Reaction of copper (II) ion reacting with sodium hydroxide

Based on an experiment, a solution that contained $\ce{Cu^2+}$ formed a precipitate with $\ce{NaOH}$. Is this how you write the equation? 


$$\ce{2NaOH + Cu^{2+} -> Cu(OH)2 + 2Na+}$$


On the experiment: solution A is mixed with NaOH (solution turned white). Using a flame test solution A is color green. I figured solution A has $\ce{Cu^{2+}}$ ion.




Yes, the reaction is correct however it may be preferable to either include *all* or *no* spectator ions.
Use either a net ionic equations (omit the Na+), molecular equation (include the copper compound full) or complete ionic equation.


For example, if the compound is copper sulphate the equations would be (without states)


**Net ionic**
$$\ce{Cu^2+}+\ce{2OH^-->Cu(OH)\_2}$$
**Molecular equation**
$$\ce{CuSO\_4}+\ce{2NaOH->Cu(OH)\_2}+\ce{Na\_2SO\_4}$$
**Complete Ionic Equation**
$$\ce{Cu^2+ +SO\_4^{2-}}+\ce{2Na^+ +2OH^-->Cu(OH)\_2}+\ce{2Na^+ +SO\_4^2-}$$


**Edit**


In the case where you don't know what copper compound it is, use a net ionic equation. As a side note, copper hydroxide is not white. You might want to check your calculations.


If you're wondering what it is, check what other compounds will emit green.


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