Transcript Chapter 4 Chemical Quantities and Aqueous Reactions

Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 4
Chemical
Quantities and
Aqueous
Reactions
Part1
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall
Reaction Stoichiometry
• the numerical relationships between chemical amounts
•
in a reaction is called stoichiometry
the coefficients in a balanced chemical equation
specify the relative amounts in moles of each of the
substances involved in the reaction
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
2 molecules of C8H18 react with 25 molecules of O2
to form 16 molecules of CO2 and 18 molecules of H2O
2 moles of C8H18 react with 25 moles of O2
to form 16 moles of CO2 and 18 moles of H2O
2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O
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Predicting Amounts from Stoichiometry
• the amounts of any other substance in a chemical
reaction can be determined from the amount of
just one substance
• How much CO2 can be made from 22.0 moles of
C8H18 in the combustion of C8H18?
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
2 moles C8H18 : 16 moles CO2
16 mol CO 2
22.0 moles C8 H18 
 176 moles CO 2
2 mol C8 H18
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Example – Estimate the mass of CO2 produced in
2004 by the combustion of 3.4 x 1015 g gasoline
• assuming that gasoline is octane, C8H18, the equation
•
for the reaction is:
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
the equation for the reaction gives the mole relationship
between amount of C8H18 and CO2, but we need to
know the mass relationship, so the Concept Plan will
be:
g C8H18
mol C8H18
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mol CO2
g CO2
4
Example – Estimate the mass of CO2 produced in
2004 by the combustion of 3.4 x 1015 g gasoline
Given:
Find:
Concept Plan:
3.4 x 1015 g C8H18
g CO2
g C8H18
mol C8H18
1 mol
114.22g
mol CO2
16 molCO 2
2 molC8H18
g CO2
44.01 g
1 mol
Relationships: 1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18 = 16 mol CO2
Solution:
1 molC8 H18
16 molCO 2 44.01g CO 2
3.4 10 g C8H18 


114.22g C8 H18 2 molC8 H18 1 molCO 2
15
 1.01016 g CO 2
Check: since 8x moles of CO as C H , but the molar mass of C H is
2
8 18
8 18
3x CO2, the number makes sense
Practice
• According to the following equation, how
many milliliters of water are made in the
combustion of 9.0 g of glucose?
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
1.
2.
3.
4.
convert 9.0 g of glucose into moles (MM 180)
convert moles of glucose into moles of water
convert moles of water into grams (MM 18.02)
convert grams of water into mL
a) How? what is the relationship between mass and
volume?
density of water = 1.00 g/mL
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Practice
According to the following equation, how many
milliliters of water are made in the combustion of
9.0 g of glucose?
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
9.0 g C 6 H12O 6 x
1 mole C 6 H12O 6
6 mole H 2 O
18.0 g H 2 O 1 mL H 2 O
x
x
x
2
1 mole C 6 H12O 6 1 mole H 2 O 1.00 g H 2 O
1.80 x 10 g
 5.4 mL H 2 O
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Limiting Reactant
• for reactions with multiple reactants, it is likely that
•
•
one of the reactants will be completely used before the
others
when this reactant is used up, the reaction stops and no
more product is made
the reactant that limits the amount of product is called
the limiting reactant
 sometimes called the limiting reagent
 the limiting reactant gets completely consumed
• reactants not completely consumed are called excess
•
reactants
the amount of product that can be made from the
limiting reactant is called the theoretical yield
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Things Don’t Always Go as Planned!
• many things can happen during the course of an
experiment that cause the loss of product
• the amount of product that is made in a reaction
is called the actual yield
generally less than the theoretical yield, never more!
• the efficiency of product recovery is generally
given as the percent yield
actual yield
Percent Yield 
100%
theoretica l yield
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Limiting and Excess Reactants in the
Combustion of Methane
•
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Our balanced equation for the combustion of methane
implies that every 1 molecule of CH4 reacts with 2
molecules of O2
H
H
C
H
H
O
+
O
+
O
O
C
+
H
H
+
O
O
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O
O
H
H
10
Limiting and Excess Reactants in the
Combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
• If we have 5 molecules of CH4 and 8 molecules
of O2, which is the limiting reactant?
H
C
H
H
H
H
H
H
H
C
H
H
H
C
C
H
H
H
H
H
H
H
C
+
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
?
H
H
11
Limiting and Excess Reactants in the
Combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
H
C
H
H
H
H
H
H
H
C
H
H
H
C
C
H
H
H
H
H
H
H
C
+
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
H
H
since less CO2
can be made
from the O2 than
the CH4, the O2
is the limiting
reactant
2 molecules CO 2
8 molecules CH 4 
 16 molecules CO 2
1 molecules CH 4
2 molecules CO 2
10 molecules O 2 
 10 molecules CO 2
2 molecules O 2
12
Example 4.4
Finding Limiting Reactant,
Theoretical Yield, and
Percent Yield
Example:
• When 28.6 kg of C are allowed to react with 88.2 kg of
TiO2 in the reaction below, 42.8 kg of Ti are obtained.
Find the Limiting Reactant, Theoretical Yield, and
Percent Yield.
T iO2(s)  2 C(s)  T i(s)  2 CO(g)
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Example:
When 28.6 kg of C reacts with 88.2
kg of TiO2, 42.8 kg of Ti are
obtained. Find the Limiting
Reactant, Theoretical Yield, and
Percent Yield.
TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
• Write down the given quantity and its units.
Given:
28.6 kg C
88.2 kg TiO2
42.8 kg Ti produced
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Information
Example:
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find the Limiting
Reactant, Theoretical
Yield, and Percent Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
• Write down the quantity to find and/or its units.
Find: limiting reactant
theoretical yield
percent yield
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Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
• Write a Concept Plan:
kg
C
kg
TiO2
smallest
mol Ti
1000 g
1 kg
1000 g
1 kg
47.87 g
1 mol Ti
g
C
g
TiO2
g Ti
1 mol C
12.01 g C
1 mol TiO 2
79.87 g TiO 2
1 kg
1 000 g
mol
C
mol
TiO2
kg Ti
T.Y.
1 mol Ti
2 mol C
1 mol Ti
1 mol TiO 2
% yield 
mol
Ti
mol
Ti
act. yield
theor. yield
}
smallest
amount is
from
limiting
reactant
% Yield
17
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
• Collect Needed Relationships:
1000 g = 1 kg
Molar Mass TiO2 = 79.87 g/mol
Molar Mass Ti = 47.87 g/mol
Molar Mass C = 12.01 g/mol
1 mole TiO2 : 1 mol Ti (from the chem. equation)
2 mole C  1 mol Ti (from the chem. equation)
18
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent
Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
• Apply the Concept Plan:
1000 g 1 mole C 1 mol Ti
28.6 kg C 


 1.1907 103 mol Ti
1 kg 12.01 g C 2 mol C
1000 g 1 mole TiO 2
1 mol Ti
88.2 kg TiO 2 


 1.1043  103 mol Ti
1 kg 79.87 g TiO 2 1 mol TiO 2
Limiting Reactant
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smallest moles of Ti
19
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent
Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
• Apply the Concept Plan:
47.87g T i 1 kg
1.104310 molT i

 52.9kg T i
1 mol
1000g
3
Theoretical Yield
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Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent
Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
• Apply the Concept Plan:
Actual Yield
100 %  Percent Yield
Theoretica l Yield
42.8kg T i
100%  80.9%
52.9kg T i
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Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent
Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
• Check the Solutions:
Limiting Reactant = TiO2
Theoretical Yield = 52.9 kg
Percent Yield = 80.9%
Since Ti has lower molar mass than TiO2, the T.Y. makes sense
The Percent Yield makes sense as it is less than 100%.
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Practice – How many grams of N2(g) can be made from
9.05 g of NH3 reacting with 45.2 g of CuO?
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
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Practice – How many grams of N2(g) can be made from 9.05 g of
NH3 reacting with 45.2 g of CuO?
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
Given:
Find:
9.05 g NH3, 45.2 g CuO
g N2
Concept Plan: g NH
3
mol NH3
1 mol N 2
2 mol NH3
1 mol
17.03g
g CuO
mol CuO
1 mol
79.55g
smallest moles N2
Relationships:
mol N2
mol N2
1 mol N 2
3 mol CuO
1 mol
28.02g
g N2
1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g
2 mol NH3 = 1 mol N2, 3 mol CuO = 1 mol N2
Practice – How many grams of N2(g) can be made from 9.05 g of
NH3 reacting with 45.2 g of CuO?
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
Solution:
1 mol NH3
1 mol N 2
9.05 g NH3 

 0.266mol N 2
17.03g NH3 2 mol NH3
1 molCuO
1 mol N 2
45.2 g CuO 

 0.189mol N 2
79.55g CuO 3 molCuO
28.02g N 2
0.189mol N 2 
 5.30 g N 2
1 mol N 2
Check:
units are correct, and since there are fewer moles
of N2 than CuO in the reaction and N2 has a
smaller mass, the number makes sense
Solutions
• when table salt is mixed with water, it seems to disappear,
or become a liquid – the mixture is homogeneous
 the salt is still there, as you can tell from the taste, or simply
boiling away the water
• homogeneous mixtures are called solutions
• the component of the solution that changes state is called
•
the solute
the component that keeps its state is called the solvent
 if both components start in the same state, the major component
is the solvent
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Describing Solutions
• since solutions are mixtures, the composition can
vary from one sample to another
pure substances have constant composition
salt water samples from different seas or lakes have
different amounts of salt
• so to describe solutions accurately, we must
describe how much of each component is present
we saw that with pure substances, we can describe
them with a single name because all samples identical
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Solution Concentration
• qualitatively, solutions are often
•
•
•
described as dilute or
concentrated
dilute solutions have a small
amount of solute compared to
solvent
concentrated solutions have a
large amount of solute
compared to solvent
quantitatively, the relative
amount of solute in the solution
is called the concentration
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Solution Concentration
Molarity
• moles of solute per 1 liter of solution
• used because it describes how many molecules
of solute in each liter of solution
amount of solute (in moles)
molarity, M 
amount of solution (in L)
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Preparing 1 L of a 1.00 M NaCl Solution
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Example 4.5 – Find the molarity of a solution that
has 25.5 g KBr dissolved in 1.75 L of solution
•
Sort
Information
•
Strategize
Given:
Find:
Concept Plan:
25.5 g KBr, 1.75 L solution
Molarity, M
g KBr
mol KBr
1 mol
119.00g
Relationships:
•
Follow the
Concept Plan
to Solve the
problem
•
Check
M
mol
L
M
L sol’n
1 mol KBr = 119.00 g,
M = moles/L
Solution:
1 mol KBr
 0.21429 mol KBr
119.00g KBr
molesKBr 0.21429 mol KBr
molarity,M 

 0.122M
L solution
1.75L
25.5 g KBr 
Check: since most solutions are between 0 and
18 M, the answer makes sense
Using Molarity in Calculations
• molarity shows the relationship between the
moles of solute and liters of solution
• If a sugar solution concentration is 2.0 M, then
1 liter of solution contains 2.0 moles of sugar
2 liters = 4.0 moles sugar
0.5 liters = 1.0 mole sugar
• 1 L solution : 2 moles sugar
2 mol sugar
1 L solution
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1 L solution
2 mol sugar
32
Example 4.6 – How many liters of 0.125 M NaOH
contains 0.255 mol NaOH?
•
Sort
Information
•
Strategize
Given: 0.125 M NaOH, 0.255 mol NaOH
Find:
liters, L
Concept Plan:
mol NaOH
L sol’n
1 L solution
0.125 mol NaOH
Relationships:
•
•
Follow the
Concept Plan
to Solve the
problem
Check
0.125 mol NaOH = 1 L solution
Solution:
0.255 mol NaOH 
1 L solution
 2.04 L solution
0.125 mol NaOH
Check: since each L has only 0.125 mol NaOH,
it makes sense that 0.255 mol should
require a little more than 2 L
Dilution
• often, solutions are stored as concentrated stock
•
solutions
to make solutions of lower concentrations from these
stock solutions, more solvent is added
 the amount of solute doesn’t change, just the volume of
solution
moles solute in solution 1 = moles solute in solution 2
• the concentrations and volumes of the stock and new
solutions are inversely proportional
M1∙V1 = M2∙V2
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Example 4.7 – To what volume should you dilute
0.200 L of 15.0 M NaOH to make 3.00 M NaOH?
•
Sort
Information
•
Strategize
Given: V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M
V2, L
Find:
Concept Plan:
V1, M1, M2
V2
M1  V1
 V2
M2
Relationships:
•
Follow the
Concept Plan
to Solve the
problem
•
Check
Solution:
M1V1 = M2V2
mol

15.0
  0.200 L 
L


 1.00L
mol

3.00


L


Check: since the solution is diluted by a factor
of 5, the volume should increase by a
factor of 5, and it does
Solution Stoichiometry
• since molarity relates the moles of solute to the
liters of solution, it can be used to convert
between amount of reactants and/or products in
a chemical reaction
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Example 4.8 – What volume of 0.150 M KCl is required to
completely react with 0.150 L of 0.175 M Pb(NO3)2 in the
reaction 2 KCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2 KNO3(aq)
•
•
Sort
Information
Strategize
Given:
0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2
Find:
L KCl
Concept Plan:
L Pb(NO3)2
mol Pb(NO3)2
0.175mol
1 L Pb(NO3 ) 2
Relationships:
•
•
Follow the
Concept
Plan to
Solve the
problem
Check
mol KCl
2 molKCl
1 mol Pb(NO3 ) 2
L KCl
1 L KCl
0.150 mol
1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol,
1 mol Pb(NO3)2 = 2 mol KCl
Solution:
0.150L Pb(NO3 ) 2 
0.175mol
2 mol KCl
1 L KCl


1 L Pb(NO3 ) 2 1 mol Pb(NO3 ) 2 0.150mol
 0.350L KCl
Check:
since need 2x moles of KCl as Pb(NO3)2, and
the molarity of Pb(NO3)2 > KCl, the volume of
KCl should be more than 2x volume Pb(NO3)2
What Happens When a Solute Dissolves?
• there are attractive forces between the solute particles
holding them together
• there are also attractive forces between the solvent
molecules
• when we mix the solute with the solvent, there are
attractive forces between the solute particles and the
solvent molecules
• if the attractions between solute and solvent are strong
enough, the solute will dissolve
38
Table Salt Dissolving in Water
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Each ion is attracted
to the surrounding
water molecules and
pulled off and away
from the crystal
When it enters the
solution, the ion is
surrounded by water
molecules, insulating
it from other ions
The result is a solution
with free moving
charged particles able
to conduct electricity 39
Electrolytes and Nonelectrolytes
• materials that dissolve
in water to form a
solution that will
conduct electricity are
called electrolytes
• materials that dissolve
in water to form a
solution that will not
conduct electricity are
called nonelectrolytes
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Molecular View of
Electrolytes and Nonelectrolytes
• in order to conduct electricity, a material must have
•
charged particles that are able to flow
electrolyte solutions all contain ions dissolved in the
water
 ionic compounds are electrolytes because they all dissociate
into their ions when they dissolve
• nonelectrolyte solutions contain whole molecules
dissolved in the water
 generally, molecular compounds do not ionize when they
dissolve in water
 the notable exception being molecular acids
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Salt vs. Sugar Dissolved in Water
ionic compounds dissociate
into ions when they dissolve
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molecular compounds do not
dissociate when they dissolve
42
Acids
• acids are molecular compounds that ionize when they
dissolve in water
 the molecules are pulled apart by their attraction for the water
 when acids ionize, they form H+ cations and anions
• the percentage of molecules that ionize varies from one
•
•
acid to another
acids that ionize virtually 100% are called strong acids
HCl(aq)  H+(aq) + Cl-(aq)
acids that only ionize a small percentage are called
weak acids
HF(aq)  H+(aq) + F-(aq)
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Strong and Weak Electrolytes
• strong electrolytes are materials that dissolve
completely as ions
 ionic compounds and strong acids
 their solutions conduct electricity well
• weak electrolytes are materials that dissolve mostly as
molecules, but partially as ions
 weak acids
 their solutions conduct electricity, but not well
• when compounds containing a polyatomic ion dissolve,
the polyatomic ion stays together
Na2SO4(aq)  2 Na+(aq) + SO42-(aq)
HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
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Classes of Dissolved Materials
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Solubility of Ionic Compounds
• some ionic compounds, like NaCl, dissolve very well in
•
•
water at room temperature
other ionic compounds, like AgCl, dissolve hardly at all
in water at room temperature
compounds that dissolve in a solvent are said to be
soluble, while those that do not are said to be insoluble
 NaCl is soluble in water, AgCl is insoluble in water
 the degree of solubility depends on the temperature
 even insoluble compounds dissolve, just not enough to be
meaningful
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When Will a Salt Dissolve?
• Predicting whether a compound will dissolve in
water is not easy
• The best way to do it is to do some experiments
to test whether a compound will dissolve in
water, then develop some rules based on those
experimental results
we call this method the empirical method
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Solubility Rules
Compounds that Are Generally Soluble in Water
Compounds Containing the Exceptions
Following Ions are Generally (when combined with ions on the
Soluble
left the compound is insoluble)
Li+, Na+, K+, NH4+
none
NO3–, C2H3O2–
none
Cl–, Br–, I–
Ag+, Hg22+, Pb2+
SO42–
Ag+, Ca2+, Sr2+, Ba2+, Pb2+
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Solubility Rules
Compounds that Are Generally Insoluble
Exceptions
Compounds Containing the (when combined with ions on the
Following Ions are Generally left the compound is soluble or
Insoluble
slightly soluble)
OH–
S2–
CO32–, PO43–
Tro, Chemistry: A Molecular Approach
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+
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Precipitation Reactions
• reactions between aqueous solutions of ionic
compounds that produce an ionic compound
that is insoluble in water are called
precipitation reactions and the insoluble
product is called a precipitate
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2 KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2
KNO3(aq)
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No Precipitate Formation =
No Reaction
KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq)
all ions still present,  no reaction
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Process for Predicting the Products of
a Precipitation Reaction
1. Determine what ions each aqueous reactant has
2. Determine formulas of possible products
 Exchange ions
 (+) ion from one reactant with (-) ion from other
 Balance charges of combined ions to get formula of each
product
3. Determine Solubility of Each Product in Water
 Use the solubility rules
 If product is insoluble or slightly soluble, it will precipitate
4. If neither product will precipitate, write no reaction
after the arrow
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Process for Predicting the Products of
a Precipitation Reaction
5. If either product is insoluble, write the formulas
for the products after the arrow – writing (s)
after the product that is insoluble and will
precipitate, and (aq) after products that are
soluble and will not precipitate
6. Balance the equation
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Example 4.10 – Write the equation for the
precipitation reaction between an aqueous solution
of potassium carbonate and an aqueous solution of
nickel(II) chloride
1. Write the formulas of the reactants
K2CO3(aq) + NiCl2(aq) 
2. Determine the possible products
a) Determine the ions present
(K+ + CO32-) + (Ni2+ + Cl-) 
b) Exchange the Ions
(K+ + CO32-) + (Ni2+ + Cl-)  (K+ + Cl-) + (Ni2+ + CO32-)
c) Write the formulas of the products

cross charges and reduce
K2CO3(aq) + NiCl2(aq)  KCl + NiCO3
Example 4.10 – Write the equation for the
precipitation reaction between an aqueous solution
of potassium carbonate and an aqueous solution of
nickel(II) chloride
3. Determine the solubility of each product
KCl is soluble
NiCO3 is insoluble
4. If both products soluble, write no reaction
does not apply since NiCO3 is insoluble
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Example 4.10 – Write the equation for the
precipitation reaction between an aqueous solution
of potassium carbonate and an aqueous solution of
nickel(II) chloride
5. Write (aq) next to soluble products and (s) next
to insoluble products
K2CO3(aq) + NiCl2(aq)  KCl(aq) + NiCO3(s)
6. Balance the Equation
K2CO3(aq) + NiCl2(aq)  2 KCl(aq) + NiCO3(s)
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