A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 600. mL of a solution that has a concentration of Na+ ions of 0.900 M?
To find out how many grams of Na3PO4 are needed to produce a solution with a concentration of Na+ ions of 0.900 M in 600 mL volume, we first need to calculate the number of moles of Na+ ions present in the solution.
From the chemical formula of tribasic sodium phosphate, Na3PO4, we can see that 1 mole of Na3PO4 produces 3 moles of Na+ ions.
Therefore, the number of moles of Na+ ions in the solution = 0.900 M * 0.600 L = 0.540 moles
Since 1 mole of Na3PO4 produces 3 moles of Na+ ions, the number of moles of Na3PO4 needed = 0.540 moles / 3 = 0.180 moles
Now, we need to convert the number of moles of Na3PO4 to grams using its molar mass.
The molar mass of Na3PO4 = 3(Na) + 1(P) + 4(O) = 3(22.99 g/mol) + 1(30.97 g/mol) + 4(16.00 g/mol) = 163.94 g/mol
Therefore, the mass of Na3PO4 needed = 0.180 moles * 163.94 g/mol = 29.51 grams
So, 29.51 grams of Na3PO4 will be needed to produce 600 mL of a solution with a concentration of Na+ ions of 0.900 M.