How do you calculate Molar Solubility in grams/100mL of Calcium iodate in water at 25 degrees Celsius? Ksp = 7.1 x 10^-7

1 Answer
anor277
Jun 9, 2016

"Solubility of calcium iodate "~=0.25*g*10^-1L^-1

Explanation:

We write out the dissolution reaction:

Ca(IO_3)_2(s) rightleftharpoonsCa^(2+) + 2IO_3^-

And then we write out the equilibrium expression, in which Ca(IO_3)_2, as a solid, does not appear:

K_(sp)=[Ca^(2+)][IO_3^-]^2=7.1xx10^-7

If we dub the solubility of calcium iodate under these conditions as S, then K_(sp)=(S)(2S)^2 = 4S^3.

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

Thus K_(sp)=(S)(2S)^2 = 4S^3 = 7.1xx10^-7

S=""^3sqrt{(7.1xx10^-7}/4)

This gives an answer in mol*L^-1. You will have to use the formula mass of calcium iodate, 389.88*g*mol^-1, to give an answer in the units required.

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