Question

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 1.30 M?

First determine the number of Na+ using the given molarity and volume. Next, use the chemical formula to determine the number of moles of Na3PO4 needed. Finally, convert moles to grams using the molecular weight of the compound.

How many moles of Na+ are in the solution? Keep in mind that M = moles/liters.

Answer 1

Concentration of Na+ ions = 1.30 M = 1.30 mol /L

Volume of solution = 750 ml = 0.75 L

No of moles of Na+ ions = (1.30 mol / L)*0.75 L = 0.975 mol

Consider a dissociation reaction of Na3PO4

Na3PO4(aq) -------> 3 Na+(aq) + PO43-

1.0 mol of Na3PO4 produces 3.0 mol of Na+. Therefore 0.975 mol of Na+(aq) need to react with(1.0mol/3.0 mol)*0.975 mol =0.325 mol of Na3PO4 (aq)

Molar mass of Na3PO4 = 163.94 g/mol

Mass of Na3PO4 needed = (163.94 g/mol)*0.325 mol=53.28 g