# Question #c648c

Nov 7, 2017

$\text{CH"_3"COOH"(aq) + "OH"^(-)(aq) -> "CH"_3"COO"^(-)(aq) + "H"_2"O} \left(l\right)$

#### Explanation:

First, check the standard equation:

$\text{CH"_3"COOH" + "NaOH" -> "H"_2"O" + "CH"_3"COONa}$

The ionic equation for the above:

$\text{CH"_3"COOH"(aq) + "Na"^(+)(aq) + "OH"^(-)(aq) -> "CH"_3"COO"^(-)(aq) + "Na"^(+)(aq) + "H"_2"O} \left(l\right)$

Therefore, the net ionic equation is (remove the ${\text{Na}}^{+}$ as it is present as both a reactant and a product):

$\text{CH"_3"COOH"(aq) + "OH"^(-)(aq) -> "CH"_3"COO"^(-)(aq) + "H"_2"O} \left(l\right)$

(note the charge on both sides has to be the same, in this case, $- 1$).