# How would you complete and balance the acid base neutralizations reaction: HBr + Ba(OH)_2 ->?

Mar 27, 2017

$2 \text{HBr" + "Ba(OH)"_2 -> "BaBr"_2 + 2 "H"_2"O}$

#### Explanation:

In a neutralization reaction, if the base is a hydroxide, then the products formed would be the salt and water.

acid + base $\to$ salt + water

The water ($\text{H"_2"O}$) is a result of the combination of the ${\text{H}}^{+}$ from the acid and the ${\text{OH}}^{-}$ from the base.

The salt (${\text{BaBr}}_{2}$) is simply the remaining anion of the acid (${\text{Br}}^{-}$) and the cation of the base (${\text{Ba}}^{2 +}$).

The unbalanced equation is therefore

$\text{HBr" + "Ba(OH)"_2 -> "BaBr"_2 + "H"_2"O}$

To balance the equation, first notice that the number of Ba on both sides are the same. Ba is balanced.

Next, notice that the LHS is short of 1 x Br. So balance it by adding 1 unit of HBr.

$\text{2HBr" + "Ba(OH)"_2 -> "BaBr"_2 + "H"_2"O}$

Now, the RHS is missing 2 x H and 1 x O. Balance that by adding 1 units of $\text{H"_2"O}$.

$\text{2HBr" + "Ba(OH)"_2 -> "BaBr"_2 + 2"H"_2"O}$

Now, all the elements have equal amounts on both sides of the equation. The chemical equation is now considered balanced.