Chapter 4 Chemical Quantities and Aqueous Reactions

1
Chapter 4Chemical Quantities and Aqueous
Reactions
Chemistry A Molecular Approach, 1st Ed.Nivaldo
Tro
Roy Kennedy Massachusetts Bay Community
College Wellesley Hills, MA
2008, Prentice Hall
2
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
3
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

4
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

5
Example Estimate the mass of CO2 produced in
2004 by the combustion of 3.4 x 1015 g gasoline
3.4 x 1015 g C8H18 g CO2
Given Find
1 mol C8H18 114.22g, 1 mol CO2 44.01g, 2
mol C8H18 16 mol CO2
Concept Plan Relationships
Solution
Check
since 8x moles of CO2 as C8H18, but the molar
mass of C8H18 is 3x CO2, the number makes sense
6
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)
• 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
• How? what is the relationship between mass and
  volume?

density of water 1.00 g/mL
7
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)
8
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

9
Things Dont 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

10
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

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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?

12
Limiting and Excess Reactants in the Combustion
of Methane
CH4(g) 2 O2(g) ? CO2(g) 2 H2O(g)
since less CO2 can be made from the O2 than the
CH4, the O2 is the limiting reactant
13
Example 4.4Finding Limiting Reactant,
Theoretical Yield, and Percent Yield
14

• 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.

15
ExampleWhen 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

16
ExampleFind 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

• Write down the quantity to find and/or its units.
  
• Find limiting reactant
• theoretical yield
• percent yield

17
ExampleFind 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

smallest mol Ti
18
ExampleFind 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)

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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 C12.01g 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
ExampleFind the Limiting Reactant, Theoretical
Yield, and Percent Yield. TiO2(s) 2 C(s) ?
Ti(s) 2 CO(g)

• Apply the Concept Plan

20
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 C12.01g 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
ExampleFind the Limiting Reactant, Theoretical
Yield, and Percent Yield. TiO2(s) 2 C(s) ?
Ti(s) 2 CO(g)

• Apply the Concept Plan

21
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 C12.01g 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
ExampleFind the Limiting Reactant, Theoretical
Yield, and Percent Yield. TiO2(s) 2 C(s) ?
Ti(s) 2 CO(g)

• Apply the Concept Plan

22
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 C12.01g 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
ExampleFind the Limiting Reactant, Theoretical
Yield, and Percent Yield. TiO2(s) 2 C(s) ?
Ti(s) 2 CO(g)

• 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.
23
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)
9.05 g NH3, 45.2 g CuO g N2
Given Find
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
Concept Plan Relationships
25
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
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
Check
26
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

27
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

28
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 ConcentrationMolarity

• moles of solute per 1 liter of solution
• used because it describes how many molecules of
  solute in each liter of solution

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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
25.5 g KBr, 1.75 L solution Molarity, M
Given Find

• Sort Information

1 mol KBr 119.00 g, M moles/L
Concept Plan Relationships

• Strategize

Solution

• Follow the Concept Plan to Solve the problem

since most solutions are between 0 and 18 M, the
answer makes sense

• Check

Check
32
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

33
Example 4.6 How many liters of 0.125 M NaOH
contains 0.255 mol NaOH?
0.125 M NaOH, 0.255 mol NaOH liters, L
Given Find

• Sort Information

0.125 mol NaOH 1 L solution
Concept Plan Relationships

• Strategize

Solution

• Follow the Concept Plan to Solve the problem

since each L has only 0.125 mol NaOH, it makes
sense that 0.255 mol should require a little more
than 2 L

• Check

Check
34
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 doesnt 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
• M1V1 M2V2

35
Example 4.7 To what volume should you dilute
0.200 L of 15.0 M NaOH to make 3.00 M NaOH?
V1 0.200L, M1 15.0 M, M2 3.00 M V2, L
Given Find

• Sort Information

M1V1 M2V2
Concept Plan Relationships

• Strategize

Solution

• Follow the Concept Plan to Solve the problem

since the solution is diluted by a factor of 5,
the volume should increase by a factor of 5, and
it does

• Check

Check
36
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

37
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)
0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2 L KCl
Given Find

• Sort Information

1 L Pb(NO3)2 0.175 mol, 1 L KCl 0.150 mol,
1 mol Pb(NO3)2 2 mol KCl
Concept Plan Relationships

• Strategize

Solution

• Follow the Concept Plan to Solve the problem

since need 2x moles of KCl as Pb(NO3)2, and the
molarity of Pb(NO3)2 gt KCl, the volume of KCl
should be more than 2x volume Pb(NO3)2

• Check

Check
38
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

39
Table Salt Dissolving in Water
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
40
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

41
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

42
Salt vs. Sugar Dissolved in Water
43
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)

44
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)

45
Classes of Dissolved Materials
46
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

47
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

48
Solubility RulesCompounds that Are Generally
Soluble in Water
49
Solubility RulesCompounds that Are Generally
Insoluble
50
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

51
2 KI(aq) Pb(NO3)2(aq) ? PbI2(s) 2 KNO3(aq)
52
No Precipitate Formation No Reaction
KI(aq) NaCl(aq) ? KCl(aq) NaI(aq) all ions
still present, ? no reaction
53
Process for Predicting the Products ofa
Precipitation Reaction

• Determine what ions each aqueous reactant has
• 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
• Determine Solubility of Each Product in Water
• Use the solubility rules
• If product is insoluble or slightly soluble, it
  will precipitate
• If neither product will precipitate, write no
  reaction after the arrow

54
Process for Predicting the Products ofa
Precipitation Reaction

• 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
• Balance the equation

55
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

• Write the formulas of the reactants
• K2CO3(aq) NiCl2(aq) ?
• Determine the possible products
• Determine the ions present
• (K CO32-) (Ni2 Cl-) ?
• Exchange the Ions
• (K CO32-) (Ni2 Cl-) ? (K Cl-) (Ni2
  CO32-)
• Write the formulas of the products
• cross charges and reduce
• K2CO3(aq) NiCl2(aq) ? KCl NiCO3

56
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

• Determine the solubility of each product
• KCl is soluble
• NiCO3 is insoluble
• If both products soluble, write no reaction
• does not apply since NiCO3 is insoluble

57
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

• Write (aq) next to soluble products and (s) next
  to insoluble products
• K2CO3(aq) NiCl2(aq) ? KCl(aq) NiCO3(s)
• Balance the Equation
• K2CO3(aq) NiCl2(aq) ? 2 KCl(aq) NiCO3(s)

58
Ionic Equations

• equations which describe the chemicals put into
  the water and the product molecules are called
  molecular equations
• 2 KOH(aq) Mg(NO3)2(aq) 2 KNO3(aq)
  Mg(OH)2(s)
• equations which describe the actual dissolved
  species are called complete ionic equations
• aqueous strong electrolytes are written as ions
• soluble salts, strong acids, strong bases
• insoluble substances, weak electrolytes, and
  nonelectrolytes written in molecule form
• solids, liquids, and gases are not dissolved,
  therefore molecule form
• 2K1(aq) 2OH-1(aq) Mg2(aq) 2NO3-1(aq)
  2K1(aq) 2NO3-1(aq) Mg(OH)2(s)

59
Ionic Equations

• ions that are both reactants and products are
  called spectator ions
• 2K1(aq) 2OH-1(aq) Mg2(aq) 2NO3-1(aq)
  2K1(aq) 2NO3-1(aq) Mg(OH)2(s)

• an ionic equation in which the spectator ions
  are
• removed is called a net ionic equation
• 2OH-1(aq) Mg2(aq) Mg(OH)2(s)

60
Acid-Base Reactions

• also called neutralization reactions because the
  acid and base neutralize each others properties
• 2 HNO3(aq) Ca(OH)2(aq) ? Ca(NO3)2(aq) 2
  H2O(l)
• the net ionic equation for an acid-base reaction
  is
• H(aq) OH?(aq) ? H2O(l)
• as long as the salt that forms is soluble in water

61
Acids and Bases in Solution

• acids ionize in water to form H ions
• more precisely, the H from the acid molecule is
  donated to a water molecule to form hydronium
  ion, H3O
• most chemists use H and H3O interchangeably
• bases dissociate in water to form OH? ions
• bases, like NH3, that do not contain OH? ions,
  produce OH? by pulling H off water molecules
• in the reaction of an acid with a base, the H
  from the acid combines with the OH? from the base
  to make water
• the cation from the base combines with the anion
  from the acid to make the salt
• acid base ???salt water

62
Common Acids
63
Common Bases
64
HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
65
Example - Write the molecular, ionic, and
net-ionic equation for the reaction of aqueous
nitric acid with aqueous calcium hydroxide

• Write the formulas of the reactants
• HNO3(aq) Ca(OH)2(aq) ?
• Determine the possible products
• Determine the ions present when each reactant
  dissociates
• (H NO3-) (Ca2 OH-) ?
• Exchange the ions, H1 combines with OH-1 to make
  H2O(l)
• (H NO3-) (Ca2 OH-) ? (Ca2 NO3-)
  H2O(l)
• Write the formula of the salt
• cross the charges
• (H NO3-) (Ca2 OH-) ? Ca(NO3)2 H2O(l)

66
Example - Write the molecular, ionic, and
net-ionic equation for the reaction of aqueous
nitric acid with aqueous calcium hydroxide

• Determine the solubility of the salt
• Ca(NO3)2 is soluble
• Write an (s) after the insoluble products and a
  (aq) after the soluble products
• HNO3(aq) Ca(OH)2(aq) ? Ca(NO3)2(aq) H2O(l)
• Balance the equation
• 2 HNO3(aq) Ca(OH)2(aq) ? Ca(NO3)2(aq) 2
  H2O(l)

67
Example - Write the molecular, ionic, and
net-ionic equation for the reaction of aqueous
nitric acid with aqueous calcium hydroxide

• Dissociate all aqueous strong electrolytes to get
  complete ionic equation
• not H2O
• 2 H(aq) 2 NO3-(aq) Ca2(aq) 2 OH-(aq) ?
  Ca2(aq) 2 NO3-(aq) H2O(l)
• Eliminate spectator ions to get net-ionic
  equation
• 2 H1(aq) 2 OH-1(aq) ? 2 H2O(l)
• H1(aq) OH-1(aq) ? H2O(l)

68
Titration

• often in the lab, a solutions concentration is
  determined by reacting it with another material
  and using stoichiometry this process is called
  titration
• in the titration, the unknown solution is added
  to a known amount of another reactant until the
  reaction is just completed, at this point, called
  the endpoint, the reactants are in their
  stoichiometric ratio
• the unknown solution is added slowly from an
  instrument called a burette
• a long glass tube with precise volume markings
  that allows small additions of solution

69
Acid-Base Titrations

• the difficulty is determining when there has been
  just enough titrant added to complete the
  reaction
• the titrant is the solution in the burette
• in acid-base titrations, because both the
  reactant and product solutions are colorless, a
  chemical is added that changes color when the
  solution undergoes large changes in
  acidity/alkalinity
• the chemical is called an indicator
• at the endpoint of an acid-base titration, the
  number of moles of H equals the number of moles
  of OH?
• aka the equivalence point

70
Titration
71
Titration
The base solution is the titrant in the burette.
As the base is added to the acid, the H reacts
with the OH to form water. But there is still
excess acid present so the color does not change.
At the titrations endpoint, just enough base has
been added to neutralize all the acid. At this
point the indicator changes color.
72
Example 4.14The titration of 10.00 mL of HCl
solution of unknown concentration requires 12.54
mL of 0.100 M NaOH solution to reach the end
point. What is the concentration of the unknown
HCl solution?

• Write down the given quantity and its units.
• Given 10.00 mL HCl
• 12.54 mL of 0.100 M NaOH

73
Example 4.14The titration of 10.00 mL of HCl
solution of unknown concentration requires 12.54
mL of 0.100 M NaOH solution to reach the end
point. What is the concentration of the unknown
HCl solution?

• Information
• Given 10.00 mL HCl
• 12.54 mL of 0.100 M NaOH

• Write down the quantity to find, and/or its
  units.
• Find concentration HCl, M

74
Example 4.14The titration of 10.00 mL of HCl
solution of unknown concentration requires 12.54
mL of 0.100 M NaOH solution to reach the end
point. What is the concentration of the unknown
HCl solution?

• Information
• Given 10.00 mL HCl
• 12.54 mL of 0.100 M NaOH
• Find M HCl

• Collect Needed Equations and Conversion Factors
• HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
• ? 1 mole HCl 1 mole NaOH
• 0.100 M NaOH ?0.100 mol NaOH ? 1 L soln

75
Example 4.14The titration of 10.00 mL of HCl
solution of unknown concentration requires 12.54
mL of 0.100 M NaOH solution to reach the end
point. What is the concentration of the unknown
HCl solution?

• Information
• Given 10.00 mL HCl
• 12.54 mL of 0.100 M NaOH
• Find M HCl
• CF 1 mol HCl 1 mol NaOH
• 0.100 mol NaOH 1 L
• M mol/L

• Write a Concept Plan

mL NaOH
L NaOH
mol NaOH
mol HCl
mL HCl
L HCl
76
ExampleThe titration of 10.00 mL of HCl
solution of unknown concentration requires 12.54
mL of 0.100 M NaOH solution to reach the end
point. What is the concentration of the unknown
HCl solution?

• Information
• Given 10.00 mL HCl
• 12.54 mL of 0.100 M NaOH
• Find M HCl
• CF 1 mol HCl 1 mol NaOH
• 0.100 mol NaOH 1 L
• M mol/L
• CP mL NaOH ? L NaOH ?
• mol NaOH ? mol HCl
• mL HCl ? L HCl mol ? M

• Apply the Solution Map

1.25 x 10-3 mol HCl
77
ExampleThe titration of 10.00 mL of HCl
solution of unknown concentration requires 12.54
mL of 0.100 M NaOH solution to reach the end
point. What is the concentration of the unknown
HCl solution?

• Information
• Given 10.00 mL HCl
• 12.54 mL NaOH
• Find M HCl
• CF 1 mol HCl 1 mol NaOH
• 0.100 mol NaOH 1 L
• M mol/L
• CP mL NaOH ? L NaOH ?
• mol NaOH ? mol HCl
• mL HCl ? L HCl mol ? M

• Apply the Concept Plan

78
ExampleThe titration of 10.00 mL of HCl
solution of unknown concentration requires 12.54
mL of 0.100 M NaOH solution to reach the end
point. What is the concentration of the unknown
HCl solution?

• Information
• Given 10.00 mL HCl
• 12.54 mL NaOH
• Find M HCl
• CF 1 mol HCl 1 mol NaOH
• 0.100 mol NaOH 1 L
• M mol/L
• CP mL NaOH ? L NaOH ?
• mol NaOH ? mol HCl
• mL HCl ? L HCl mol ? M

• Check the Solution

HCl solution 0.125 M
The units of the answer, M, are correct. The
magnitude of the answer makes sense since the
neutralization takes less HCl solution than NaOH
solution, so the HCl should be more concentrated.
79
Gas Evolving Reactions

• Some reactions form a gas directly from the ion
  exchange
• K2S(aq) H2SO4(aq) ? K2SO4(aq) H2S(g)
• Other reactions form a gas by the decomposition
  of one of the ion exchange products into a gas
  and water
• K2SO3(aq) H2SO4(aq) ? K2SO4(aq) H2SO3(aq)
• H2SO3 ? H2O(l) SO2(g)

80
NaHCO3(aq) HCl(aq) ? NaCl(aq) CO2(g) H2O(l)
81
Compounds that UndergoGas Evolving Reactions
82
Example 4.15 - When an aqueous solution of sodium
carbonate is added to an aqueous solution of
nitric acid, a gas evolves

• Write the formulas of the reactants
• Na2CO3(aq) HNO3(aq) ?
• Determine the possible products
• Determine the ions present when each reactant
  dissociates
• (Na1 CO3-2) (H1 NO3-1) ?
• Exchange the anions
• (Na1 CO3-2) (H1 NO3-1) ? (Na1 NO3-1)
  (H1 CO3-2)
• Write the formula of compounds
• cross the charges
• Na2CO3(aq) HNO3(aq) ? NaNO3 H2CO3

83
Example 4.15 - When an aqueous solution of sodium
carbonate is added to an aqueous solution of
nitric acid, a gas evolves

• Check to see either product H2S - No
• Check to see if either product decomposes Yes
• H2CO3 decomposes into CO2(g) H2O(l)
• Na2CO3(aq) HNO3(aq) ? NaNO3 CO2(g) H2O(l)

84
Example 4.15 - When an aqueous solution of sodium
carbonate is added to an aqueous solution of
nitric acid, a gas evolves

• Determine the solubility of other product
• NaNO3 is soluble
• Write an (s) after the insoluble products and a
  (aq) after the soluble products
• Na2CO3(aq) 2 HNO3(aq) ? 2 NaNO3(aq) CO2(g)
  H2O(l)
• Balance the equation
• Na2CO3(aq) 2 HNO3(aq) ? 2 NaNO3 CO2(g)
  H2O(l)

85
Other Patterns in Reactions

• the precipitation, acid-base, and gas evolving
  reactions all involved exchanging the ions in the
  solution
• other kinds of reactions involve transferring
  electrons from one atom to another these are
  called oxidation-reduction reactions
• also known as redox reactions
• many involve the reaction of a substance with
  O2(g)
• 4 Fe(s) 3 O2(g) ? 2 Fe2O3(s)

86
Combustion as Redox2 H2(g) O2(g) ? 2 H2O(g)
87
Redox without Combustion2 Na(s) Cl2(g) ? 2
NaCl(s)
2 Na ? 2 Na 2 e?
Cl2 2 e? ? 2 Cl?
88
Reactions of Metals with Nonmetals

• consider the following reactions
• 4 Na(s) O2(g) ? 2 Na2O(s)
• 2 Na(s) Cl2(g) ? 2 NaCl(s)
• the reaction involves a metal reacting with a
  nonmetal
• in addition, both reactions involve the
  conversion of free elements into ions
• 4 Na(s) O2(g) ? 2 Na2O (s)
• 2 Na(s) Cl2(g) ? 2 NaCl(s)

89
Oxidation and Reduction

• in order to convert a free element into an ion,
  the atoms must gain or lose electrons
• of course, if one atom loses electrons, another
  must accept them
• reactions where electrons are transferred from
  one atom to another are redox reactions
• atoms that lose electrons are being oxidized,
  atoms that gain electrons are being reduced

2 Na(s) Cl2(g) ? 2 NaCl(s) Na ? Na 1 e
oxidation Cl2 2 e ? 2 Cl reduction
90
Electron Bookkeeping

• for reactions that are not metal nonmetal, or
  do not involve O2, we need a method for
  determining how the electrons are transferred
• chemists assign a number to each element in a
  reaction called an oxidation state that allows
  them to determine the electron flow in the
  reaction
• even though they look like them, oxidation states
  are not ion charges!
• oxidation states are imaginary charges assigned
  based on a set of rules
• ion charges are real, measurable charges

91
Rules for Assigning Oxidation States

• rules are in order of priority
• free elements have an oxidation state 0
• Na 0 and Cl2 0 in 2 Na(s) Cl2(g)
• monatomic ions have an oxidation state equal to
  their charge
• Na 1 and Cl -1 in NaCl
• (a) the sum of the oxidation states of all the
  atoms in a compound is 0
• Na 1 and Cl -1 in NaCl, (1) (-1) 0

92
Rules for Assigning Oxidation States

• (b) the sum of the oxidation states of all the
  atoms in a polyatomic ion equals the charge on
  the ion
• N 5 and O -2 in NO3, (5) 3(-2) -1
• (a) Group I metals have an oxidation state of 1
  in all their compounds
• Na 1 in NaCl
• (b) Group II metals have an oxidation state of
  2 in all their compounds
• Mg 2 in MgCl2

93
Rules for Assigning Oxidation States

• in their compounds, nonmetals have oxidation
  states according to the table below
• nonmetals higher on the table take priority

94
Practice Assign an Oxidation State to Each
Element in the following

• Br2
• K
• LiF
• CO2
• SO42-
• Na2O2

95
Practice Assign an Oxidation State to Each
Element in the following

• Br2 Br 0, (Rule 1)
• K K 1, (Rule 2)
• LiF Li 1, (Rule 4a) F -1, (Rule 5)
• CO2 O -2, (Rule 5) C 4, (Rule 3a)
• SO42- O -2, (Rule 5) S 6, (Rule 3b)
• Na2O2 Na 1, (Rule 4a) O -1, (Rule 3a)

96
Oxidation and ReductionAnother Definition

• oxidation occurs when an atoms oxidation state
  increases during a reaction
• reduction occurs when an atoms oxidation state
  decreases during a reaction

CH4 2 O2 ? CO2 2 H2O -4 1
0 4 2 1 -2
97
OxidationReduction

• oxidation and reduction must occur simultaneously
  
• if an atom loses electrons another atom must take
  them
• the reactant that reduces an element in another
  reactant is called the reducing agent
• the reducing agent contains the element that is
  oxidized
• the reactant that oxidizes an element in another
  reactant is called the oxidizing agent
• the oxidizing agent contains the element that is
  reduced

2 Na(s) Cl2(g) ? 2 NaCl(s) Na is oxidized, Cl
is reduced Na is the reducing agent, Cl2 is the
oxidizing agent
98
Identify the Oxidizing and Reducing Agents in
Each of the Following

• 3 H2S 2 NO3 2 H 3 S 2 NO 4 H2O
• MnO2 4 HBr MnBr2 Br2 2 H2O

99
Identify the Oxidizing and Reducing Agents in
Each of the Following
ox ag
red ag

• 3 H2S 2 NO3 2 H 3 S 2 NO 4 H2O
• MnO2 4 HBr MnBr2 Br2 2 H2O

1 -2 5 -2 1 0
2 -2 1 -2
red ag
ox ag
4 -2 1 -1 2 -1 0
1 -2
100
Combustion Reactions

• Reactions in which O2(g) is a reactant are called
  combustion reactions
• Combustion reactions release lots of energy
• Combustion reactions are a subclass of
  oxidation-reduction reactions

2 C8H18(g) 25 O2(g) ? 16 CO2(g) 18 H2O(g)
101
Combustion Products

• to predict the products of a combustion reaction,
  combine each element in the other reactant with
  oxygen

102
Practice Complete the Reactions

• combustion of C3H7OH(l)
• combustion of CH3NH2(g)

103
Practice Complete the Reactions

• C3H7OH(l) 5 O2(g) ? 3 CO2(g) 4 H2O(g)
• CH3NH2(g) 3 O2(g) ? CO2(g) 2 H2O(g) NO2(g)