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 750 mL of a solution that has a concentration of Na^+ ions of 1.3 M?

Express your answer numerically in grams.

M = moles/L

1.3 M = moles/0.750
Solve for moles.
Then moles = grams/molar mass.
Solve for grams. That will give you g Na3PO4 to make 1.3 M Na3PO4 so divide that number by 3 to give 1.3 M sodium ions and not Na3PO4.

111.797

159.85

To find the number of grams of Na3PO4 needed, we need to first calculate the number of moles of Na^+ ions in 750 mL of the solution. Then, we can use the molar ratio between Na3PO4 and Na^+ ions to find the number of moles of Na3PO4. Finally, we can multiply the moles of Na3PO4 by its molar mass to get the mass in grams.

To calculate the number of moles of Na^+ ions, we'll use the equation:

Molarity = Moles/Volume

Since we know the molarity (1.3 M) and the volume (750 mL = 0.75 L), we can rearrange the equation to solve for moles:

Moles = Molarity * Volume

Moles of Na^+ ions = 1.3 M * 0.75 L
Moles of Na^+ ions = 0.975 moles (rounded to 3 decimal places)

Now, we'll use the molar ratio between Na3PO4 and Na^+ ions. Looking at the chemical formula, we can see that there are three Na^+ ions for every one Na3PO4 molecule. So, the number of moles of Na3PO4 will be three times the number of moles of Na^+ ions:

Moles of Na3PO4 = 3 * 0.975 moles
Moles of Na3PO4 = 2.925 moles (rounded to 3 decimal places)

Finally, we'll use the molar mass of Na3PO4 to convert the number of moles to grams. The molar mass of Na3PO4 is the sum of the molar masses of its individual elements:

Molar mass of Na = 22.99 g/mol
Molar mass of P = 30.97 g/mol
Molar mass of O = 16.00 g/mol (there are four oxygen atoms in Na3PO4)

Molar mass of Na3PO4 = (3 * 22.99 g/mol) + 30.97 g/mol + (4 * 16.00 g/mol)
Molar mass of Na3PO4 = 163.94 g/mol

Mass of Na3PO4 = Moles of Na3PO4 * Molar mass of Na3PO4
Mass of Na3PO4 = 2.925 moles * 163.94 g/mol
Mass of Na3PO4 = 479.0055 grams

So, 479.0055 grams (rounded to 4 decimal places) of Na3PO4 will be needed to produce 750 mL of a solution with a concentration of Na^+ ions of 1.3 M.