chm1045 chapter 4b
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Chapter 4Chemical
Quantities and Aqueous Reactions
2008, Prentice Hall
Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro
Tro, Chemistry: A Molecular Approach 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 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O
We can come up with lots of unit factors
8 18(l) 8 18(l) 2(g)
2(g) 2(g) 2 (g)
2 C H 2 C H 16 CO 1
25 O 16 CO 18 H O= = =
Tro, Chemistry: A Molecular Approach 3
Predicting Amounts from Stoichiometry• 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
2188
2188 CO moles 176
HC mol 2CO mol 16HC moles 22.0 =×
2
Tro, Chemistry: A Molecular Approach 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)
Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline
since 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense
1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18 = 16 mol CO2
3.4 x 1015 g C8H18g CO2
Check:
Solution:
Concept Plan:
Relationships:
Given:Find:
g 114.22mol 1
216
2
2
188
2
188
188188
15
CO g 101.0
CO mol 1CO g 44.01
HC mol 2CO mol 16
HC g 114.22HC mol 1HC g 10.43
×=
××××
188
2HC mol 2
CO mol 61
g C8H18 mol CO2 g CO2mol C8H18
mol 1g 44.01
Tro, Chemistry: A Molecular Approach 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)
density of water = 1.00 g/mL
3
Tro, Chemistry: A Molecular Approach 7
Practice
OH mL 5.4
OH g 1.00OH mL 1 x
OH mole 1OH g 18.0 x
OHC mole 1 OH mole 6 x
g 10 x 80.1OHC mole 1 x OHC g 0.9
2
2
2
2
2
6126
22
61266126
=
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)
Tro, Chemistry: A Molecular Approach 8
• Limiting Reagents: The extent to which a reaction takes place depends on the reactant that is present in limiting amounts—the limiting reagent.
The limiting reagent is the one that runs out first. It is the one that limits the extent of the
reaction.
•the limiting reactant gets completely consumed
•reactants not completely consumed are called excess reactants
Tro, Chemistry: A Molecular Approach 9
Limiting Reactant• the reactant that limits the amount of product is
called the limiting reactantsometimes called the limiting reagentthe limiting reactant gets completely consumed
• reactants not completely consumed are called excess reactants
4
Tro, Chemistry: A Molecular Approach 10
Limiting Reactant• the amount of product that can be made from
the limiting reactant is called the theoretical yield
• the amount of product that is made in a reaction is called the actual yield
• the efficiency of product recovery is generally given as the percent yield
%100yield ltheoretica
yield actualYieldPercent ×=
Tro, Chemistry: A Molecular Approach 11
• Limiting Reagent Calculation: Lithium oxide is a drying agent used on the space shuttle. If 80.0 kg of water is to be removed and 65 kg of lithium oxide is available, which reactant is limiting?
Li2O(s) + H2O(l) → 2 LiOH(s)
• MM(Li2O) = 29.88 g/mol
• MM(H2O) = 18.02 g/mol
•The best way to calculate this is to convert all reactants to moles.
•Divide by the coefficient of that reagent in the balanced chemical equation
•The smallest quotient is the limiting reagent.
All the calculations are then based on the limiting reagent.
Tro, Chemistry: A Molecular Approach 12
• 25.0 g of Lead (II) nitrate and 18.2 g of sodium chloride are mixed in 250.00 mL of solution. How many grams of Lead (II) chloride are made.? We first need a balanced chemical equation.
Pb(NO3)2 (aq) + 2NaCl (aq) PbCl2 (s) + 2 NaNO3 (aq)
Which reactant do we base the calculation on?
The Limiting Reactant.
mass Limiting Reactant
moles limiting reactant
moles Desired product
mass Desired product
5
Tro, Chemistry: A Molecular Approach 13
mass Limiting Reactant
moles limiting reactant
moles Desired product
mass Desired product
1. Make sure have a balanced chemical reaction
2. Determine limiting reactant
3. Every subsequent calculation is dependent on the limiting reactant
Tro, Chemistry: A Molecular Approach 14
• Limiting Reagent Calculation: Cisplatin is an anti-cancer agent prepared as follows:
•K2PtCl4 + 2 NH3 → Pt(NH3)2Cl2 + 2 KCl
• If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react:
(a) which is the limiting reagent? (b) How many grams of cisplatin are formed?
• MM(K2PtCl4) = 415.08 g/mol MM(NH3) = 17.04 g/mol
Tro, Chemistry: A Molecular Approach 15
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)
6
Tro, Chemistry: A Molecular Approach 16
Solutions• when table salt is mixed with water, it seems to disappear,
or become a liquid – the mixture is homogeneousthe 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
Tro, Chemistry: A Molecular Approach 17
Describing Solutions• since solutions are mixtures, the composition can
vary from one sample to anotherpure substances have constant compositionsalt 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
Tro, Chemistry: A Molecular Approach 18
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
7
Tro, Chemistry: A Molecular Approach 19
Solution ConcentrationMolarity
• moles of solute per 1 liter of solution• used because it describes how many molecules
of solute in each liter of solution
L)(in solution ofamount moles) (in solute ofamount M molarity, =
Tro, Chemistry: A Molecular Approach 20
Preparing 1 L of a 1.00 M NaCl Solution
Example 4.5 – Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution
since most solutions are between 0 and 18 M, the answer makes sense
1 mol KBr = 119.00 g, M = moles/L
25.5 g KBr, 1.75 L solutionMolarity, M
Check:• Check
Solution:• Follow the Concept Plan to Solve the problem
Concept Plan:
Relationships:
• Strategize
Given:Find:
• Sort Information
g 119.00mol 1
M 0.122L 1.75
KBr mol 29421.0solution L
KBr moles M molarity,
KBr mol 2940.21 KBr g 119.00
KBr mol 1KBr g 5.52
===
=×
g KBr mol KBr
L sol’nML
mol M =
8
Tro, Chemistry: A Molecular Approach 22
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
solution L 1sugar mol 2
sugar mol 2solution L 1
Example 4.6 – How many liters of 0.125 M NaOH contains 0.255 mol NaOH?
since each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should
require a little more than 2 L
0.125 mol NaOH = 1 L solution
0.125 M NaOH, 0.255 mol NaOHliters, L
Check:• Check
Solution:• Follow the Concept Plan to Solve the problem
Concept Plan:
Relationships:
• Strategize
Given:Find:
• Sort Information
NaOH mol 0.125solution L 1
solution L 2.04 NaOH mol 0.125
solution L 1NaOH mol 552.0 =×
mol NaOH L sol’n
Tro, Chemistry: A Molecular Approach 24
Dilution• often, solutions are stored as concentrated stock
solutions• to make solutions of lower concentrations from these
stock solutions, more solvent is addedthe 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 proportionalM1·V1 = M2·V2
9
Example 4.7 – To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH?
since the solution is diluted by a factor of 5, the volume should increase by a
factor of 5, and it does
M1V1 = M2V2
V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 MV2, L
Check:• Check
Solution:• Follow the Concept Plan to Solve the problem
Concept Plan:
Relationships:
• Strategize
Given:Find:
• Sort Information
22
11 VM
VM=
•
( )L 1.00
Lmol3.00
L 200.0L
mol15.0=
⎟⎠⎞
⎜⎝⎛
•⎟⎠⎞
⎜⎝⎛
V1, M1, M2 V2
Tro, Chemistry: A Molecular Approach 26
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
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)
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
1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 = 2 mol KCl
0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2
L KCl
Check:• Check
Solution:• Follow the Concept Plan to Solve the problem
Concept Plan:
Relationships:
• Strategize
Given:
Find:
• Sort Information
23)Pb(NO L 1mol 0.175
KCl L .3500 mol .1500
KCl L 1)Pb(NO mol 1
KCl mol 2)Pb(NO L 1
mol 0.175)Pb(NO L .15002323
23
=
×××
23)Pb(NO mol 1KCl mol 2
L Pb(NO3)2 mol KCl L KClmol Pb(NO3)2
mol0.150KCl L 1
10
28
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
Tro, Chemistry: A Molecular Approach 29
Table Salt Dissolving in WaterEach ion is attracted to the surrounding water molecules and pulled off and away from the crystalWhen 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
Tro, Chemistry: A Molecular Approach 30
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
11
Tro, Chemistry: A Molecular Approach 31
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
Tro, Chemistry: A Molecular Approach 32
Salt vs. Sugar Dissolved in Water
ionic compounds dissociate into ions when they dissolve
molecular compounds do not dissociate when they dissolve
Tro, Chemistry: A Molecular Approach 33
Acids• acids are molecular compounds that ionize when they
dissolve in waterthe molecules are pulled apart by their attraction for the waterwhen 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 acidsHCl(aq) → H+(aq) + Cl-(aq)
• acids that only ionize a small percentage are called weak acids
HF(aq) ⇔ H+(aq) + F-(aq)
12
Tro, Chemistry: A Molecular Approach 34
Strong and Weak Electrolytes• strong electrolytes are materials that dissolve
completely as ionsionic compounds and strong acidstheir solutions conduct electricity well
• weak electrolytes are materials that dissolve mostly as molecules, but partially as ions
weak acidstheir 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)
Tro, Chemistry: A Molecular Approach 35
Classes of Dissolved Materials
Tro, Chemistry: A Molecular Approach 36
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 insolubleNaCl is soluble in water, AgCl is insoluble in waterthe degree of solubility depends on the temperatureeven insoluble compounds dissolve, just not enough to be meaningful
13
Tro, Chemistry: A Molecular Approach 37
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
Tro, Chemistry: A Molecular Approach 38
Ag+, Ca2+, Sr2+, Ba2+, Pb2+SO42–
Ag+, Hg22+, Pb2+Cl–, Br–, I–
noneNO3–, C2H3O2
–
noneLi+, Na+, K+, NH4+
Exceptions(when combined with ions on the left the compound is insoluble)
Compounds Containing the Following Ions are Generally Soluble
Solubility RulesCompounds that Are Generally Soluble in Water
Tro, Chemistry: A Molecular Approach 39
Li+, Na+, K+, NH4+CO3
2–, PO43–
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
S2–
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
OH–
Exceptions(when combined with ions on the left the compound is soluble or slightly soluble)
Compounds Containing the Following Ions are Generally Insoluble
Solubility RulesCompounds that Are Generally Insoluble
14
Tro, Chemistry: A Molecular Approach 40
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
41
2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)
Tro, Chemistry: A Molecular Approach 42
No Precipitate Formation = No Reaction
KI(aq) + NaCl(aq) → KCl(aq) + NaI(aq)all ions still present, ∴ no reaction
15
Tro, Chemistry: A Molecular Approach 43
Process for Predicting the Products ofa Precipitation Reaction
1. Determine what ions each aqueous reactant has2. 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 WaterUse the solubility rulesIf product is insoluble or slightly soluble, it will precipitate
4. If neither product will precipitate, write no reaction after the arrow
Tro, Chemistry: A Molecular Approach 44
Process for Predicting the Products ofa 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
Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of
nickel(II) chloride1. 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+ + CO3
2-) + (Ni2+ + Cl-) → (K+ + Cl-) + (Ni2+ + CO32-)
c) Write the formulas of the productscross charges and reduce
K2CO3(aq) + NiCl2(aq) → KCl + NiCO3
16
Tro, Chemistry: A Molecular Approach 46
3. Determine the solubility of each productKCl is soluble
NiCO3 is insoluble4. If both products soluble, write no reaction
does not apply since NiCO3 is insoluble
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
Tro, Chemistry: A Molecular Approach 47
5. Write (aq) next to soluble products and (s) next to insoluble products
K2CO3(aq) + NiCl2(aq) → KCl(aq) + NiCO3(s)6. Balance the EquationK2CO3(aq) + NiCl2(aq) → 2 KCl(aq) + NiCO3(s)
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
Tro, Chemistry: A Molecular Approach 48
Ionic Equations• equations which describe the chemicals put into the water
and the product molecules are called molecular equations2 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 ionssoluble salts, strong acids, strong bases
insoluble substances, weak electrolytes, and nonelectrolyteswritten in molecule form
solids, liquids, and gases are not dissolved, therefore molecule form2K+1
(aq) + 2OH-1(aq) + Mg+2
(aq) + 2NO3-1
(aq) → 2K+1(aq) + 2NO3
-1(aq) + Mg(OH)2(s)
17
Tro, Chemistry: A Molecular Approach 49
Ionic Equations• ions that are both reactants and products are called
spectator ions
2K+1(aq) + 2OH-1
(aq) + Mg+2(aq) + 2NO3
-1(aq) → 2K+1
(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) + Mg+2
(aq) → Mg(OH)2(s)
Tro, Chemistry: A Molecular Approach 50
Acid-Base Reactions• also called neutralization reactions because the
acid and base neutralize each other’s properties2 HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2 H2O(l)
• the net ionic equation for an acid-base reaction isH+(aq) + OH−(aq) → H2O(l)
as long as the salt that forms is soluble in water
Tro, Chemistry: A Molecular Approach 51
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
18
Common AcidsChemical Name Formula Uses Strength Perchloric Acid HClO4 explosives, catalyst Strong
Nitric Acid HNO3 explosives, fertilizer, dye, glue Strong
Sulfuric Acid H2SO4 explosives, fertilizer, dye, glue,
batteries Strong
Hydrochloric Acid HCl metal cleaning, food prep, ore refining, stomach acid Strong
Phosphoric Acid H3PO4 fertilizer, plastics & rubber,
food preservation Moderate
Chloric Acid HClO3 explosives Moderate
Acetic Acid HC2H3O2 plastics & rubber, food preservation, vinegar Weak
Hydrofluoric Acid HF metal cleaning, glass etching Weak Carbonic Acid H2CO3 soda water Weak
Hypochlorous Acid HClO sanitizer Weak Boric Acid H3BO3 eye wash Weak
Tro, Chemistry: A Molecular Approach 53
Common BasesChemical
Name Formula Common Name Uses Strength
sodium hydroxide NaOH lye,
caustic soda soap, plastic,
petrol refining Strong
potassium hydroxide KOH caustic potash soap, cotton,
electroplating Strong
calcium hydroxide Ca(OH)2 slaked lime cement Strong
sodium bicarbonate NaHCO3 baking soda cooking, antacid Weak
magnesium hydroxide Mg(OH)2
milk of magnesia antacid Weak
ammonium hydroxide
NH4OH, {NH3(aq)}
ammonia water
detergent, fertilizer,
explosives, fibers Weak
Tro, Chemistry: A Molecular Approach 54
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
19
Tro, Chemistry: A Molecular Approach 55
Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric
acid with aqueous calcium hydroxide1. Write the formulas of the reactants
HNO3(aq) + Ca(OH)2(aq) →2. Determine the possible products
a) Determine the ions present when each reactant dissociates(H+ + NO3
-) + (Ca+2 + OH-) →b) Exchange the ions, H+1 combines with OH-1 to make H2O(l)
(H+ + NO3-) + (Ca+2 + OH-) → (Ca+2 + NO3
-) + H2O(l)c) Write the formula of the salt
cross the charges (H+ + NO3
-) + (Ca+2 + OH-) → Ca(NO3)2 + H2O(l)
Tro, Chemistry: A Molecular Approach 56
3. Determine the solubility of the saltCa(NO3)2 is soluble
4. 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)5. Balance the equation2 HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2 H2O(l)
Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric
acid with aqueous calcium hydroxide
Tro, Chemistry: A Molecular Approach 57
Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric
acid with aqueous calcium hydroxide6. Dissociate all aqueous strong electrolytes to
get complete ionic equationnot H2O
2 H+(aq) + 2 NO3-(aq) + Ca+2(aq) + 2 OH-(aq) →
Ca+2(aq) + 2 NO3-(aq) + H2O(l)
7. Eliminate spectator ions to get net-ionic equation
2 H+1(aq) + 2 OH-1(aq) → 2 H2O(l)H+1(aq) + OH-1(aq) → H2O(l)
20
Tro, Chemistry: A Molecular Approach 58
Titration• often in the lab, a solution’s 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
Tro, Chemistry: A Molecular Approach 59
Acid-Base Titrations• the difficulty is determining when there has been just
enough titrant added to complete the reactionthe 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
Tro, Chemistry: A Molecular Approach 60
Titration
21
Tro, Chemistry: A Molecular Approach 61
TitrationThe base solution is thetitrant in the burette.As the base is added tothe acid, the H+ reacts withthe OH– to form water. But there is still excess acid present so the colordoes not change.
At the titration’s endpoint,just enough base has been added to neutralize all theacid. At this point the indicator changes color.
Tro, Chemistry: A Molecular Approach 62
Example 4.14:The titration of 10.00 mL of HClsolution 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
Tro, Chemistry: A Molecular Approach 63
• Write down the quantity to find, and/or its units.Find: concentration HCl, M
InformationGiven: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Example 4.14:The titration of 10.00 mL of HClsolution 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?
22
Tro, Chemistry: A Molecular Approach 64
• Collect Needed Equations and Conversion Factors:HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
∴ 1 mole HCl = 1 mole NaOH0.100 M NaOH ∴0.100 mol NaOH ≡ 1 L sol’n
InformationGiven: 10.00 mL HCl
12.54 mL of 0.100 M NaOHFind: M HCl
Example 4.14:The titration of 10.00 mL of HClsolution 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?
solution literssolute molesMolarity =
Tro, Chemistry: A Molecular Approach 65
• Write a Concept Plan:
mLNaOH
LNaOH
molNaOH
NaOH L 1NaOH mol 1000.
mL 1L 0010.
molHCl
NaOH mol 1HCl mol 1
InformationGiven: 10.00 mL HCl
12.54 mL of 0.100 M NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 LM = mol/L
Example 4.14:The titration of 10.00 mL of HClsolution 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?
mLHCl
LHCl
mL 1L 0010. HClliters
HCl molesMolarity =
Tro, Chemistry: A Molecular Approach 66
NaOH mole 1HCl mol 1
L 1NaOH mol 1000
mL 1L 0.001NaOH mL 2.541 ×××
.
• Apply the Solution Map:
= 1.25 x 10-3 mol HCl
InformationGiven: 10.00 mL HCl
12.54 mL of 0.100 M NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 LM = mol/L
CP: mL NaOH → L NaOH → mol NaOH → mol HCl;mL HCl → L HCl & mol ⇒ M
Example:The titration of 10.00 mL of HClsolution 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?
23
Tro, Chemistry: A Molecular Approach 67
HCl L 010000mL 1
L 0.001NaOH mL 0.001 .=×
• Apply the Concept Plan:
InformationGiven: 10.00 mL HCl
12.54 mL NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 LM = mol/L
CP: mL NaOH → L NaOH → mol NaOH → mol HCl;mL HCl → L HCl & mol ⇒ M
Example:The titration of 10.00 mL of HClsolution 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?
M 1250HCl L 0.01000
HCl moles 10 x 1.25Molarity-3
.==
Tro, Chemistry: A Molecular Approach 68
• 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 thanNaOH solution, so the HCl should be more concentrated.
InformationGiven: 10.00 mL HCl
12.54 mL NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 LM = mol/L
CP: mL NaOH → L NaOH → mol NaOH → mol HCl;mL HCl → L HCl & mol ⇒ M
Example:The titration of 10.00 mL of HClsolution 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?
Tro, Chemistry: A Molecular Approach 69
Gas Evolving Reactions• Some reactions form a gas directly from the ion
exchangeK2S(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)
24
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NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l)
Tro, Chemistry: A Molecular Approach 71
Compounds that UndergoGas Evolving Reactions
K2SO3(aq) + 2HCl(aq) →2KCl(aq) + SO2(g) + H2O(l)
SO2yesH2SO3acidmetalnSO3
metal HSO3
K2CO3(aq) + 2HCl(aq) →2KCl(aq) + CO2(g) + H2O(l)
CO2yesH2CO3acidmetalnCO3,metal HCO3
KOH(aq) + NH4Cl(aq) →KCl(aq) + NH3(g) + H2O(l)
NH3yesNH4OHbase(NH4)nanion
no
Decom-pose?
K2S(aq) + 2HCl(aq) →2KCl(aq) + H2S(g)
H2SH2SacidmetalnS,metal HS
ExampleGasFormed
Ion ExchangeProduct
ReactingWith
ReactantType
Tro, Chemistry: A Molecular Approach 72
Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution
of nitric acid, a gas evolves1. Write the formulas of the reactants
Na2CO3(aq) + HNO3(aq) →2. Determine the possible products
a) Determine the ions present when each reactant dissociates(Na+1 + CO3
-2) + (H+1 + NO3-1) →
b) Exchange the anions(Na+1 + CO3
-2) + (H+1 + NO3-1) → (Na+1 + NO3
-1) + (H+1 + CO3-2)
c) Write the formula of compoundscross the charges
Na2CO3(aq) + HNO3(aq) → NaNO3 + H2CO3
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Tro, Chemistry: A Molecular Approach 73
3. Check to see either product H2S - No4. Check to see if either product decomposes –
YesH2CO3 decomposes into CO2(g) + H2O(l)
Na2CO3(aq) + HNO3(aq) → NaNO3 + CO2(g) + H2O(l)
Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution
of nitric acid, a gas evolves
Tro, Chemistry: A Molecular Approach 74
5. Determine the solubility of other productNaNO3 is soluble
6. 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)
7. Balance the equationNa2CO3(aq) + 2 HNO3(aq) → 2 NaNO3 + CO2(g) + H2O(l)
Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution
of nitric acid, a gas evolves
Tro, Chemistry: A Molecular Approach 75
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 reactionsmany involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
26
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Combustion as Redox2 H2(g) + O2(g) → 2 H2O(g)
Tro, Chemistry: A Molecular Approach 77
Redox without Combustion2 Na(s) + Cl2(g) → 2 NaCl(s)
2 Na → 2 Na+ + 2 e−
Cl2 + 2 e− → 2 Cl−
Tro, Chemistry: A Molecular Approach 78
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 Na+
2O– (s)2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
27
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Oxidation and Reduction• in order to convert a free element into an ion, the
atoms must gain or lose electronsof 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 Na+Cl–(s)Na → Na+ + 1 e– oxidationCl2 + 2 e– → 2 Cl– reduction
Leo
Ger
Tro, Chemistry: A Molecular Approach 80
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
Tro, Chemistry: A Molecular Approach 81
Rules for Assigning Oxidation States• rules are in order of priority1. free elements have an oxidation state = 0
Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)2. monatomic ions have an oxidation state equal
to their chargeNa = +1 and Cl = -1 in NaCl
3. (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
28
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Rules for Assigning Oxidation States3. (b) the sum of the oxidation states of all the atoms in
a polyatomic ion equals the charge on the ionN = +5 and O = -2 in NO3
–, (+5) + 3(-2) = -1
4. (a) Group I metals have an oxidation state of +1 in all their compounds
Na = +1 in NaCl
4. (b) Group II metals have an oxidation state of +2 in all their compounds
Mg = +2 in MgCl2
Tro, Chemistry: A Molecular Approach 83
Rules for Assigning Oxidation States5. in their compounds, nonmetals have oxidation
states according to the table belownonmetals higher on the table take priority
NH3-3Group 5ACS2-2Group 6ACCl4-1Group 7ACO2-2OCH4+1HCF4-1F
ExampleOxidation StateNonmetal
Tro, Chemistry: A Molecular Approach 84
Practice – Assign an Oxidation State to Each Element in the following
• Br2
• K+
• LiF
• CO2
• SO42-
• Na2O2
29
Tro, Chemistry: A Molecular Approach 85
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)
Tro, Chemistry: A Molecular Approach 86
Oxidation and ReductionAnother Definition
• oxidation occurs when an atom’s oxidation state increases during a reaction
• reduction occurs when an atom’s oxidation state decreases during a reaction
CH4 + 2 O2 → CO2 + 2 H2O-4 +1 0 +4 –2 +1 -2
oxidationreduction
Tro, Chemistry: A Molecular Approach 87
Oxidation–Reduction• 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 agentthe 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 Na+Cl–(s)Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent
30
Tro, Chemistry: A Molecular Approach 88
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
Tro, Chemistry: A Molecular Approach 89
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
+1 -2 +5 -2 +1 0 +2 -2 +1 -2
ox agred ag
+4 -2 +1 -1 +2 -1 0 +1 -2
oxidationreduction
oxidationreduction
red agox ag
Tro, Chemistry: A Molecular Approach 90
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)
31
Tro, Chemistry: A Molecular Approach 91
Combustion Products• to predict the products of a combustion
reaction, combine each element in the other reactant with oxygen
M2On(s)contains metal
NO(g) or NO2(g)contains N
SO2(g)contains S
H2O(g)contains H
CO2(g)contains C
Combustion ProductReactant
Tro, Chemistry: A Molecular Approach 92
Practice – Complete the Reactions
• combustion of C3H7OH(l)
• combustion of CH3NH2(g)
Tro, Chemistry: A Molecular Approach 93
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)
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