Question 20 of 25

What is the net ionic equation for this reaction?

[tex]\[
2 \text{KOH (aq)} + \text{H}_2 \text{SO}_4 \text{(aq)} \rightarrow 2 \text{H}_2 \text{O (l)} + \text{K}_2 \text{SO}_4 \text{(aq)}
\][/tex]

A. [tex]\[\text{OH}^{-} + \text{H}^{+} \rightarrow \text{H}_2 \text{O (l)}\][/tex]

B. [tex]\[2 \text{K}^{+} + \text{OH}^{-} + \text{H}^{+} + \text{SO}_4^{2-} \rightarrow 2 \text{H}_2 \text{O (l)} + 2 \text{K}^{+} + \text{SO}_4^{2-}\][/tex]

C. [tex]\[2 \text{KOH} + \text{H}^{+} + \text{SO}_4^{2-} \rightarrow 2 \text{H}_2 \text{O (l)} + 2 \text{K}^{+} + \text{SO}_4^{2-}\][/tex]

D. [tex]\[2 \text{KOH} + \text{H}_2 \text{SO}_4 \rightarrow 2 \text{H}_2 \text{O} + \text{K}_2 \text{SO}_4\][/tex]



Answer :

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. H₂SO₄ (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. K₂SO₄ (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]