0 mL.
x = 3.07 x 10-6
1 mol TlIO3 = 474.3 g
x = 3.07 x 10-6 mol/L
x = 474.3 g/L x 3.07 x 10-6 mol/L
x = 1.45 x 10-3 g/100.0 mL
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.
1 answer
1 answer