To solve this problem, we need to write the complete ionic equation and then eliminate the spectator ions to get the net ionic equation. Here are the steps:
1. Write the Given Molecular Equation:
[tex]\[
HNO_2(aq) + Li_3PO_4(aq) \rightarrow Li_2HPO_4(aq) + LiNO_2(aq)
\][/tex]
2. Write the Complete Ionic Equation:
- Nitrous acid ([tex]\(HNO_2\)[/tex]) is a weak acid, so it remains as [tex]\(HNO_2(aq)\)[/tex].
- Lithium phosphate ([tex]\(Li_3PO_4\)[/tex]) dissociates completely in water into [tex]\(3Li^{+}(aq)\)[/tex] and [tex]\(PO_4^{3-}(aq)\)[/tex].
- Lithium hydrogen phosphate ([tex]\(Li_2HPO_4\)[/tex]) dissociates partially because [tex]\(HPO_4^{2-}\)[/tex] is a weak base so it remains as [tex]\(Li_2HPO_4(aq)\)[/tex] but contributes to [tex]\(2Li^{+}(aq)\)[/tex] and [tex]\(HPO_4^{2-}(aq)\)[/tex].
- Lithium nitrite ([tex]\(LiNO_2\)[/tex]) dissociates completely into [tex]\(Li^{+}(aq)\)[/tex] and [tex]\(NO_2^{-}(aq)\)[/tex].
The complete ionic equation becomes:
[tex]\[
HNO_2(aq) + 3Li^{+}(aq) + PO_4^{3-}(aq) \rightarrow 2Li^{+}(aq) + HPO_4^{2-}(aq) + Li^{+}(aq) + NO_2^{-}(aq)
\][/tex]
3. Identify and Cancel the Spectator Ions:
The spectator ions in this equation are [tex]\(Li^+(aq)\)[/tex], which appear on both the reactant and product sides.
4. Write the Net Ionic Equation:
After canceling the spectator ions, the net ionic equation is:
[tex]\[
HNO_2(aq) + PO_4^{3-}(aq) \rightarrow HPO_4^{2-}(aq) + NO_2^{-}(aq)
\][/tex]
Thus, the net ionic equation representing the reaction is:
[tex]\[
HNO_2(aq) + PO_4^{3-}(aq) \rightarrow HPO_4^{2-}(aq) + NO_2^{-}(aq)
\][/tex]
Now, apply this to the provided groups and drag the appropriate chemical formulas to their respective columns:
- Group 1 (Reactants):
[tex]\[
HNO_2(aq), PO_4^{3-}(aq)
\][/tex]
- Group 2 (Products):
[tex]\[
HPO_4^{2-}(aq), NO_2^{-}(aq)
\][/tex]
