Net Ionic Equations - AP Chemistry

1. Aqueous sodium hydroxide is mixed with aqueous hydroiodic acid. The balanced molecular equation is: $\mathrm{NaOH(aq) + HI(aq) \rightarrow NaI(aq) + H_2O(l)}$ Which is the correct net ionic equation?
   Answer: $\mathrm{H^{+}(aq) + OH^{-}(aq) \rightarrow H_2O(l)}$ This question tests the skill of writing net ionic equations for acid-base neutralizations. The molecular equation shows sodium hydroxide and hydroiodic acid forming water and aqueous sodium iodide. Ions are Na⁺, OH⁻, H⁺, and I⁻, with H⁺ and OH⁻ combining to form H₂O. Spectators Na⁺ and I⁻ are removed, yielding H⁺(aq) + OH⁻(aq) → H₂O(l). A tempting distractor is B, which shows Na⁺ + I⁻ → NaI, incorrectly highlighting salt formation due to ignoring the neutralization. A transferable strategy is to simplify strong acid-base reactions to the core water-forming equation.
2. Aqueous sodium bromide is added to aqueous mercury(I) nitrate, forming a precipitate. The balanced molecular equation is: $\mathrm{2\,NaBr(aq) + Hg_2(NO_3)_2(aq) \rightarrow Hg_2Br_2(s) + 2\,NaNO_3(aq)}$ Which is the correct net ionic equation?
   Answer: $\mathrm{Hg_2^{2+}(aq) + 2\,Br^{-}(aq) \rightarrow Hg_2Br_2(s)}$ This question tests the skill of writing net ionic equations for precipitation with polyatomic ions. The molecular equation involves sodium bromide and mercury(I) nitrate forming solid mercury(I) bromide and aqueous sodium nitrate. Ions are 2Na⁺, 2Br⁻, Hg₂²⁺, and 2NO₃⁻, with Hg₂²⁺ and 2Br⁻ forming Hg₂Br₂. Spectators Na⁺ and NO₃⁻ are eliminated, resulting in Hg₂²⁺(aq) + 2Br⁻(aq) → Hg₂Br₂(s). A tempting distractor is D, which uses Hg⁺, incorrectly splitting the dimer ion due to misunderstanding mercury(I) structure. A transferable strategy is to preserve special ion formulas like Hg₂²⁺ in equations.
3. When aqueous solutions of lead(II) nitrate and potassium iodide are mixed, a yellow precipitate forms. The balanced molecular equation is: Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) Which of the following is the correct net ionic equation for this reaction?
   Answer: Pb2+(aq) + 2 I−(aq) → PbI2(s) This question tests the skill of writing net ionic equations for precipitation reactions. In the molecular equation, Pb(NO₃)₂ and KI are both soluble ionic compounds that dissociate completely in water, giving Pb²⁺, NO₃⁻, K⁺, and I⁻ ions. When these solutions mix, Pb²⁺ and I⁻ ions combine to form the insoluble precipitate PbI₂(s), while K⁺ and NO₃⁻ remain as spectator ions in solution. The net ionic equation includes only the ions that participate in forming the precipitate: Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s). Choice E incorrectly shows I₂²⁻ instead of 2 I⁻, demonstrating the misconception that polyatomic ions can have multiple charges on a single atom rather than understanding that two separate iodide ions are needed. To write net ionic equations correctly, first identify all spectator ions (those appearing unchanged on both sides) and remove them, leaving only the species that undergo chemical change.
4. Hydrochloric acid reacts with aqueous sodium hydroxide. The balanced molecular equation is: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Which of the following is the correct net ionic equation for the reaction?
   Answer: H+(aq) + OH−(aq) → H2O(l) This question tests the skill of writing net ionic equations for acid-base neutralization reactions. HCl is a strong acid that completely dissociates into H⁺ and Cl⁻ ions, while NaOH is a strong base that dissociates into Na⁺ and OH⁻ ions. When these solutions react, H⁺ and OH⁻ combine to form water molecules, while Na⁺ and Cl⁻ remain as spectator ions in solution. The net ionic equation shows only the actual chemical change: H⁺(aq) + OH⁻(aq) → H₂O(l). Choice E incorrectly includes all ions including the spectators Na⁺ and Cl⁻, demonstrating the misconception that net ionic equations must show all ions present rather than only those participating in the reaction. To write net ionic equations for acid-base reactions, focus on the H⁺ and OH⁻ ions that combine to form water, and eliminate spectator ions.
5. Aqueous sodium hydroxide is added to aqueous nickel(II) chloride, forming a precipitate. The balanced molecular equation is: $\mathrm{NiCl_2(aq) + 2\,NaOH(aq) \rightarrow Ni(OH)_2(s) + 2\,NaCl(aq)}$ Which is the correct net ionic equation?
   Answer: $\mathrm{Ni^{2+}(aq) + 2\,OH^{-}(aq) \rightarrow Ni(OH)_2(s)}$ This question tests the skill of writing net ionic equations for hydroxide precipitation. The molecular equation shows nickel(II) chloride and sodium hydroxide forming solid nickel(II) hydroxide and aqueous sodium chloride. Ions include Ni²⁺, 2Cl⁻, 2Na⁺, and 2OH⁻, with Ni²⁺ and 2OH⁻ producing Ni(OH)₂. Spectators Na⁺ and Cl⁻ are eliminated, resulting in Ni²⁺(aq) + 2OH⁻(aq) → Ni(OH)₂(s). A tempting distractor is C, which includes Na⁺, mistakenly retaining it due to not identifying spectators properly. A transferable strategy is to check if ions appear identically on both sides before removal.
6. A student mixes aqueous solutions of barium chloride and sodium sulfate. The balanced molecular equation is: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq) Which of the following is the correct net ionic equation for the reaction?
   Answer: Ba2+(aq) + SO42−(aq) → BaSO4(s) This question tests the skill of writing net ionic equations for precipitation reactions. The molecular equation shows BaCl₂ and Na₂SO₄ reacting to form BaSO₄(s) and NaCl. Since BaCl₂, Na₂SO₄, and NaCl are all soluble ionic compounds, they exist as dissociated ions in solution: Ba²⁺, Cl⁻, Na⁺, and SO₄²⁻. Only Ba²⁺ and SO₄²⁻ combine to form the insoluble precipitate BaSO₄, while Na⁺ and Cl⁻ remain as spectator ions throughout the reaction. The net ionic equation therefore includes only the precipitate-forming ions: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). Choice D incorrectly includes all ions on both sides, showing the misconception that spectator ions must be written in net ionic equations even when they don't participate in the reaction. When writing net ionic equations, systematically identify and remove all spectator ions that appear identically on both sides of the equation.
7. A student mixes aqueous hydrochloric acid with aqueous sodium hydroxide. Balanced molecular equation (with states): HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) What is the correct net ionic equation for the reaction?
   Answer: H+(aq) + OH−(aq) → H2O(l) This question tests the skill of writing net ionic equations for acid-base neutralization reactions. The molecular equation shows HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l), a classic neutralization reaction. To write the net ionic equation, we dissociate all strong electrolytes: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l). The spectator ions are Na⁺ and Cl⁻, which appear unchanged on both sides. Removing these gives us H⁺(aq) + OH⁻(aq) → H₂O(l), which is choice A. Choice B incorrectly shows HCl as a molecule rather than dissociated ions, reflecting the misconception that weak acids remain molecular in net ionic equations—but HCl is a strong acid and fully dissociates. For acid-base neutralizations involving strong acids and bases, the net ionic equation is always H⁺(aq) + OH⁻(aq) → H₂O(l), regardless of which specific strong acid or base is used.
8. Aqueous calcium chloride is mixed with aqueous sodium carbonate, producing a precipitate. The balanced molecular equation is: $\mathrm{CaCl_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2\,NaCl(aq)}$ Which is the correct net ionic equation?
   Answer: $\mathrm{Ca^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CaCO_3(s)}$ This question tests the skill of writing net ionic equations for precipitation reactions by identifying participating ions. The molecular equation shows calcium chloride and sodium carbonate forming solid calcium carbonate and aqueous sodium chloride. Ions are Ca²⁺, 2Cl⁻, 2Na⁺, and CO₃²⁻, with Ca²⁺ and CO₃²⁻ producing CaCO₃. Spectators Na⁺ and Cl⁻ are eliminated, leaving Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s). A tempting distractor is B, which includes all ions but omits Cl⁻ on the right, incorrectly partially canceling due to miscounting spectators. A transferable strategy is to verify solubility rules to confirm precipitates and ensure only reacting ions remain.
9. Aqueous solutions of barium chloride and sodium sulfate are mixed, producing a white precipitate. The balanced molecular equation is: $\mathrm{BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2\,NaCl(aq)}$ Which is the correct net ionic equation?
   Answer: $\mathrm{Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)}$ This question tests the skill of writing net ionic equations for precipitation reactions by identifying and removing spectator ions. The molecular equation shows barium chloride and sodium sulfate forming solid barium sulfate and aqueous sodium chloride. Dissociating the reactants yields Ba²⁺, 2Cl⁻, 2Na⁺, and SO₄²⁻ ions, with Ba²⁺ and SO₄²⁻ forming the BaSO₄ precipitate. Spectator ions Na⁺ and Cl⁻ are unchanged and removed, leaving Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). A tempting distractor is A, which includes Cl⁻ on the right, mistakenly keeping spectators due to confusing the complete ionic with the net ionic. A transferable strategy is to always dissociate soluble ionic compounds fully and eliminate ions that do not participate in the reaction.
10. Aqueous solutions of iron(III) chloride and sodium hydroxide are mixed, forming a precipitate. The balanced molecular equation is: FeCl3(aq) + 3 NaOH(aq) → Fe(OH)3(s) + 3 NaCl(aq) Which of the following is the correct net ionic equation?
    Answer: Fe3+(aq) + 3 OH−(aq) → Fe(OH)3(s) This question tests the skill of writing net ionic equations for precipitation reactions. FeCl₃ dissociates into Fe³⁺ and 3 Cl⁻ ions, while NaOH dissociates into Na⁺ and OH⁻ ions. When mixed, Fe³⁺ combines with 3 OH⁻ ions to form the insoluble precipitate Fe(OH)₃(s), while Na⁺ and Cl⁻ remain as spectator ions in solution. The net ionic equation shows only the precipitate formation: Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s). Choice E incorrectly shows Fe²⁺ forming Fe(OH)₂, demonstrating the misconception of confusing iron(III) with iron(II) compounds, which have different charges and form different hydroxide precipitates. Pay careful attention to Roman numerals in compound names as they indicate the specific oxidation state of the metal ion.
11. Aqueous solutions of acetic acid and sodium hydroxide react. The balanced molecular equation is: $\text{HC}_2\text{H}_3\text{O}_2(aq) + \text{NaOH}(aq) \rightarrow \text{NaC}_2\text{H}_3\text{O}_2(aq) + \text{H}_2\text{O}(l)$ Which of the following is the correct net ionic equation after dissociation and cancellation (treating acetic acid as a weak acid that remains mostly molecular)?
    Answer: $\text{HC}_2\text{H}_3\text{O}_2(aq) + \text{OH}^-(aq) \rightarrow \text{C}_2\text{H}_3\text{O}_2^-(aq) + \text{H}_2\text{O}(l)$ This question tests the skill of writing net ionic equations for weak acid-strong base reactions. Acetic acid (HC₂H₃O₂) is a weak acid that remains mostly molecular in solution, while NaOH is a strong base that completely dissociates into Na⁺ and OH⁻ ions. The OH⁻ ion removes a proton from the molecular acetic acid, forming the acetate ion (C₂H₃O₂⁻) and water, while Na⁺ remains as a spectator ion. The net ionic equation is: HC₂H₃O₂(aq) + OH⁻(aq) → C₂H₃O₂⁻(aq) + H₂O(l). Choice C incorrectly shows H⁺ + OH⁻ → H₂O, demonstrating the misconception that all acid-base reactions have the same net ionic equation, failing to recognize that weak acids remain molecular. Remember that weak acids and bases remain largely molecular in net ionic equations, unlike strong acids and bases which fully dissociate.
12. When aqueous solutions of lead(II) nitrate and potassium iodide are mixed, a yellow precipitate forms. The balanced molecular equation is: $\mathrm{Pb(NO_3)_2(aq) + 2\,KI(aq) \rightarrow PbI_2(s) + 2\,KNO_3(aq)}$ Which of the following is the correct net ionic equation?
    Answer: $\mathrm{Pb^{2+}(aq) + 2\,I^{-}(aq) \rightarrow PbI_2(s)}$ This question tests the skill of writing net ionic equations for precipitation reactions by identifying and removing spectator ions. The molecular equation involves aqueous lead(II) nitrate and potassium iodide forming solid lead(II) iodide and aqueous potassium nitrate. Dissociating the soluble compounds gives Pb²⁺, 2NO₃⁻, 2K⁺, and 2I⁻ ions, with Pb²⁺ and 2I⁻ combining to form the PbI₂ precipitate. The spectator ions K⁺ and NO₃⁻ appear unchanged on both sides and are eliminated, resulting in the net ionic equation Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s). A tempting distractor is B, which includes NO₃⁻ ions, incorrectly retaining spectator ions due to the misconception that all anions must be shown. A transferable strategy is to write the complete ionic equation and cancel identical spectator ions on both sides.
13. Aqueous sodium bicarbonate reacts with aqueous hydrobromic acid, producing carbon dioxide gas. The balanced molecular equation is: $\mathrm{NaHCO_3(aq) + HBr(aq) \rightarrow NaBr(aq) + CO_2(g) + H_2O(l)}$ Which is the correct net ionic equation?
    Answer: $\mathrm{H^{+}(aq) + HCO_3^{-}(aq) \rightarrow CO_2(g) + H_2O(l)}$ This question tests the skill of writing net ionic equations for gas-evolving reactions. The molecular equation involves sodium bicarbonate and hydrobromic acid forming carbon dioxide, water, and aqueous sodium bromide. Ions are Na⁺, HCO₃⁻, H⁺, and Br⁻, with H⁺ and HCO₃⁻ reacting to form CO₂ and H₂O. Spectators Na⁺ and Br⁻ are removed, yielding H⁺(aq) + HCO₃⁻(aq) → CO₂(g) + H₂O(l). A tempting distractor is E, which uses CO₃²⁻, incorrectly assuming full deprotonation due to confusing bicarbonate with carbonate. A transferable strategy is to represent intermediate ions accurately in stepwise reactions.
14. Aqueous aluminum sulfate is mixed with aqueous barium hydroxide, forming a precipitate. The balanced molecular equation is: $\mathrm{Al_2(SO_4)_3(aq) + 3\,Ba(OH)_2(aq) \rightarrow 2\,Al(OH)_3(s) + 3\,BaSO_4(s)}$ Which is the correct net ionic equation?
    Answer: $\mathrm{2\,Al^{3+}(aq) + 3\,SO_4^{2-}(aq) + 3\,Ba^{2+}(aq) + 6\,OH^{-}(aq) \rightarrow 2\,Al(OH)_3(s) + 3\,BaSO_4(s)}$ This question tests the skill of writing net ionic equations for reactions with multiple precipitates and no spectators. The molecular equation shows aluminum sulfate and barium hydroxide forming solid aluminum hydroxide and barium sulfate. All compounds dissociate to 2Al³⁺, 3SO₄²⁻, 3Ba²⁺, and 6OH⁻, with Al³⁺ pairing with OH⁻ and Ba²⁺ with SO₄²⁻ to form precipitates. Since no ions remain unchanged, the net equation is 2Al³⁺(aq) + 3SO₄²⁻(aq) + 3Ba²⁺(aq) + 6OH⁻(aq) → 2Al(OH)₃(s) + 3BaSO₄(s). A tempting distractor is A, which shows only one precipitate, incorrectly omitting others due to focusing on one product. A transferable strategy is to include all reacting species when there are no true spectators.
15. Aqueous strontium nitrate is mixed with aqueous sodium sulfate, producing a precipitate. The balanced molecular equation is: $\mathrm{Sr(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow SrSO_4(s) + 2\,NaNO_3(aq)}$ Which is the correct net ionic equation?
    Answer: $\mathrm{Sr^{2+}(aq) + SO_4^{2-}(aq) \rightarrow SrSO_4(s)}$ This question tests the skill of writing net ionic equations for precipitation reactions. The molecular equation depicts strontium nitrate and sodium sulfate forming solid strontium sulfate and aqueous sodium nitrate. Ions are Sr²⁺, 2NO₃⁻, 2Na⁺, and SO₄²⁻, with Sr²⁺ and SO₄²⁻ forming SrSO₄. Spectators Na⁺ and NO₃⁻ are eliminated, resulting in Sr²⁺(aq) + SO₄²⁻(aq) → SrSO₄(s). A tempting distractor is C, which forms Na₂SO₄ solid, incorrectly assuming insolubility due to misapplying solubility rules. A transferable strategy is to consult solubility tables to identify precipitates correctly.
16. Aqueous ammonium chloride is mixed with aqueous sodium hydroxide, producing ammonia gas. The balanced molecular equation is: $\mathrm{NH_4Cl(aq) + NaOH(aq) \rightarrow NH_3(g) + H_2O(l) + NaCl(aq)}$ Which is the correct net ionic equation?
    Answer: $\mathrm{NH_4^{+}(aq) + OH^{-}(aq) \rightarrow NH_3(g) + H_2O(l)}$ This question tests the skill of writing net ionic equations for reactions producing gases by removing spectators. The molecular equation involves ammonium chloride and sodium hydroxide forming ammonia gas, water, and aqueous sodium chloride. Ions include NH₄⁺, Cl⁻, Na⁺, and OH⁻, with NH₄⁺ and OH⁻ reacting to form NH₃ and H₂O. Spectators Na⁺ and Cl⁻ are removed, yielding NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l). A tempting distractor is E, which shows NH₄OH formation, incorrectly assuming a stable compound due to outdated notation misconceptions. A transferable strategy is to recognize weak bases or gases as undissociated products in net equations.
17. Aqueous solutions of calcium nitrate and sodium carbonate are mixed. The balanced molecular equation is: Ca(NO3)2(aq) + Na2CO3(aq) → CaCO3(s) + 2 NaNO3(aq) Which of the following is the correct net ionic equation for the reaction?
    Answer: Ca2+(aq) + CO32−(aq) → CaCO3(s) This question tests the skill of writing net ionic equations for precipitation reactions. Ca(NO₃)₂ and Na₂CO₃ are soluble ionic compounds that dissociate in water to give Ca²⁺, NO₃⁻, Na⁺, and CO₃²⁻ ions. When mixed, Ca²⁺ and CO₃²⁻ ions combine to form the insoluble precipitate CaCO₃(s), while Na⁺ and NO₃⁻ remain as spectator ions in solution. The net ionic equation shows only the precipitate formation: Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s). Choice E incorrectly shows Ca⁺ and CO₃⁻, demonstrating the misconception about ion charges where calcium has a +2 charge and carbonate has a -2 charge, not +1 and -1. To write accurate net ionic equations, ensure you know the correct charges of common ions and include only the species that undergo chemical change.
18. Aqueous potassium carbonate reacts with aqueous nitric acid, producing carbon dioxide gas. The balanced molecular equation is: $\mathrm{K_2CO_3(aq) + 2\,HNO_3(aq) \rightarrow 2\,KNO_3(aq) + CO_2(g) + H_2O(l)}$ Which is the correct net ionic equation?
    Answer: $\mathrm{CO_3^{2-}(aq) + 2\,H^{+}(aq) \rightarrow CO_2(g) + H_2O(l)}$ This question tests the skill of writing net ionic equations for gas-producing acid-carbonate reactions. The molecular equation involves potassium carbonate and nitric acid forming carbon dioxide, water, and aqueous potassium nitrate. Ions are 2K⁺, CO₃²⁻, 2H⁺, and 2NO₃⁻, with CO₃²⁻ and 2H⁺ reacting to form CO₂ and H₂O. Spectators K⁺ and NO₃⁻ are eliminated, resulting in CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l). A tempting distractor is E, which forms HCO₃⁻, incorrectly stopping at one protonation due to stepwise reaction misconception. A transferable strategy is to account for complete reaction products like gases in net equations.
19. Aqueous sodium acetate is mixed with aqueous hydrochloric acid. The balanced molecular equation is: $\mathrm{NaC_2H_3O_2(aq) + HCl(aq) \rightarrow HC_2H_3O_2(aq) + NaCl(aq)}$ Which is the correct net ionic equation? (Assume $\mathrm{HC_2H_3O_2}$ is a weak acid and remains mostly undissociated.)
    Answer: $\mathrm{H^{+}(aq) + C_2H_3O_2^{-}(aq) \rightarrow HC_2H_3O_2(aq)}$ This question tests the skill of writing net ionic equations for weak acid formation in solutions. The molecular equation shows sodium acetate and hydrochloric acid forming acetic acid and aqueous sodium chloride. Assuming HC₂H₃O₂ is weak and undissociated, ions are Na⁺, C₂H₃O₂⁻, H⁺, and Cl⁻, with H⁺ and C₂H₃O₂⁻ forming HC₂H₃O₂. Spectators Na⁺ and Cl⁻ are removed, yielding H⁺(aq) + C₂H₃O₂⁻(aq) → HC₂H₃O₂(aq). A tempting distractor is E, which forms NaH²⁺, incorrectly inventing species due to misunderstanding ion combinations. A transferable strategy is to represent weak electrolytes as molecular in net equations.
20. Aqueous silver nitrate is added to aqueous sodium chloride, forming a white precipitate. The balanced molecular equation is: $\mathrm{AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)}$ Which is the correct net ionic equation?
    Answer: $\mathrm{Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)}$ This question tests the skill of writing net ionic equations for precipitation reactions by eliminating spectator ions. The molecular equation involves silver nitrate and sodium chloride forming solid silver chloride and aqueous sodium nitrate. Dissociating gives Ag⁺, NO₃⁻, Na⁺, and Cl⁻, with Ag⁺ and Cl⁻ forming AgCl precipitate. Spectators Na⁺ and NO₃⁻ are removed, resulting in Ag⁺(aq) + Cl⁻(aq) → AgCl(s). A tempting distractor is A, which includes all ions, mistakenly presenting the complete ionic equation due to not canceling spectators. A transferable strategy is to cross out identical ions on both sides of the complete ionic equation to derive the net version.

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