[Chz]

How can I determine from concentrations and then activites the molar solubility of La(IO3)3 in 0.100 M NaNO3, 0.100 M NaIO3, and 0.100 M La(NO3)3?

I can find the Activiy coefficents for the ions with the extended Debye-Huckel equation and a table. I realize how the activity coeffiecent applies to the equilibrium constant, but I fail to see how to solve this. Does it have something to do with Ksp?

Thanks for your *delete me*

[Borek]

Does it have something to do with Ksp?

Yes.

[Chz]

The Ksp of La(IO3)3 is 1.0X10^-11, so without using activities, do i set this equal to (X)(.10) to solve from the concentrations?
Without taking the activities into consideration, would the solubility be the same for all of them?

[AWK]

NaNO3 gives you only pure ionic strenght effect, but the other salts also the common ion effect.
Moreover, extended Debye-Huckel method has some limits for ionic strenght.





two printing error corrected

[Chz]

I am still a bit confused on how to approach this...

Could you set up the equilibrium expressions for the Ksp?

[Borek]

Write Ksp for La(IO₃)₃.

[Chz]

La(IO3)3 <--> La(3+) + (IO3)3(3-)  Ksp= 1.0X10^-11

I am still not making the connection on how to apply this to the problem.

[Borek]

La(IO3)3 <--> La(3+) + (IO3)3(3-)  Ksp= 1.0X10^-11

This is not Ksp formula, sorry for not being clear.

Write Ksp formula, show how you will use it to calculate molar solubility of solid La(IO₃)₃ in pure water.

[AWK]

http://www.coolschool.ca/lor/CH12/unit3/U03L01.htm
http://www.chem.purdue.edu/gchelp/howtosolveit/equilibrium/solubility_products.htm
http://jan.ucc.nau.edu/~doetqp-p/courses/env440/env440_2/lectures/lec8/lec8.html

[Chz]

La(IO3)3 <--> La(3+) + (IO3)3(3-)  Ksp= 1.0X10^-11

This is not Ksp formula, sorry for not being clear.

Write Ksp formula, show how you will use it to calculate molar solubility of solid La(IO₃)₃ in pure water.

So Ksp= [La][IO3]^3= 1.0X10^-11

so in pure water the solubility would be 1.0X10^-11= (X)(3X)^3, x=.00014938 M

Would this be the same molar solubility in .100 M NaNO3 because there is no common ion?

[Borek]

That's where the activities kick in.

Note, that other cases will require you to take care of more effects.