What is the Na+ concentration of 2.85 M sodium sulfate?
Sodium sulfate is Na2SO4; there are 2 Na^+ in 1 molecule of Na2SO4.
2.85 moles/L = Na2SO4.
Na^+ is twice that.
To find the Na+ concentration in 2.85 M sodium sulfate, we need to consider the dissociation of the compound in water. Sodium sulfate (Na2SO4) completely dissociates into two sodium ions (2 Na+) and one sulfate ion (SO4^2-).
Since the compound dissociates completely, the concentration of sodium ions (Na+) is equal to twice the molarity of sodium sulfate (Na2SO4).
Therefore, the Na+ concentration in a 2.85 M sodium sulfate solution would be 2 times the molarity:
Na+ concentration = 2 * 2.85 M = 5.7 M
So, the Na+ concentration in a 2.85 M sodium sulfate solution is 5.7 M.