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Net Ionic Equations

Net Ionic Equations. An Application of Double Replacement Reactions. Introduction. We know that double replacement reactions result in the formation of either - a precipitate , or an insoluble gas , or water Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq)

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Net Ionic Equations

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  1. Net Ionic Equations • An Application of Double Replacement Reactions

  2. Introduction • We know that double replacement reactions result in the formation of either - • a precipitate, or • an insoluble gas, or • water Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • “An aqueous solution of lead(II) nitrate is mixed with an aqueous solution of potassium iodide and results in the formation of solid lead(II) iodide and an aqueous solution of potassium nitrate.”

  3. Ions in Solution Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Let’s look at what happens when we make the two starting solutions - Pb(NO3)2(s) → Pb2+(aq) + 2 NO3−(aq) KI(s) → K+(aq) + I−(aq) • Our solutions are actually composed of the ionsin solution. • When we write “Pb(NO3)2(aq)” we really mean “Pb2+(aq) + 2 NO3−(aq)” H2O H2O

  4. Ions in Solution Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions – Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 KNO3(aq) • The PbI2(s) is a solidand is not in solution - • we don’t have separated ions • The KNO3(aq) is in solution and represents solvated ions – KNO3(aq) → K+(aq) + NO3−(aq) H2O

  5. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq)

  6. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) • This is called the “complete ionic equation” • We have all the ionic species on both sides of the arrow

  7. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) • This is called the “complete ionic equation” • If we look carefully at the equation, we will see compounds that are the same on both sides

  8. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) • This is called the “complete ionic equation” • If we look carefully at the equation, we will see compounds that are the same on both sides • 2 NO3−(aq)

  9. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) • This is called the “complete ionic equation” • If we look carefully at the equation, we will see compounds that are the same on both sides • 2 NO3−(aq)

  10. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) • This is called the “complete ionic equation” • If we look carefully at the equation, we will see compounds that are the same on both sides • 2 NO3−(aq) and2 K+(aq)

  11. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) • This is called the “complete ionic equation” • If we look carefully at the equation, we will see compounds that are the same on both sides • 2 NO3−(aq) and2 K+(aq)

  12. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) • This is called the “complete ionic equation” • If we look carefully at the equation, we will see compounds that are the same on both sides • 2 NO3−(aq) and2 K+(aq) • These are called “spectatorions”

  13. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) • This is called the “complete ionic equation” • Spectator ions don’t participate in the reaction • They hang around and watch

  14. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) • If we remove the spectator ions from the equation ...

  15. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 I−(aq) → PbI2(s) • If we remove the spectator ions from the equation ...

  16. Ionic Equations Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) • Now, we can write the equation as a mixture of solvated ions - Pb2+(aq) + 2 I−(aq) → PbI2(s) • If we remove the spectator ions from the equation, we end up with an equation that has only the reacting species. • This is called the “netionicequation”

  17. Applications Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq)

  18. Applications Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq) • Ions in solution: • Ba2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + SO42−(aq)

  19. Applications Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq) • Ions in solution: • Ba2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + SO42−(aq) • Ions on both sides of the arrow: • 2 Cl−(aq) + 2 Na+(aq)

  20. Applications Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq) • Ions in solution: • Ba2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + SO42−(aq) • Ions on both sides of the arrow: • 2 Cl−(aq) + 2 Na+(aq) These are the spectator ions

  21. Applications Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq) • Complete Ionic Equation: • Ba2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + SO42−(aq) → BaSO4(s) + 2 Na+(aq) +2 Cl−(aq)

  22. Applications Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq) • Complete Ionic Equation: • Ba2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + SO42−(aq) → BaSO4(s) + 2 Na+(aq) +2 Cl−(aq) • Net Ionic Equation: • Ba2+(aq) + SO42−(aq) → BaSO4(s)

  23. Applications Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq) • Complete Ionic Equation: • Ba2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + SO42−(aq) → BaSO4(s) + 2 Na+(aq) +2 Cl−(aq) • Net Ionic Equation: • Ba2+(aq) + SO42−(aq) → BaSO4(s) • Spectator Ions: • Na+(aq) and Cl−(aq)

  24. Applications Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO4(aq) + NaCl(aq) → AgCl(s) + NaClO4(aq)

  25. Applications Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO4(aq) + NaCl(aq) → AgCl(s) + NaClO4(aq) • Ions in solution: • Ag+(aq) + ClO4−(aq) + Na+(aq) + Cl−(aq)

  26. Applications Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO4(aq) + NaCl(aq) → AgCl(s) + NaClO4(aq) • Ions in solution: • Ag+(aq) + ClO4−(aq) + Na+(aq) + Cl−(aq) • Ions on both sides of the arrow: • ClO4−(aq) + Na+(aq)

  27. Applications Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO4(aq) + NaCl(aq) → AgCl(s) + NaClO4(aq) • Ions in solution: • Ag+(aq) + ClO4−(aq) + Na+(aq) + Cl−(aq) • Ions on both sides of the arrow: • ClO4−(aq) + Na+(aq) These are the spectator ions

  28. Applications Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO4(aq) + NaCl(aq) → AgCl(s) + NaClO4(aq) • Complete Ionic Equation: • Ag+(aq) + ClO4−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + ClO4−(aq)

  29. Applications Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO4(aq) + NaCl(aq) → AgCl(s) + NaClO4(aq) • Complete Ionic Equation: • Ag+(aq) + ClO4−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + ClO4−(aq) • Net Ionic Equation: • Ag+(aq) + Cl−(aq) → AgCl(s)

  30. Applications Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO4(aq) + NaCl(aq) → AgCl(s) + NaClO4(aq) • Complete Ionic Equation: • Ag+(aq) + ClO4−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + ClO4−(aq) • Net Ionic Equation: • Ag+(aq) + Cl−(aq) → AgCl(s) • Spectator Ions: • Na+(aq) and ClO4−(aq)

  31. Summary • To write the complete ionic equation - • separate all aqueous ionic compounds into their aqueous ions • keep all solids, insoluble gases, and water together

  32. Summary • To write the complete ionic equation - • separate all aqueous ionic compounds into their aqueous ions • keep all solids, insoluble gases, and water together • To find the spectator ions - • find the aqueous ions that are the same on both sides of the arrow

  33. Summary • To write the complete ionic equation - • separate all aqueous ionic compounds into their aqueous ions • keep all solids, insoluble gases, and water together • To find the spectator ions - • find the aqueous ions that are the same on both sides of the arrow • To write the net ionic equation - • remove the spectator ions from the complete ionic equation

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