Question

Question #c648c

Answer 1

`"CH"_3"COOH"(aq) + "OH"^(-)(aq) -> "CH"_3"COO"^(-)(aq) + "H"_2"O"(l)`

Explanation:

First, check the standard equation:

`"CH"_3"COOH" + "NaOH" -> "H"_2"O" + "CH"_3"COONa"`

The ionic equation for the above:

`"CH"_3"COOH"(aq) + "Na"^(+)(aq) + "OH"^(-)(aq) -> "CH"_3"COO"^(-)(aq) + "Na"^(+)(aq) + "H"_2"O"(l)`

Therefore, the net ionic equation is (remove the `"Na"^(+)` as it is present as both a reactant and a product):

`"CH"_3"COOH"(aq) + "OH"^(-)(aq) -> "CH"_3"COO"^(-)(aq) + "H"_2"O"(l)`

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