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(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (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(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Let’s look at what happens when we make the two starting solutions - Pb(NO 3 ) 2 (s) → Pb 2+ (aq) + 2 NO 3 − (aq) KI(s) → K + (aq) + I − (aq) Our solutions are actually composed of the ions in solution. When we write “Pb(NO 3 ) 2 (aq)” we really mean “Pb 2+ (aq) + 2 NO 3 − (aq)” H2OH2O H2OH2O

5. Ions in Solution Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 KNO 3 (aq)

6. Ions in Solution Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 KNO 3 (aq) The PbI 2 (s) is a solid and is not in solution - we don’t have separated ions

7. Ions in Solution Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 KNO 3 (aq) The PbI 2 (s) is a solid and is not in solution - we don’t have separated ions The KNO 3 (aq) is in solution and represents solvated ions - KNO 3 (aq) → K + (aq) + NO 3 − (aq) H2OH2O

8. Ions in Solution Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (aq) The PbI 2 (s) is a solid and is not in solution - we don’t have separated ions The KNO 3 (aq) is in solution and represents solvated ions - KNO 3 (aq) → K + (aq) + NO 3 − (aq) H2OH2O

9. Ionic Equations Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (aq)

10. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (aq) This is called the “complete ionic equation” We have all the ionic species on both sides of the arrow Ionic Equations

11. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (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 Ionic Equations

12. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (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 NO 3 − (aq) Ionic Equations

13. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (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 NO 3 − (aq) Ionic Equations

14. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (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 NO 3 − (aq) and 2 K + (aq) Ionic Equations

15. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (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 NO 3 − (aq) and 2 K + (aq) Ionic Equations

16. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (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 NO 3 − (aq) and 2 K + (aq) These are called “spectator ions” Ionic Equations

17. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (aq) This is called the “complete ionic equation” Spectator ions don’t participate in the reaction They hang around and watch Ionic Equations

18. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 NO 3 − (aq) + 2 K + (aq) + 2 I − (aq) → PbI 2 (s) + 2 K + (aq) + 2 NO 3 − (aq) If we remove the spectator ions from the equation... Ionic Equations

19. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 I − (aq) → PbI 2 (s) If we remove the spectator ions from the equation... Ionic Equations

20. Pb(NO 3 ) 2 (aq) + 2 KI(aq) → PbI 2 (s) + 2 KNO 3 (aq) Now, we can write the equation as a mixture of solvated ions - Pb 2+ (aq) + 2 I − (aq) → PbI 2 (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 “net ionic equation” Ionic Equations

21. Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2 NaCl(aq) Applications

22. Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2 NaCl(aq) Ions in solution: Ba 2+ (aq) + 2 Cl − (aq) + 2 Na + (aq) + SO 4 2− (aq) Applications

23. Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2 NaCl(aq) Ions in solution: Ba 2+ (aq) + 2 Cl − (aq) + 2 Na + (aq) + SO 4 2− (aq) Ions on both sides of the arrow: 2 Cl − (aq) + 2 Na + (aq) Applications

24. Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2 NaCl(aq) Ions in solution: Ba 2+ (aq) + 2 Cl − (aq) + 2 Na + (aq) + SO 4 2− (aq) Ions on both sides of the arrow: 2 Cl − (aq) + 2 Na + (aq) Applications These are the spectator ions

25. Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2 NaCl(aq) Complete Ionic Equation: Ba 2+ (aq) + 2 Cl − (aq) + 2 Na + (aq) + SO 4 2− (aq) → BaSO 4 (s) + 2 Na + (aq) +2 Cl − (aq) Applications

26. Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2 NaCl(aq) Complete Ionic Equation: Ba 2+ (aq) + 2 Cl − (aq) + 2 Na + (aq) + SO 4 2− (aq) → BaSO 4 (s) + 2 Na + (aq) +2 Cl − (aq) Net Ionic Equation: Ba 2+ (aq) + SO 4 2− (aq) → BaSO 4 (s) Applications

27. Example 1: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2 NaCl(aq) Complete Ionic Equation: Ba 2+ (aq) + 2 Cl − (aq) + 2 Na + (aq) + SO 4 2− (aq) → BaSO 4 (s) + 2 Na + (aq) +2 Cl − (aq) Net Ionic Equation: Ba 2+ (aq) + SO 4 2− (aq) → BaSO 4 (s) Spectator Ions: Na + (aq) and Cl − (aq) Applications

28. Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO 4 (aq) + NaCl(aq) → AgCl(s) + NaClO 4 (aq) Applications

29. Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO 4 (aq) + NaCl(aq) → AgCl(s) + NaClO 4 (aq) Ions in solution: Ag + (aq) + ClO 4 − (aq) + Na + (aq) + Cl − (aq) Applications

30. Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO 4 (aq) + NaCl(aq) → AgCl(s) + NaClO 4 (aq) Ions in solution: Ag + (aq) + ClO 4 − (aq) + Na + (aq) + Cl − (aq) Ions on both sides of the arrow: ClO 4 − (aq) + Na + (aq) Applications

31. Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO 4 (aq) + NaCl(aq) → AgCl(s) + NaClO 4 (aq) Ions in solution: Ag + (aq) + ClO 4 − (aq) + Na + (aq) + Cl − (aq) Ions on both sides of the arrow: ClO 4 − (aq) + Na + (aq) Applications These are the spectator ions

32. Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO 4 (aq) + NaCl(aq) → AgCl(s) + NaClO 4 (aq) Complete Ionic Equation: Ag + (aq) + ClO 4 − (aq) + Na + (aq) + Cl − (aq) → AgCl(s) + Na + (aq) + ClO 4 − (aq) Applications

33. Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO 4 (aq) + NaCl(aq) → AgCl(s) + NaClO 4 (aq) Complete Ionic Equation: Ag + (aq) + ClO 4 − (aq) + Na + (aq) + Cl − (aq) → AgCl(s) + Na + (aq) + ClO 4 − (aq) Net Ionic Equation: Ag + (aq) + Cl − (aq) → AgCl(s) Applications

34. Example 2: Write the complete ionic equation, the net ionic equation, and determine the spectator ions for the following equation: AgClO 4 (aq) + NaCl(aq) → AgCl(s) + NaClO 4 (aq) Complete Ionic Equation: Ag + (aq) + ClO 4 − (aq) + Na + (aq) + Cl − (aq) → AgCl(s) + Na + (aq) + ClO 4 − (aq) Net Ionic Equation: Ag + (aq) + Cl − (aq) → AgCl(s) Spectator Ions: Na + (aq) and ClO 4 − (aq) Applications

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

36. 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 Summary

37. 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 Summary