# What is the net ionic equation for the reaction "NaOH" + "Cu"("NO"_3)_2 -> "Cu"("OH")_2 + "NaNO"_3?

## I keep rereading what's in my textbook over and over again, but I'm just lost! Could someone please help and explain! Many thanks!

Mar 22, 2016

2"OH"_text((aq])^(-) + "Cu"_text((aq])^(2+) -> "Cu"("OH")_text(2(s]) darr

#### Explanation:

You'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 your case, sodium hydroxide, $\text{NaOH}$, and copper(II) nitrate, "Cu"("NO"_3)_2, will dissociate completely in aqueous solution to form cations and anions

${\text{NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

${\text{Cu"("NO"_3)_text(2(aq]) -> "Cu"_text((aq])^(2+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

The reaction will produce copper(II) hydroxide, "Cu"("OH")_2, an insoluble ionic compound that precipitates out of solution, and aqueous sodium nitrate, ${\text{NaNO}}_{3}$, another soluble ionic compound.

The balanced chemical equation for this reaction would look like this

$2 {\text{NaOH"_text((aq]) + "Cu"("NO"_3)_text(2(aq]) -> "Cu"("OH")_text(2(s]) darr + 2"NaNO}}_{\textrm{3 \left(a q\right]}}$

Now, notice that you need $2$ moles of sodium hydroxide for every $1$ mole of copper(II) nitrate that takes part in the reaction.

To get the complete ionic equation, rewrite the soluble ionic compounds as cations and anions

2 xx overbrace(("Na"_text((aq])^(+) + "OH"_text((aq])^(-)))^(color(purple)("NaOH")) + overbrace(("Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)))^(color(red)("Cu"("NO"_3)_2)) -> "Cu"("OH")_text(2(s]) $\downarrow$ + 2 xx underbrace(("Na"_text((aq])^(+) + "NO"_text(3(aq])^(-)))_color(blue)("NaNO"_3)

This is equivalent to

$2 {\text{Na"_text((aq])^(+) + 2"OH"_text((aq])^(-) + "Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) -> "Cu"("OH")_text(2(s]) darr + 2"Na"_text((aq])^(+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

Now, in order to get the net ionic equation, you must eliminate spectator ions, i.e. ions that are present on both sides of the equation

In this case, you would have

$\textcolor{red}{\cancel{\textcolor{b l a c k}{2 {\text{Na"_text((aq])^(+)))) + 2"OH"_text((aq])^(-) + "Cu"_text((aq])^(2+) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-)))) -> "Cu"("OH")_text(2(s]) darr + color(red)(cancel(color(black)(2"Na"_text((aq])^(+)))) + color(red)(cancel(color(black)(2"NO}}_{\textrm{3 \left(a q\right]}}^{-}}}}$

which is equivalent to

2"OH"_text((aq])^(-) + "Cu"_text((aq])^(2+) -> "Cu"("OH")_text(2(s]) darr

Copper(II) hydroxide is a blue insoluble solid that precipitates out of solution. 