What is the net ionic equation of the reaction of ##FeCl_2## with ##NaOH##?

##"Fe"^(2+) (aq) + "2OH"^(-)(aq)## → ##"Fe(OH)"_2(s) ##

Step 1. Write the word equation for the reaction.

##"FeCl"_2## reacts with ##"NaOH"## to produce ##"Fe(OH)"_2## and ##"NaCl"##

Step 2. Write the molecular equation for the reaction.

You use the Solubility Rules to determine if any of the products is a solid.

1. All of Group 1 metals are soluble.
2. All hydroxides are insoluble except those of Group 1 metals and ##"Ba"^(2+)##, ##"Ca"^(2+)##, and ##"Sr"^(2+)##.

This tells you that ##"Fe(OH)"_2## is a solid and ##"NaCl"## is soluble. The molecular equation is then

##"FeCl"_2"(aq)"+ "2NaOH(aq)" → "Fe(OH)"_2"(s)" + "2 NaCl(aq)"##

Step 3. Write the ionic equation.

Iron(II) chloride when dissolved produces ##"Fe"^(2+)## ions and ##"Cl"^-## ions.

Sodium hydroxide produces ##"Na"^+## ions and ##"OH"^-## ions.

You do not write solids as ions.

The ionic equation is

##"Fe"^(2+)(aq) + "2Cl"^(-)(aq) +"2Na"^(+)(aq) + "2OH"^(-)(aq) → "Fe(OH)"_2(s) + "2Na"^(+)(aq) + "2Cl"^(-)(aq)##

Step 4. Cancel items that appear on each side of the equation (##"2Cl"^-## and ##"2Na"^+##) to get the net ionic equation.

##"Fe"^(2+)(aq) + color(red)(cancel(color(black)("2Cl"^(-)(aq))) + color(red)(cancel(color(black)("2Na"^(+)(aq))))) + "2OH"^(-)(aq) → "Fe(OH)"_2(s) + color(red)(cancel(color(black)("2Na"^(+)(aq)))) + color(red)(cancel(color(black)("2Cl"^(-)(aq)))##

The net ionic equation is

##"Fe"^(2+)(aq) + "2OH"^(-)(aq) → "Fe(OH)"_2(s) ##

And there you have it.

Here's another example, this one using ##"FeCl"_3 + "NaOH"##:

video from: Noel Pauller