Question #c648c

1 Answer

Answer:

#"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#).