Al(H_2O)_6^{3+}]} = \frac{x^2}{0.050 - x} \] Assuming x is significantly smaller than 0.050, we can simplify the equation as: \[ 1.4 \times 10^{-5} = \frac{x^2}{0.050} \]

Calculate the value of x

Solving for x, which represents the [H₃O⁺] concentration: \[ x = \sqrt{1.4 \times 10^{-5} \times 0.050} \approx 5.29 \times 10^{-4} \]

Calculate the pH

Use the pH formula: \[ pH = -\log[H_3O^+] = -\log(5.29 \times 10^{-4}) \]

Find the final pH value

Calculate the pH: \[ pH \approx 3.28 \] So, the pH of the 0.050 M Al(NO₃)₃ solution is approximately 3.28.

Key concepts

Hydrolysis Reaction

Hydrolysis reactions are a fundamental concept when studying how water impacts chemical compounds. In chemistry, hydrolysis involves the reaction of a compound with water, resulting in the breaking of a chemical bond. For the given example, we focus on the hydrolysis of the aluminum ion complex, specifically \[ Al(H_2O)_6^{3+} + H_2O \rightleftharpoons H_3O^+ + Al(H_2O)_5(OH)^{2+} \] In this reaction, the metal complex interacts with water to produce hydronium ions ([H₃O⁺]) and a hydroxo complex. This process causes an increase in the acidity of the solution, which can be measured using pH. When calculating the pH of solutions like this, understanding the nature of such reactions is key.- **Importance**: Hydrolysis significantly affects pH, especially with metal ions like aluminum, which often possess high charges that attract electron pairs from water.- **Result**: This process leads to the formation of acidic solutions because of the production of hydronium ions.

ICE Table

An ICE table is a simple, yet powerful tool used in chemistry to track the concentration changes in different species during a reaction. ICE stands for Initial, Change, Equilibrium, representing the three stages in the table.**How to Use an ICE Table:**- **Initial**: Start with the initial concentrations of reactants and products. For this reaction, we begin with \([Al(H_2O)_6^{3+}] = 0.050 \) mol. \([H_3O^+] \) and \([Al(H_2O)_5(OH)^{2+}] \) start at 0 since they are products formed in the reaction.- **Change**: Indicate how concentrations change. The reactants decrease by \(x\), and each product increases by \(x\).- **Equilibrium**: Show the concentrations at equilibrium. Use the expressions from the Change step to write the final concentrations. This acts as the basis for further calculations like finding the pH.

Equilibrium Constant

The equilibrium constant (\( K_a \)) is crucial for understanding how far a reaction proceeds before reaching equilibrium. For the hydrolysis reaction, \[ K_a = \frac{[H_3O^+][Al(H_2O)_5(OH)^{2+}]}{[Al(H_2O)_6^{3+}]} \] **Key Aspects of Equilibrium Constants:**- **Magnitude**: A small value like \(K_a = 1.4 \times 10^{-5}\) indicates that the reaction doesn't proceed far, meaning not much of the \([Al(H_2O)_6^{3+}]\) is converted into products at equilibrium.- **Calculation**: Solving for \(x\) using the equilibrium constant provides the concentration of hydronium ions, a vital step in determining pH.- **Assumptions**: Often, we assume \(x\) is small relative to the starting concentration, simplifying the math: \( \frac{x^2}{0.050} \approx 1.4 \times 10^{-5} \). This makes it manageable to solve.

Al(NO3)3 Solution

Aluminum nitrate solution, noted as \(Al(NO_3)_3\), is an aqueous solution critical in this type of pH calculation. When we dissolve \(Al(NO_3)_3\), the aluminum ion complex \(Al(H_2O)_6^{3+}\) forms and undergoes hydrolysis.**Properties of the Solution:**- **Acidity**: The hydrolysis of \(Al(H_2O)_6^{3+}\) in water contributes to the acidity of the solution, where a low pH value, such as 3.28, signifies an acidic environment.- **Concentration**: The concentration of \(Al(NO_3)_3\), given as \(0.050 M\) influences the position of equilibrium and the resultant pH.- **Applications**: Such calculations are not only academic; they also have practical applications in understanding chemical behaviors in various contexts, from industrial processes to environmental studies.