Question

If you were to filter and dry the barium sulfate precipitate that fromed in the titration in part A, what mass would you expect to obtain based on your molarity of the barium hydroxide solution and the volume used in trial 2?

molarity of barium hydroxide: 0.1M

volume of barium hydroxide dispensed: 10.00 mL/0.01 L

Please help! Thank you!

Answer 1

Ba(OH)2 + H2SO4 ---------------------> BaSO4 + 2H2O

moles of Ba(OH)2 = moles of BaSO4

moles of Ba(OH)2 = molarity x volume

                              = 0.1 x 0.01

                              = 10^-3

moles of BaSO4 is also = 10^-3

BaSO4 molar mass = 233. 3896 g/mol

moles = mass / molar mass

10^-3 = mass / 233.3896

mass = 0.233 g

mass of BaSO4 formed = 0.233 g