A solution is prepared by dissolving 15.6 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00 mL sample of this stock is added to 52.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.
I know concentration is Molarity = moles/L I need help starting the problem.
15.6 g (NH4)2SO4 = ?? moles.
moles = 15.6/molar mass (NH4)2SO4.
M (NH4)2SO4 = moles/0.100 L for the stock solution = xx M
Technically you can't go from this solution to the final solution BECAUSE 10 mL of the stock is ADDED to 52 mL water and the final volume may or may not be 62 mL. We will assume it is 62 mL.
final concn (NH4)2SO4 = xxM x (10/62) = yy M. The sulfate will be yy M (1 mole sulfate/1 mole ammonium sulfate) and the NH4 concn will be twice that (2 NH4^+ in 1 mole (NH4)2SO4).
To solve this problem, we need to find the concentration of ammonium ions (NH4+) and sulfate ions (SO42-) in the final solution.
Let's start by calculating the moles of ammonium sulfate in the initial stock solution and the moles of ammonium sulfate in the final solution.
1. Moles of ammonium sulfate in stock solution:
First, we need to convert the mass of ammonium sulfate (15.6 g) to moles. The molar mass of ammonium sulfate (NH4)2SO4 is:
(1 × 2 + 4 × 1 + 1 × 32) + (32 + 4 × 16) = 132.14 g/mol.
Moles of ammonium sulfate = mass / molar mass
Moles of ammonium sulfate = 15.6 g / 132.14 g/mol
2. Volume of the stock solution:
The volume of the stock solution is given as 100.0 mL.
Next, let's calculate the concentration of ammonium and sulfate ions in the stock solution.
3. Concentration of ammonium and sulfate ions in stock solution:
The stock solution is prepared by dissolving 15.6 g of ammonium sulfate in enough water to make 100.0 mL of solution.
To find the concentration, we need to determine how many moles of ammonium and sulfate ions are present in 100.0 mL of stock solution.
Since ammonium sulfate dissociates into two ammonium ions and one sulfate ion, we need to multiply the moles of ammonium sulfate by 2 (for ammonium ions) and 1 (for sulfate ions).
Concentration of ammonium ions in stock solution = (2 × moles of ammonium sulfate) / volume of stock solution
Concentration of sulfate ions in stock solution = (1 × moles of ammonium sulfate) / volume of stock solution
Now we need to calculate the moles of ammonium sulfate in the final solution.
4. Moles of ammonium sulfate in the final solution:
10mL of the stock solution is added to 52mL of water.
Since there is no information provided about the concentration of ammonium sulfate in the stock solution, we assume that the solution is dilute enough to be considered as pure water.
Therefore, we can assume that the moles of ammonium sulfate added to the final solution remain the same as in the 10.00 mL sample of the stock solution.
Finally, let's calculate the concentration of ammonium and sulfate ions in the final solution.
5. Concentration of ammonium and sulfate ions in the final solution:
In the final solution, 10.00 mL of the stock solution is added to 52.00 mL of water.
To determine the concentration, we need to adjust the volume of the final solution.
Total volume of the final solution = volume of stock solution + volume of water
Total volume of the final solution = 10.00 mL + 52.00 mL
Concentration of ammonium ions in the final solution = (2 × moles of ammonium sulfate) / total volume of final solution
Concentration of sulfate ions in the final solution = (1 × moles of ammonium sulfate) / total volume of final solution
You can now proceed with plugging the values into these equations to find the concentration of ammonium and sulfate ions in the final solution.