1. UNIT 2: CHEMICAL REACTIONS

2. d+ d- solution solute solvent UNIT 2: CHEMICAL REACTIONS
4.1, 4.2 Aqueous Solutions
_____________ = _____________ + _______________
Water is polar!!!
solution
solute
solvent
must be soluble!! d+
O - 3.5
H - 2.1
1.4 EN Diff
so VERY Polar bond d-
H2O strong dipole

3. © 2012 Pearson Education, Inc.
Solutions
Solutions are defined as homogeneous mixtures of two or more pure substances.
The solvent is present in greatest abundance.
All other substances are solutes.
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4. Solubility of a substance in water is determined by
the AFFINITY of the water molecules for the solute particles
For INSOLUBLE cmpds:
the attraction of the (+) and (-) ions
for each other
is stronger than
their attraction for
POLAR water molecules

5. An ELECTROLYTE is a substance
whose aqueous solution conducts electric current.
The current is carried by IONS (+) and (-)
Key: The more ions in solution, the more current is carried.
Thus if NO ions in a solution (just neutral molecules)
then NO current can be conducted.

6. 8 STRONG ELECTROLYTES: are excellent “conductors” 7
because ___ solute particle are in the form of ______
all
ions
1) soluble ionics
NaCl
NH4NO3 KI
See Solubility Table!
2) strong acids
neutral molecules that “ionize”(react with) H2O 7
“Is Bright and Clear, No Snow, 3,4”
HI, HBr, HCl,
HClO3, HClO4, HNO3, H2SO4
all end in “ic”
3) strong bases
soluble hydroxides 8
backwards “J”
“Can, Sara, Bat”
LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2

7. © 2012 Pearson Education, Inc.
Acids
There are only seven strong acids:
Hydrochloric (HCl)
Hydrobromic (HBr)
Hydroiodic (HI)
Nitric (HNO3)
Sulfuric (H2SO4)
Chloric (HClO3)
Perchloric (HClO4)
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8. © 2012 Pearson Education, Inc.
Bases
The strong bases are the soluble metal salts of hydroxide ion:
Alkali metals
Calcium
Strontium
Barium
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Dissociation
An electrolyte is a substances that dissociates into ions when dissolved in water.
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10. © 2012 Pearson Education, Inc.
Solutions
An electrolyte is a substance that dissociates into ions when dissolved in water.
A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.
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11. Electrolytes and Nonelectrolytes
Soluble ionic compounds tend to be electrolytes.
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12. WEAK ELECTROLYTES: are poor “conductors” because
________ solute particles are in the form of _____.
very few
ions
(Almost all solute remains as neutral molecules)
1) weak acids HF
HC2H3O2
HNO2
2) weak (molecular) bases
NH3
CH3NH2

13. NaCl(s) --------> Na+(aq) + Cl-(aq)
H2O
NaCl(s) > Na+(aq) + Cl-(aq)
“hydrated” or (“solvated”) ions
IowaState visual

14. strong electrolyte weak electrolyte non- electrolyte ALL ions
VERY FEW ions
NO ions + - - + + + + - + - +
--- -
Str Acids
Str Bases
Soluble Ionics
Weak Acids
Weak Bases
POLAR
molecular
compounds

15. Electrolytes and Nonelectrolytes
Soluble ionic compounds tend to be electrolytes.
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16. Electrolytes and Nonelectrolytes
Molecular compounds tend to be nonelectrolytes, except for acids and bases.
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17. © 2012 Pearson Education, Inc.
Electrolytes
A strong electrolyte dissociates completely when dissolved in water.
A weak electrolyte only dissociates partially when dissolved in water.
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18. Strong Electrolytes Are…
Strong acids
Strong bases
Soluble ionic salts
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19. BaCl2(s) ------> Ba2+(aq) + 2Cl-(aq)
SOLUBLE
IONIC
BaCl2(s)
dissolves
solvated
ions
ionic solid
H2O
BaCl2(s) > Ba2+(aq) Cl-(aq)
“DISSOCIATION” of an ionic compound in water:
breaking bonds between ions in the solid then
making bonds between each ion and
polar H2O molecules that surround them

20. © 2012 Pearson Education, Inc.
Dissociation
When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them.
This process is called dissociation.
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21. HCl(g) + H2O(l) ---> H3O+(aq) + Cl-(aq)
STRONG
ACID
HCl solution
ALL solute is ions.
Str ACID ionizes 100% in H2O
HCl(g) + H2O(l) ---> H3O+(aq) + Cl-(aq)
“IONIZATION” of an ACID in H2O.

22. NaOH(s) -------> Na+(aq) + OH-(aq)
STRONG BASE
NaOH solution
ALL solute is ions
Str BASE dissociates 100% in H2O
H2O
NaOH(s) > Na+(aq) + OH-(aq)
(ionic)

23. HC2H3O2(l) + H2O(l) <---> H3O+(aq) + C2H3O2-(aq)
acetic acid
solution
WEAK ACID
mostly molecules
& very few ions
Weak ACID ionizes
only slightly in H2O
HC2H3O2(l) + H2O(l) <---> H3O+(aq) + C2H3O2-(aq)

24. Methods of Expressing Solution Concentration
1) Mass % = mass of solute x 100
(mass of solute + mass of solvent)
Mass %A = mass of component A ratio x
total mass of solution
“10% MgSO4(aq)”
means g MgSO g H2O = 100.g solution
ratio: g solute
100.g solution

25. OR you could find %H2O first, then do a proportion:
Problem: How many grams H2O are needed to prepare g of 6.80%NaCl(aq) ?
525g x 6.80% = 35.7g NaCl solute so
525g solution g NaCl = g H2O
solute
solvent
OR you could find %H2O first, then do a proportion:
93.20g = x g
100g g
100% % = 93.20%
x = 489.3g H2O

26. 2) Volume% = volume of solute x 100
(volume of solute + volume of solvent)
Volume%A = volume of component A x
total volume of solution
again it is a ratio
mL of A
100mL solution

27. Problem: How many liters of water are needed to prepare
1.00 liter of 23% by volume CH3OH(aq)?
1.00L x 23% = 0.23L CH3OH so
1.00L solution L CH3OH = 0.77L H2O
OR % % = 77%
77L = x L
100L L
“Proof” is a volume %.
Proof = 2 x volume % alcohol C2H5OH) also called “EtOH”
so “180 proof” whiskey means 90% ethanol by volume!!
solute
solvent

28. 3) Parts per million or parts per billion (ppm) (ppb)
used for very small solute concentrations
(for mass or volume)
ppm(mass) = mass of solute x 106 ppb….x mass of solution OR
ppm = mg solute ppb = µg solute
kg solution kg solution
“0.1ppm by mass of Pb2+ ions”
in drinking water means
0.1mg Pb2+ in 1kg water

29. Problem: Find ppm of 0.5g of Ca2+ ions in 2500g tapwater
ppm = 0.5g x 106
2500g
= 200ppm Ca2+

30. 4) MOLARITY (M) = moles of solute
Liter of solution
mol L M

31. Molarity
Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different.
Molarity is one way to measure the concentration of a solution:
moles of solute
volume of solution in liters
Molarity (M) =
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32. © 2012 Pearson Education, Inc.
Mixing a Solution
To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute.
The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.
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33. Problem: Find Molarity when 53. 3grams FeCl3 are
Problem: Find Molarity when 53.3grams FeCl3 are dissolved in 755mL of solution.
Fe –
Cl – 3(35.45)
g/mol
to convert
to moles M
= 53.3g x
1 mol
162.20g
0.755L
…….
_ _ _ 3
= M (means mol/L FeCl3)
Now find Molarity of chloride ions in this solution:
1 to 1 ratio
FeCl Fe Cl-
1 to 3 ratio
3(0.435M) = M Cl-

34. moles = Molarity x VolumeLiters
Problem: How many moles of K+ ions are present in mL of 2.50M K2O?
moles = Molarity x VolumeLiters
= (2.50M) (0.0250L)
= mol K2O
2 mol K+ ions
1 mol K2O
0.625mol K2O x
= mol K+ ions

35. mg mol kg L 1kg = 1L 1 L 1kg solution 2.0mg x g x 1 mol = mol
Calculate Molarity of 2.0ppm Cr6+ ions in well water. (find mol/L)
(in this very dilute solution, 1mL has a mass of ~1g)
2.0ppm means _______________
  ______ = ______
______ = ______
Convert: numerator ______----> ________
denominator _____----> ________
2.0mg Cr6+ ions
1kg solution
1kg H2O
1L H2O
1g H2O
1mL H2O mg
mol kg L
2.0mg
x g
x 1 mol
= mol
10-3 1
52.00
3.846 x 10-5 mg g
3.8 x 10-5mol
1kg = 1L
= 3.8 x 10-5M
1 L

36. 1.20mol LiF x mol 1 L 0.450L 0.540mol x 25.94 g 1 mol = 14.0g LiF 0.54
2. How much LiF is needed to prepare 450.mL of a 1.20M sol’n. (M is ratio of mol solute to L solution so find mol…then mol-->g)
_____________ = _______ or mol = M x L
1.20mol LiF
x mol
1 L
0.450L
Li – 6.94
F – 19.00
25.94 g/mol
X = 0.540mol
0.54
0.540mol
x g
1 mol =
14.0g LiF

37. g mol g L 0.0151mol Pb(NO3)2 0.0952L sol’n 5.00g x 1 mol = 331.20g
3. Find Molarity of a 5.00% Pb(NO3)2 solution by mass.
(find mol/L)
D = 1.05g/mL
5.00% Pb(NO3)2 means ______________
Convert: numerator _____----> ______
denominator _____----> ______
5.00g Pb(NO3)2
100g solution g
mol
Pb –
N – 2(14.00)
O - 6(16.00)
331.20g/mol g L
5.00g
x 1 mol =
331.20g
0.0151mol Pb(NO3)2
100g x mL
1.05g
x L =
1000mL
0.0952L sol’n

38. 0.0151mol
M =
0.0952L
= M
Pb(NO3)2(s) Pb2+(aq) NO3-(aq)
NONE of this ALL of this
in solution in solution
[Pb2+] = _____________
[NO3-] = _____________
[total ion] = _____________
H2O
0.159M M
(x 2)
0.477M
(x 3)

39. 1) Find total moles of iodide ions in 250.mL of a 0.750M BaI2 sol’n.
____________ = _________
0.750mol
x mol
1 L
0.250L
0.1875
x = 0.188mol BaI2
H2O
BaI2(s) > Ba2+(aq) I-(aq)
0.188mol
0.188mol
2(0.188)mol
0.376mol of I- ions

40. 9.82g 1 mol 0.600L 134.45g 2 mol Cl- 0.244mol Cl- 1 mol CuCl2 0.12173
2. Find [Cl-] when 9.82g CuCl2 are dissolved in 600.mL of sol’n.
M = ________ X __________
9.82g
1 mol
0.600L
134.45g
Cu –
Cl – 2(35.45)
134.45g/mol
= M CuCl2
0.122M CuCl2
2 mol Cl- =
0.244mol Cl-
x __________
1 mol CuCl2

41. 3. Which solution of strong electrolytes contains the
largest # of chloride ions?
A or B
50.0mL of 0.60M MgCl2
200.mL of 0.40M NaCl
mol = M x L
mol = M x L
= (0.60M)(0.0500L)
= (0.40M)(0.200L)
= 0.030mol MgCl2
= 0.080mol NaCl
MgCl2(s)---> Mg2+(aq) + 2Cl-(aq)
NaCl(s)---> Na+(aq) + Cl-(aq)
1 to 2
1 to 1
0.030mol
2(0.030)mol
0.080mol
0.080mol
0.060mol Cl-
0.080mol Cl-

42. (16M) (x mL) (2.5M) (1000.00mL) x = 160mL) add 160mL conc acid
4. How would you prepare:
1.00L of a 2.5M HNO3 solution from conc. (16M) nitric acid?
dilution formula:
Mconc x Vconc = Mdil x Vdil
(16M)
(x mL)
(2.5M)
( mL) =
x = 160mL)
add 160mL
conc acid
2) add H2O
to line
3) cap & mix
1000mL

43. © 2012 Pearson Education, Inc.
Dilution
One can also dilute a more concentrated solution by
Using a pipet to deliver a volume of the solution to a new volumetric flask, and
Adding solvent to the line on the neck of the new flask.
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44. © 2012 Pearson Education, Inc.
Dilution
The molarity of the new solution can be determined from the equation
Mc  Vc = Md  Vd,
where Mc and Md are the molarity of the concentrated and dilute solutions, respectively, and Vc and Vd are the volumes of the two solutions.
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45. 21.3g 21.25 0.250M is 0.250mol NaNO3 in 1Liter sol’n 0.250mol x 85.00g
How would you prepare:
1.00L of a 0.250M NaNO3 solution from the pure solid?
Na –
N –
O - 3(16.00)
85.00g/mol
0.250M is 0.250mol NaNO3 in 1Liter sol’n
0.250mol
x g
1mol =
21.3g
21.25

46. 5. Find [Li+] when 20.0mL of 1.0M LiCl is added to
80.0mL of 2.0M LiBr
1) find moles of each cmpd (solute)
2) find total moles of solute ions
20.0mL of 1.0M LiCl
(0.0200L)(1.0M) = 0.020mol LiCl
0.020mol Li+
80.0mL of 2.0M LiBr
0.16mol Li+
(0.0800L)(2.0M) = 0.16mol LiBr
100.0mL
total volume
0.18mol Li+
total
[Li+] = mol L
= 0.18mol
0.100L
= 1.8M

47. AD CB (s) ppt (l) H2O or other molec cmpd (WA) (g) GAS
TYPES OF REACTIONS IN AQUEOUS SOLUTION
precipitation (formation of insoluble solid product)
acid-base (formation of water, a neutralization reaction)
oxidation-reduction reaction (changing oxidation #s)
METATHESIS reaction = DOUBLE REPLACEMENT
AB CD > _____( ) + _____( )
make it go! AD CB
(s) ppt
(l) H2O or other
molec cmpd (WA)
(g) GAS
“spectator ions”
are (aq) product

48. Metathesis (Exchange) Reactions
Metathesis comes from a Greek word that means “to transpose.”
AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)
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49. Metathesis (Exchange) Reactions
Metathesis comes from a Greek word that means “to transpose.”
It appears as though the ions in the reactant compounds exchange, or transpose, ions:
AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)
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50. Precipitation Reactions
When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.
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51. 2 2 NaNO3 PbCl2 s Na+ NO3- Pb2+ Cl- Writing equations:
The “MOLECULAR” equation is a balanced equation with all formulas & state of matter symbols
Ex:
________( ) + _______ ( ) 2 2
NaNO3
PbCl2
1) NaCl(aq) + Pb(NO3)2(aq)----> aq s
Na+
NO3-
Pb2+
Cl-

52. Molecular Equation
The molecular equation lists the reactants and products in their molecular form:
AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)
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53. 2Na+(aq) + 2Cl-(aq) + Pb2+(aq) + 2NO3-(aq) -----> 2Na+(aq)
2) The “COMPLETE IONIC” equation shows all reactants and products separated into individual ions as they exist in solution:
strong acids, strong bases and soluble ionics
(remember SA,SB,SI)
Note: Never separate the ppt, H2O or gas
2Na+(aq)
+ 2Cl-(aq)
+ Pb2+(aq)
+ 2NO3-(aq)
----->
2Na+(aq)
+ 2NO3-(aq)
+ PbCl2(s)

54. Ionic Equation
In the ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions.
This more accurately reflects the species that are found in the reaction mixture:
Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq) 
AgCl(s) + K+(aq) + NO3−(aq)
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55. Pb2+(aq) + 2Cl-(aq) ----> PbCl2(s)
3) The “NET IONIC EQUATION” is missing the spectator ions, (cancelled)
showing only the particles involved in the reaction that made PRODUCTS. (made it go!!)
Net ionic equation: 3)
Pb2+(aq)
+ 2Cl-(aq)
----> PbCl2(s)

56. Net Ionic Equation
To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right:
Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq) 
AgCl(s) + K+(aq) + NO3−(aq)
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57. Net Ionic Equation
To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.
The only things left in the equation are those things that change (i.e., react) during the course of the reaction:
Ag+(aq) + Cl−(aq)  AgCl(s)
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58. Net Ionic Equation
To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.
The only things left in the equation are those things that change (i.e., react) during the course of the reaction.
Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions:
Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq) 
AgCl(s) + K+(aq) + NO3−(aq)
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59. Writing Net Ionic Equations
Write a balanced molecular equation.
Dissociate all strong electrolytes.
Cross out anything that remains unchanged from the left side to the right side of the equation.
Write the net ionic equation with the species that remain.
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60. BaSO4(s) BaSO4(s) Ba2+(aq) + SO42-(aq)
SHORTCUT TO NET IONIC EQUATIONS:
insoluble
BaSO4(s)
BaCl2(aq) + K2SO4(aq)
Net ionic equation:
______________________________________
soluble
spect ions
Ba2+(aq)
+ SO42-(aq)
BaSO4(s)

61. Fe3(PO4)2 3Fe2+(aq) + 2PO43-(aq) Fe3(PO4)2 insoluble
2) FeCl2 (aq) + Na3PO4(aq)
Net ionic equation:
______________________________________
soluble
spect ions
3Fe2+(aq)
+ 2PO43-(aq)
Fe3(PO4)2

62. H2O(l) H+(aq) + OH-(aq) H2O(l) water HCl(aq) + KOH(aq) soluble
Net ionic equation:
______________________________________
soluble
spect ions
H+(aq)
+ OH-(aq)
H2O(l)

63. Any COMPOUNDS containing these ions are SOLUBLE: “CLAAAN” ! Cl A A A N
SOLUBILITY RULES
Any COMPOUNDS containing these ions are SOLUBLE: “CLAAAN” !
Cl A A A N
Chlorates Acetates Ammonium Alkali metals Nitrates
ClO C2H3O NH Li NO3-
or K+
ClO4- CH3COO- Na+
Rb+
Cs+
Also all chlorides, bromides and iodides are SOLUBLE
except those of Pb2+, Ag+, Hg “PAH” !
All 8 strong bases (hydroxides) are SOLUBLE.

64. Precipitation Reactions in Aqueous Solution
Steps for Determining the Mass of Product (ppt) Formed:
1) Identify the substances present in the mixture of solutions
and determine what reaction occurs.
Write the balanced molecular equation with
state of matter symbols.
Calculate moles of each reactant given.(moles = M x VL )
Solve to find the “Limiting Reactant”.
5) Convert moles of product to grams of product.

65. 3 3 Fe(NO3)3 KOH Fe(OH)3 KNO3 Problem:
What mass of iron(III) hydroxide is produced when
35.0mL of 0.250M Fe(NO3)3 is mixed with
55.0mL of 0.180M KOH?
BAL. EQ:
__________ ________ > _________ +________ 3 3
Fe(NO3)3
KOH
Fe(OH)3
KNO3
(aq)
(aq)
(aq)
(s)
(0.250M)(0.0350L)
(0.180M)(0.0550L)
mol
mol
3 mol KOH
mol Fe(NO3)3
= mol KOH
1 mol Fe(NO3)3
Need this much KOH, but don’t have so Fe(NO3)3 in excess
And KOH is LR

66. ppt = 0.00330mol Fe(OH)3 x 106.88g = 0.353g Fe(OH)3 0.00330mol KOH
106.88g/mol
1 mol Fe(OH)3
= mol Fe(OH)3
mol mol KOH
3 mol KOH
x g =
1 mol
0.353g Fe(OH)3
mol

67. Now, calculate the Molarity of all ions left in solution:
Fe(NO3) KOH Fe(OH) KNO3
ppt!
mol mol mol
subscripts coefficients
BEFORE AFTER
Fe NO K OH Fe(OH) K NO3-
1 to 3
1 to 1
disregard!!
mol
mol
mol
mol
mol
mol
mol
spectator ions No CHANGE
Since KOH is LR, then only OH- ions got used up as
they formed Fe(OH)3 ppt. Some Fe3+ ions will be left over.

68. Now to make solving clearer, write “net ionic equation”:
& work backwards!!
Fe OH > Fe(OH)3
mol ppt formed
[Note: 1:1 ratio between
Fe3+mol used & Fe(OH)3 made]
Fe3+ given: Total Vol. Sol’n: mL
Fe3+ used: mL
Fe3+ left: mL =
______L
mol
used up to
make ppt
3( )mol
mol
ALL USED UP
mol
35.0
mol
55.0
mol
90.0
0.0900

69. Ions remaining in Sol’n:
K _________mol [K+] = _______
NO ________mol [NO3-] = _______
Fe _________mol [Fe3+] = ________
0.110M
0.11
0.0900L
0.292M
0.0900L
0.0606M
0.0900L

70. 3 3 When 200.0mL of 1.0M NaOH is mixed with
300.0mL of 0.50M AlBr3, find mass of ppt
& concentration of all ions left in solution. 3 3
NaOH
AlBr3
NaBr
Al(OH)3
__________ ________ > _________ +________
(aq)
(aq)
(aq)
(s)
(1.0M)(0.2000L)
(0.50M)(0.3000L)
0.20mol
0.15mol
NaOH ---> ppt
AlBr3 ---> ppt
3 mol = 1 mol
0.20mol x mol
1 mol = 1 mol
0.15mol x mol
x = mol ppt
x = 0.15mol ppt
LR = NaOH

71. Mass of ppt: 0.067mol x 78.01g = 5.2g Al(OH)3 Al(OH)3 5.22667 1 mol
78.01g/mol

72. LR Al3+ + 3OH- ------> Al(OH)3 BEFORE AFTER
Na OH Al Br Al(OH) Na Br-
0.20
mol
0.20
mol
0.15
mol
0.45
mol
0.067
mol
0.20
mol
0.45
mol
spectator ions LR
Al OH > Al(OH)3
0.067
mol
3(0.067)
0.20mol all used up
0.067
mol
Al3+ given: Total Vol. Sol’n: mL
Al3+ used: mL
Al3+ left: mL = ______L
0.15mol
200.0
mol
300.0
0.08mol
0.083
500.0
0.5000

73. Ions Conc left::
Na+ = _____mol [Na+] = _______
Br- = _____mol [Br-] = _______
Al3+ = ____mol [Al3+] = ________
0.40M
0.20
0.4
0.5000L
0.90M
0.45
0.9
0.5000L
0.16
0.2M
0.08
0.5000L

74. 15-30-15 means it contains “at least” 15% urea (NH2)2CO 30% P2O5
15% K2O

75. In this experiment, phosphorus in P2O5
will be determined by precipitation of the
sparingly soluble salt
magnesium ammonium phosphate hexahydrate according to the reactions:
1) P2O5(s) H2O(l) > 2 H3PO4 (aq)
2) H3PO4(aq) ----> 2H+(aq) HPO42-(aq)
3) 5H2O(l) + Mg2+(aq) + NH4+(aq)+ OH-(aq) + HPO42-(aq) > MgNH4PO4•6H2O(s)
we add water
non-metallic oxide
phosphoric acid
ionizes, the 1st H+ then the 2nd H+
We add NH3(aq) and MgSO4(aq)
PPT has captured all P atoms

76. If a 10.00g sample of soluble plant food yields
Sample Problem:
If a 10.00g sample of soluble plant food yields
10.22g of ppt. MgNH4PO4•6H2O(s),
calculate the %P and %P2O5 in the sample?
solve for grams of P in the sample:
using Law of Definite Proportions
245.44g MgNH4PO4•6H2O = 30.97g P
10.22g ppt x gP
molar mass ppt = molar mass P
actual mass ppt x grams P
X = g P

77. using Law of Definite Proportions
Thus %P = grams P x 100
g plant food
1.290g P_____ x = %P Ans!
10.00g plant food
2) solve for grams of P2O5:
using Law of Definite Proportions
2(30.97)g P = g P2O5
1.290 g P X g
X = g P2O5

78. Pre-Lab Questions: (on loose leaf & copy questions)
%P2O5 = g P2O5 x
g plant food
2.956g P2O5__ x = % P2O5 Ans!
10.00g plant food
Pre-Lab Questions: (on loose leaf & copy questions)
1. The label on a plant food reads Explain what this means.
2. What is %age of phosphorus element in the plant food in Question 1?
3. What is %age of potassium element in the plant food in Question 1?
4. What is the molar mass of MgSO4•7H2O known as Epsom salts?

79. neutralization HA BOH BA H2O ACID-BASE REACTIONS formation of water
are ______________________reactions.
The “driving force” behind neutralization reactions is the
___________________ or a__________________. !!!!
ACID BASE > SALT H2O
_____ _____ > _____ _____
neutralization
formation of water
weak electrolyte HA
BOH BA
H2O

80. produces H3O+ ions in water
Definitions:
Arrhenius acid _____________________________
Acid increases _________________________________
Ex: recall: HCl(g) H2O(l) > H3O+(aq) Cl-(aq)
simplified:
Arrhenius base _____________________________
Ex: recall: Ca(OH)2(s) > Ca2+(aq) OH- (aq)
produces H3O+ ions in water
[H+] (molarity of H+ ions) in solution
HCl(aq) > H+(aq) Cl-(aq)
releases OH- ions in water
H2O

81. is a proton acceptor (H+)
is a proton donor (H+)
Bronsted-Lowry acid ______________________________
Bronsted-Lowry base _____________________________
Ex: recall:
NH3(g) H2O(l) > NH4+(aq) OH- (aq)
_____ ______ ______ _______
is a proton acceptor (H+)
WEAK
BASE
BLB
BLA
Conj Acid
Conj Base
Conj Base

82. © 2012 Pearson Education, Inc.
Acids
The Swedish physicist and chemist S. A. Arrhenius defined acids as substances that increase the concentration of H+ when dissolved in water.
Both the Danish chemist J. N. Brønsted and the British chemist T. M. Lowry defined them as proton donors.
© 2012 Pearson Education, Inc.

83. © 2012 Pearson Education, Inc.
Bases
Arrhenius defined bases as substances that increase the concentration of OH− when dissolved in water.
Brønsted and Lowry defined them as proton acceptors.
© 2012 Pearson Education, Inc.

84. H2O KBr H+(aq) + Br-(aq) + K+(aq) + OH-(aq) ---> H2O(l) + K+(aq)
STRONG ACID-STRONG BASE
Molec. EQ:
Complete Ionic EQ:
Net Ionic EQ:
(spect will cancel)
H2O
________ + _____
HBr(aq) + KOH(aq)---->
(l)
KBr
(aq)
SA,SB,SI get pulled apart
H+(aq)
+ Br-(aq)
+ K+(aq)
+ OH-(aq)
--->
H2O(l)
+ K+(aq)
+ Br-(aq)
SAME for every SA/SB reaction.

85. © 2012 Pearson Education, Inc.
Acid-Base Reactions
In an acid–base reaction, the acid donates a proton (H+) to the base.
© 2012 Pearson Education, Inc.

86. Neutralization Reactions
Generally, when solutions of an acid and a base are combined, the products are a salt and water:
CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)
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87. H2O NaF HF(aq) + Na+(aq) + OH-(aq) ---> H2O(l) + Na+(aq) + F-(aq)
WEAK ACID-STRONG BASE
Molec. EQ:
Complete Ionic EQ:
Net Ionic EQ:
H2O
NaF
HF(aq) + NaOH(aq)---->
________ _____
(l)
(aq)
HF(aq)
+ Na+(aq)
+ OH-(aq)
--->
H2O(l)
+ Na+(aq)
+ F-(aq) WA
HF(aq)
+ OH-(aq)
--->
H2O(l)
+ F-(aq)

88. 2 2 H2O Mg(NO3)2 ---> 2H2O(l) 2H+(aq) + 2NO3-(aq) + Mg(OH)2(aq)
STRONG ACID-WEAK BASE
Molec. EQ:
Complete Ionic EQ:
Net Ionic EQ: 2 2
H2O
Mg(NO3)2
HNO3(aq) + Mg(OH)2(aq)---->
________ _________
(l)
(aq)
--->
2H2O(l)
2H+(aq)
+ 2NO3-(aq)
+ Mg(OH)2(aq)
+ Mg2+(aq)
+ 2NO3-(aq)
2H+(aq)
+ Mg(OH)2(aq)
--->
2H2O(l)
+ Mg2+(aq)

89. Ba(s) 2 H2(g) Ba(OH)2(aq) H2O(l) x 1 mol = 137.33g
FINDING MOLARITY OF IONS
5.25g of Ba metal is placed in enough water to make 45.0mL Find [OH-]
_________ + _______ > ______ _________
Find moles Ba reacted:
Find moles OH-:
Find Molarity of OH-: M = mol L
Ba(s) 2
H2(g)
H2O(l)
Ba(OH)2(aq)
1 mol > 1 mol Ba2+ ions
2 mol OH- ions
5.25g Ba
x 1 mol =
137.33g
mol Ba
mol Ba
x 2 mol OH- =
1 mol Ba
mol OH-
0.0764
[OH- ] = 1.70M
0.0450

90. TITRATION: is an analytical technique whereby a solution of known concentration and volume (titrant) is reacted with a solution of known volume (analyte) to determine its concentration.

91. © 2012 Pearson Education, Inc.
Titration
Titration is an analytical technique in which one can calculate the concentration of a solute in a solution.
© 2012 Pearson Education, Inc.

92. Neutralization equation:
MOLARITYacid ion x VOLUMEacid = MOLARITYbase ion x VOLUMEbase
MH+ x Vacid = MOH- x Vbase
moles of H+ = moles of OH-
H+(aq) + OH-(aq) > H2O(l)

93. (x) x = (0.472M)(43.2mL) x = 0.556M = [H+] = [HCl]
1. Find molarity of HCl if 36.7mL of acid is needed to titrate
43.2mL of 0.236M Ca(OH)2
Note: HCl ---> H Cl Ca(OH) > Ca OH-
MH+ x Vacid = MOH- x Vbase
[H+] = [HCl]
[OH-] = 2[Ca(OH)2]
[OH-] = 2(0.236M) = 0.472M
(x)
(36.7mL) =
(0.472M)
(43.2mL)
x = (0.472M)(43.2mL)
(36.7mL)
x = M = [H+] = [HCl]

94. MH+ x Vacid = MOH- x Vbase
2. What volume of 0.80M Ba(OH)2 is needed to neutralize
50.0mL of a 0.20M H3PO4 solution?
Note: H3PO4 ----> 3H+ + PO Ba(OH)2 ----> Ba OH-
MH+ x Vacid = MOH- x Vbase
[H+] = 3[H3PO4]
[OH-] = 2[Ba(OH)2]
[H+] = 3(0.20M) = 0.60M
[OH-] = 2(0.80M) = 1.60M
(1.60M)
(x)
(0.60M)
(50.0mL) =
18.75
x = 19mL Ba(OH)2

95. FINDING MOLARITY OF ALL IONS LEFT IN SOLUTION
When 12.5g KOH is added to 450.0mL of 0.250M H2SO4, find concentration of all ions left in solution following this neutralization reaction. Is solution acidic? basic/ or neutral?
1) Initially, find moles of base added & moles of acid present:
H2SO4(aq) KOH(s) > H2O(l) K2SO4(aq)
( )( )
moles moles
K –
O – H
56.11g/mol
12.5g KOH
x 1 mol =
56.11g
0.223mol
spect 2 2
0.250M
0.4500L
0.113
0.223

96. 1 mol = 2 mol 0.113mol x mol 2 mol = 2 mol 0.223mol x mol
2) Find LR and moles H2O(l) formed:
H2SO H2O
KOH H2O
1 mol = 2 mol
0.113mol x mol
2 mol = 2 mol
0.223mol x mol
x = 0.226mol H2O
x = 0.223mol H2O
LR = KOH

97. H+(aq) + OH-(aq) H2O(l) H+ SO42- K+ OH- -----> H2O K+ SO42-
3) Before & After moles of all ions:
H SO K OH > H2O K SO42-
4) Net Ionic EQ:
2(0.113)
0.226
mol
0.113
mol
0.223
mol
0.223
mol
0.223
mol
0.223
mol
0.113
mol
H+(aq)
+ OH-(aq)
H2O(l)
0.223
mol ALL
used up LR
0.223
mol
0.223
mol
H+ given: Total Vol. Sol’n: mL =
H+ used:
H+ left:
0.226 mol
450.0
0.223 mol
0.4500L
0.003 mol

98. 5) now find Molarity of all ions left is solution spectators:
  K _____mol = [K+] = L
SO _____mol = [SO42-] =
 excess ion when forming water:
  H ______mol = [H+] =
 Since acid ion is left over , the solution is therefore ______
  Formula: pH = - log [H+] so pH = ____
Acidic solutions: pH < 7
Basic solutions: pH > 7
Neutral solutions: pH = 7
0.223
0.496M
0.4500
0.113
0.251M
0.4500
0.003
0.007M
0.4500
acidic
2.2
= - log 0.007M _ 1
2.1549 _ 1

99. Oxidation-Reduction Reactions
An oxidation occurs when an atom or ion loses electrons.
A reduction occurs when an atom or ion gains electrons.
One cannot occur without the other.
© 2012 Pearson Education, Inc.

100. © 2012 Pearson Education, Inc.
Oxidation Numbers
To determine if an oxidation–reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.
© 2012 Pearson Education, Inc.

101. o o o o +1 +4 -3 -1 OXIDATION-REDUCTION REACTIONS “REDOX” H2 Fe C P4
OIL-RIG
OXIDATION is
REDUCTION is
A “redox” reaction always involves a
RULES for ASSIGNING OXIDATION NUMBERS:
1) Any free element is __
2) Monatomic ions is
loss of electrons & an increase in oxidation #.
gain of electrons & a decrease in oxidation #.
change in oxidation #s.
o o o o
H2 Fe C P4
the charge on the ion
Na+ Pb N Br-

102. IBr NF3 BrCl +1 -1 +1 -1 +3 -1 -2 -1 +1 -1 +1 -1 -1 +1 -1 +3 -1 +1 -1
3) H in a molecular cmpd is ___ H in a metal hydride is ___
4) O in a compound is ___ except peroxides is ___   
NaH FeH3 -2 -1
Na2O2 H2O2 -1
5) F in a compound is always ____
In ionic compounds, it is the ion’s charge anyway,
but in a molecular compound, F atom gets a because it has the highest EN of all atoms
(ie greatest attraction for electrons in a shared pair)
highest EN
halogen
gets
the -1
+1 -1
+3 -1
+1 -1
IBr
NF3
BrCl

103. Cr = +6 N= +3 -8 = -2 -4 = -1 charge CrO4 2- NO2 -
6) sum of oxidation #s in a compound is ___
sum of oxidation #s in a polyatomic ion is its _________
charge -2
Cr = +6
CrO4 2- -8
= -2 -2
N= +3
NO2 - -4
= -1

104. REACTION TYPES. Yes, ONLY IF oxid #s change
after the reaction!
1) Synthesis:
E E ---> C E C ---> C C C ---> C
__________ ___________ ___________
2) Decomposition:
C ---> E E C ---> E C C ---> C C
__________ __________ ____________
3) Single Replacement:
E C ----> E C _______________
Yes
Yes
Usually Not
Yes
Yes
Usually Not
Yes always

105. NO -?+1 0 +4-2 +1-2 Yes always 4) Double Replacement (metathesis):
C C ----> C + C ____________
5) Combustion: CxHy + O > CO H2O
__________________ NO
(2 ions switching places) -?
Yes always

106. reduction oxidation lose oxidize 0 0 +1 -2
A REDUCING AGENT causes __________ in another reactant,
but itself undergoes __________.
Metal elements love to _____ electrons (________) and
form (+) ions, so they are good reducing agents.
2Na(s) O2(g) > 2Na2O(s)
oxidation
lose
oxidize
Oxid. Agent: O2
oxid
Red. Agent: Na
red

107. oxidation reduction gain reduce
An OXIDIZING AGENT causes __________ in another reactant,
but itself undergoes __________.
Non-metal elements love to _____ electrons (_________)
and form (-) ions, so they are good oxidizing agents.
reduction
gain
reduce

108. Note: Atoms in POLYATOMIC IONS with HIGH oxidation #s are
______________________________________________
good reducers therefore strong oxidizing agents +1 -2
KMnO4
Mn = +7 +1 -8
= 0

109. It is possible to get a “non-integer” oxidation #:
Means
Fe3O4
+8/3 -2 4 +8 -8
= 0 -8 +8

110. +2 -2 +2-2 0 +4 -2 gain of oxygen loss of oxygen
Broader Definition: OXIDATION is the ______________
REDUCTION is the _______________
Ex: PbO(s) CO(g) > Pb(s) CO2(g)
PbO _______________ and _______________
CO _______________ and _______________
Common method for reducing metal oxides to get the metal element isolated.
gain of oxygen
loss of oxygen
reduces from +2 to 0
loses oxygen
oxidizes from +2 to +4
gains oxygen

111. (+12) + (+6) + (-1) = (+18) + (-1) + (0)
6FeCl2(aq) + 6HCl(aq) + NaClO3(aq) --> 6FeCl3(aq) + NaCl(aq) +3H2O(l)
6Fe2+(aq) + 12Cl-(aq) + 6H+(aq) + 6Cl-(aq) + Na+(aq) + ClO3-(aq) >
6Fe3+(aq) Cl-(aq) + Na+(aq) + Cl-(aq) H2O(l)
Net ionic EQ:
6Fe2+(aq)
+ 6H+(aq)
+ ClO3-(aq) --->
6Fe3+(aq)
+ Cl-(aq)
+ 3H2O(l)
(+12) (+6) (-1) = (+18) (-1) (0)
(+17) = (+17)
Oxidation:
6Fe2+
Everything balances:
elements, e-s & charges
---> 6Fe3+
+ 6e-
Reduction:
ClO3-
+ 6e-
--> Cl-

112. We can consider a balanced net ionic equation
as the sum of
an oxidation ½ reaction + a reduction ½ reaction

113. today
tomorrow

114. BALANCE by the Half-Reaction Method (see steps!)
MnO C2O > Mn CO2
oxidation reduction
(in acidic solution) 5 2
+ 2e- 2
C2O42- --> CO2
5e- +
8H+ +
MnO4- -->Mn2+
+ 4H2O
(-2) = (0)
(+8) + (-1) = (+2) (0)
(+7) = (+2)
5C2O42- --> 10CO2 + 10e-
10e- + 16H+ + 2MnO4- --> 2Mn2+ + 8H2O
5C2O H+ + 2MnO4- --> 10CO Mn2+ + 8H2O

115. 1) Redox Balancing in BASIC solution 0 +7 -2 +4 -2 +3 -2+1 + 3e- + 3e- 1)
Al MnO > MnO Al(OH)4-
oxidation reduction
4OH-+
4H2O+
Al-->Al(OH)4-
+4H+
4H++
MnO4- -->MnO2
+ 2H2O
+4OH-
+4OH-
+4OH-
(-4) + (0) + (0) = (-1) + (0)
(0) + (-1) = (0) +(0) + (-4)
4H2O
4H2O
+ 3e-
+ 3e-
4OH-+
4H2O+ Al
+ 4H2O
+ MnO4- -->
Al(OH)4- +4H2O+
MnO2
+ 2H2O
+ 2H2O +4OH-
Al + 2H2O + MnO > Al(OH)4- + MnO2

116. 2) 0 -1 -2+1 2 2 hypochlorite ion 2e- + + 2e- Cl2 ----> Cl- + OCl- 2)
Cl > Cl OCl-
oxidation reduction 2 2
4OH-+
2H2O+
Cl2---> OCl-
+4H++
2e- +
Cl > Cl-
4OH-
(0) = (-2)
(-4) + (0) + (0) = (-2) + (0)
4H2O
+ 2e-
4OH-+
2H2O +
Cl2
+ Cl > 2OCl-
+ 4H2O
+ 2Cl-
2H2O
4OH- +
2Cl > 2OCl-
+ 2H2O
+ 2Cl-

117. B- M AB A MB A+ N AB B AN 0 + - 0 + - 0 + - 0 + -
SINGLE-REPLACEMENT REACTIONS all are redox!!!
E C > E C
When Element is a METAL:
spect. ion_____
____ + _____ ----> ____ _____ B- M AB A MB
M oxidizes ↑ while metal ion A+ reduces↓
The metal element _________________________________
When Element is a NON-METAL:
spect. ion_____
____ _____ ----> ____ _____
The non-metal element ______________________________ _ A+ N AB B AN
N reduces ↓ while nonmetal ion B-↑
oxidizes

118. Problem: Zn(s) + AgNO3(aq) ---> _____ + _________
Eooxid = _____
Eored = _____
Eocell = _____ 2 2
Ag(s)
Zn(NO3)2(aq)
Spect ion: _____
NO3-
0.76V
0.80V
1.56V
yes
Complete ionic EQ:
____________________________________________________
Zn(s) + 2Ag+(aq) + 2NO3-(aq) -----> 2Ag(s) + Zn2+(aq) + 2NO3-(aq)
Net Ionic EQ:
Zn(s) + 2Ag+(aq) -----> 2Ag(s) + Zn2+(aq)