Problem 81 - Chemistry: An Atoms-Focused Approach

Question

What is the pH of \(0.25 M \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} ?\)

Step-by-Step Solution

Answer

Answer: The pH of a 0.25 M Al(NO3)3 solution is 0.

Step 1: Write the chemical equation of Al3+ reacting with water

Al3+ will react with water to produce hydrogen ions (H+) and the aluminum hydroxide complex ion (Al(OH)4-). The balanced equation for this reaction is: Al3+ (aq) + 4H2O (l) <=> Al(OH)4- (aq) + 4H+ (aq)

Step 2: Determine the equilibrium expression

Now, we will write the equilibrium expression for the reaction. The equilibrium constant (K) for this reaction can be calculated as: K = [Al(OH)4-][H+]^4 / [Al3+] Since the concentration of water is constant, it is not included in the equilibrium expression.

Step 3: Find the concentration of H+ ions

The 0.25 M Al(NO3)3 solution means that the initial concentration of Al3+ ([Al3+]_initial) is 0.25 M. From the balanced equation, for every Al3+ that reacts, 4 H+ ions are produced. Therefore, we can set up the following expressions: [Al3+] = 0.25 - x [Al(OH)4-] = x [H+] = 4x Since K is significantly large, we can assume that all of the Al3+ ions have reacted. Therefore, the concentration of H+ ions [H+] will be: [H+] = 4 * (0.25) = 1

Step 4: Calculate the pH

Once we have the concentration of H+ ions in the solution, we can then calculate the pH of the solution. The pH is defined as the negative logarithm of the hydrogen ion concentration: pH = -log[H+] For our 0.25 M Al(NO3)3 solution: pH = -log(1) = 0 The pH of the 0.25 M Al(NO3)3 solution is 0.