Posted by
**Jason** on
.

Thallium (I) iodate is only slightly soluble in water. Its Ksp at 25°C is

3.07 x 10-6. Estimate the solubility of thallium iodate in units of grams per

100.0 ml of water.

TlIO3(s) <==> Tl^+ + IO3^-

Let x = molar solubility of TlIO3.

The x is molar solubility of Tl^+ and x is molar solubility of IO3^-

Ksp = (Tl^+)(IO3^-) = 3.07E-6

Plug in and solve for x.

Post your work if you get stuck.

A small correction here.

x is in mols/L. You want to change that to grams/100.