# Equilibrium constants and

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!

• chemistry.stackexchange.com/a/40586/16683 – orthocresol May 12 '16 at 15:35
• 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