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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

3. 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.”

4. 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 ions in solution. • When we write “Pb(NO3)2(aq)” we really mean “Pb2+(aq) + 2 NO3−(aq)” H2O H2O

5. 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 solvatedions - • Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 KNO3(aq)

6. 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 solid and is not in solution - • we don’t have separated ions

7. 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 solid and 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

8. 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 K+(aq) + 2 NO3−(aq) • The PbI2(s) is a solid and 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

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)

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” • We have all the ionic species on both sides of the arrow

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

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)

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” • If we look carefully at the equation, we will see compounds that are the same on both sides • 2 NO3−(aq)

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) • 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)

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 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)

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 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”

17. 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

18. 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 ...

19. 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 ...

20. 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”

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)

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) • Ions in solution: • Ba2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + SO42−(aq)

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) • 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)

24. 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

25. 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)

26. 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)

27. 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)

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)

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) • Ions in solution: • Ag+(aq) + ClO4−(aq) + Na+(aq) + Cl−(aq)

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) • Ions in solution: • Ag+(aq) + ClO4−(aq) + Na+(aq) + Cl−(aq) • Ions on both sides of the arrow: • ClO4−(aq) + Na+(aq)

31. 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

32. 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)

33. 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)

34. 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)

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

36. 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

37. 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