The ionic compound, (NH4)2SO4, will fully dissolve in water. If 13.01 g is dissolved in water to give a final volume of 546.0 mL, then find the following ion concentrations.
The concentration of the NH4+ is: _____ M
The concentration of the SO42- is: _____ M
13.01 grams (NH4)2SO4 in 546.0 mL
I will do this with figures that don't show the precision you are using so you will need to recalculate using the number of significant figures you want to use.
molar mass (NH4)2SO4 is approximately 132 so you should recalculate that number to four significant figures.
mols (NH4)2SO4 = g/molar mass = about 13.01/132 = about 0.986
M (NH4)2SO4 = mols/L = about 0.986 moles/0.546 L = about 0.180 M.
So (NH4)^+ = 0.180 M x (2 mols NH4^+ ions/1 mol (NH4)2SO4) = ?
(SO4^2-) = 0.180 x (1 mol SO4^2- ions/1 mol (NH4)2SO4 = ?
It may look complicated but look at it logically. If the concn of NH4)2SO4 is 0.180 M, the sulfate is the same since there is 1 mol sulfate ions in the molecule and there is twice the concn for NH4^+ since there are two NH4+ ions/molecule.
Well, well, well! Looks like we have some ions looking to make some noise in the water! Let's see what we've got here.
First, let's calculate the molarity of NH4+. To do that, we need to figure out how many moles of NH4+ are present in 13.01 g of (NH4)2SO4.
The molar mass of (NH4)2SO4 is:
(2 x molar mass of NH4+) + molar mass of SO4
(2 x 18.03 g/mol) + 96.06 g/mol
= 36.06 g/mol + 96.06 g/mol
= 132.12 g/mol
Now, let's calculate the number of moles of NH4+:
Number of moles = mass / molar mass
= 13.01 g / 132.12 g/mol
= 0.098 g/mol (approximately)
Since (NH4)2SO4 contains 2 NH4+ ions, the number of moles of NH4+ is twice that of (NH4)2SO4. So, the number of moles of NH4+ is:
0.098 mol x 2
= 0.196 mol
Now, let's calculate the concentration of NH4+ in the final volume of 546.0 mL.
Concentration (in M) = Number of moles / Volume (in L)
= 0.196 mol / 0.546 L
= 0.359 M (approximately)
So, the concentration of NH4+ is approximately 0.359 M. Those NH4+ ions are ready to party in the water!
Now, let's move on to calculate the concentration of SO42-. The process is similar.
First, let's calculate the number of moles of SO42-. Using the molar mass of SO42- (96.06 g/mol), we get:
Number of moles = mass / molar mass
= 13.01 g / 96.06 g/mol
= 0.135 mol (approximately)
Since (NH4)2SO4 contains 1 SO42- ion, the number of moles of SO42- is the same as that of (NH4)2SO4. So, the number of moles of SO42- is approximately 0.135 mol.
Now, let's calculate the concentration of SO42- in the final volume of 546.0 mL.
Concentration (in M) = Number of moles / Volume (in L)
= 0.135 mol / 0.546 L
= 0.247 M (approximately)
So, the concentration of SO42- is approximately 0.247 M. Those SO42- ions are ready to go swimming in the water!
I hope I didn't dissolve your brain with all these calculations. Molarity can be a bit intense sometimes. But hey, now you know the concentrations of NH4+ and SO42- ions in the solution. Enjoy the chemistry party!
To find the ion concentrations, we need to use the formula for molarity, which is:
Molarity (M) = moles of solute / volume of solution in liters
First, we need to find the number of moles of (NH4)2SO4:
Given:
Mass of (NH4)2SO4 = 13.01 g
Molar mass of (NH4)2SO4 = 132.14 g/mol (2 x 14.01 g/mol for N + 8 x 1.01 g/mol for H + 32.06 g/mol for S + 4 x 16.00 g/mol for O)
Number of moles of (NH4)2SO4 = mass / molar mass
Number of moles of (NH4)2SO4 = 13.01 g / 132.14 g/mol
Next, we need to convert the volume of the solution to liters:
Given:
Volume of solution = 546.0 mL
Volume of solution in liters = 546.0 mL / 1000 mL/L
Now we can calculate the concentration of NH4+ ion:
The concentration of NH4+ = moles of NH4+ / volume of solution in liters
Since (NH4)2SO4 has two NH4+ ions per formula unit, the moles of NH4+ ion will be twice the moles of (NH4)2SO4:
Moles of NH4+ = 2 x (moles of (NH4)2SO4)
Finally, plug in the values:
Concentration of NH4+ = (2 x moles of (NH4)2SO4) / volume of solution in liters
To find the concentration of SO42- ion, we use the same formula:
Concentration of SO42- = (moles of (NH4)2SO4) / volume of solution in liters
Now we can calculate the ion concentrations step-by-step.
To find the ion concentrations, we need to determine the number of moles of NH4+ and SO42- ions in the given mass of (NH4)2SO4 and then divide by the final volume of the solution. Here's how you can calculate it step by step:
1. Calculate the molar mass of (NH4)2SO4:
(NH4)2SO4 = (2 x NH4+) + SO4²-
Molar mass of NH4+ = 14.01 g/mol (1 nitrogen + 4 hydrogens)
Molar mass of SO42- = 32.07 g/mol (1 sulfur + 4 oxygens)
Molar mass of (NH4)2SO4 = (2 x 14.01 g/mol) + 32.07 g/mol = 132.14 g/mol
2. Convert the given mass of (NH4)2SO4 to moles:
moles = mass / molar mass
moles = 13.01 g / 132.14 g/mol
moles = 0.0985 mol
3. Calculate the ion concentrations:
ion concentration (M) = moles / volume (L)
The concentration of NH4+:
There are 2 moles of NH4+ ions in every mole of (NH4)2SO4, so the moles of NH4+ = 2 x 0.0985 mol = 0.197 mol
concentration of NH4+ = 0.197 mol / 0.546 L = 0.36 M
The concentration of SO42-:
There is 1 mole of SO42- ions in every mole of (NH4)2SO4, so the moles of SO42- = 1 x 0.0985 mol = 0.0985 mol
concentration of SO42- = 0.0985 mol / 0.546 L = 0.180 M
Therefore, the concentration of NH4+ is 0.36 M, and the concentration of SO42- is 0.180 M.