What is the total concentration of cations in a solution made by combining 700.0 mL of 3.0 M (NH4)3PO4 with 300.0 mL of 2.0 M Na2SO4? numbers next to compounds are subscripts
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To find the total concentration of cations in the solution, we need to calculate the concentration of each cation present, taking into account the volumes and molarities of the two solutions.
Step 1: Calculate the number of moles of (NH4)3PO4:
Volume of (NH4)3PO4 solution = 700.0 mL = 0.700 L
Molarity of (NH4)3PO4 solution = 3.0 M
Number of moles of (NH4)3PO4 = 0.700 L x 3.0 mol/L = 2.1 mol
Step 2: Calculate the number of moles of Na2SO4:
Volume of Na2SO4 solution = 300.0 mL = 0.300 L
Molarity of Na2SO4 solution = 2.0 M
Number of moles of Na2SO4 = 0.300 L x 2.0 mol/L = 0.6 mol
Step 3: Calculate the total number of moles of cations:
Each formula unit of (NH4)3PO4 contains 3 ammonium ions (NH4+) and each formula unit of Na2SO4 contains 2 sodium ions (Na+).
Number of moles of NH4+ ions = 2.1 mol x 3 = 6.3 mol
Number of moles of Na+ ions = 0.6 mol x 2 = 1.2 mol
Step 4: Calculate the total concentration of cations:
Total concentration of cations = moles of cations / total volume of the solution
Total moles of cations = moles of NH4+ ions + moles of Na+ ions = 6.3 mol + 1.2 mol = 7.5 mol
Total volume of the solution = volume of (NH4)3PO4 solution + volume of Na2SO4 solution = 0.700 L + 0.300 L = 1.0 L
Total concentration of cations = 7.5 mol / 1.0 L = 7.5 M
Therefore, the total concentration of cations in the solution is 7.5 M.
To find the total concentration of cations in the solution, we need to identify the cations present in the two compounds, and then calculate their combined concentration.
In the compound (NH4)3PO4, the cations are NH4+ and Na+, while in Na2SO4, the cations are Na+ and SO4^2-. However, since SO4^2- is an anion, we only need to consider Na+ as the cation in Na2SO4.
First, let's calculate the moles of each compound:
For (NH4)3PO4:
Molarity = moles/volume
moles = Molarity * volume
moles of (NH4)3PO4 = 3.0 M * 0.700 L = 2.1 mol
For Na2SO4:
moles of Na2SO4 = 2.0 M * 0.300 L = 0.6 mol
Next, let's calculate the total moles of cations (NH4+ and Na+):
moles of NH4+ = moles of (NH4)3PO4 * 3
moles of NH4+ = 2.1 mol * 3 = 6.3 mol
moles of Na+ = moles of (NH4)3PO4 + moles of Na2SO4
moles of Na+ = 2.1 mol + 0.6 mol = 2.7 mol
Finally, let's calculate the combined concentration of cations:
Total concentration of cations = moles of cations / combined volume
combined volume = volume of (NH4)3PO4 + volume of Na2SO4
combined volume = 0.700 L + 0.300 L = 1.000 L
Total concentration of cations = (moles of NH4+ + moles of Na+) / combined volume
Total concentration of cations = (6.3 mol + 2.7 mol) / 1.000 L
Total concentration of cations = 9.0 mol / 1.000 L
Total concentration of cations = 9.0 M
Therefore, the total concentration of cations in the solution is 9.0 M.
millimols NH4^= = 3 x mL x M = ?
millimols Na^+ = 2 x mL x M = ?
Total millimols cations = ?
Concn cations in M = millimols/total mL.