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

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.”

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

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)

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

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

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

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)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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