Aqueous reactions and solution stoichiometry

1
Chapter 4

• Aqueous reactions and solution stoichiometry

2
Aqueous Chemistry

• Virtually all chemistry that makes life possible
  occurs in solution
• Common tests for sugar, cholesterol, and iron are
  all done in solution.

3
Solution Vocabulary

• Solution Homogeneous mix of 2 substances
• Solvent dissolving medium (water, oil, liquid
  nitrogen), present in larger qty.
• Solute what is being dissolved (salt, sugar,
  sodium hydroxide), present in smaller qty.
• Aqueous Solution When water is the dissolving
  medium (Making Kool-Aid)
• Electrolytes substance that dissolves in water
  to yield a solution that conducts electricity.
  (salt water)

4
Soft drink (l)
H2O
Sugar, CO2
Air (g)
N2
O2, Ar, CH4
Soft Solder (s)
Pb
Sn
4.1
5
4.1 General Properties of Solutions

• One of the properties of water is its ability to
  dissolve MANY different substances.
• Polar unequal distribution of charges makes
  water polar, allowing it to dissolve solutes.

6
Ions dissolving in water

• Ionic substances such as salts dissolve in water
  to release cations () and anions
• (-).
• Ex NaCl -gt Na , Cl-
• The dissociation of NaCl is called hydration.

7

• Solvation- The process in which an ion or a
  molecule is surrounded by solvent molecules
  arranged in a specific manner.
• Hydration solvation with water as solvent

8
A Note On Dissociation

• You must memorize the charges or all the ions and
  poly atomic ions in order to predict how
  molecules will dissociate.
• If you dont know them by now you are in
  trouble. Make flash cards and study!

9
Solubility

• The amount of a substance that dissolves in a
  given volume of solvent at a given temperature.
• If ionic compounds are not greatly attracted to
  the ions in water then that compound will be less
  soluble in water.

10
What is not soluble in water?

• Like dissolves like
• In general polar (unequal distribution of
  charges) and ionic substances are expected to be
  more soluble in water than nonpolar substances.

11
Solubility of alcohol

• Hint (-anol) suffix alcohol. Ethanol (C2H5OH),
  methanol, isopropanol.

12
Electrolyte Vocabulary

• Ionization the process of adding or removing
  electrons from an atom or molecule. Which gives
  the atom a net charge.
• Complete Ionization
• substances that only exist as ions in solution

13
Strong and Weak electrolytes

• Strong electrolyte ionize completely in water
  (single arrow). Are good conductors of
  electricity
• Ex Strong acids, strong bases and salts.

14
Weak Electrolytes

• Ionize partially in water (double arrows). Poor
  conductors of electricity
• Ex weak acids (acetic acid shown), weak bases,
  partially soluble salts.

And thus The reaction is reversible
15
Non-electrolytes

• Do not IONIZE in water or conduct electricity.
  They are covalent compounds that are not acids or
  bases.
• Dissolve in water as molecules instead of ions.
  Sugar (C12H22O11)

16
Identify the strong and weak electrolytes.
17

• A simple device to demonstrate the electrical
  conductivity of an ionic solution.
• (a) A solution of NaCl conducts electricity
  because of the movement of charged particles
  (ions), thereby completing the circuit and
  allowing the bulb to light.
• (b) A solution of sucrose does not conduct
  electricity or complete the circuit because it
  has no charged particles. The bulb therefore
  remains dark.

18
An electrolyte is a substance that, when
dissolved in water, results in a solution that
can conduct electricity.
A nonelectrolyte is a substance that, when
dissolved, results in a solution that does not
conduct electricity.
4.1
19
Molecular Compounds
20
Homework

• Chang pg 157 s 1 2 4 7 11
• BL Pg 145 s 3, 5, 6, 7, 8

21
4.2 Precipitate Reactions

• Reactions that result in the formation of an
  insoluble substance.
• (-) anion
• () cation

Ag(NO3)2 (aq) 2NaI (aq) ? AgI2 (s) 2NaNO3 (aq)
22
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23
Question

• How would each molecule dissociate? Label cations
  and anions.
• NiSO4 Ca(NO3)2 Na3PO4 Al2(SO4)3

24
Answer

• Ni 2 SO4 2-
• Ca2 (NO3)2 -
• Na3 PO43-
• Al23 (SO4)32-

25
Solubility Rules to Memorize

• You must memorize the following rules.
• You will have a pop quiz on solubility rules in
  the next week.
• If you do not pass with 85 you must write all
  the rules out 5 times each.

26
Solubility Rules (memorize)

• Solubility Rules (memorize)
• 1. NH4 and alkali metal (group IA) salts are
  soluble.
• 2. Nitrate, NO3-, acetate C2H3O2-, chlorate,
  ClO3-, perchlorate ClO4 salts are soluble.
• 3. Chloride, Cl-, bromide, Br-,iodide, I-, salts
  are soluble.
• EXCEPT Ag, Hg22, Pb2 (AgBr, Hg2I2, PbCl2)
• 4. Sulfate, SO42- , salts are soluble.
• EXCEPT PbSO4, HgSO4, CaSO4, BaSO4, AgSO4, SrSO4
• 5. Most Hydroxide, OH- ,salts are slightly
  soluble.
• Hydroxide salts of Group I elements are soluble
  (Li, Na, K, Rb, Cs, Fr).
• Hydroxide salts of Group II elements (Ca, Sr, and
  Ba) are slightly soluble.
• Hydroxide salts of transition metals and Al3 are
  insoluble. Thus, Fe(OH)3, Al(OH)3, Co(OH)2 are
  not soluble.

27
Use Solubility Rules to Classify as Soluble or
Insoluble

• Sodium carbonate Na2CO3
• Lead Sulfate PbSO4
• Cobalt (II) hydroxide
• Barium nitrate
• Ammonium phosphate

28
Answer

• Sodium carbonate Na2CO3 Soluble (1)
• Lead Sulfate PbSO4 Insoluble (4)
• Co(OH)2 Insoluble (5)
• Ba(NO3)2 Soluble (2)
• Ammonium phosphate (NH4)3PO4
• New Polyatomic PO4-3 Phosphate

29
Goal

• Use solubility rules to predict whether a
  precipitate will form when electrolytic solutions
  are mixed.
• Hint find products that have insoluble salt(s).
  This implies a precipitate reaction.

30
Example

• Predict what will happen when the following pairs
  of solutions are mixed. (Hint break it into
  ions)
• KNO3(aq) and BaCl2 (aq)
• Na2SO4(aq) and Pb(NO3)2(aq)

31
Answer

• KNO3(aq) and BaCl2 (aq)
• Reactants K NO3- Ba2 Cl- ?
• Products KCl Ba (NO3)2
• Both are soluble according to the rules and thus
  no precipitate forms.
• Rule 1 Rule 2

32

• Na2SO4(aq) and Pb(NO3)2(aq)
• Reactants Na SO42- Pb2 NO3- ?
• Products NaNO3 PbSO4
• NaNO3 Soluble according to rule 1.
• PbSO4 Insoluble according to rule 4

33
Predicting Precipitates

• Predict the precipitate that forms when solutions
  mix and write a balanced chemical equation.
• BaCl2 (aq) K2SO4 (aq) ?
• Fe2(SO4)3 (aq) LiOH (aq) ?

34
Answer

• BaCl2 (aq) K2SO4 (aq) ? BaSO4 (s) 2KCl (aq)
• Fe2(SO4)3 (aq) 6LiOH (aq)? 2Fe(OH)3 (s)
  3Li2SO4 (aq)

35
Describing reactions

• In this section we will talk about the types of
  equations used to represent reactions in
  solution.
• Three types of equations
• Molecular
• Complete ion
• Net ionic

36
Precipitation Reactions
Precipitate insoluble solid that separates from
solution
molecular equation
ionic equation
net ionic equation
Na and NO3- are spectator ions
4.2
37
Molecular Equations/Chemical equation

• Shows the complete chemical formula for reactants
  and products
• HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
• Strong acid Strong Base
  Soluble Salt weak electrolyte
• Strong elect. Strong elect.

38
Complete Ionic Equation

• Shows the formula of cations and anions for ionic
  compounds.
• H(aq) Cl-(aq) Na(aq) OH-(aq) ? Na(aq)
  Cl-(aq) H2O(l)
• Water (l) and solid (s) precipitates do not break
  down into ions
• Solids on reactant side can break down in water
  (i.e adding salt to water)

39
Net Ionic Equation

• Includes only those solutions components directly
  involved in the reaction.
• H(aq) OH-(aq) ? H2O(l)
• Spectator ions ions that appear in identical
  forms in reactants and product side of a chemical
  rxn that do not participate in the rxn directly.
• NOTE The net ionic equation of any strong
  acid-base neutralization rxn is always like the
  above rxn.

40
Write the three types of equations for the
following rxn.

• Pb(NO3)2 (aq) 2KI (aq) ?

41

• Molecular
• Pb(NO3)2 (aq) 2KI (aq) ? PbI2 (s) 2KNO3
  (aq)
• Complete Ionic
• Pb 2 (aq) 2NO3-1 (aq) 2K (aq) 2I- (aq)?
  PbI2 (s) 2K (aq) 2NO3- (aq)
• Note subscripts that a re multiplied through
  become coefficients
• Coefficients apply to all atoms in a compound
  
  
• Net Ionic
• Pb 2 (aq) 2I- (aq) ? PbI2 (s)

42
Writing Net Ionic Equations

• Write the balanced molecular equation.
• Write the ionic equation showing the strong
  electrolytes completely dissociated into cations
  and anions.
• Cancel the spectator ions on both sides of the
  ionic equation
• Check that charges and number of atoms are
  balanced in the net ionic equation

4.2
43
4.2 Homework

• Chang pg 156-157 s 9,10,12,15,18,19,21, 22,
• BL Pg 145 s 11, 12, 14, 15, 16, 19

44
4.3 Acids

• Substances that are able to donate a hydrogen ion
  (H) and increase H in aqueous solutions.

45
Properties of acids

• Sour taste
• Acids neutralize bases
• Acids corrode active metals
• Acids release a hydrogen ion into water (aqueous)
  solution
• Strong acids conduct electricity

46
Properties of Bases

• Bitter taste
• Slippery feel
• Bases denature protein
• Bases neutralize acids
• Bases release a hydroxide ion into water solution
  
• Strong bases conduct electricity

47
Bases

• Are soluble ionic compounds containing hydroxide
  ion (OH-).
• When dissolved in water the cations and OH- ions
  separate and move independently.

48
Strong electrolytes

• All of the strong bases and acids that you
  memorized are strong electrolytes because they
  ionize completely.
• All weak acids and bases are weak electrolytes
  because they only partially ionize.

49
Salts

• Another name for an ionic compound. When a salt
  dissolves in water, it breaks up into its ions,
  which move about independently.
• ionic substances are electrolytes
• Result from acid base neutralizations

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50

• Classify the following dissolved substances as
  strong or weak electrolytes
• CaCl2
• HNO3
• C2H5OH
• HCHO2

51

• CaCl2 ionic strong electrolyte
• HNO3 strong acid strong electrolyte
• C2H5OH molecular nonelectrolyte
• HCHO2 molecular nonelectrolyte

52
Neutralization Rxn

• Neutralization when an acid and a base mix and
  their products share no characteristics with
  their reactants. (i.e acid base qualities)

53
Neutralization Rxn

• Neutralization reaction between an acid and a
  metal hydroxide produce water and a salt.
• HCl (aq) NaOH (aq) ? H2O (l) NaCl (aq)
• (acid) (base) (water)
  (salt)

54

• Molecular equation
• HCl (aq) NaOH (aq) ? H2O (l) NaCl (aq)
• (acid) (base) (water)
  (salt)
• Complete Ionic Equation
• H Cl- Na OH- ? H2O Na Cl-
• Net Ionic Equation
• H OH- ? H2O

55
Neutralization Reaction
4.3
56
Question

• Write a balanced complete chemical equation for
  the reaction between aqueous solutions of acetic
  acid and barium hydroxide.

57
Answer

• Complete chemical equation
• We are given an acid and a base (metal hydroxide)
  so the result should be water and a salt
• 2HC2H3O2 (aq) Ba(OH)2 (aq) ? 2H2O (l)
  Ba(C2H3O2)2 (aq)

58

• Now Write the complete ionic equation
• Steps
• Determine if (aq) solutions are strong or weak
  electrolytes to see how they will dissociate.
• 2HC2H3O2 (aq) Ba(OH)2 (aq) ? 2H2O (l)
  Ba(C2H3O2)2 (aq)
• (weak elect) (strong Base/elect
  (strong elect
  
• weak acid Rule 5 exception)
  ionic salt)

59
Complete Ionic Equation

• 2HC2H3O2 (aq) Ba(OH)2 (aq) ? 2H2O (l)
  Ba(C2H3O2)2 (aq)
• 2HC2H3O2 (aq) Ba2 (aq) 2OH- (aq) ? 2H2O
  (l) Ba 2 (aq) 2C2H3O2- (aq)
• Note subscripts become coefficients

60
Now write the Net Ionic Equation

• 2HC2H3O2 (aq) Ba2 (aq) 2OH- (aq) ? 2H2O
  (l) Ba 2 (aq) 2C2H3O2- (aq)
• Net Ionic Equation
• 2HC2H3O2 (aq) 2OH- (aq) ? 2H2O (l) 2C2H3O2-
  (aq)
• Simplify coefficients for final answer
• HC2H3O2 (aq) OH- (aq) ? H2O (l) C2H3O2- (aq)

61
Acid Base reactions forming Gases

• Sulfide ion and carbonate ion react with acids
  to form gases with low solubility's in water.
• 2HCl (aq) Na2S (aq) ? H2S (g) 2NaCl (aq)
• HCl (aq) NaHCO3 (aq) ? NaCl (aq) H2O (l)
  CO2 (g)

62
Homework

• BL Pg 146 s 23, 24, 25, 27, 29, 31

63
4.4 Oxidation Reduction Reactions

• Rxns in which one or more electrons are
  transferred between reactants.
• Ex 2 Na Cl2 ? 2NaCl
• neutral neutral Na
  Cl-

64
Symantics

• Charges are written sign 1- 3
• Oxidation numbers are written sign -1 3

65

• Photosynthesis is a redox rxn
• Most energy producing rxns are redox rxns. Such
  as combustion rxn (fuel)
• Rusting of iron is a redox rxn
• Ca (s) 2H (aq) ? Ca2 (aq) H2 (g)

66
Oxidation

• An atom ion, molecule, becomes more positive it
  has lost electrons.
• We say that that atom, ion, molecule has been
  oxidized.
• Ca (s) 2H (aq) ? Ca2 (aq) H2 (g)

67
Reduction

• When an atom, ion, molecule has become more
  negatively charged its has gained electrons.
• We say this atom, ion, molecule has been reduced.
• Ca (s) 2H (aq) ? Ca2 (aq) H2 (g)

68
Oxidation reduction

• When one atom loses electrons it is gained by the
  other atom involved in the reaction.
• Ca (s) 2H (aq) ? Ca2 (aq) H2 (g)

69
Oxidation half-reaction (lose e-)
Reduction half-reaction (gain e-)
4.4
70
Oxidation numbers

• Helps us keep track of the electrons being gained
  and lost.
• Oxidation number is the actual charge of the atom
  if it is a mono-atomic ion, other wise it is the
  hypothetical charged assigned to the atom using a
  set of rules.

71
The Rules

• 1. The rule is that the cation is written first
  in a formula, followed by the anion.
• Example in NaH, the H is H- in HCl, the H is
  H.
• -
  -
• 2. The oxidation number of a free element is
  always 0.
• Example The atoms in He and N2, for
  example, have oxidation numbers of 0.
• 3. The oxidation number of a monatomic ion equals
  the charge of the ion.
• Example oxidation number of Na is 1 the
  oxidation number of N3- is -3.
• 4. The oxidation number of oxygen in compounds is
  usually
• -2.

72

• 5. The oxidation number of a Group 1 element in a
  compound is 1.
• 6. The oxidation number of a Group 2 element in a
  compound is 2.
• 7. The oxidation number of a Group 3 element in a
  compound is 3.
• 8. The oxidation number of a Group 7 element in
  a compound is -1, except when that element is
  combined with one having a higher
  electronegativity.

73

• 9. The sum of the oxidation numbers of all of the
  atoms in a neutral compound is 0.
• EX CO2 we know O -2 and there are 2 Os and
  CO2 is neutral so
• C 2(-2) O
• C 4
• 10. The sum of the oxidation numbers in a
  polyatomic ion is equal to the charge of the ion.
  
• EX the sum of the oxidation numbers for SO4 2-
  is -2.
• S 4(-2) - 2
• S 6

74
Oxidation number
The charge the atom would have in a molecule (or
an ionic compound) if electrons were completely
transferred.

• Free elements (uncombined state) have an
  oxidation number of zero.

Na, Be, K, Pb, H2, O2, P4 0

• In monatomic ions, the oxidation number is equal
  to the charge on the ion.

Li, Li 1 Fe3, Fe 3 O2-, O -2

• The oxidation number of oxygen is usually 2. In
  H2O2 and O22- it is 1.

4.4
75

• The oxidation number of hydrogen is 1 except
  when it is bonded to metals in binary compounds.
  In these cases, its oxidation number is 1.

• Group IA metals are 1, IIA metals are 2 and
  fluorine is always 1.

6. The sum of the oxidation numbers of all the
atoms in a molecule or ion is equal to the charge
on the molecule or ion.
7. Oxidation numbers do not have to be integers.
Oxidation number of oxygen in the superoxide ion,
O2-, is -½.
HCO3-
O -2
H 1
3x(-2) 1 ? -1
C 4
4.4
76

• 10. The sum of the oxidation numbers of all of
  the atoms in a neutral compound is 0.
• EX CO2 we know O -2 and there are 2 and CO2
  is
• neutral so
• C 2(-2) O
• C 4
• 11. The sum of the oxidation numbers in a
  polyatomic ion is equal to the charge of the ion.
  
• EX the sum of the oxidation numbers for SO4 2-
  is -2.
• S 4(-2) - 2
• S 6

77
The oxidation numbers of elements in their
compounds
4.4
78
ExamplesHINT start with what you know for
sure

• H2S
• Neutral molecule so all oxidation numbers must
  add up to 0.
• Let X the oxidation number of S. H has an
  oxidation number of (1)2
• X 2(1) 0 so charge of S -2

79

• S8
• In elemental form so oxidation number is 0 (rule
  1)

80

• SCl2
• This is a binary compound. We expect Cl to have
  an oxidation number or -1.
• The sum of the ox s must equal zero because
  this is a neutral compound.
• X 2(-1) 0 X 2

81

• Na2SO3
• Oxidation numbers of Alkali metals always have an
  oxidation number of 1 in compounds. Oxygen has a
  common oxidation state of -2.
• Let x number of S
• 2(1) X 3(-2) 0
• X 4

82
Assign Oxidation numbers for

• CO2
• SF6-
• NO3-

83

• CO2 C 4 O
  -2(2)
• SF6- S 5
  F -1(6)
• NO3- N 5 O
  -2(3)
• 5 -6 -1

84
Oxidations of metals by acids and salts

• General pattern
• A BX ? AX B
• Zn (s) 2HBr (aq) ? ZnBr2 (aq) H2 (g)
• 0 1 -1 2 -1(2) 0
• Gain e- reduced
• Lose e- oxidized
• Stays the same

85
Fe 3(2) 6 O -2(3) -6
Oxidation for free elements is zero
86

• L - E -O
• Lose electrons Oxidized
• The Lion Says
• G E - R
• Gain Electrons Reduced

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87
Leo the Lion Says Ger!!!
OXIDATION
REDUCTION
88

• In this reaction, oxygen maintains a -2 oxidation
  number throughout.
• Iron becomes reduced from 3 to 0 oxidation
  number, while carbon becomes oxidized from 0 to
  4 state.
• The species that becomes reduced is called the
  oxidizing agent since it is accepting electrons
  from some other species. Conversely, the species
  that becomes oxidized is called the reducing
  agent since it is giving up electrons.

89
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90

• Identify the atoms being oxidized and reduced as
  well as the oxidizing and reducing agents.
• 2Al 3I2 ? 2AlI3

91

• 2Al 3I2 ? 2AlI3
• 0 0 3 -1
• Since each Al atom changes from a 0 ? 3
• Al is Oxidized
• Iodine is reduced 0 -1
• Al donates the electrons so it is the reducing
  agent
• I2 accepts the electrons and it is the
  oxidizing agent

92
Zn is the reducing agent
Zn is oxidized
Cu2 is reduced
Cu2 is the oxidizing agent
Ag is reduced
Ag is the oxidizing agent
4.4
93
Activity Series

• We need to know what metals are most likely to
  oxidize others.
• Example We cant store nickel nitrate in an iron
  container because the solution would eat through
  the container.

94
Activity Series

• A list of metals arranged in order of decreasing
  ease of oxidation.
• Page 139 table

95
Using activity series

• Any metal on the list can be oxidized by the
  metal below it.
• Give FeCl2 Mg
• Find will iron oxidize
• Magnesium metal?
• I finger on Fe
• 1 finger on Mg
• Is the bound chemical below
• Yes Fe is below Mg.
• Then complete the reaction

96

• Give NaCl2 Mg
• Find will sodium oxidize
• Magnesium metal?
• I finger on Na
• 1 finger on Mg
• Is the bound chemical below
• no
• Then the reaction is not possible

97
The Activity Series for Halogens
F2 gt Cl2 gt Br2 gt I2
Halogen Displacement Reaction
0
-1
-1
0
4.4
98
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99
Homework

• Chang Pg 158-159 s 43,54,55,56
• BL146-147
• s 35, 39, 41, 42, 44, 45,

100
Concentrations of Solutions 4.5

• A solution is a homogeneous mixture of two or
  more substances. One of these substances is a
  solvent the other is the solute.
• Solvent component in greater quantity
• Solute component in lesser quantity.
• Solution solvent solute
• Concentration amount of solute dissolved in a
  given amount of solution. Units vary.

101
Molarity (M)

• Number of moles of solute in one liter of
  solution
• Molarity moles of solute
• liter of solution
• Units mol/L M

102
A B
C D

• To make 250 mL (0.250 L) of 1.00 M CuSO4
• Use the formula grams needed Molecular weight
  x Volume x Molarity
• g
  159.6 x 0.250 x 1 39.9 g CUSO4
• Place chemical into flask and add a small
  quantity of water to dissolve.
• Mix solution
• Bring total volume up to 0.250 L

103
Question

• Calculate the molarity of a solution made by
  dissolving 5.00g of C6H12O6 (MW 180 amu) in
  sufficient water to form 100 ml solution.
• Recall M mole/ L

104
Answer

• 5.0 g 1mol 0.027 mol
• 180g
• M 0.027 0.27M
• 0.1

105
Question

• How many grams of Na2SO4 are there in 5ml of 0.50
  M Na2SO4.

106
Answer

• 0.05 M X 0.0075 mol
• 0.015
• 0.0075 mol 142 g Na2SO4 1.07 g Na2SO4
• 1 mol
• Or use g M.W x V x L

107
Finding Concentration of one type of atom

• We can find the concentration of one type of atom
  in a molecule by multiplying the molarity of the
  solution by that number of atoms.

108
Question

• Which of the following solutions of strong
  electrolytes contains the largest concentration
  of chloride ions
• A.0.30 M AlCl3
• B. 0.60M MgCl2
• C. 0.40 NaCl

109
Answer

• 0.30(3 Cl) 0.90 M Cl
• 0.60 (2Cl) 1.2 M Cl
• 0.40 ( 1 Cl) 0.4 M Cl
• MgCl2 gt AlCl3 gt NaCl

110
Dilutions

• Moles before dilution moles after dilution
• M1V1 M2V2
• How many milliliters of 5.0 M K2Cr2O7 solution
  must be diluted in order to prepare a 250 mL of
  0.10M solution.
• NOTE All Volumes must be in L to satisfy units
  of Molarity!!!!!!

111
Answer

• 250 mL .250 L
• 5.0 M stock ( V1) stock (0.10 M)want 0.250 L
  want
• V1 .005 L stock
• To make this solution we will add 0.005 L of
  stock to 0.245 L of water to make a 0.10M
  solution.

112
Homework Sections

• Chang pg 159 s 59,60,61,63,69, 74
• Page 147
• s 49, 51, 52, 54, 56, 59, 60, 61

113
4.7 Stoichiometry of precipitation reactions

• 2 differences
• its hard to predict products in solution, so we
  need to think and remember the rules.
• To obtain moles of reactants we must use the
  volume of the solution and its molarity. Recall M
  mol
• 1L

114
Question

• How many moles are in 1 L of 0.3M solution?
• How many moles are in 6.9L of 0.45M solution?

115
Example

• What is the mass of NaCl solid that must be added
  to 1.50L of a 0.100M AgNO3 solution to
  precipitate all the Ag ions in the form of AgCl
  according to the balanced equation below?
• NaCl AgNO3 ? AgCl NaNO3

116

• Apply solubility rules
• Products AgCl NaNO3
• NaNO3 is soluble (rule 1) AgCl is insoluble (rule
  3)
• Forming a solid.
• SO
• Lets add enough Cl- ions (form NaCl) to react
  with all of the Ag (from AgNO3) to form a
  precipitate of AgCl. But how many moles of AgNO3
  do we have?

117

• Given
• 1.50L 0.100M AgNO3
• we know from our homework that that there is
• 0.1M(1 Ag) 0.1M Ag
• 0.1M( 1 NO3) 0.1M NO3
• We can use our molarity (moles per one liter) and
  how many Liters we are given to find how many
  moles we have in that volume.
• 1.5 L AgNO3 0.100 mol AgNO3 0.150 mol
  AgNO3in 1.5L
• 1L AgNO3

118

• Because AgNo3 and NaCl react in a 11 ratio
  (see rxn)
• NaCl AgNO3 ? AgCl NaNO3
• 0.150 mol AgNO3 are present (we just solved for
  that) thus 0.150 mol NaCl are needed. But we need
  to know grams not moles.
• 0.150 mol NaCl 58.45g NaCl 8.77g NaCl
• 1 mol NaCl

119
Question

• When aqueous solutions of Na2SO4 and Pb(NO3)2 are
  mixed, PbSO4 precipitates. Calculate the mass of
  PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and
  2.00L of 0.0250 M Na2SO4 are mixed?

120

• Step 1
• Write a balanced equation and apply solubility
  rules
• Na2SO4 (aq) Pb(NO3)2 (aq) ? PbSO4 (s)
  2NaNO3 (aq)
• We need to find the mass of PbSO4 (s) formed.
  Where is the Pb and SO4 coming from?
• Which one is going to limit us?
• How do I find this answer?

121

• Step 2
• Find limiting reactant g ? mol ? mol ?g
• Given
• 1.25 L of 0.0500 M Pb(NO3)2
• 2.00L of 0.0250 M Na2SO4
• Pb2(aq) SO42-(aq) ? PbSO4 (s)
• 1.25 L Pb(NO3) x 0.0500 mol Pb(NO3) 0.0625
  mol Pb2 formed
• 1L
• 2.00 L Na2SO4 x 0.0250 mol Na2SO4 0.0500
  mol SO42- formed
• 1 L

122

• Step 3
• calculate moles of product PbSO4 that can be
  formed according to how much LR we have been
  given. ( g-mol-mol-g)
• .05 molNa2SO4 x 1 mol PbSO4 0.05mol PbSO4
• 1 mol Na2SO4
• only 0.0500 mol of solid PbSO4 will be formed
  since we only have 0.0500 mole of S04 available
  to us because it is the limiting recatant.

123

• Step 4
• convert moles to grams because that is the units
  the question wants us to report our answers in
• The mass of PbSO4 formed can be calculated using
  the molar mass of PbSO4 (303.3g/mol)
• 0.0500 mol PbSO4 303.3g PbSO4 15.2 gPbSO4
• 1mol PbSO4

124
With your partner solve

• What mass of NaCl is needed to precipitate all of
  the silver ions from 20.0 mL of 0.100 M AgNO3
  solution?

125
Example neutralization rxn

• What volume of 0.100 M HCL solution is needed to
  neutralize 25.0 ml of 0.350 M NaOH?
• Step 1 list reactants
• HCL NaOH ?
• H Cl- Na OH- ?

126

• Step 2
• What possible rxns will occur
• HCl (aq) NaOH (aq) ? NaCl (aq) H2O (l)
• Na Cl- ? NaCl (soluble and cant neutralize)
• H OH- ?H2O (insoluble and can neutralize)

127

• Step 3 write a balanced net ionic equation and
  calculate moles of reactant needed.
• H OH- ?H2O
• 0.025 L NaOH 0.035 mol OH 8.75 x 10-1 mol OH-
• 1 L NaOH
• We do not need to determine limiting reactant
  since the addition of H ions react exactly with
  the OH- present in a11 ratio. Thus 8.75 x 10-1
  mol H is required to neutralize the solution.

128

• Convert to volume required to neutralize the rxn.
  
• Volume X 0.100 mol H 8.75 x 10-1 mol H
  
• 1L
• 8.75 x 10-2 L of .100M HCL is required to
  neutralize 25.0 mL of NaOH

129
Titrations

• A procedure used for determining the
  concentration of an acid or a base in a solution
  by addition of a base or an acid of a known
  concentration.
• We know the solution is at its end point or
  stoichiometric point when the indicator changes
  color.

130
Titration vocabulary

• Standard solution Solution with a known
  concentration.
• Equivalence point when unknown solution and
  standard solution are at the same concentration.
• Indicators help establish equivalence point by a
  color change.

131
Equivalence point the point in the titration
when exactly enough base is added to neutralize
the acid.
132
Indicators

• Organic dyes that change color as they go from an
  acidic solution to basic solution.

133
Titration Question

• What volume of 0.25M HNO3 is required to titrate
  (neutralize) a solution containing 0.200 g of
  KOH.

134

• Given
• 0.200 g of KOH.
• 0.25M HNO3
• Find volume of 0.25M HNO3 required to
  neutralize 0.200 g KOH.
• KOH HNO3 ? KNO3 H2O
• Grams to moles to moles to liters
• 0.2 g KOH 1 mol KOH x 1 mol HNO3 x 1 L HNO3
  0.014 L HNO3
• 56.1 g KOH 1 mol KOH 0.25
  mol HNO3

135
Homework

• Pg 148
• s 65, 67, 69, 73
• Bonus 3 pts number 76 must be turned in to the
  box tomorrow first thing. On a piece of a paper
  with all units clearly worked out.
• S.O.S