Double Displacement Rxns

1. Double Displacement Rxns • 2 soluble ionic compounds trade cations (+ve ions) • General Equation AB + CD  CB +AD • 3 possible products • Precipitate • Gas • Water

2. Will a precipitate form? • If the product is insoluble in water , a precipitate will form (the compound is solid). • If the product is soluble in water, no precipitate forms (the compound is aqueous) • How do we know which compounds are soluble? • Use a solubility table

4. Example • Aqueous solution of lead(II) nitrate is added an aqueous solution of potassium iodide. Predict the products, balance the chemical equation and predict the solubility of the products. • Products Pb2+ NO3- K+ I-

5. Example Con’t • Skeleton Equation Pb (NO3)2 + KI  PbI2 + KNO3 • Balanced Equation: • Add states to the equation: • Reactants are both aqueous • Go to the table • Lead (II) iodide • Potassium nitrate Insoluble due to lead(II) ion (precipitate forms) Soluble – all nitrates are soluble

6. Determine the products, balance and predict state of the products • Silver(I) nitrate (aq) + potassium sulfate (aq)  • Li2CO3(aq) + NaCl (aq) • Mercury(II) nitrate(aq) + Lithium bromide(aq) 

7. Double Displacement Rxns Producing Gases • 3 situations where gases are produced: • Carbonates react with acids • Sulfites react with acids • Ammonium salts react with bases • In all cases the product containing the carbonate, sulfite or ammonium ion breaks down into H2O and a gas.

8. Examples • CaCO3(s) + HCl(aq) CaCl2(aq) + H2CO3(aq) CO2(g) + H2O(l) • The final equation is: CaCO3(s) + HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) Decomposes

9. Examples Con’t • NH4Cl(aq) + NaOH(aq) NaCl(aq) + NH4OH(aq) NH3(g) + H2O(l) • The final equation is: NH4Cl(aq) + NaOH(aq) NaCl(aq) + NH3(g) + H2O(l) Decomposes

10. Examples Con’t • Na2SO3(s) + HCl(aq) + H2O(l) • The final equation is: • Na2SO3(s) + HCl(aq) Decomposes

11. How do we know when a gas will form? • If one of the products is: • H2CO3 (ag)  CO2(g) + H2O(l) • H2SO3(aq)  SO2(g) + H2O(l) • NH4OH(aq)  NH3(g) + H2O(l) • = decomposes into Remember, we still need to show the other product of the double displacement rxn.

12. Practice: Determine the products, states and balance the equation • Na2CO3 (aq) + HBr(aq)

13. Neutralization Reactions • Double displacement rxn involving an acid and a base as reactants • Products are a salt and water • Eg. HNO3(aq) + KOH(aq)  KNO3(aq) + H2O(l)

14. For the following neutralization rxn: Predict the products, states and balance the equation HCl(aq) + NaOH(aq) H2SO4(aq) + CaOH(aq) 

15. Wrap-up of Double Displacement Rxns Remember!!! • Double displacement rxns only happen when: • A precipitate is formed • A gas is formed • Water is formed • In all other cases, the ions stay in solution!!! • No rxn happens!!!

16. Animations for Double Displacement Rxns • http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/clm2s3_4.swf • http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/crm3s2_3.swf

17. Net Ionic Equations • Used when a precipitate is formed in a double displacement rxn. • Soluble ionic compounds are present in solution as aqueous ions. disolves in water • NaCl(s)  Na+ + Cl- disolves in water • K2SO4(s)  2K+ + SO42-

18. Why use a Net Ionic Equation • It is the ions that will join to create the precipitate, by looking at the ions we get a better idea of what is happening in the solution. Eg. For the reaction: K2CO3 (aq) + BaCl2 (aq) What are the potential products? Are they soluble?

19. Total Ionic Equations • K2CO3(aq) + BaCl2(aq) 2KCl(aq) + BaCO3(s) • KCl  soluble From a • BaCO3  insoluble table of solubility • We can write the equation using ions for all aqueous compounds • K2CO3(aq) + BaCl2(aq) 2KCl(aq) + BaCO3(s) Becomes, 2K+ (aq) + CO32- (aq) + Ba2+ (aq) + 2Cl- (aq) 2K+ (aq) + 2Cl- (aq) + BaCO3(s)

20. From Total  Net Ionic Equations • Total ionic equation – contains all ions/compounds 2K+(aq) + CO32- (aq) + Ba2+ (aq) + 2Cl-(aq) 2K+ (aq) + 2Cl-(aq) + BaCO3(s) • The potassium and chloride ions we unchanged by the rxn (in red) • Because they do not take part in the rxn, we call them spectator ions • If we show only the ions involved in the rxn, we get a net ionic equation

21. Net Ionic Equations • Total ionic equation – contains all ions/compounds 2K+(aq) + CO32- (aq) + Ba2+ (aq) + 2Cl-(aq) 2K+ (aq) + 2Cl-(aq) + BaCO3(s) • Net ionic equation – contains only ions involved in the reaction and the solid they form. • For the above reaction the net ionic equation is: CO32- (aq) + Ba2+ (aq)  BaCO3(s)

22. Down side to the Net Ionic Equation • Do not show what aqueous were mixed Ex. CO32- (aq) + Ba2+ (aq)  BaCO3(s) • We don’t know which barium compound and which carbonate compound were mixed. • We do know that any reaction involving the combination of barium ions and carbonate ions will produce a precipitate! Try one out.

23. Rules for Net Ionic Equations • Solutions of soluble salts (KCl, NaNO3 etc.) All written as ions • Solutions of strong acids (HCl, HNO3, H2SO4 and HClO4) • Solutions of strong bases (NaOH, KOH, Ba(OH)2) • All covalent compounds left in molecular form, states are written beside them. • Including all gases, liquids (eg. H2O) and soluble covalent compounds (eg. ethanol)

24. Steps for Writing a Net Ionic Equation • Identify Type of Reaction and Possible Products • Look Up Solubility of Both Products • Indicate States of Reactants and Products • Write Chemical Equation for Reaction • Balance Equation • Write Total Ionic Equation • Write Net Ionic Equation

25. Ex. Write the total ionic equation and net ionic equation for the reaction of barium sulfide with sodium sulfate.