1. Section 6.4—Solubility & Precipitation
How can we make sure everything that’s added to the sports drink will dissolve?

2. A Review of Double-Replacement Reactions

3. Double Replacement Reactions
The cations from two compounds replace each other.
NaCl + AgNO3  AgCl + NaNO3 Cl Ag Cl Na Ag O N Na O N
Two ionic compounds switch ions

4. Double Replacement Reactions
General format of a double replacement reaction:

5. Products of a Double Replacement1
Combine the cation of the first reactant with the anion of the second reactant Ca
Cl2 + Ag
NO3

6. Products of a Double Replacement2
Combine the cation of the second reactant with the anion of the first reactant Ca
Cl2 + Ag
NO3

7. Products of a Double Replacement3
Remember to write cations first …
& balance charges with subscripts when writing formulas
Only leave subscripts that are in the original compound there if they are a part of a polyatomic ion! Ca
Cl2 + Ag
NO3 Ca
Cl2 + Ag
NO3
Ca(NO3)2 +
AgCl

8. Precipitation Reactions

9. Precipitation Reactions
A precipitation reaction is when 2 soluble substances are mixed together and they form an insoluble substance. This is called a precipitate.
Reactants
2 soluble
chemicals:
NaOH and Cu(NO3)2
NaOH
Cu(NO3)2

10. Precipitation Reactions(DR Rxns)
Na+1
OH-1
Cu+2
NO3 -1
Cu(OH)2(S)
Products: 1 soluble chemical: NaNO3
1 insoluble chemical (the precipitate): Cu(OH)2
Na+1
NO3 -1

11. Solubility Rules

12. Solubility Rules Table
Use the table on the reference sheet!
Insoluble = Precipitate

13. Decide whether each is soluble or not
Let’s Practice #1
NaNO3
Fe(C2H3O2)2
CaBr2
Ba(OH)2
Cu(OH)2
Example:
Decide whether each is soluble or not

14. Decide whether each is soluble or not
Let’s Practice #1
NaNO3
Fe(C2H3O2)2
CaBr2
Ba(OH)2
Cu(OH)2
Soluble
Not Soluble
Example:
Decide whether each is soluble or not

15. Write the products for this reaction & predict the precipitate
Let’s Practice #2
Example:
Write the products for this reaction & predict the precipitate
Remember to indicate compounds that dissolve with “aq” for “aqueous” and compounds that don’t dissolve with “s” for “solid”
AgNO3 (aq) + NaCl (aq) 

16. Write the products for this reaction
Let’s Practice #2
Example:
Write the products for this reaction
AgNO3 (aq) + NaCl (aq) 
AgCl (s) + NaNO3 (aq)
precipitate

17. Write the products for this reaction & identify the precipitate
Let’s Practice #3
Example:
Write the products for this reaction & identify the precipitate
Remember to indicate compounds that dissolve with “aq” for “aqueous” and compounds that don’t dissolve with “s” for “solid”
BaCl2 (aq) + K2CO3 (aq) 

18. Write the products for this reaction
Let’s Practice #3
Example:
Write the products for this reaction
BaCl2 (aq) + K2CO3 (aq) 
KCl (aq) + BaCO3 (s)
precipitate

19. Net Ionic Reactions
Shows the details of aqueous reactions that involve ions in aqueous solution
Molecular Equation: the typical equation you are use to writing keeping all molecules together
Complete Ionic Equation: shows all the particles in a solution as they really exist, as IONS or MOLECULES.
Anything aqueous needs to be split apart into the cation and anion
Anything solid stays intact
Coefficients need to be multiplied by subscripts to determine the exact amount of each cation and anion.
Spectator ions: ions that do not participate in a reaction; they are identical on both sides of the equation & are crossed out!
Net Ionic Equation: the final equation showing the major players. All spectator ions have been removed.

20. NET IONIC REACTIONS for Precipitation Reactions
Molecular equation:
KI(aq) + AgNO3(aq)  AgI(s) + KNO3(aq)
Complete Ionic equation:
K+1 + I-1 + Ag+1+ NO3-1  AgI K NO3-1
Spectator ions: ions that do not participate in a reaction; they are identical on both sides of the equation & are crossed out!
Net Ionic equation:
I-1 + Ag+1  AgI

21. NET IONIC REACTIONS for Precipitation Reactions
Molecular equation:
2 NaOH(aq) + CuCl2(aq)  2 NaCl(aq) + Cu(OH)2(s)
Complete Ionic equation:
2 Na OH-1 + Cu Cl-1  2 Na Cl-1 + Cu(OH)2
Net Ionic equation:
2 OH-1 + Cu+2  Cu(OH)2

22. Take Home Practice: Predict products and balance Iron (III) chloride reacts with sodium hydroxide
Molecular equation:
1 FeCl3(aq) + 3 NaOH(aq)  1 Fe(OH)3(s) +3 NaCl(aq)
Complete Ionic equation:
3 Na OH-1 + Fe Cl-1  3 Na Cl-1 + Fe(OH)3
Net Ionic equation:
3 OH-1 + Fe+3  Fe(OH)3

23. Section 6.5—Stoichiometry
How can we determine in a lab the concentration of electrolytes?

24. What do those coefficients really mean?
The coefficient of the balanced chemical equation tells how many moles of each substance is used in the reaction.
For every 2 moles of H2…
and 2 moles of H2O are produced 2 2
2 H2 + O2  2 H2O
No coefficient = 1
1 mole of O2 is need to react…

25. Mole Ratio
Is a conversion factor that relates 2 substances in moles; must use a balanced chemical equation to create it
2 H2 + O2  2 H2O
Examples of Mole Ratio’s
2mol H mol O mol H2O
1 mol O mol H2O mol H2

26. What is stoichiometry?
Stoichiometry – Calculations using the mole ratio from the balanced equation and information about one compound in the reaction to determine information about another compound in the equation.

27. Example: What is the mole ratio of chlorine to sodium?
2 Na + Cl2  2 NaCl
2mol Na mol Cl mol Na
1 mol Cl mol NaCl mol NaCl

28. Stoich (Mole-Mole) : 1 step problem using the mole ratio
Example:
If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are needed?
2 H2 + O2  2 H2O

29. Stoichiometry with Moles
Example:
If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are needed?
2 H2 + O2  2 H2O
From balanced equation:
2 mole H2  1 mole O2
4.2 mole H2 1
mole O2
= ________ mole O2
2.1 2
mole H2

30. Stoich (Mole-Mole) Example:
If 0.67 moles of potassium nitrate reacts, how many moles of oxygen are produced?
2KNO3  2KNO2 + O2
From balanced equation:
2 mole KNO3  1 mole O2
0.67 mole KNO3 1
mole O2
= ________ mole O2
0.34 2
Mole KNO3

31. But we can’t measure moles in lab!
We can’t go to the lab and count or measure moles…so we need a way to work in measurable units, such as grams and liters!
Molecular mass gives the grams = 1 mole of a compound!

32. Stoich( Mole-Mass): 2 step problem use mole ratio & then molar mass conversion factors
Example:
How many grams of AgCl will be precipitated if
0.45 mole AgNO3 is reacted as follows:
2 AgNO3 + CaCl2  2 AgCl + Ca(NO3)2

33. Stoichiometry with Moles & Mass
Example:
How many grams of AgCl will be precipitated if
0.45 mole AgNO3 is reacted as follows:
2 AgNO3 + CaCl2  2 AgCl + Ca(NO3)2
From balanced equation:
2 mole AgNO3  2 mole AgCl
Molar Mass of AgCl:
1 mole AgCl = g
0.45 mole AgNO3 2
mole AgCl
143.32
g AgCl
= ________ g AgCl 64 2
mole AgNO3 1
mole AgCl

34. If 4.42 g of H2 reacts, how many moles of NH3 are produced?
Stoich( Mass- Mol): 2 step problem use molar mass & then mole ratio conversion factors
Example:
If 4.42 g of H2 reacts, how many moles of NH3 are produced?
N2 + 3H2  2NH3
From balanced equation:
3 mole H2  2 mole N2
4.42 g H2 1
Mole H2
2 mole NH3
=1.46 mole NH3
2.02
g H2
3 mole H2

35. 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Stoich( Mass-Mass): 3 step problem use molar mass, then mole ratio & then molar mass conversion factors (Honors Only)
Example:
How many grams Ba(OH)2 are precipitated from 14.5 g of NaOH in the following reaction:
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl

36. Stoichiometry with Mass (Honors)
Example:
How many grams Ba(OH)2 are precipitated from 14.5 g of NaOH in the following reaction:
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Molar Mass of NaOH:
1 mole NaCl = g
From balanced equation:
2 mole NaOH  1 mole Ba(OH)2
Molar Mass of Ba(OH)2:
1 mole Ba(OH)2 = g
14.5 g NaOH 1
mole NaOH 1
mole Ba(OH)2
171.35
g Ba(OH)2
40.00
g NaOH 2
mole NaOH 1
mole Ba(OH)2
= ________ g Ba(OH)2
31.1

37. Stoichiometry with Mass (Honors)
Example:
How many grams of HCl are needed to produce 65.0 g of magnesium chloride:
__Mg + ____HCl  ____MgCl2 + __H2
Balance the equation and fill in the missing information:
1 mole MgCl2
mole HCl
g HCl
65 g MgCl2
= g HCl
g MgCl2
mole MgCl2
mole HCl

38. What about the stoichiometry of gases? Recall
Molar Volume of a Gas – at STP
1 mole of any gas = 22.4 liters

39. Stoichiometry with Gases 1 mol= 22.4 L @STP
Example:
If you need react 1.5 g of zinc completely, what volume of hydrogen gas will be produced at STP?
2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)

40. Stoichiometry with Gases
Example:
If you need react 1.5 g of zinc completely, what volume of hydrogen gas will be produced at STP?
2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)
Molar volume of a gas:
1 mole H2 = 22.4 L
From balanced equation:
1 mole Zn  1 mole H2
Molar Mass of Zn:
1 mole Zn = g
1.5 g Zn 1
mole Zn 1
mole H2
22.4
L H2
65.39
g Zn 1
mole Zn 1
mole H2
= ________ L H2
0.51

41. Stoichiometry with Gases
Example: How many moles of water will be produced from the complete combustion of 7.3 L of oxygen gas? Assume STP
C3H O2 → 3CO H2O
Molar volume of a gas:
1 mole O2 = 22.4 L
From balanced equation:
4 mole H2O  5 mole O2
Molar Mass of H:
1 mole H2O = 18 g
7.3 L O2 1
Mole O2 4
mole H2 O
22.4 L
L O2 5
Mole O2
= _
0.26 mole H2O

42. Keeping all these equalities straight!
TO GO BETWEEN
USE THE EQUALITY
Grams & moles
Molar mass (g)= 1 mole
Particles & Moles
1 mol = 6.02 x 1023 particles
Moles & liters of a gas at STP
1 mole = 22.4 L at STP
2 different chemicals in a reaction
Coefficient ratio(MOLE RATIO) from balanced equation

43. You Try!
Given the UNBALANCED EQUATION: __MgCO3  __MgO + __CO2, how many liters of CO2 gas are produced from the reaction of 15 grams of MgCO3? Assume STP!

44. Section 6.5b
Percent Yield

45. Percent Yield A “Yield” is a product
Actual Yield(A): the actual amount of product you produce in the lab
Theoretical Yield(T): the amount of product you should produce if nothing went wrong; use the balanced chemical equation to calculate this amount.
Percent Yield: ratio of actual yield to theoretical yield

46. %yield = A x 100 T Percent Yield Lets Practice in steps:
1a. If 4.20 moles H2 reacts completely with oxygen, how many grams of H2O are produced?
2 H2 + O2  2 H2O
This is a mol-mass problem.
Your answer is the theoretical yield of water?

47. If 4.20 moles H2 react completely with oxygen how many grams of H2O are produced? 2 H2 + O2  2 H2O
From balanced equation:
2 mole H2O  1 mole O2
Molar Mass of O2:
1 mole H2O = 18.02g
18.02 grams H2O
2 moleH2O
75.7 g
4.2 mol H2
= ______g H2O
1 mole H2O
2 mol H2
This is the theoretical yield.

48. What is the percent yield if 60.0 grams of H2O are produced?
A= 60.0 g
T= 75.7 g
%yield = A x 100 T
%yield = x 100
75.7
79.3% yield

49. You have precipitated 8. 50 g of Ba(OH)2. If you start with 4
You have precipitated 8.50 g of Ba(OH)2. If you start with 4.57 grams of NaOH, what is the % yield NaOH + BaCl2  Ba(OH)2 + 2 NaCl
From balanced equation:
2 mole NaOH  1 mole Ba(OH)2
Molar Mass of Ba(OH)2: g
Molar Mass of NaOH: g
1 molBa(OH)2
171.35g Ba(OH)2
1 molNaOH
4.57 g NaOH
40.00g NaOH
2 mol NaOH
1mol Ba(OH)2
= g Ba(OH)2
This is the theoretical yield.

50. If 9.78 grams are obtained in the experiment, what is the percent yield?
A= 8.50 g
T= 9.79 g
%yield = A x 100 T
%yield = x 100
9.79
86.8% yield

51. Section 6.5c
Titrations

52. Titrations—Using Stoichiometry
Titration – A technique where the addition of a known volume of a known concentration solution to a known volume of unknown concentration solution to determine the concentration.
Use a buret to titrate unknown concentration of solutions.

53. Titrations—Using Stoichiometry
The titrant is the known concentration in the buret and the analyte is the unknown concentration in the flask.
Formula: nMaVa = nMbVb
na= number of H+ in the acid
nb= number of OH- in the base
Ma= molarity of acid
Mb= molarity of base
V= volume

54. End Point vs. Equivalence Point
Equivalence Point (or Stoichiometric Point)
– When there are no reactants left over—they have all been reacted and the solution contains only products
the point where the acid and the base are equal in equal moles
moles acid = moles base

55. Importance of Indicators
Indicators – Paper or liquid that change color based on pH level.
End Point: point at which the indicator in the solution changes color
It signals the equivalence point and the stop of the titration
Always select an indicator that has a pH value close to that of the pH of the equivalence point of the titration.

57. Titration Process

58. Titration Problem #1
How many liters of 0.10 M NaOH is needed to react with L of 0.25 M HCl?
NaOH + HCl  H2O + NaCl

59. Titration Problem #2
What is the molarity of a Ca(OH)2 solution if 30.0 ml of the solution is neutralized by 20.0 ml of a 0.50 M solution of HCl?
  Ca(OH) 2 + 2HCl  2H2O + CaCl2

60. Titration Problem #3
What volume of 2.0M solution of NH4OH is needed to neutralize 50.0 ml of a 0.50M solution of H2SO4?
2 NH4OH + H2SO4  2H2O + (NH4) 2SO4

61. Titration Curves Shows the changes of pH during a titration
Strong Base - Strong Acid
Weak Base - Strong Acid
Shows the changes of pH during a titration
Identifies the pH of the equivalence point
Strong Base - Weak Acid
Weak Base - Weak Acid

63. Titration curve for Titrating a strong acid with a strong base pH is always = 7
The titration curve graph shows the pH of the equivalence point. Take the vertical region and cut the length in half and then look to what pH value aligns to that point.

64. Titration curve for Titrating a strong base with an strong acid pH is always = 7

65. Titration curve for Titrating a weak acid with an strong base pH is >7