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.

Related questions
See all questions in Ksp
Impact of this question
9659 views around the world
You can reuse this answer
Creative Commons License