Question

Barium hydroxide reacts with phosphoric acid to form barium phosphate and water. How many mL of 0.2195 M barium hydroxide would be required to titrate 12.57 mL of phosphoric acid? (Density of ohosphoric acid is 1.834 g/mL.) >> answer should be 1608 mL

Answer 1

The balanced equation is

3Ba(OH)2 + 2 H3PO4 ----> Ba3(PO4)2 + 6 H2O

density = mass / volume

1.834 g/mL = mass of phosphoric acid / 12.57 mL

Therefore, the  mass of phosphoric acid = 23.0534 g

number moles of phosphoric acid = 23.0534g / 97.995 g/mol = 0.23525 mole

from the balanced equation we can say that

2 mole of H3PO4 requires 3 mole of barium hydroxide so

0.23525 mole of H3PO4 will require 0.3529 mole of barium hydroxide

molarity of barium hydroxide = number of moles of barium hydroxide / volume of solution in L

volume of solution in L = 0.3529 / 0.2195 = 1.608 L

1 L = 1000 mL

1.608 L = 1608 mL

Therfore, the volume of solution = 1608 mL