# Concerning Solubility Product Equations and Phases? [closed]

NaOH(?) <---> $$Na^+$$(aq) + $$OH^-$$(aq) What is the phase of the NaOH, I originally thought it was Aqueous due to it being readily soluble in water but I also heard that it was common practice to always write Solubility Product Eq'ns putting the substance on the Left hand side as solid as Solubility is a continuum and every substance is soluble to a certain degree. Thank you very much.

Even in $$\ce{NaOH(s)}$$, there are no $$\ce{NaOH}$$ molecules, but ions $$\ce{Na+}$$ and $$\ce{OH-}$$, as sodium hydroxide is a ionic substance.

When it is being dissolved, water molecules pull ions from the solid ionic lattice and wrap them into their hydrated forms.

So it is written

$$\ce{NaOH(s) ->[\ce{H2O}]Na+(aq) + OH-(aq)}$$

If it were e.g. a benzoic acid, its crystals contain molecules, that are dissolved and then it dissociates:

$$\ce{HA(s) <=>[\ce{H2O}]HA(aq)<=>H+(aq) + A-(aq)}$$

• Thank you very much, I get it now. – Kyro Apr 25 '20 at 6:22