I’m really confused about this. If we consider $\ce{Ca3(PO4)2}$, it can be seen it undergoes an equilibrium. But why it’s an equilibrium?

This is the equilibrium it undergoes $$\ce{Ca3(PO4)2_{(s)}<=> 3Ca^2+_{(aq)} + 2 PO4_{(aq)}^3-}$$

By the way I’m having a doubt of why some are using backward arrow for a dissociation of a particular precipitate while some are using a solid forward arrow for the same reaction. For an example let’s take $\ce{AgCl(s)}$

Now this is a place where a backward arrow is used, $$\ce{AgCl_{(s)}<=> Ag+_{(aq)} + Cl_{(aq)}-}$$

But I have seen in some books where there is a solid forward arrow instead of back-ward arrow for the dissociation of $\ce{AgCl(s)}$.


In the two examples you mentioned, we have an equilibrium between the solid and the small quantity of the solid that is soluble in water and completely dissociated: $$\ce{Ca3(PO4)2_{(s)}<=> 3Ca^2+_{(aq)} + 2 PO4^{3-}_{(aq)} }$$ $$\ce{AgCl_{(s)}<=> Ag^+(aq)_{(aq)} + Cl^{-}_{(aq)} }$$

I think it's wrong to write a solid forward reaction. It's an heterogeneous equilibrium between the solid state (the precipitate) and the aqueous state (ions constituents of the precipitate). The constant of this equilibrium is the solubility product $K_{\mathrm{sp}}$ of the precipitate.

  • $\begingroup$ @ Youmen Then if some one wrote Ag+ + Cl^- --->(solid forward arrow)AgCl , is it correct? Could you explain it also? $\endgroup$ May 5 '15 at 13:04
  • $\begingroup$ @On the way to success Please see my answer (I have added some explanations). $\endgroup$ May 5 '15 at 13:43
  • $\begingroup$ @ Youmen How can you say constant of Ag+(aq)+Cl−(aq)<---> (AgCl(s) is the inverse of solubility product of AgCl(s)<-->.Ag+(aq)+Cl−(aq) $\endgroup$ May 5 '15 at 13:54
  • $\begingroup$ @ On the way to success It's the definition of Ksp =[Ag+][Cl-]. Ok? $\endgroup$ May 5 '15 at 14:09
  • $\begingroup$ @ Youmen Yes I do know Ksp =[Ag+][Cl-] but I'm confused with about the thing you said " constant of Ag+(aq)+Cl−(aq)<---> (AgCl(s) is the inverse of solubility product of AgCl(s)<-->.Ag+(aq)+Cl−(aq)" $\endgroup$ May 5 '15 at 14:55

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