How do you calculate Molar Solubility in grams/100mL of Calcium iodate in water at 25 degrees Celsius? Ksp = 7.1x107

1 Answer
Jun 9, 2016

Solubility of calcium iodate 0.25g101L1

Explanation:

We write out the dissolution reaction:

Ca(IO3)2(s)Ca2++2IO3

And then we write out the equilibrium expression, in which Ca(IO3)2, as a solid, does not appear:

Ksp=[Ca2+][IO3]2=7.1×107

If we dub the solubility of calcium iodate under these conditions as S, then Ksp=(S)(2S)2 = 4S3.

This expression follows the equilibrium equation: each equiv of salt that dissolves gives 1 equiv of Ca2+, but 2 equiv iodate ion.

Thus Ksp=(S)(2S)2 = 4S3 = 7.1×107

S=37.1×1074

This gives an answer in molL1. You will have to use the formula mass of calcium iodate, 389.88gmol1, to give an answer in the units required.