Question

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!

Answer 1

`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, `"NaOH"`, and copper(II) nitrate, `"Cu"("NO"_3)_2`, will dissociate completely in aqueous solution to form cations and anions

`"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)`

`"Cu"("NO"_3)_text(2(aq]) -> "Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)`

The reaction will produce copper(II) hydroxide, `"Cu"("OH")_2`, an insoluble ionic compound that precipitates out of solution, and aqueous sodium nitrate, `"NaNO"_3`, another soluble ionic compound.

The balanced chemical equation for this reaction would look like this

`2"NaOH"_text((aq]) + "Cu"("NO"_3)_text(2(aq]) -> "Cu"("OH")_text(2(s]) darr + 2"NaNO"_text(3(aq])`

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])` `darr` `+ 2 xx underbrace(("Na"_text((aq])^(+) + "NO"_text(3(aq])^(-)))_color(blue)("NaNO"_3)`

This is equivalent to

`2"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"_text(3(aq])^(-)`

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

`color(red)(cancel(color(black)(2"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"_text(3(aq])^(-))))`

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.