# Which are the spectator ions when a solution of copper(II) nitrate reacts with a solution of potassium hydroxide?

May 25, 2017

${\text{K"^(+) and "NO}}_{3}^{-}$

#### Explanation:

Copper(II) nitrate and potassium hydroxide are soluble ionic compounds, which implies that they dissociate completely when dissolved in water to produce ions.

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

${\text{KOH"_ ((aq)) -> "K"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

When you mix these two solutions, the copper(II) cations and the hydroxide anions will combine to form copper(II) hydroxide, an insoluble solid that precipitates out of solution.

The other product will be aqueous potassium nitrate, which, as its name suggests, will exist as ions in aqueous solution.

You can thus say that the balanced chemical equation that describes this double replacement reaction looks like this

${\text{Cu"("NO"_ 3)_ (2(aq)) + color(red)(2)"KOH"_ ((aq)) -> "Cu"("OH")_ (2(s)) darr + color(red)(2)"KNO}}_{3 \left(a q\right)}$

The complete ionic equation looks like this

${\text{Cu"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + color(red)(2)"K"_ ((aq))^(+) + color(red)(2)"OH"_ ((aq))^(-) -> "Cu"("OH")_ (2(s)) darr + color(red)(2)"K"_ ((aq))^(+) + color(red)(2)"NO}}_{3 \left(a q\right)}^{-}$

To get the net ionic equation, eliminate the spectator ions, i.e. the ions that are present on both sides of the chemical equation.

You will have

"Cu"_ ((aq))^(2+) + color(blue)(cancel(color(black)(2"NO"_ (3(aq))^(-)))) + color(blue)(cancel(color(black)(color(red)(2)"K"_ ((aq))^(+)))) + color(red)(2)"OH"_ ((aq))^(-) -> "Cu"("OH")_ (2(s)) darr + color(blue)(cancel(color(black)(color(red)(2)"K"_ ((aq))^(+)))) + color(blue)(cancel(color(black)(color(red)(2)"NO"_ (3(aq))^(-))))

This is equivalent to

"Cu"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-) -> "Cu"("OH")_ (2(s)) darr