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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