Answer :
To determine the net ionic equation for the given chemical 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]
we follow these steps:
1. Write the balanced molecular equation:
The given equation is already balanced:
[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]
2. Write the complete ionic equation:
Strong electrolytes (such as KOH and H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) dissociate completely in aqueous solution. Therefore:
[tex]\[ 2 \text{K}^+(\text{aq}) + 2 \text{OH}^-(\text{aq}) + 2 \text{H}^+(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow 2 \text{H}_2\text{O(l)} + 2 \text{K}^+(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \][/tex]
3. Identify and cancel the spectator ions:
The ions that appear on both sides of the equation and do not change during the reaction are called spectator ions. In this case, the spectator ions are [tex]\( \text{K}^+ \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex]. Canceling them out, we get:
[tex]\[ 2 \text{OH}^-(\text{aq}) + 2 \text{H}^+(\text{aq}) \rightarrow 2 \text{H}_2\text{O(l)} \][/tex]
4. Simplify the net ionic equation if necessary:
Since all coefficients are even, we can simplify by dividing through by 2:
[tex]\[ \text{OH}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{H}_2\text{O(l)} \][/tex]
Thus, the correct net ionic equation and the correct choice from the given options are:
D. [tex]\(\text{OH}^- + 2 \text{H}^+ \rightarrow 2 \text{H}_2\text{O(l)}\)[/tex]
[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]
we follow these steps:
1. Write the balanced molecular equation:
The given equation is already balanced:
[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]
2. Write the complete ionic equation:
Strong electrolytes (such as KOH and H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) dissociate completely in aqueous solution. Therefore:
[tex]\[ 2 \text{K}^+(\text{aq}) + 2 \text{OH}^-(\text{aq}) + 2 \text{H}^+(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow 2 \text{H}_2\text{O(l)} + 2 \text{K}^+(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \][/tex]
3. Identify and cancel the spectator ions:
The ions that appear on both sides of the equation and do not change during the reaction are called spectator ions. In this case, the spectator ions are [tex]\( \text{K}^+ \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex]. Canceling them out, we get:
[tex]\[ 2 \text{OH}^-(\text{aq}) + 2 \text{H}^+(\text{aq}) \rightarrow 2 \text{H}_2\text{O(l)} \][/tex]
4. Simplify the net ionic equation if necessary:
Since all coefficients are even, we can simplify by dividing through by 2:
[tex]\[ \text{OH}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{H}_2\text{O(l)} \][/tex]
Thus, the correct net ionic equation and the correct choice from the given options are:
D. [tex]\(\text{OH}^- + 2 \text{H}^+ \rightarrow 2 \text{H}_2\text{O(l)}\)[/tex]
