Question

Solution A is prepared by dissolving 89.4 g of Na₃PO₄ in enough water to make 1.50 L of solution. Solution B is 2.5 L of 0.696 M Na₂SO₄.

a) What is the molar concentration of Na₃PO₄ in Solution A?  

0.36 M

b) How many millilitres of Solution A will give 2.50 g of Na₃PO₄?

41.95 mL

c) A 50.0 mL sample of Solution B is mixed with a 75.00 mL sample of Solution A. Calculate the concentration of Na⁺ ions in the final solution.

______M

Answer 1

Answer

1. Moles of Na3PO4 = mass/molar mass

= 89.4/163.94 = 0.5453 mol

Molar concentration of Na3PO4 in solution A = moles of Na3PO4/volume

= 0.5453/1.50 = 0.363 M

2. Volume of solution A = 2.50/89.4 x 1.50

= 0.0419 L = 41.9 mL

3. Na2SO4 => 2 Na+ + SO42-

Moles of Na2SO4 = volume x concentration of solution B

= 50.0/1000 x 0.696 = 0.034 mol

Moles of Na+ = 2 x moles of Na2SO4

= 2 x 0.034 = 0.068 mol

Na3PO4 => 3 Na+ + PO43-

Moles of Na3PO4 = volume x concentration of solution A

= 75.00/1000 x 0.363 = 0.02722 mol

Moles of Na+ = 3 x moles of Na3PO4

= 0.02722 x 3 = 0.08167 mol

Total moles of Na+ = 0.068 + 0.08167 = 0.14967 mol

Total volume of solution = 50.0 + 75.00 = 125 mL = 0.125 L

Concentration of Na+ = total moles/total volume

= 0.14967/0.125

= 1.19 M