Transcript Chapter 4 Notes

Chapter 4
Aqueous reactions and
solution stoichiometry
1
Aqueous Chemistry

Virtually all chemistry that makes life
possible occurs in solution

Common tests for sugar, cholesterol, and
iron are all done in solution.
2
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)
3
Solution
Solvent
Solute
H2O
Sugar, CO2
Air (g)
N2
O2, Ar, CH4
Soft Solder (s)
Pb
Sn
Soft drink (l)
4
4.1
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.

5
Ions dissolving in water
Ionic substances such as
salts dissolve in water to
release cations (+) and
anions
(-).
Ex: NaCl -> Na+ , Cl-
The dissociation of NaCl is
called hydration.
6

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
7
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 don’t know them by now you are in
trouble. Make flash cards and study!
8
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.
9
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.
10
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

12
Strong and Weak electrolytes

Strong electrolyte: ionize completely in
water (single arrow). Are good conductors
of electricity
 Ex:
Strong acids, strong bases and salts.
13
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
14
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)

15
Identify the strong and weak
electrolytes.
16
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.
nonelectrolyte
weak electrolyte
strong electrolyte
18
4.1
Molecular Compounds
19
Homework

Chang pg 157 # ‘s 1 2 4 7 11

BL Pg 145 #’s
3, 5, 6, 7, 8
20
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)
21
22
Question
How would each molecule dissociate? Label
cations and anions.
NiSO4
Ca(NO3)2
Na3PO4
Al2(SO4)3
23
Answer
Ni 2+ SO4 2Ca2+ (NO3)2 Na3+ PO43Al23+ (SO4)3224
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.
25
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.
26
Use Solubility Rules to Classify as
Soluble or Insoluble
Sodium carbonate Na2CO3
 Lead Sulfate PbSO4
 Cobalt (II) hydroxide
 Barium nitrate
 Ammonium phosphate

27
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
28
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.
29
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)
30
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
31
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
32
Demonstrations
Aqueous iron (III) nitrate reacts with sodium
hydroxide
33
Demonstrations
CoCl2(aq) + Ca(OH)2(aq) 
34
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) 
35
Answer
BaCl2 (aq)+ K2SO4 (aq)  BaSO4 (s) + 2KCl
Fe2(SO4)3 (aq) + 6LiOH (aq) 2Fe(OH)3 (s) + 3Li2SO4 (aq)
36
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
 Net
ion
ionic
37
Precipitation Reactions
Precipitate – insoluble solid that separates from solution
precipitate
Pb(NO3)2 (aq) + 2NaI (aq)
PbI2 (s) + 2NaNO3 (aq)
molecular equation
Pb2+ (aq) + 2NO3- (aq) + 2Na+ (aq) + 2I- (
aq)
PbI2 (s) + 2Na+ (aq) + 2NO3- (aq)
ionic equation
Pb2+aq + 2I- aq
PbI2 (s)
net ionic equation
P
bI
Na+ and NO3- are spectator ions
38
4.2
Molecular Equations/Chemical
equation

Shows the complete chemical formula for
reactants and products
HCl(aq) + NaOH(aq)  NaCl(aq) +
Strong acid
Strong elect.
Strong Base
Strong elect.
Soluble Salt
H2O(l)
weak electrolyte
39
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)
40
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.
41
Write the three types of equations
for the following rxn.
Pb(NO3)2 (aq) + KI (aq) 
42
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)
43
Writing Net Ionic Equations
1. Write the balanced molecular equation.
2. Write the ionic equation showing the strong electrolytes
completely dissociated into cations and anions.
3. Cancel the spectator ions on both sides of the ionic equation
4. Check that charges and number of atoms are balanced in the
net ionic equation
Write the net ionic equation for the reaction of silver
nitrate with sodium chloride.
AgNO3 (aq) + NaCl (aq)
AgCl (s) + NaNO3 (aq)
Ag+ + NO3- + Na+ + Cl-
AgCl (s) + Na+ + NO3-
Ag+ + Cl-
AgCl (s)
44
4.2
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
45
4.3 Acids
Substances that are
able to donate a
hydrogen ion (H+) and
increase [H+] in
aqueous solutions.
MEMORIZE
Nitric acid
HNO3
chloric acid
HClO3
perchloric acid HClO4
sulfuric acid
H2SO4
hydrochloric
acid
HCl
hydrobromic
acid
HBr
hydroiodic
acid
HI
46
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
47
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
48
Bases

Are soluble ionic
compounds containing
hydroxide ion (OH-).

When dissolved in water
the cations and OH- ions
separate and move
independently.

MEMORIZE /LEARN
Hydroxides NaOH
of 1A
LiOH
KOH
RbOH
CsOH
FrOH
Hydroxides Be (OH)2
of 2A
Mg(OH)2
Ca(OH)2
Sr(OH)2
Ba(OH)2
Ra(OH)2
49
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.
50
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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51
Classify the following dissolved substances
as strong or weak electrolytes:
CaCl2
HNO3
C2H5OH
HCHO2
52
CaCl2
= ionic = strong electrolyte
HNO3 = strong acid = strong electrolyte
C2H5OH = molecular = nonelectrolyte
HCHO2 = molecular = nonelectrolyte
53
Neutralization Rxn

Neutralization: when an acid and a base
mix and their products share no
characteristics with their reactants. (i.e
acid base qualities)
54
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)
55
Molecular equation
HCl (aq) + NaOH (aq)  H2O (l) + NaCl (aq)
(acid)
(base)
(water)
(salt)
Complete Ionic Equation
H+ + Cl- + Na+ + OH-  H2O + Na+ + ClNet Ionic Equation
H+ + OH-  H2O
56
Neutralization Reaction
acid + base
HCl (aq) + NaOH (aq)
H+ + Cl- + Na+ + OH-
H+ + OH-
salt + water
NaCl (aq) + H2O
Na+ + Cl- + H2O
H2O
57
4.3
Question

Write a balanced complete chemical
equation for the reaction between aqueous
solutions of acetic acid and barium
hydroxide.
58
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)
59
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)
weak acid
(strong Base/elect
Rule 5 exception)
(strong elect
ionic salt)
60
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
61
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)
62
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)
63
Homework

BL Pg 146 #’s : 23, 24, 25, 27, 29, 31
64
4.4 Oxidation –Reduction
Reactions

Rxn’s in which one or more electrons are
transferred between reactants.

Ex: 2 Na
neutral
+
Cl2 
neutral
2NaCl
Na+ Cl-
65
Symantics
Charges are written: # sign :
1- 3+
Oxidation numbers are written: sign # : -1 +3
66

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)
67
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)
68
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)

69
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)
70
2Mg
O2 + 4e-
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2O2Reduction half-reaction (gain e-)
2Mg + O2 + 4e2Mg2+ + 2O2- + 4e2Mg + O2
2MgO
71 4.4
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.
72
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.
73
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
74
electronegativity.
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 O’s 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
75
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
78
The oxidation numbers of elements in their compounds
79
4.4
Examples
HINT: 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
80

S8
In elemental form so oxidation number is 0
(rule 1)
81

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
82

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
83
Assign Oxidation numbers for
CO2
SF6NO3-
84
CO2
C = +4
O= -2(2)
SF6-
S = +5
F = -1(6)
NO3-
N = +5
O = -2(3)
5+ -6 = -1
85
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
86
Fe = +3(2) = +6
Oxidation for
free elements
is zero
O = -2(3) = -6
87
L - E -O
Lose electrons Oxidized

I think you
are pretty
and smart!!!
The Lion Says
G–E-R
Gain Electrons Reduced
88
Leo the Lion Says Ger!!!
OXIDATION
REDUCTION
89
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.
90
91
Identify the atoms being oxidized and
reduced as well as the oxidizing and
reducing agents.
2Al + 3I2  2AlI3
92
2Al
+
0
3I2
0

2AlI3
+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
93
Zn (s) + CuSO4 (aq)
Zn
ZnSO4 (aq) + Cu (s)
Zn2+ + 2e- Zn is oxidized
Cu2+ + 2e-
Zn is the reducing agent
Cu Cu2+ is reduced Cu2+ is the oxidizing agent
Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu (s) + 2AgNO3 (aq)
Cu
Ag+ + 1e-
Cu(NO3)2 (aq) + 2Ag (s)
Cu2+ + 2eAg Ag+ is reduced
Ag+ is the oxidizing agent
94
4.4
Activity Series

We need to know what metals are most
likely to oxidize others.

Example: We can’t store nickel nitrate in
an iron container because the solution
would eat through the container.
95
Activity Series

A list of metals
arranged in order of
decreasing ease of
oxidation.

Page 139 table
96
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?

1.
2.
3.
4.
5.
I finger on Fe
1 finger on Mg
Is the bound chemical below
Yes Fe is below Mg.
Then complete the reaction
97
Give: NaCl2 + Mg

Find: will sodium oxidize
Magnesium metal?

1.
2.
3.
4.
5.
I finger on Na
1 finger on Mg
Is the bound chemical below
no
Then the reaction is not
possible
98
The Activity Series for Halogens
F2 > Cl2 > Br2 > I2
Halogen Displacement Reaction
0
-1
Cl2 + 2KBr
I2 + 2KBr
-1
0
2KCl + Br2
2KI + Br2
99
4.4
100
Homework

Chang Pg 158-159 #’s 46,48, 50, 54,56 (id
what is red and ox)
BL146-147
#’s : 35, 39, 41, 42, 44, 45,
101
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.



102
Molarity (M)

Number of moles of solute in one liter of
solution

Molarity = moles of solute
liter of solution
Units = mol/L = M
103
A
B
C
D
To make 250 mL (0.250 L) of 1.00 M CuSO4
A. Use the formula grams needed = Molecular weight x Volume x Molarity
g = 159.6 x 0.250 x 1 = 39.9 g CUSO4
B. Place chemical into flask and add a small quantity of water to dissolve.
C. Mix solution
D. Bring total volume up to 0.250 L
104
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
105
Answer
5.0 g 1mol = 0.027 mol
180g
M = 0.027 = 0.27M
0.1
106
Question

How many grams of Na2SO4 are there in
5ml of 0.50 M Na2SO4.
107
Answer
g = M.W x V x L
g = 142 x 5/1000 x .5
108
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.
109
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
110
Answer
A.
B.
C.
0.30(3 Cl) = 0.90 M Cl
0.60 (2Cl) = 1.2 M Cl
0.40 ( 1 Cl) = 0.4 M Cl
MgCl2 > AlCl3 > NaCl
111
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!!!!!!
112
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.
113
Homework Sections

Chang pg 159 #’s 59,60,61,63,69, 74

Page 147
#’s 49, 51, 52, 54, 56, 59, 60, 61
114
4.7 Stoichiometry of precipitation
reactions

1.
2.
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
115
Question

How many moles are in 1 L of 0.3M
solution?

How many moles are in 6.9L of 0.45M
solution?
116
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
117

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?
118
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
119
Because AgNo3 and NaCl react in a 1:1 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
1 mol NaCl
= 8.77g NaCl
120
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?
121
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?
122
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
1L
123
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.
124
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
125
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?
126
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- 
127
Step 2:
What possible rxn’s will occur
HCl (aq)+ NaOH (aq)  NaCl (aq)+ H2O (l)
Na+ + Cl-  NaCl (soluble and cant
neutralize)
H+ + OH- H2O (insoluble and can
neutralize)
128
Step 3: write a balanced net ionic equation and
calculate moles of reactant needed.
H+ + OH- H2O
0.025 L NaOH 0.35 mol OH = 8.75 x 10-3 mol OH1 L NaOH
We do not need to determine limiting reactant since
the addition of H+ ions react exactly with the OHpresent in a1:1 ratio. Thus 8.75 x 10-1 mol H+ is
required to neutralize the solution.
129
Convert to volume required to neutralize the rxn.
Volume X 0.100 mol H+ = 8.75 x 10-3 mol H+
1L
= 8.75 x 10-2 L of .100M HCL is required to neutralize
25.0 mL of NaOH
130
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.
131
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.
132
Equivalence point: the point in the titration when exactly
enough base is added to neutralize the acid.
133
Indicators
Organic dyes that
change color as they
go from an acidic
solution to basic
solution.
134
Titration Question

What volume of 0.25M HNO3 is required to
titrate (neutralize) a solution containing
0.200 g of KOH.
135
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
136
Homework
Pg 148
 #’s 65, 67, 69, 73


Bonus 2 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
137