Question

What will the molarity be of the resulting solutions made by mixing the following? Assume that volumes are additive.

a)116 mL of 3.6 M H₃PO₄ with 776 mL of H₂O

(b) 219 mL of 0.39 M Na₂SO₄ with 737 mL of H₂O

(c) 75 mL of 0.50 M HNO₃ with 75 mL of 1.5 M HNO₃

Answer 1

a) Total volume of solution = Volume of H₃PO₄ + Volume of H₂O = 116 mL + 776 mL = 892 mL = 0.892 L

Molarity = No. of moles of solute / Volume of solution in Litres

Concentration of H₃PO₄ = 3.6 M = 3.6 moles of solute / Volume of solution

Volume of H₃PO₄ solution = 116 mL = 0.116 L

Moles of H₃PO₄ solute = 3.6 0.116 = 0.4176 moles of H₃PO₄ solute

Molarity of solution = Moles of solute / Total volume of solution in Litres

                               = 0.4176 moles of H₃PO₄ solute/0.892 L = 0.468 M H₃PO₄

b) Total volume of solution = Volume of Na₂SO₄ + Volume of H₂O = 219 mL + 737 mL = 956 mL = 0.956 L

Molarity = No. of moles of solute / Volume of solution in Litres

Concentration of Na₂SO₄ = 0.39 M = 0.39 moles of solute / Volume of solution

Volume of Na₂SO₄ solution = 219 mL = 0.219L

Moles of Na₂SO₄ solute = 0.39 0.219 = 0.0854 moles of Na₂SO₄ solute

Molarity of solution = Moles of solute / Total volume of solution in Litres

                               = 0.0854 moles of Na₂SO₄ solute/0.956 L = 0.0893 M Na₂SO₄

c) Total volume of solution = Volume of 0.5 M HNO₃ + Volume of 1.5 M HNO₃ = 75 mL + 75 mL = 150 mL = 0.150 L

Molarity = No. of moles of solute / Volume of solution in Litres

Moles of HNO₃ solute in 0.5 M HNO₃= 0.5 0.075 = 0.0375 moles of HNO₃ solute

Moles of HNO₃ solute in 1.5 M HNO₃= 1.5 0.075 = 0.1125 moles of HNO₃ solute

Total number of moles of HNO₃ solute = 0.15 moles

Molarity of solution = Moles of solute / Total volume of solution in Litres

                               = 0.15 moles of HNO₃ solute/0.150 L = 1 M HNO₃