Can anyone help me solve a chemistry problem involving Ksp..?

Question

here is the question:
what compound if any, will precipitate when 150mL .004M barium hydroxide is mixed with 50mL .003M iron(III) sulfate? determine concentration of all the ions when the mixture reaches equilibrium. Ksp for BaSO4 is 1.5x10-9 and Fe(OH)3 is 4x10-38.

Answer 1

Moles Ba(OH)2 = 150 x 0.004 / 1000 = 0.0006

Ba(OH)2 is a strong base so we get 0.0006 mole Ba2+ and 0.0012 mole OH-

Moles Fe2(SO4)3 = 50 x 0.003 / 1000 = 0.00015

Fe2(SO4)3 is a strong electrolyte so we get

0.00015 x 2 = 0.00030 mole Fe3+ and 0.00015 x 3 = 0.00045 mole SO42-

mole Ba2+ =0.0006 change - 0.00045 after 0.00015
mole OH- = 0.0012 change - 0.0009 after 0.0003
mole Fe3+ = 0.00030 change - 0.00030 after 0
mole SO42- = 0.00045 change - 0.00045 after 0

Total volume = 200 mL = 0.200 L

[ Ba2+ ] = 0.00015 / 0.200 = 0.00075 M
[ OH- ] = 0.0003 / 0.200 = 0.0015 M

[ SO42- ] = 1.5 x 10^-9 / 0.00075 = 0.000002 M

[Fe3+] = 4 x 10^-38 /( 0.0015)^3 = 1.18 x 10^-29 M

Answer 2

BaSO4 will precipitate out.

moles of Ba(OH)2 = .6 millimol
moles of Fe2(SO4)3 = .15 millimol

[Fe3+] = 2*4x10^-38/(27x.6^3)

[Ba2+] = [Fe3+]/2

Answer 3

You should do your own hw or ask your teacher if you don't know. But, I'll feeling nice and I've got extra time so I'll just do a couple. 1) Technically, it wouldn't even be solvable b/c of significant digit issues but I'll just ignore that. So pH= -log[H], right? So plug 8 in for pH and solve for [H]. You get 1x10^-8. But we want [OH], and since [H]x[OH] = 1x10^-14, do (1x10^-14)/(1x10^-8) to get [OH]. 2) For the reverse reaction, K is 1/K of the forward reaction, so just do 1/(2x10^5).