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What is the net ionic equation of sodium hydroxide when it dissolves in water?

For the net ionic equation I got $$\ce{NaOH(s) + H2O(l) -> NaH+(aq) + OH- (aq) + H2O(l)}$$ but it was wrong. Then I tried: $$\ce{NaOH(aq) -> Na+(aq) + OH- (aq)}$$ and it was wrong as well. Can someone please explain?

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    $\begingroup$ Your second answer is close but think about the state of sodium hydroxide before dissolving in water. $\endgroup$ – Ian Fang Apr 16 '15 at 2:19
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    $\begingroup$ Got it! NaOH is a solid! $\endgroup$ – jeterpaul Apr 16 '15 at 2:24
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When you write down the net ionic equation, you have to consider that the educt and the products are in different phases, a solid and a liquid phase. Undissolved $\ce{NaOH}$ is a solid, denoted by the index $\ce{(s)}$, while the dissolved $\ce{Na+}$ and $\ce{OH-}$ ions are solvated in the aqueous phase (index $\ce{(aq)}$ in the equation):

$$\ce{NaOH(s) ~->~ Na+(aq) + OH^{-}(aq)}$$

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You don't need the H ion with the Na ion. You were correct the second time. and if there is two of the same product or reactant of on either side of the equation it is eliminated because it is just a spectator.

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    $\begingroup$ "NaOH (aq)" is not correct. See Jannis Andreska's answer. Identical products and reactants does not make them a spectator either, acid catalysis is a thing even in secondary school chemistry. $\endgroup$ – Nij Jan 11 '17 at 0:56

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