A scientist wants to make a solution of tribasic sodium phosphate, Na_3PO_4, for a laboratory experiment. How many grams of Na_3PO_4 will be needed to produce 550 mL of a solution that has a concentration of Na^+ ions of 0.500 it M?
You want 0.5 M in Na^+. That will be 0.5/3 in Na3PO4.
M = moles/L soln
Substitute M and L and solve for moles.
moles Na3PO4 = grams/molar mass.
Solve for grams.
163.97
To determine the grams of Na3PO4 needed, we need to calculate the number of moles of Na+ ions required for the desired concentration in the solution.
Concentration is defined as the number of moles of a solute per liter of solution (molarity, M). Here, the concentration is given as 0.500 M.
Molarity (M) = moles of solute / liters of solution
Rearranging the equation, we can calculate moles of solute:
moles of solute = Molarity × liters of solution
Liters of solution is given as 550 mL. To convert it to liters, we divide by 1000:
Liters of solution = 550 mL / 1000 = 0.55 L
Now, let's calculate the moles of Na+ ions:
moles of Na+ ions = 0.500 M × 0.55 L
Next, we need to convert moles of Na+ ions to moles of Na3PO4. From the chemical formula, we can see that there are three Na+ ions for every one Na3PO4 molecule, so we multiply the moles of Na+ ions by 1/3:
moles of Na3PO4 = (0.500 M × 0.55 L) / 3
Finally, we can calculate the grams of Na3PO4 needed using the molar mass of Na3PO4. The molar mass of Na is approximately 23 g/mol, and the molar mass of PO4 is approximately 95 g/mol. Adding them up, we get:
molar mass of Na3PO4 = (3 × 23 g/mol) + 95 g/mol = 164 g/mol
Now, we can calculate the grams of Na3PO4:
grams of Na3PO4 = moles of Na3PO4 × molar mass of Na3PO4
Substituting the previously calculated values:
grams of Na3PO4 = [(0.500 M × 0.55 L) / 3] × 164 g/mol
Performing the calculation will give you the answer in grams of Na3PO4 needed to produce a 550 mL solution with a concentration of Na+ ions of 0.500 M.