Question

A 200 mL saturadted solution of barium nitrate was obtaied by dissolving 5.50 g of salt water.

a) determine the molar solubility of barium nitrate,

b) Calculate the solubility product constant of barium nitrate,

c) determine of precipitate will form after addition of 0.011 mol of barium sulfate to the solution without a chnage in volume.

Answer 1

a)

moles of barium nitrate = mass / molar mass

                                     = 5.50 / 261.33

                                     = 0.02105

molar solubility = 0.02105 / 0.2

                      = 0.105 M

molar solubility = 0.105 M

b) soluility product = 4S^3

                               = 4 (0.105)^3

                               = 4.66 x 10^-3

solubility product constant= 4.66 x 10^-3

b)

Qsp = [Ba+2] [SO4-2]

Qsp = 0.011^2

Qsp = 1.21 x 10^-4

Ksp < Qsp

so there is no precipitate form