Final answer:
To find the mass of sodium sulfate required to prepare 350 mL of a 0.125 M sodium ion solution, you need to calculate the moles of Na2SO4 based on its molar mass and the volume and concentration of the desired solution.
Step-by-step explanation:
To calculate the mass of sodium sulfate (Na2SO4) needed to prepare a solution with a specific sodium ion concentration, we use stoichiometry and molarity concepts.
First, we need to recognize that sodium sulfate dissociates into two sodium ions (2 Na+) and one sulfate ion (SO42-) in solution. For a solution with a sodium ion concentration of 0.125 M, each liter of solution contains 0.125 moles of Na+ ions. Since Na2SO4 provides two Na+ ions per formula unit, we only need half of that amount in moles of sodium sulfate to get the desired concentration of sodium ions.
The molar mass of Na2SO4 is approximately 142 g/mol. To find the mass of Na2SO4, we multiply the number of moles needed by the molar mass of Na2SO4.
To prepare 350 mL of a 0.125 M Na+ solution, we calculate the moles of Na2SO4 required:
- V = 0.350 L (350 mL)
- MNa+ = 0.125 M
- Moles of Na2SO4 = (0.350 L) × (0.125 M / 2)
- Mass of Na2SO4 = Moles × Molar mass
The calculation will show the mass of Na2SO4 needed to be weighed and dissolved in water to prepare the desired solution.