Question

What is the net ionic equation for the reaction $ NaOH+Cu{{(N{{O}_{3}})}_{2}}\to Cu{{(OH)}_{2}}+NaN{{O}_{3}} $ ?

Answer 1

Hint: We know that the net ionic equation is a chemical equation for a reaction that lists only those species participating in the reaction. In other words, the net ionic equation applies to reactions that are strong electrolytes in water. During a chemical reaction net charge (either positive or negative) remains the same before & after the reaction. Check the number of atoms and oxidation states on both sides.

Complete answer:
During a chemical reaction, the sum of charge on all cations or anions on reactant is equal to the sum of charge of all cations or anions on the product side. That is, net positive or negative charge remains the same before and after the reaction in a balanced chemical equation.
As copper nitrate solution is added to sodium hydroxide solution, forming products that are copper hydroxide and sodium nitrate. As copper solutions have cupric cation and nitrate anion. On the other hand, sodium hydroxide solution has sodium cation and hydroxide anion. During mixing, cation of one solution mixes with anion of another solution and vice versa. So in our case, cupric ion from copper nitrate solution reacts with hydroxide ion of sodium hydroxide solution and results into formation of copper hydroxide in order to make compound, two hydroxide ion net $ charge\text{ }-\text{ }2 $ combine with one cupric ion to keep neutrality of compound. Similarly sodium ion from sodium hydroxide will react with nitrate ion of copper nitrate and results in formation of sodium nitrate as two sodium ion will combine with one nitrate ion in order to keep neutrality of species
Here we’re dealing with a double replacement reaction in which two soluble ionic compounds react to form an insoluble solid that precipitates out of the aqueous solution. In this case, sodium hydroxide, $ NaOH, $ and copper $ \left( II \right) $ nitrate, $ Cu{{\left( N{{O}_{3}} \right)}_{2}}, $ will dissociate completely in aqueous solution to form cations and anions.
 $ NaOH_{(aq)}^{{}}\to Na_{(aq)}^{+}+OH_{(aq)}^{-} $ and $ Cu{{(N{{O}_{3}})}_{2}}_{(aq)}^{{}}\to Cu_{(aq)}^{2+}+2NO_{3(aq)}^{-} $
The reaction will produce copper $ \left( II \right) $ hydroxide, $ Cu{{\left( OH \right)}_{2}}, $ an insoluble ionic compound that precipitates out of solution, and aqueous sodium nitrate, $ NaN{{O}_{3}}, $ another soluble ionic compound, which is equivalent to
 $ 2OH_{(aq)}^{-}+Cu_{(aq)}^{2+}\to Cu{{(OH)}_{2(s)}}\downarrow $ Copper(II) hydroxide is a blue insoluble solid that precipitates out of solution.

Note:
Remember that the reaction which involves exchange of ions does not involve change in oxidation state of ions but oxidation state of ion changes in redox reactions. Reduction causes decrease in oxidation number and oxidation causes increase in oxidation number. But still net charge remains the same even in redox reactions.