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Answer 1 · 2017-11-07T19:18:41

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

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, $-1$ ).