# Question #d2974

Apr 19, 2017

$N {H}_{4} O H + C {H}_{3} C O O H \to C {H}_{3} C O O N {H}_{4} + {H}_{2} O$

#### Explanation:

The basic chemical equation would be:
$N {H}_{4} O H + C {H}_{3} C O O H \to C {H}_{3} C O O N {H}_{4} + {H}_{2} O$

The net ionic equation would be:
$O {H}^{-} + {H}^{+} \to {H}_{2} O$

But personally, when doing calculations, I find this equation to be the most helpful:
$N {H}_{4} O H \left(a q\right) + C {H}_{3} C O O H \left(a q\right) r i g h t \le f t h a r p \infty n s C {H}_{3} C O {O}^{-} \left(a q\right) + N {H}_{4}^{+} \left(a q\right) + {H}_{2} O \left(l\right)$

Because acetic acid is a weak acid, it will not fully dissociate. That means that the solution will reach an equilibrium between the reactants and products.