# Question 75f48

May 19, 2017

A double replacement reaction.

#### Explanation:

Calcium chloride, ${\text{CaCl}}_{2}$, and sodium hydroxide, $\text{NaOH}$, are soluble in water, which implies that they exist as ions in aqueous solution.

${\text{CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + 2"Cl}}_{\left(a q\right)}^{-}$

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

When you mix aqueous solutions of calcium chloride and sodium hydroxide, the calcium cations will combine with the hydroxide anions to form calcium hydroxide, "Ca"("OH")_2, an insoluble solid that precipitates out of solution. The other product is aqueous sodium chloride, itself a soluble ionic compound that exists as ions in solution.

So the two cations are exchanging partners, which is why this reaction is a double replacement reaction.

The calcium cations start combined with the chloride anions in solid calcium chloride and end up combined with the hydroxide anions.

Similarly, the sodium cations start combined with the hydroxide anions in solid sodium hydroxide and end up combined with the chloride anions--keep in mind that if we were to evaporate all the water, we would get solid sodium chloride as the second product of the reaction.

So, put all this together to get

${\text{CaCl"_ (2(aq)) + 2"NaOH"_ ((aq)) -> "Ca"("OH")_ (2(s)) darr + 2"NaCl}}_{\left(a q\right)}$

The complete ionic equation looks like this

${\text{Ca"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + 2"Na"_ ((aq))^(+) + 2"OH"_ ((aq))^(-) -> "Ca"("OH")_ (2(s)) darr + 2"Na"_ ((aq))^(+) + 2"Cl}}_{\left(a q\right)}^{-}$

The net ionic equation, which you get by eliminating the spectator ions

"Ca"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + 2"OH"_ ((aq))^(-) -> "Ca"("OH")_ (2(s)) darr + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-))))

looks like this

"Ca"_ ((aq))^(2+) + 2"OH"_ ((aq))^(-) -> "Ca"("OH") _(2(s)) darr#