# How do you balance NaOH + Cu(NO_3)_2 -> Cu(OH)_2 + NaNO_3?

Mar 12, 2016

You have almost done it:

$2 N a O H \left(a q\right) + C u {\left(N {O}_{3}\right)}_{2} \left(a q\right) \rightarrow C u {\left(O H\right)}_{2} \downarrow + 2 N a N {O}_{3} \left(a q\right)$

#### Explanation:

The equation is stoiciometrically balanced. It is balanced with respect to MASS and to charge. We could write the net ionic equation as :

$C {u}^{2 +} + 2 O {H}^{-} \rightarrow C u {\left(O H\right)}_{2} \left(s\right) \downarrow$

This represents the macroscopic, observable chemical change. All hydroxides are soluble (save for those of the alkali metals). Thus, the sodium counterion is simply along for the ride in the reaction.