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Net Ionic Equations An Application of Double Replacement Reactions.

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Presentation on theme: "Net Ionic Equations An Application of Double Replacement Reactions."— Presentation transcript:

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(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.”

3 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

4 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

5 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) Ionic Equations

6 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

7 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

8 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

9 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

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” 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

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 2 NO 3 − (aq) and 2 K + (aq) 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) and 2 K + (aq) These are called “spectator ions” 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” Spectator ions don’t participate in the reaction They hang around and watch 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) If we remove the spectator ions from the equation... 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 I − (aq) → PbI 2 (s) If we remove the spectator ions from the equation... 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 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

17 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

18 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

19 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

20 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

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) 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

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) 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

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) 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

24 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

25 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

26 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

27 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

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) Complete Ionic Equation: Ag + (aq) + ClO 4 − (aq) + Na + (aq) + Cl − (aq) → AgCl(s) + Na + (aq) + ClO 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) 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

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) 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

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

32 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

33 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

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