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 450 mL of a solution that has a concentration of Na+ ions of 0.700 M
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To calculate the number of grams of Na3PO4 needed, we need to have the molar mass of Na3PO4 and use the molarity and volume of the solution.
Step 1: Find the molar mass of Na3PO4.
The molar mass of Na (sodium) is 22.99 g/mol, and the molar mass of P (phosphorus) is 30.97 g/mol. The molar mass of O (oxygen) is 15.99 g/mol. Since there are three sodium atoms, one phosphorus atom, and four oxygen atoms in Na3PO4, we can calculate its molar mass as follows:
(3 x 22.99 g/mol) + (1 x 30.97 g/mol) + (4 x 15.99 g/mol) = 163.94 g/mol
Step 2: Calculate the moles of Na+ ions in the solution.
The molarity (M) of a solution is given in moles of solute per liter of solution. We are given that the concentration of Na+ ions (solute) is 0.700 M, which means there are 0.700 moles of Na+ ions per liter of solution. Since we want to make 450 mL of solution (0.450 L), we can calculate the number of moles of Na+ ions as follows:
0.700 moles/L x 0.450 L = 0.315 moles
Step 3: Calculate the grams of Na3PO4 needed.
Using the stoichiometry of the chemical formula, we know that one mole of Na3PO4 contains 3 moles of Na+ ions. Therefore, the number of moles of Na3PO4 needed is one-third of the moles of Na+ ions:
0.315 moles / 3 = 0.105 moles
Finally, we can calculate the grams of Na3PO4 needed by multiplying the number of moles by the molar mass:
0.105 moles x 163.94 g/mol ≈ 17.21 grams
Therefore, approximately 17.21 grams of Na3PO4 are needed to produce 450 mL of a 0.700 M solution of Na+ ions.