A solution is prepared by dissolving 16.5 g ammonium sulfate in enough water to make 145.0 mL of stock solution. A 10.50 mL sample of this stock solution is added to 57.70 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.
NH4+
Suppose 15.o mL of 0.315 M NH4CL is combine with 10.0mL OF 0.430 M (NH4)3PO4. What is the total concentration of ammonium ions in this mixture?
To calculate the concentration of ammonium ions (NH4+) and sulfate ions (SO4^2-) in the final solution, we need to consider the principles of stoichiometry and dilution.
First, let's calculate the number of moles of ammonium sulfate (NH4)2SO4 in the stock solution.
1. Calculate the molar mass of (NH4)2SO4:
The molar mass of ammonium sulfate (NH4)2SO4 is calculated by summing the molar masses of the individual atoms:
NH4 = 14.01 g/mol (ammonium)
SO4 = 32.07 g/mol (sulfate)
Molar mass of (NH4)2SO4 = (2 * NH4) + SO4 = (2 * 14.01 g/mol) + 32.07 g/mol = 132.14 g/mol
2. Calculate the number of moles of (NH4)2SO4 in the stock solution:
Moles of (NH4)2SO4 = Mass of (NH4)2SO4 / Molar mass of (NH4)2SO4
Moles of (NH4)2SO4 = 16.5 g / 132.14 g/mol
Next, let's calculate the concentration of ammonium ions and sulfate ions in the stock solution.
3. Calculate the molarity of the stock solution:
Molarity (M) = Moles / Volume in liters
Volume in liters = Volume in mL / 1000 (since 1 L = 1000 mL)
Molarity of stock solution = Moles of (NH4)2SO4 / Volume of stock solution in L
Molarity of stock solution = (16.5 g / 132.14 g/mol) / (145.0 mL / 1000)
Now, we have the molarity of the stock solution. Next, we need to consider the dilution of the stock solution.
4. Calculate the number of moles of (NH4)2SO4 in the diluted solution:
According to the dilution formula (M1V1 = M2V2), the moles of (NH4)2SO4 remain constant before and after dilution. Therefore, the number of moles of (NH4)2SO4 in the 10.50 mL sample is the same as in the stock solution.
5. Calculate the volume of the diluted solution:
The volume of the diluted solution is the sum of the volumes of the stock solution and water added.
Volume of diluted solution = Volume of stock solution + Volume of water added
Volume of diluted solution = 10.50 mL + 57.70 mL
Finally, let's calculate the concentration of ammonium ions and sulfate ions in the diluted solution.
6. Calculate the concentration of ammonium ions in the diluted solution:
Concentration of NH4+ in diluted solution = Moles of (NH4)2SO4 / Volume of diluted solution in L
Concentration of NH4+ in diluted solution = Moles of (NH4)2SO4 / (Volume of diluted solution in mL / 1000)
7. Calculate the concentration of sulfate ions in the diluted solution:
Since ammonium sulfate dissociates into two ammonium ions (NH4+) and one sulfate ion (SO4^2-), the concentration of sulfate ions is half the concentration of ammonium ions.
Concentration of SO4^2- in diluted solution = Concentration of NH4+ in diluted solution / 2
By following these steps, you can calculate the concentration of ammonium ions and sulfate ions in the final solution.
16.5 g (NH4)2SO4 in 145.0 mL solution =
16.5 g/145 mL = xx g/mL and use this below.
Concn of diluted solution is
xx g/mL x 10.50 mL/(57.70+10.50) = yy g (NH4)2SO4/mL.
How is the concentration to be measured? g/mL is what the yy is above.
(SO4^=) = same as (NH4)2SO4 = zz g/mL.
(NH4^+) = 2 x [(NH4)2SO4] = ww g/mL.
If you want these numbers in molarity, just multiply zz or ww g/mL x 1000 and divide by molar mass (NH4(2SO4.
Check my thinking. Check my work.