Molar Solubility Calculation - Common Ion Effect?

Question

Calculate the molar solublity of Ca(IO3)2 in 0.305 M Ca(NO3)2.
(Ksp = 7.1×10-7 for Ca(IO3)2.)

I tried a bunch of different methods. I though it was the ksp/M^2 but the answer is wrong...Help please

Answer 1

Here you are with my try:

1.- First we use the known expression of solubility product:

Kps = [Ca++] [IO3-]²

2.- We apply the above expression:

7.1 x 10^-7 = [Ca++] [IO3-]²

we assume that "x" is the unknown concentration of the ions, but we also see, the presence of Ca++ coming from Calcium nitrate. As we know that concentration of Ca(NO3)2 is 0.305 M the same is for the ions Ca++ coming from that salt, we add it to the Ca++ coming from iodate.

Hence, the expression of Kps would be:

7.1 x 10^-7 = (x+0.305) * x² = x³ + 0.305x²

this a cubic equation. Solving it would be a very hard task. I will assume that third grade term is too small to affect result, so I will drop it and we get:

0.305x² = 7.1 x 10^-7

can solve this one for x:

x = 1.52x 10^-3 mol/L (molar solubility of Calcium Iodate)

Hope it is the correct result.

Good luck!

Answer 2

Solve using common ion effect of Ca(NO3)2 on Ca(IO3)2.