At $\pu{25^\circ C}$ I used the free energy table for $G$ for the reaction and got $K_\mathrm{sp}= 10^{-9.02}$.

I now need to find out what the solubility is at $10$, $20$ and $\pu{30^\circ C}$. Could you also show what equation you used and how to work out the solubility by one of these other temperatures?

  • $\begingroup$ Could you elaborate on how you might approach the problem first, rather than asking for a solution, I think it will be a lot more likely to receive an answer that way. $\endgroup$
    – jonsca
    Oct 16, 2012 at 6:48

1 Answer 1


It depends what you have access to… but if we suppose that you have access to decent thermodynamic tables, here's a hint:

You know how solubility is linked to dissolution reaction (Gibbs) free energy. Now, how does the free energy depend on temperature? In other words, what is $\displaystyle\frac{\partial G}{\partial T}$?

  • $\begingroup$ I do have access to thermodynamic tables. I tried using the van 't Hoff equation, because there is the relationship between temperature and the K solubility product. However, I am not 100% sure that I am using the right equation and I don't have any way to check if my answers are correct. The answers I get from using the van 't Hoff equation are a little bit counterintuitive. For 10° Celsius, I get a solubility product of Ksp= 10^−9.02 For 20° Celsius, I get a solubility product of Ksp= 10^−9.03 For 30° Celsius, I get a solubility product of Ksp= 10^−9.00 This just seems incorrect to me. $\endgroup$
    – Erin
    Oct 16, 2012 at 11:59
  • $\begingroup$ I think you misread the results. The 20 C reading is the smallest. Assuming that Erin used a constant $\Delta H$, that's not possible. I'd guess that there is a math error involved. $\endgroup$ Nov 16, 2012 at 19:04

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