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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.

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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:

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

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

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