A solution is prepared by adding 45.0g of Na2SO4*6H2O to pure water to make 150mL of solution. What are molar concentrations of these ions in the resulting solution?
Sodium Ion? Sulfate Ion? Phosphate Ion?
I've determined the solution is 1.2M
45/250.16g/mol Na2SO4*6H2O = .180 mol
.180 mol Na2SO4*6H2O/.150 L=1.2M
Where do I go now?
You've done the hard part.
(Na2SO4.6H2O) = 1.2 M.
Now there is 1 sulfate/1 Na2SO4 so (SO4^2-) must the same as (Na2SO4). Right? Right.
Now there are 2 Na/1 Na2SO4 so (Na^+) = 2 x (Na2SO4) = 2*1.2 = 2.4. Right?
Now that you have determined the molarity of the Na2SO4*6H2O solution to be 1.2M, you can use this information to calculate the molar concentrations of the sodium ions (Na+), sulfate ions (SO42-), and phosphate ions (PO43-) in the resulting solution.
To find the molar concentration of each ion, you need to consider the stoichiometry of the compound Na2SO4 and understand that each molecule dissociates into two sodium ions (2Na+) and one sulfate ion (SO42-).
To calculate the molar concentration of the sodium ions (Na+), you can multiply the molarity of the Na2SO4*6H2O solution by the number of sodium ions per formula unit (2Na+).
1.2M Na2SO4*6H2O × 2 Na+ / 1 Na2SO4*6H2O = 2.4M Na+
Similarly, to find the molar concentration of sulfate ions (SO42-), you can multiply the molarity of the Na2SO4*6H2O solution by the number of sulfate ions per formula unit (1SO42-).
1.2M Na2SO4*6H2O × 1 SO42- / 1 Na2SO4*6H2O = 1.2M SO42-
Finally, for the phosphate ions (PO43-), you need to know if the Na2SO4*6H2O compound is a source of phosphate. However, Na2SO4*6H2O does not contain phosphate (PO43-). Therefore, the molar concentration of phosphate ions in the resulting solution would be zero.
So, the molar concentrations of the ions in the resulting solution are:
- Sodium ion (Na+): 2.4M
- Sulfate ion (SO42-): 1.2M
- Phosphate ion (PO43-): 0M (not present in the compound Na2SO4*6H2O)