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I have a few questions about this two relations:

We know that:

$K_p$ = $e^{\frac{-\Delta G^0}{RT}}$ $(1)$

and

$K_c = K_p (RT)^{\Delta n}$ $(2)$

Ok now I'm solving this problem about solubility:

The solubility product $K_s$ of $Ni(OH)_2$ is $6.0\times10^{-16}$ at $T=25ºC$

Calculate $\Delta G^0$ of this reaction at $25ºC$

I checked my teacher's resolution and he is using $(1)$ to directly calculate $\Delta G^0$? But is that correct since $K_s$ is equivalent to $K_c$ and not $K_p$?

But I also thought about converting $K_s$ to $K_p$ using $(2)$ but does that make sense since we are not dealing with gases?

Can someone please clarify me what's the best process to follow here?

Thanks!

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  • http://chemistry.stackexchange.com/a/40586/16683 – orthocresol May 12 '16 at 15:35
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    Also your equation for $K_c$ depends on the ideal gas law. It isn't true for non-ideal gases. (And solubility equilibria happen in liquids, and liquids are definitely not ideal gases...) – Curt F. May 12 '16 at 17:19

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