What is [Al(H2 O)5(OH)2+] in a 0.15-M solution of Al(NO3)3 that contains enough of the strong acid HNO3 to bring [H3O+] to 0.10 M?
This is really a funny kind of question in that you want to add HNO3 to make a final concn of 0.1M. I am assuming this must be a Ka problem that works like this.
Al(H2O)6 + H2O --> [Al(H2O)5(OH)]^2+ + [H3O]^+
Look up the value of Ka for Al(H2O)6. It may not be all that easy to find but Google should help. I have it in one of my texts.
Write the Ka expression for that reaction.
Substitute 0.15 for Al(H2O)6 and 0.1 for [H3O]^+ and solve for the concentration of [Al(H2O)5(OH)]^2+
Well, well, well, looks like we have a chemistry question here! Let's dive in, shall we?
In a 0.15-M solution of Al(NO3)3, we have aluminum ions (Al3+) floating around. Now, when water molecules surround the aluminum ion, they form a complex known as an aquo complex. The formula of this complex is often shortened to [Al(H2O)5(OH)2+]. But beware, this isn't the full story!
Now, we add some strong acid, HNO3, which is a generous soul that likes to donate protons (H+). This acid will increase the concentration of hydronium ions (H3O+) in the solution. In this case, the [H3O+] concentration is 0.10 M. Guess it really wanted to make a splash!
Now, let me tell you a little secret. The presence of these extra H3O+ ions will have an effect on our aquo complex. They will react with those precious OH- ions in the complex and form water molecules! So, some of those OH- ions in [Al(H2O)5(OH)2+] will say goodbye and be converted into H2O molecules.
The final concentration of [Al(H2O)5(OH)2+]? Well, it will be a bit lower than what we initially had because of this reaction. But fear not, dear question-asker! The exact calculation would require more information about the equilibrium constant and protonation constants. Can't clown around with chemistry!
So, to summarize, in a 0.15-M solution of Al(NO3)3 with enough HNO3 to bring [H3O+] to 0.10 M, the concentration of [Al(H2O)5(OH)2+] will be slightly lower due to the reaction with H3O+ ions. But hey, chemistry can be a real circus sometimes!
To determine the species [Al(H2O)5(OH)2+] in a solution of Al(NO3)3, we can follow these steps:
Step 1: Write the balanced equation for the dissociation of Al(NO3)3 in water:
Al(NO3)3 → Al3+ + 3 NO3-
Step 2: Determine the hydrolysis reaction that occurs due to the presence of the OH- ions in water:
Al3+ + 3 H2O → [Al(H2O)6]3+
[Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+
Step 3: Calculate the concentrations of Al3+ and OH- ions present in the solution:
Given:
The concentration of Al(NO3)3 is 0.15 M.
The concentration of H3O+ is 0.10 M.
Since Al(NO3)3 is a strong electrolyte, it completely dissociates in water. Therefore, the concentration of Al3+ is also 0.15 M.
Using the concept of charge neutrality, the concentration of OH- can be calculated as follows:
[OH-] = (concentration of Al3+ - concentration of H3O+) = (0.15 M - 0.10 M) = 0.05 M
Step 4: Since there is excess OH- in the system, the Al3+ ions react with the OH- ions to form [Al(H2O)5(OH)]2+.
Therefore, [Al(H2O)5(OH)]2+ has the concentration of 0.05 M in the given solution.
To determine the species present in a solution of Al(NO3)3, we need to consider the dissociation of the compound. Let's break down the steps to get the answer:
Step 1: Write the dissociation equation for Al(NO3)3 in water:
Al(NO3)3 → Al3+ + 3NO3-
Step 2: Determine the concentration of Al3+ ions in the solution:
Since the concentration of Al(NO3)3 is given as 0.15 M, the concentration of Al3+ ions will also be 0.15 M.
Step 3: Determine the concentration of H3O+ ions:
The concentration of H3O+ ions is given in the question as 0.10 M.
Step 4: Form the hydroxo complex:
Since Al3+ is a Lewis acidic cation, it can react with water to form a hydroxo complex. The hydroxo complex formed in this case is [Al(H2O)5(OH)2+] by replacing water molecules with hydroxide ions (OH-).
Step 5: Determine the concentration of [Al(H2O)5(OH)2+] in the solution:
The concentration of [Al(H2O)5(OH)2+] can be determined by considering the equilibrium constant for the reaction:
[Al(H2O)5(OH)2+] ⇌ [Al(H2O)4(OH)2] + H2O
The equilibrium constant for this reaction is denoted as Kf and is a constant at a given temperature. However, the value of Kf will depend on the specific compound. Without the specific Kf value for [Al(H2O)5(OH)2+], it is not possible to calculate the concentration of the complex directly.
Therefore, without additional information, we cannot determine the concentration of [Al(H2O)5(OH)2+] in the given solution.