# What is the net ionic equation of the reaction of FeCl2 with NaOH? Express you answer as a chemical equation including phases.

Mar 4, 2018

$F {e}^{2 +} \left(a q\right) + 2 O {H}^{-} \left(a q\right) \to F e {\left(O H\right)}_{2} \left(s\right) \downarrow$

#### Explanation:

We first write down the balanced equation for this reaction:

$F e C {l}_{2} \left(a q\right) + 2 N a O H \left(a q\right) \to 2 N a C l \left(a q\right) + F e {\left(O H\right)}_{2} \left(s\right) \downarrow$

This is a precipitation reaction, where two soluble salts are mixed together in solution forms to make a precipitate.

So, we can now write the ionic equation for this reaction, that is:

$F {e}^{2 +} \left(a q\right) + 2 C {l}^{-} \left(a q\right) + 2 N {a}^{+} \left(a q\right) + 2 O {H}^{-} \left(a q\right) \to 2 N {a}^{+} \left(a q\right) + 2 C {l}^{-} \left(a q\right) + F e {\left(O H\right)}_{2} \left(s\right) \downarrow$

Now, we need to cancel the spectator ions, which are ions that are present on both sides in the equation. So, the equation becomes

$F {e}^{2 +} \left(a q\right) + \textcolor{red}{\cancel{\textcolor{b l a c k}{2 C {l}^{-} \left(a q\right)}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{2 N {a}^{+} \left(a q\right)}}} + 2 O {H}^{-} \left(a q\right) \to \textcolor{red}{\cancel{\textcolor{b l a c k}{2 N {a}^{+} \left(a q\right)}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{2 C {l}^{-} \left(a q\right)}}} + F e {\left(O H\right)}_{2} \left(s\right) \downarrow$

The final net ionic equation is now:

$F {e}^{2 +} \left(a q\right) + 2 O {H}^{-} \left(a q\right) \to F e {\left(O H\right)}_{2} \left(s\right) \downarrow$

The iron(II) hydroxide formed is easily oxidized when exposed to air and can easily become iron(III) hydroxide $\left(F e {\left(O H\right)}_{3}\right)$.