To determine the net ionic equation for the given reaction:
[tex]\[
2 \text{KOH} \, (\text{aq}) + \text{H}_2\text{SO}_4 \, (\text{aq}) \rightarrow 2 \text{H}_2\text{O} \, (\ell) + \text{K}_2\text{SO}_4 (\text{aq}),
\][/tex]
we first identify the dissociation of each compound in the aqueous phase:
1. KOH (aq): This dissociates completely into its ions:
[tex]\[
2 \text{KOH} \rightarrow 2 \text{K}^+ + 2 \text{OH}^-
\][/tex]
2. Hâ‚‚SOâ‚„ (aq): This dissociates completely into its ions:
[tex]\[
\text{H}_2\text{SO}_4 \rightarrow 2 \text{H}^+ + \text{SO}_4^{2-}
\][/tex]
3. The products include water in the liquid state, which does not dissociate:
[tex]\[
2 \text{H}_2\text{O} (\ell)
\][/tex]
4. Kâ‚‚SOâ‚„ (aq): This dissociates completely into its ions:
[tex]\[
\text{K}_2\text{SO}_4 \rightarrow 2 \text{K}^+ + \text{SO}_4^{2-}
\][/tex]
Next, we write all the ions present before the reaction takes place:
[tex]\[
2 \text{K}^+ + 2 \text{OH}^- + 2 \text{H}^+ + \text{SO}_4^{2-}
\][/tex]
After the reaction, the ions present are:
[tex]\[
2 \text{K}^+ + \text{SO}_4^{2-} + 2 \text{H}_2\text{O} (\ell)
\][/tex]
We can now eliminate the spectator ions, which are ions that do not change during the reaction. In this case, the spectator ions are \( \text{K}^+ \) and \( \text{SO}_4^{2-} \), as they appear unchanged on both sides of the reaction.
After removing the spectator ions, we are left with:
[tex]\[
2 \text{OH}^- + 2 \text{H}^+ \rightarrow 2 \text{H}_2\text{O} (\ell)
\][/tex]
We simplify the equation by dividing through by 2:
[tex]\[
\text{OH}^- + 2 \text{H}^+ \rightarrow 2 \text{H}_2\text{O} (\ell)
\][/tex]
Thus, the net ionic equation is:
[tex]\[
\text{OH}^- + 2 \text{H}^+ \rightarrow 2 \text{H}_2\text{O} (\ell)
\][/tex]
This matches option A. Therefore, the correct answer is:
A. [tex]\( \text{OH}^- + 2 \text{H}^+ \rightarrow 2 \text{H}_2\text{O} (\ell) \)[/tex]
