Help!!! I have done this a million times... Solubility question?

Question

What is the solubility of Ba(IO3)2 in g/100mL (Ksp = 1.5E-9)?

Here is what I did (a million times)
Ba(IO3)2 (s)  Ba+ +2(IO3)-

1.5E-9 = [x] [2x]^2

1.5E-9= [4(x^3)]

(1.5E-9 / 4) ^(1/3) =x

x = (7.211E-4 M)(487.1 g of Ba(IO3)2 / 1 mol ) (0.1/100) = 3.51E-4

(a.) .033 g/100mL
(b.) .100 g/100mL
(c.) .058 g/100mL
(d.) .047 g/100mL
(e.) .244 g/100mL

Here is what I did (a million times)

the rxn did not come out properly it should be:

Ba(IO3)2 (s) <----> Ba+ + 2(IO3)-

oops... just received and email from tutor.. I did it right.. Don't worry vnav_in, you will still get points for being brave enough to try...thank you :)

Answer 1

Ba(IO3)2 --> Ba2+ + 2(IO3)-

1.5E-9 = [x][x]^2 ( I dont know why u used 2x)

1.5E-9 = x^3

X = 1.14E-3 M

Weight/100ml = 1.14E-3 * 487.1/10 = 0.056