Question

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4Na3PO4, for a laboratory experiment. How many grams of Na3PO4Na3PO4 will be needed to produce 625 mLmL of a solution that has a concentration of Na+Na+ ions of 1.00 MM ?

Answer 1

To make a solution with a concentration of 1.00 M Na+ ions in 625 mL of water, you would need 341.7 grams of tribasic sodium phosphate (Na3PO4).

Step-by-step explanation:

To calculate the mass of Na3PO4 needed, we need to use the relationship between molarity, volume, and molar mass. First, we convert the given concentration of Na+ ions (1.00 M) to moles using the equation:

moles of Na+ ions = concentration(molar) x volume(L)
moles of Na+ ions = 1.00 M x 0.625 L = 0.625 moles

Since tribasic sodium phosphate (Na3PO4) has 3 Na+ ions per formula unit, we need to multiply the moles of Na+ ions by the molar ratio:

moles of Na3PO4 = 0.625 moles x (1 mole Na3PO4 / 3 moles Na+ ions) = 0.2083 moles Na3PO4

Finally, we can calculate the mass of Na3PO4 using its molar mass:

mass of Na3PO4 = moles of Na3PO4 x molar mass = 0.2083 moles x 163.94 g/mol = 341.7 grams