1. Double Replacement Rxn
Chemistry I

2. A closer look at Double Replacement Reactions
Instead of FIGHTING (Single Replacement)
Double Rep. Are all about LOVE
Evidence of the DR is either:
A precipitate forming
an acid/base (two ionic) reaction forming covalent water molecules
neutralization
If neither happens, No Reaction
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3. Aqueous Solutions
Many times, the chemicals we are reacting together are dissolved in water.
Mixtures of a chemical dissolved in water are called aqueous solutions.
The chemicals must first be dissolved in water
Compounds separate into free ions.

4. When some ionic compounds dissolve in water,
The anions and cations are separated from each other. Dissociation
Because they are more attracted to polar water molecules.
NaCl(s)  Na+(aq) +Cl- (aq)
Ions are FREE to move
throughout the solution.
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5. When compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion.
AgNO3 (s)  Ag+ (aq) + NO3-(aq)

6. Dissociation, Continued
Sodium chloride dissociates in water
NaCl(s) → Na+1(aq) + Cl-1(aq)
Silver nitrate dissociates in water
AgNO3(s) → Ag+1(aq) + NO3-1(aq)
Na+1
Cl-1 Na Cl
NO3
NO3-1 Ag
Ag+1

7. A cation and anion combine to form an insoluble precipitate.
DR Reactions
Occurs if
A cation and anion combine to form an insoluble precipitate.
If no precipitate forms, then NR (no reaction).

8. Why did a precipitate form?
Ions in AgCl are more attracted to each other than to the polar water molecules.
Fall madly deeply in love, wrap their arms around each other, and drop out of the solution as a solid.
AgCl(s)

9. What about the other couple, NaNO3?
The ions of NaNO3 are more attracted to the polar water molecules and stay separate (dissociated)
Soluble product is shown as (aq)
Na+ (aq) + NO3- (aq)
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10. So, how do you know if a new product is (s) or (aq)?
Gross equation:
NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)
Net equation:
Cl- (aq) + Ag+ (aq)  AgCl (s)
So, how do you know if a new product is (s) or (aq)?
Solubility Rules
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11. Solubility of Ionic Salts
General Rules
1.    Sodium (Na+1), potassium (K+1), and ammonium (NH4+1)
compounds are soluble in water.
2. Nitrates (NO3-1), acetates (C2H3O2-1), and chlorates (ClO3-1)
are soluble.
3. Chlorides (Cl-1) are soluble except with silver (Ag+1),
mercury(I) (Hg2+2), and lead(II) (Pb+2).
[Lead(II) chloride is soluble in hot water.]
4. Sulfates (SO4-2) are soluble except with calcium (Ca+2),
barium (Ba+2), strontium (Sr+2), and lead(II) (Pb+2).
5. Carbonates (CO3-2), phosphates(PO4-3), and
silicates (SiO3-2) are insoluble except with sodium (Na+1),
potassium (K+1), and ammonium (NH4+1).
6. Sulfides (S-2) are insoluble except with calcium (Ca+2),
barium (Ba+2), strontium (Sr+2), magnesium (Mg+2),
sodium (Na+1), potassium (K+1), and
ammonium (NH4+1).
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12. Solubility Patterns… Generally Soluble Generally Insoluble Cations:
NH4+, Li+, Na+, K+
Ag+, Pb2+, Hg2+
Anions:
Cl-, Br-, I-, NO3-, ClO3-, C2H3O2-
OH-, S2-, CO32-, PO43-, CrO42-

13. Determine whether each of the salts is soluble or insoluble:
AgBr
CaCl2
Pb(NO3)2
PbSO4
CuS
Fe2(SO4 )3
PbCO3
NH4Cl
AgBr insoluble
CaCl2 soluble
Pb(NO3)2 soluble
PbSO4 insoluble
CuS insoluble
Fe2(SO4 )3 soluble
PbCO3 insoluble
NH4Cl soluble
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15. Precipitation Reactions, Continued
2 KI(aq) + Pb(NO3)2(aq)  2 KNO3(aq) + PbI2(s)

16. DR Precipitation Reaction
2 KI(aq) + Pb(NO3)2(aq)  2 KNO3(aq) + PbI2(s)

17. No Precipitate Formation = No Reaction
KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq)
All ions still present,  no reaction.

18. Process for Predicting the Products of a Precipitation Reaction
Write the formula for the reactants.
Exchange ions. (Trade dance partners)
(+) ion combines with (-) ion
Criss-cross ionic charges of new dance couples to get formulas. (Don’t worry about balancing the equation yet.)
Use the solubility rules to determine solubility of each product in water.
If product is insoluble or slightly soluble, it will precipitate. Write (s) behind it.
If the product is soluble, write (aq) behind it.
If neither product will precipitate, no reaction.
5. Eliminate the spectator ions (those that are (aq) on both sides)
6. Write a balanced net ionic equation.

19. Write the formulas of the reactants Exchange the ions.
An aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Copper(II) Chloride, a White Solid Forms.
Write the formulas of the reactants
Na2CO3(aq) + CuCl2(aq) 
Exchange the ions.
Na2CO3 (aq) + CuCl2 (aq)  Na+ + Cl- + Cu2+ + CO32-
3. Criss-Cross charges and reduce
Na2CO3(aq) + CuCl2(aq)  NaCl + CuCO3

20. 4. Determine solubility of products
NaCl is soluble
CuCO3 is insoluble.
Na2CO3(aq) + CuCl2(aq)  2 NaCl(aq) + CuCO3(s)
5. Na+ and Cl- are (aq) on both sides. Spectators. Eliminate.
Net ionic equation:
CO32-(aq) + Cu2+(aq)  CuCO3(s)
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21. Consider: AlCl3(aq) + AgNO3(aq) 
Write the dissociation equation of each reactant:
AlCl3(s)  Al+3(aq) + 3 Cl-1(aq)
Why 3 Cl-1(aq) and NOT Cl3-1 ?
AgNO3(s)  Ag+1(aq) + NO3-1(aq)
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22. One of two things can happen:
All ions remain “free”
No reaction happens.
Two oppositely charged ions are attracted to each other strongly, bond
Form a precipitate.
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23. What are the products (possible new dance couples?
Al+3 combines with NO3- : Al(NO3)3
Ag+1 combines with Cl-1 : AgCl
Why NOT AlAg? Or NO3Cl ?
Does one, both, or neither product form a precipitate?
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24. AgCl insoluble Al(NO3)3 soluble AgCl (s) Al+3(aq) + 3NO3-1 (aq)
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25. Gross and Net Equations
Al+3 (aq) + 3 Cl-1(aq) + 3Ag+1(aq) + 3NO3-1(aq)
3 AgCl (s) + Al+3 (aq) +3 NO3- (aq)
Eliminate spectators.
Net Ionic equation:
Ag+(aq) + Cl_ (aq)  AgCl (s)
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26. Pb(NO3)2 (aq) + KI (aq) 
Write the dissociation equations for each salt.
Pb+2(aq) + NO3-(aq) + K+ (aq) + I-(aq)
Determine what the products are.
PbI2 and KNO3
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27. Pb(NO3)2 (aq) + KI (aq)  Which is the precipitate? Gross Equation:
PbI2 is insoluble (forms a precipitate)
Gross Equation:
Pb+2(aq) + NO3-(aq) + K+ (aq) + I-(aq)  PbI2 (s) + KNO3 (aq)
Net Ionic Equation:
Pb2+ (aq) + 2 I-(aq)  PbI2 (s)
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29. FeBr2(aq) + AgNO3(aq)  Dissociation of reactants?
Fe+2(aq) + 2 Br –1(aq) + Ag+1(aq) + NO3 -1(aq)
New Dance Couples?
Fe (NO3)2 and AgBr
Check table for precipitate
Fe(NO3)2 is soluble or (aq)
AgBr is insoluble or (s)
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30. Ag+1(aq) + Br-1(aq)  2AgBr (s)
Fe+2(aq) + 2 Br –1(aq) + Ag+1(aq) + NO3 -1(aq)
Fe+2(aq) + NO3 -1(aq) + AgBr (s)
Net ionic Equation:
Ag+1(aq) + Br-1(aq)  2AgBr (s)
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