Wouldn't it be 3*0.234 instead of 2* 0.234? since....2Al3+ +3SO4 2- ----> Al2(SO4)3
4.) A solution is prepared by dissolving 25.0 g of aluminum sulfate in enough water to make 125.0 mL of stock solution. A 10.0 mL aliquot (portion) of this stock solution is then removed and added to 15.0 mL of water. Calculate the concentrations of all ions in the final solution.
My answer:
[Al2(SO4)3]= 0.234M
[Al3+]= 0.468M
[SO4 2-]= 0.702M
I would have use 0.2338 x 2 = 0.0701 for SO4^2-. The others look ok.
Of course, that's the only way to get 0.701. You should check all of these closely, especially for typos. My eyesight is not all that good. I did proof each answer before posting but sometimes I miss them the second time through.
To determine the concentrations of all ions in the final solution, let's break down the process step by step:
1. First, we have a stock solution consisting of 25.0 g of aluminum sulfate dissolved in enough water to make 125.0 mL of solution.
2. To find the molarity (M) of this solution, we need to calculate the number of moles of aluminum sulfate.
Moles = Mass / Molar Mass
The molar mass of aluminum sulfate (Al2(SO4)3) is calculated as:
2(Al) + 3(S) + 12(O) = 54.0 g/mol + 96.0 g/mol + (12.0 g/mol x 4) = 342.0 g/mol
Moles = 25.0 g / 342.0 g/mol = 0.0731 moles
3. Now, we can calculate the molarity of the stock solution using the formula:
Molarity (M) = Moles / Volume (in L)
Volume (in L) = 125.0 mL / 1000 mL/L = 0.125 L
Molarity (M) = 0.0731 moles / 0.125 L = 0.584 M
4. Next, we take a 10.0 mL aliquot of the stock solution and add it to 15.0 mL of water. This gives us a final solution with a total volume of 25.0 mL.
To calculate the concentrations of ions in the final solution, we need to consider the dilutions that have occurred.
For the final volume of 25.0 mL:
- The concentration of Al2(SO4)3 will remain the same at 0.584 M (as it is undiluted).
- The concentration of Al3+ ion will also remain the same at 0.584 M (as it doesn't change during dilution).
- The concentration of SO4^2- ion will be diluted by a factor of 2 since we use a 10.0 mL aliquot and dilute it to 25.0 mL.
So, the concentration of SO4^2- ion can be calculated as:
[SO4^2-] = (Concentration of Stock Solution) x (Dilution Factor)
Dilution Factor = (Final Volume) / (Initial Volume)
Dilution Factor = 25.0 mL / 10.0 mL = 2.5
[SO4^2-] = 0.584 M x 2.5 = 1.46 M
Therefore, the concentrations of ions in the final solution are as follows:
[Al2(SO4)3] = 0.584 M
[Al3+] = 0.584 M
[SO4^2-] = 1.46 M
Note: Your initial calculation of [SO4^2-] as 0.702 M seems incorrect. It should be 1.46 M based on the dilution factor.