Given the aqueous solubility rules presented in this module please answer the following questions based on the information presented below:

You are mixing exactly 95.0 mL of 1.5 x 10^-4 M Ba(OH)₂ with 35.5 mL of 2.8 x 10^-5 M Fe₂(SO₄)₃.

K_sp (BaSO₄) = 1.1 x 10^-8 K_sp (Fe(OH)₃) = 2.6 x 10^-23

What are the possible products in this reaction mixture?

What are the possible IP expressions to be studied?

What is the concentration of SO₄^2- that is being used in the appropriate IP expression?

Is a precipitate formed from this mixing of the two solutions, and if so, which precipitate would it be?

Select 5 correct answer(s)

	IP = [Ba2+][SO42-]
	Possible precipitate products are BaSO4 and Fe(OH)3
	No, there is no precipitate from the mixing of these two solutions.
	[SO42-] = 2.3 x 10-5 M
	IP = [Ba2+][OH1-]2
	Yes there is a precipitate and it is BaSO4.
	[SO42-] = 1.5 x 10-5 M
	Yes there is a precipitate and it is Fe(OH)3.
	Yes there is a precipitate and it is Fe2(SO4)3.
	IP = [Ba2+][OH1-]2
	IP = [Fe3+]2[SO42-]3
	IP = [Fe3+][OH1-]3
	Possible precipitate products are Ba(OH)2 and Fe2(SO4)3
	[SO42-] = 7.6 x 10-6 M
	Possible precipitate products are Ba(OH)2 and BaSO4
	Possible precipitate products are Ba(OH)2 and Fe(OH)3
	[SO42-] = 2.2 x 10-4 M
	Yes there is a precipitate and it is Ba(OH)2.
	Possible precipitate products are BaSO4 and Fe2(SO4)3
	[SO42-] = 1.1 x 10-4 M