To determine the net ionic equation that depicts the neutralization of KOH by KHP, let’s go through the steps carefully:
1. Write the balanced molecular equation:
[tex]\[
\text{KHP (aq)} + \text{KOH (aq)} \rightarrow \text{K}_2\text{P (aq)} + \text{H}_2\text{O (l)}
\][/tex]
2. Identify and dissociate all the strong electrolytes (soluble ionic compounds and strong acids/bases):
- KHP dissociates in water to produce [tex]\( \text{K}^+ \)[/tex] and [tex]\( \text{HP}^- \)[/tex].
- KOH dissociates in water to produce [tex]\( \text{K}^+ \)[/tex] and [tex]\( \text{OH}^- \)[/tex].
- [tex]\( \text{K}_2\text{P} \)[/tex] dissociates in water to produce [tex]\( 2\text{K}^+ \)[/tex] and [tex]\( \text{P}^{2-} \)[/tex].
So, the dissociated form of the equation becomes:
[tex]\[
\text{K}^+ (aq) + \text{HP}^- (aq) + \text{K}^+ (aq) + \text{OH}^- (aq) \rightarrow 2\text{K}^+ (aq) + \text{P}^{2-} (aq) + \text{H}_2\text{O (l)}
\][/tex]
3. Cancel out the spectator ions:
Spectator ions are ions that appear on both the reactant and product side of the equation without undergoing any change. Here, the K[tex]\(^+\)[/tex] ions are spectator ions.
After canceling the spectator ions [tex]\( \text{K}^+ \)[/tex], the net ionic equation is:
[tex]\[
\text{HP}^- (aq) + \text{OH}^- (aq) \rightarrow \text{P}^{2-} (aq) + \text{H}_2\text{O (l)}
\][/tex]
Thus, the net ionic equation depicting the neutralization of the base KOH by KHP is:
[tex]\[
\text{HP}^- (aq) + \text{OH}^- (aq) \rightarrow \text{P}^{2-} (aq) + \text{H}_2\text{O (l)}
\][/tex]
