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 375 mL of a solution that has a concentration of Na+ ions of 1.50 M ?

Answer 1

Solution :-

Dissociation equation for the Na3PO4 is as follows

Na3PO4 --- > 3Na^+ + PO4^3-

We want 1.50 M solution with Na^+ ions with total volume 375 ml (0.375 L)

Mass of Na3PO4 = ?

Lets first calculate the moles of Na^+ using the molarity and volume

Moles = molarity x volume in liter

Moles of Na^+ = 1.50 mol per L * 0.375 L

                           = 0.5625 mol Na^+

Now using the moles of Na^+ we can find the mass of Na3PO4 using the mole ratio of the Na^+ and Na3PO4

(0.5625 mol Na^+ * 1 mol Na3PO4 / 3 mol Na^+)*(163.94 g / 1 mol Na3PO4) = 30.7 g Na3PO4

Therefore mass of Na3PO4 needed to make the solution is 30.7 g