Question

for the chemical reaction
mathrm{hclo}_{4}(mathrm{aq})+mathrm{naoh}(mathrm{aq})
ightarrow mathrm{h}_{2} mathrm{o}(mathrm{l})+mathrm{naclo}_{4}(mathrm{aq})
write the net ionic equation, including the phases.

net ionic equation:

which ions are considered spectator ions for this reaction?
(square mathrm{clo}_{4}^{-})
(square mathrm{na}^{+})
(square mathrm{h}^{+})
(square mathrm{oh}^{-})

Explanation:

Step1: Write the complete ionic equation

$\ce{HClO4(aq)}$ dissociates into $\ce{H+(aq) + ClO4-(aq)}$, $\ce{NaOH(aq)}$ dissociates into $\ce{Na+(aq) + OH-(aq)}$, and $\ce{NaClO4(aq)}$ dissociates into $\ce{Na+(aq) + ClO4-(aq)}$.
So the complete ionic equation is:
$\ce{H+(aq) + ClO4-(aq) + Na+(aq) + OH-(aq) \to H2O(l) + Na+(aq) + ClO4-(aq)}$

Step2: Cancel out spectator ions

Spectator ions are ions that do not participate in the reaction. In this case, $\ce{Na+(aq)}$ and $\ce{ClO4-(aq)}$ are on both sides of the equation.
After canceling them, the net ionic equation is:
$\ce{H+(aq) + OH-(aq) \to H2O(l)}$

Answer:

net ionic equation: $\ce{H+(aq) + OH-(aq) \to H2O(l)}$
Spectator ions: $\ce{ClO4-}$, $\ce{Na+}$