# Question #e8338

Jul 4, 2015

You're indeed dealing with a neutralization reaction.

#### Explanation:

Barium hydroxide, $B a {\left(O H\right)}_{2}$, is a strong base, so it will react with hydrochloric acid, $H C l$, a strong acid, to produce water and barium chloride, a soluble salt.

Keep in mind that barium hydroxide is not very soluble in aqueous solution, but what does dissolve will dissociate completely to form barium cations, $B {a}^{2 +}$, and hydroxide anions, $O {H}^{-}$.

$B a {\left(O H\right)}_{2 \left(a q\right)} \to B {a}_{\textrm{\left(a q\right]}}^{2 +} + 2 O {H}_{\left(a q\right)}^{-}$

Hydrochloric acid will dissociate completely in aqueous solution to give ${H}^{+}$ cations and $C {l}^{-}$ anions.

$H C {l}_{\left(a q\right)} \to {H}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

The chemical reaction between these two compounds will be

$B a {\left(O H\right)}_{2 \left(a q\right)} + 2 H C {l}_{\textrm{\left(a q\right]}} \to B a C {l}_{2 \left(a q\right)} + 2 {H}_{2} {O}_{\left(l\right)}$

The complete ionic equation will be

$B {a}_{\left(a q\right)}^{2 +} + 2 O {H}_{\left(a q\right)}^{-} + 2 {H}_{\left(a q\right)}^{+} + 2 C {l}_{\left(a q\right)}^{-} \to B {a}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-} + 2 {H}_{2} {O}_{\left(l\right)}$

The net ionic equation, which you get if you remove spectator ions, will be

$\cancel{B {a}_{\left(a q\right)}^{2 +}} + 2 O {H}_{\left(a q\right)}^{-} + 2 {H}_{\left(a q\right)}^{+} + \cancel{2 C {l}_{\left(a q\right)}^{-}} \to \cancel{B {a}_{\left(a q\right)}^{2 +}} + \cancel{2 C {l}_{\left(a q\right)}^{-}} + 2 {H}_{2} {O}_{\left(l\right)}$

$2 O {H}_{\left(a q\right)}^{-} + 2 {H}_{\left(a q\right)}^{+} \to 2 {H}_{2} {O}_{\left(l\right)}$

which is equivalent to

$O {H}_{\left(a q\right)}^{-} + {H}_{\left(a q\right)}^{+} \to {H}_{2} {O}_{\left(l\right)}$