Transcript Introductory Chemistry, 2nd Edition Nivaldo Tro

Introductory Chemistry, 3rd Edition
Nivaldo Tro
Chapter 14
Acids and
Bases
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2009, Prentice Hall
Types of Electrolytes
• Salts are water-soluble ionic compounds.
All strong electrolytes.
• Acids form H+1 ions in water solution.
• Bases combine with H+1 ions in water solution.
Increases the OH-1 concentration.
May either directly release OH-1 or pull H+1 off H2O.
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Properties of Acids
• Sour taste.
• React with “active” metals.
 I.e., Al, Zn, Fe, but not Cu, Ag or Au.
2 Al + 6 HCl AlCl3 + 3 H2
 Corrosive.
• React with carbonates, producing CO2.
 Marble, baking soda, chalk, limestone.
CaCO3 + 2 HCl CaCl2 + CO2 + H2O
• Change color of vegetable dyes.
 Blue litmus turns red.
• React with bases to form ionic salts.
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Common Acids
Chemical name
Formula
Uses
Strength
Nitric acid
HNO3
Explosive, fertilizer, dye, glue
Strong
Explosive, fertilizer, dye, glue,
Strong
batteries
Metal cleaning, food prep, ore
Strong
refining, stomach acid
Fertilizer, plastics and rubber,
Moderate
food preservation
Plastics and rubber, food
Weak
preservation, vinegar
Sulfuric acid
H2SO4
Hydrochloric acid
HCl
Phosphoric acid
H3PO4
Acetic acid
HC2H3O2
Hydrofluoric acid
HF
Metal cleaning, glass etching
Weak
Carbonic acid
H2CO3
Soda water
Weak
Boric acid
H3BO3
Eye wash
Weak
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Structures of Acids
• Binary acids have acid
hydrogens attached to a
nonmetal atom.
HCl, HF
Hydrofluoric acid
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Structure of Acids
• Oxyacids have acid
hydrogens attached to
an oxygen atom.
H2SO4, HNO3
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Structure of Acids
• Carboxylic acids have
COOH group.
 HC2H3O2, H3C6H5O3
• Only the first H in the
formula is acidic.
 The H is on the COOH.
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Properties of Bases
• Also known as alkalis.
• Taste bitter.
 Alkaloids = Plant product that is
alkaline.
 Often poisonous.
• Solutions feel slippery.
• Change color of vegetable dyes.
 Different color than acid.
 Red litmus turns blue.
• React with acids to form ionic salts.
 Neutralization.
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Common Bases
Chemical
name
Sodium
hydroxide
Potassium
hydroxide
Calcium
hydroxide
Sodium
bicarbonate
Magnesium
hydroxide
Common
name
Lye,
caustic soda
Caustic
potash
Soap, plastic,
petrol refining
Soap, cotton,
electroplating
Ca(OH)2
Slaked lime
Cement
Strong
NaHCO3
Baking soda
Cooking, antacid
Weak
Mg(OH)2
Milk of
magnesia
Antacid
Weak
Ammonium NH4OH,
hydroxide {NH3(aq)}
Ammonia
water
Detergent,
fertilizer,
explosives, fibers
Weak
Formula
NaOH
KOH
Uses
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Strength
Strong
Strong
9
Structure of Bases
• Most ionic bases contain
OH ions.
 NaOH, Ca(OH)2
• Some contain CO32- ions.
 CaCO3 NaHCO3
• Molecular bases contain
structures that react with
H+.
 Mostly amine groups.
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Arrhenius Theory
• Bases dissociate in water to produce OH- ions and
cations.
 Ionic substances dissociate in water.
NaOH(aq) → Na+(aq) + OH–(aq)
• Acids ionize in water to produce H+ ions and
anions.
 Because molecular acids are not made of ions, they
cannot dissociate.
 They must be pulled apart, or ionized, by the water.
HCl(aq) → H+(aq) + Cl–(aq)
 In formula, ionizable H is written in front.
HC2H3O2(aq) → H+(aq) + C2H3O2–(aq)
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Arrhenius Theory, Continued
HCl ionizes in water,
producing H+ and Cl– ions.
NaOH dissociates in water,
producing Na+ and OH– ions.
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Arrhenius Acid–Base Reactions
• The H+ from the acid combines with the
OH- from the base to make a molecule of
H2O.
It is often helpful to think of H2O as H—OH.
• The cation from the base combines with the
anion from the acid to make a salt.
acid + base → salt + water
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
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Problems with Arrhenius Theory
• Does not explain why molecular substances, like NH3,
dissolve in water to form basic solutions—even though
they do not contain OH– ions.
• Does not explain acid–base reactions that do not take
place in aqueous solution.
• H+ ions do not exist in water. Acid solutions contain
H3O+ ions.
 H+ = a proton!
 H3O+ = hydronium ions.
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Arrow Conventions
• Chemists commonly use two
kinds of arrows in reactions to
indicate the degree of completion
of the reactions.
• A single arrow indicates that all
the reactant molecules are
converted to product molecules at
the end.
• A double arrow indicates that the
reaction stops when only some of
the reactant molecules have been
converted into products.
  in these notes.
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Brønsted–Lowry Theory
•
A Brønsted-Lowry acid–base reaction is any
reaction in which an H+ is transferred.
 Does not have to take place in aqueous solution.
 Broader definition than Arrhenius.
•
Acid is H+ donor; base is H+ acceptor.
 Since H+ is a proton, acid is a proton donor and
base is a proton acceptor.
 Base structure must contain an atom with an
unshared pair of electrons to bond to H+.
•
In the reaction, the acid molecule gives an H+
to the base molecule.
H–A + :B  :A– + H–B+
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Comparing Arrhenius Theory and
Brønsted–Lowry Theory
•
• Arrhenius theory
HCl(aq) 
H+(aq) + Cl−(aq)
HF(aq) 
H+(aq) + F−(aq)
NaOH(aq) 
Na+(aq) + OH−(aq)
NH4OH(aq) 
NH4+(aq) + OH−(aq)
Brønsted–Lowry theory
HCl(aq) + H2O(l) 
Cl−(aq) + H3O+(aq)
HF(aq) + H2O(l) 
F−(aq) + H3O+(aq)
NaOH(aq) + H2O(l) 
Na+(aq) + OH−(aq) + H2O(l)
NH3(aq) + H2O(l) 
NH4+(aq) + OH−(aq)
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Amphoteric Substances
• Amphoteric substances can act as either an acid or
a base.
 They have both transferable H and an atom with a lone
pair.
• HCl(aq) is acidic because HCl transfers an H+ to
H2O, forming H3O+ ions.
 Water acts as base, accepting H+.
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
• NH3(aq) is basic because NH3 accepts an H+ from
H2O, forming OH–(aq).
 Water acts as acid, donating H+.
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
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Brønsted–Lowry
Acid–Base Reactions
• One of the advantages of the Brønsted–Lowry
theory is that it allows reactions to be reversible.
H–A + :B → :A– + H–B+
• The original base has an extra H+ after the
reaction, so it can act as an acid in the reverse
process.
• The original acid has a lone pair of electrons after
the reaction, so it can act as a base in the reverse
process.
:A– + H–B+ → H–A + :B
• A double arrow, , is usually used to indicate a
process that is reversible.
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Conjugate Pairs
• In a Brønsted-Lowry acid-base reaction, the
original base becomes an acid in the reverse
reaction, and the original acid becomes a base in
the reverse process.
• Each reactant and the product it becomes is
called a conjugate pair.
• The original base becomes the conjugate acid;
the original acid becomes the conjugate base.
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Brønsted–Lowry
Acid–Base Reactions
H–A
Acid
+
HCHO2
Acid
H2O +
Acid
:B
Base

+ H2O 
Base
NH3 
Base
:A–
Conjugate
base
CHO2–
Conjugate
base
+
HO–
Conjugate
base
+
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H–B+
Conjugate
acid
+
H3O+
Conjugate
acid
NH4+
Conjugate
acid
21
Conjugate Pairs
In the reaction H2O + NH3  HO– + NH4+:
H2O and HO– constitute an
acid/conjugate–base pair.
NH3 and NH4+ constitute a
base/conjugate–acid pair.
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Example—Identify the Brønsted–Lowry Acids and
Bases and Their Conjugates in the Reaction.
H2SO4 + H2O 
HSO4–
+
H3O+
When the H2SO4 becomes HSO4, it loses an H+, so
H2SO4 must be the acid and HSO4 its conjugate base.
When the H2O becomes H3O+, it accepts an H+, so
H2O must be the base and H3O+ its conjugate acid.
H2SO4
Acid
+
H2O 
Base
HSO4–
Conjugate
base
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+
H3O+
Conjugate
acid
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Example—Identify the Brønsted-Lowry Acids and
Bases and Their Conjugates in the Reaction,
Continued.
HCO3– +
H2O

H2CO3
+
HO–
When the HCO3 becomes H2CO3, it accepts an H+, so
HCO3 must be the base and H2CO3 its conjugate acid.
When the H2O becomes OH, it donates an H+, so
H2O must be the acid and OH its conjugate base.
HCO3– +
Base
H2O 
Acid
H2CO3
Conjugate
acid
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+
HO–
Conjugate
base
24
Practice—Write the Formula for the
Conjugate Acid of the Following:
•H2O
•NH3
•CO32−
•H2PO41−
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Practice—Write the Formula for the
Conjugate Acid of the Following,
Continued:
•H2O
H3O+
•NH3
NH4+
•CO32−
HCO3−
•H2PO41−
H3PO4
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Practice—Write the Formula for the
Conjugate Base of the Following:
•H2O
•NH3
•CO32−
•H2PO41−
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Practice—Write the Formula for the
Conjugate Base of the Following,
Continued:
•H2O
HO−
•NH3
NH2−
•CO32−
Since CO32− does not have an H, it cannot be an acid.
•H2PO41−
HPO42−
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Practice—Write the Equations for the Following
Reacting with Water and Acting as a Monoprotic Acid.
Label the Conjugate Acid and Base.
• HBr
• HSO4-1
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Practice—Write the Equations for the Following
Reacting with Water and Acting as a Monoprotic Acid.
Label the Conjugate Acid and Base, Continued.
• HBr
HBr + H2O  Br-1 + H3O+1
Acid
Base
Conjugate Conjugate
base
acid
• HSO4-1 HSO4-1 + H2O  SO4-2 + H3O+1
Acid
Base
Conjugate Conjugate
base
acid
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Practice—Write the Equations for the Following
Reacting with Water and Acting as a MonoproticAccepting Base.
Label the Conjugate Acid and Base.
• I−
• CO32−
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Practice—Write the Equations for the Following
Reacting with Water and Acting as a MonoproticAccepting Base.
Label the Conjugate Acid and Base, Continued.
•
I−
I− + H2O  HI + OH−
Base Acid Conjugate Conjugate
acid
base
• CO32−
CO32− + H2O  HCO3− + OH−
Base
Acid
Conjugate Conjugate
acid
base
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Neutralization Reactions
• H+ + OH- H2O
• Acid + base salt + water
• Double-displacement reactions.
 Salt = cation from base + anion from
acid.
 Cation and anion charges stay constant.
H2SO4 + Ca(OH)2 → CaSO4 + 2 H2O
• Some neutralization reactions are gas
evolving, where H2CO3 decomposes
into CO2 and H2O.
H2SO4 + 2 NaHCO3 → Na2SO4 + 2 H2O + 2 CO2
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Example 14.2—Write the Equation for the
Reaction of Aqueous Nitric Acid with Aqueous
Calcium Hydroxide.
1. Write the formulas of the reactants.
HNO3(aq) + Ca(OH)2(aq) 
2. Determine the ions present when each reactant
dissociates.
(H+ + NO3−) + (Ca2+ + OH−) 
3. Exchange the ions. H+1 combines with OH-1 to
make H2O(l).
(H+ + NO3−) + (Ca2+ + OH−)  (Ca2+ + NO3−) + H2O(l)
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Example 14.2—Write the Equation for the
Reaction of Aqueous Nitric Acid with Aqueous
Calcium Hydroxide, Continued.
4. Write the formulas of the products.
 Cross charges and reduce.
HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2 + H2O(l)
5. Balance the equation.
 May be quickly balanced by matching the numbers
of H and OH to make H2O.
 Coefficient of the salt is always 1.
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2 + 2 H2O(l)
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Example 14.2—Write the Equation for the
Reaction of Aqueous Nitric Acid with Aqueous
Calcium Hydroxide, Continued.
6. Determine the solubility of the salt.
Ca(NO3)2 is soluble (all NO3- are soluble).
7. Write an (s) after the insoluble products and an
(aq) after the soluble products.
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
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Example—When an Aqueous Solution of
Sodium Carbonate Is Added to an Aqueous
Solution of Nitric Acid, a Gas Evolves.
1. Write the formulas of the reactants.
Na2CO3(aq) + HNO3(aq) 
2. Determine the ions present when each
reactant dissociates.
(Na+ + CO32−) + (H1 + NO3−) 
3. Exchange the ions.
(Na+ + CO32−) + (H+ + NO3−)  (Na+ + NO3−) + (H+ + CO32−)
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Example—When an Aqueous Solution of Sodium
Carbonate Is Added to an Aqueous Solution of Nitric
Acid, a Gas Evolves, Continued.
4. Write the formulas of the products.
 Cross charges and reduce.
Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3
5. Check to see of product decomposes – Yes
 H2CO3 decomposes into CO2(g) + H2O(l)
Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l)
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Example—When an Aqueous Solution of Sodium
Carbonate Is Added to an Aqueous Solution of Nitric
Acid, a Gas Evolves, Continued.
6. Balance the equation.
Na2CO3(aq) + 2 HNO3(aq)  NaNO3 + CO2(g) + H2O(l)
7. Determine the solubility of other product.
NaNO3 is soluble (all Na+ are soluble).
8. Write an (s) after the insoluble products and an
(aq) after the soluble products.
Na2CO3(aq) + 2 HNO3(aq)  NaNO3(aq) + CO2(g) +
H2O(l)
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Practice–Complete Each Reaction.
• Ca(OH)2(s) + H2SO3(aq) 
• HClO3(aq) + Pb(OH)4(s) 
• CaCO3(s) + HNO3(aq) 
• Mg(HCO3)2(aq) + HC2H3O2(aq) 
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Practice–Complete Each Reaction,
Continued.
• Ca(OH)2(s) + H2SO3(aq)  CaSO3(s) + 2 H2O(l)
• 4 HClO3(aq) + Pb(OH)4(s)  Pb(ClO3)4(s) + 4 H2O(l)
• CaCO3(s) + 2 HNO3(aq)  Ca(NO3)2(aq) + CO2(g) + 2 H2O(l)
• Mg(HCO3)2(aq) + 2 HC2H3O2(aq)  Mg(C2H3O2)2(aq) + 2 CO2(g) + 2 H2O(l)
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Acid Reactions:
Acids React with Metals
• Acids react with many metals.
But not all!!
• When acids react with metals, they
produce a salt and hydrogen gas.
3 H2SO4(aq) + 2 Al(s) → Al2(SO4)3(aq) + 3 H2(g)
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Acid Reactions:
Acids React with Metal Oxides
• When acids react with metal oxides, they
produce a salt and water.
3 H2SO4 + Al2O3 → Al2(SO4)3 + 3 H2O
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Example 14.3a—Write an Equation for the
Reaction of Hydroiodic Acid with Potassium
Metal.
1. Write formulas of reactants.
 HI(aq) + K(s) 
2. Identify the type of reaction and predict the
pattern.
 acid + metal  salt + H2(g)
3. Determine the charge on the cation.
 K  K+
4. Determine the formula of the salt.
 K+ + I−  KI
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Example 14.3a—Write an Equation for the
Reaction of Hydroiodic Acid with Potassium
Metal, Continued.
5. Write the skeletal equation.
 HI(aq) + K(s)  KI + H2(g)
6. Check the solubility of the salt.
 KI is soluble.
7. Write (aq) if soluble; (s) if insoluble.
 HI(aq) + K(s)  KI(aq) + H2(g)
8. Balance the equation.
 2 HI(aq) + 2 K(s)  2 KI(aq) + H2(g)
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Example 14.3b—Write an Equation for the
Reaction of Hydrobromic Acid with Sodium
Oxide(s).
1. Write formulas of reactants.
 HBr(aq) + Na2O(s) 
2. Identify the type of reaction and predict the
pattern.
 Acid + metal oxide  salt + H2O(l)
3. Determine the charge on the cation.
 Na+
4. Determine the formula of the salt.
 Na+ + Br−  NaBr
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Example 14.3b—Write an Equation for the
Reaction of Hydrobromic Acid with Sodium
Oxide(s), Continued.
5. Write the skeletal equation.
 HBr(aq) + Na2O(s)  NaBr + H2O(l)
6. Check the solubility of the salt.
 NaBr is soluble.
7. Write (aq) if soluble; (s) if insoluble.
 HBr(aq) + Na2O(s)  NaBr(aq) + H2O(l)
8. Balance the equation.
 2 HBr(aq) + Na2O(s)  2 NaBr(aq) + H2O(l)
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Practice—Complete and Balance the
Following Reactions:
• HCl(aq) + CaO(s) 
• HCl(aq) + Ca(s) 
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Practice—Complete and Balance the
Following Reactions, Continued:
2 HCl(aq) + CaO(s)  CaCl2(aq) + H2O(l)
2 HCl(aq) + Ca(s)  CaCl2(aq) + H2(g)
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Base Reactions
• The reaction all bases have in common is
neutralization of acids.
• Strong bases will react with Al metal to form
sodium aluminate and hydrogen gas.
2 NaOH + 2 Al + 6 H2O → 2 NaAl(OH)4 + 3 H2
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Titration
• Titration is a technique that
uses reaction stoichiometry to
determine the concentration of
an unknown solution.
• Titrant (unknown solution) is
added from a buret.
• Indicators are chemicals that
are added to help determine
when a reaction is complete.
• The endpoint of the titration
occurs when the reaction is
complete.
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Titration, Continued
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Acid–Base Titration
The base solution is the
titrant in the buret.
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 titration’s endpoint,
just enough base has been
added to neutralize all the
acid. At this point, the
indicator changes color.
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Example 14.4—What Is the Molarity of an HCl Solution if
10.00 mL Is required to Titrate 12.54 mL of 0.100 M NaOH?
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
Given: 12.54 mL NaOH, 10.00 mL HCl
Find: M HCl
Solution Map: mL L NaOH mol NaOH mol HCl M  mol
L
M
mL HCl
L HCl
Relationships: M = mol/L, 1 mol NaOH = 1 mol HCl, 1 mL = 0.001L
Solve:
-3
1.2
5
4

10
molHCl
HCl
0.001 L 0.100 mol
NaOH
1 mol
12.54 mL NaOH 
 M

 1.254 103 mol HCl
2 NaOH
1 mL
1L
1.000  101-mol
mol HCl
0.001 L
2
10.00 mLM
HCl


1
.
000

10
L HCl
 0.125
M
HCl
1 mL
Check: The unit is correct, the magnitude is reasonable.
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Practice—What Is the Molarity of a Ba(OH)2 Solution if
37.6 mL Is Required to Titrate 43.8 mL of 0.107 M HCl?
Ba(OH)2(aq) + 2 HCl(aq)  BaCl2(aq) + 2 H2O(l)
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Practice—What Is the Molarity of a Ba(OH)2 Solution if 37.6 mL
Is Required to Titrate 43.8 mL of 0.107 M HCl?
Ba(OH)2(aq) + 2 HCl(aq)  BaCl2(aq) + 2 H2O(l), Continued
Given: 37.6 mL Ba(OH)2, 43.8 mL HCl
Find: M Ba(OH)2
Solution Map: mL L HCl mol HCl mol Ba(OH)2 M  mol
L
mL Ba(OH)2
M
L Ba(OH)2
Relationships: M = mol/L, 1 mol Ba(OH)2= 2 mol HCl, 1 mL = 0.001L
Solve:
2.343  10 -3 mol Ba(OH) 2
43.8 mL HCl 
0.001 L M
0.107
Ba(OH) 2
 mol HCl  1 mol

 2.343 103 mol Ba(OH) 2
-2
1 mL
13
L .76  102 mol
HClBa(OH) 2
mol
0.001 L
2
37.6 mL M
Ba(OH)


3
.
76

10
L Ba(OH) 2
 0.0623
M
Ba(OH)
2
2
1 mL
Check: The unit is correct, the magnitude is reasonable.
Tro's Introductory Chemistry, Chapter
14
64
Strong or Weak
• A strong acid is a strong electrolyte.
 Practically all the acid molecules ionize, →.
• A strong base is a strong electrolyte.
 Practically all the base molecules form OH– ions, either
through dissociation or reaction with water, →.
• A weak acid is a weak electrolyte.
 Only a small percentage of the molecules ionize, .
• A weak base is a weak electrolyte.
 Only a small percentage of the base molecules form OH–
ions, either through dissociation or reaction with water, .
Tro's Introductory Chemistry, Chapter
14
65
Strong Acids
• The stronger the acid, the
more willing it is to donate
H.
HCl  H+ + ClHCl + H2O H3O+ + Cl-
 Use water as the standard
base.
• Strong acids donate
practically all their Hs.
 100% ionized in water.
 Strong electrolyte.
• [H3O+] = [strong acid].
 [ ] = molarity.
Tro's Introductory Chemistry, Chapter
14
66
Strong Acids, Continued
Hydrochloric acid
HCl
Hydrobromic acid
HBr
Hydroiodic acid
HI
Nitric acid
HNO3
Perchloric acid
HClO4
Sulfuric acid
H2SO4
Tro's Introductory Chemistry, Chapter
14
67
Strong Acids, Continued
Pure water
HCl solution
Tro's Introductory Chemistry, Chapter
14
68
Weak Acids
• Weak acids donate a small
fraction of their Hs.
HF  H+ + FHF + H2O  H3O+ + F-
Most of the weak acid
molecules do not donate H
to water.
Much less than 1% ionized
in water.
• [H3O+] << [weak acid].
Tro's Introductory Chemistry, Chapter
14
69
Weak Acids, Continued
Hydrofluoric acid
HF
Acetic acid
HC2H3O2
Formic acid
HCHO2
Sulfurous acid
H2SO3
Carbonic acid
H2CO3
Phosphoric acid
H3PO4
Tro's Introductory Chemistry, Chapter
14
70
Weak Acids, Continued
Pure water
HF solution
Tro's Introductory Chemistry, Chapter
14
71
Degree of Ionization
• The extent to which an acid ionizes in water
depends in part on the strength of the bond
between the acid H+ and anion compared to
the strength of the bond between the acid H+
and the O of water.
HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
Tro's Introductory Chemistry, Chapter
14
72
Relationship Between Strengths of
Acids and Their Conjugate Bases
• The stronger an acid is, the weaker the attraction of the
ionizable H for the rest of the molecule is.
• The better the acid is at donating H, the worse its
conjugate base will be at accepting an H.
Strong acid HCl + H2O → Cl– + H3O+ Weak conjugate base
Weak acid HF + H2O  F– + H3O+ Strong conjugate base
73
Example 14.5—Determine the [H3O+] in the
Following Solutions:
•
1.5 M HCl
 Since HCl is a strong acid, [H3O+] = [HCl] = 1.5 M.
•
3.0 M HC2H3O2
 Since HC2H3O2 is a weak acid, [H3O+] << [HC2H3O2].
 Therefore, [H3O+] << 3.0 M.
Tro's Introductory Chemistry, Chapter
14
74
Practice—Determine the [H3O+] in the
Following Solutions:
Strong
•
•
0.5 M HI
0.1 M
Hydrochloric
acid
HCl
Hydrobromic
HBr
acid
Hydroiodic
HI
HCHO2 acid
Nitric
HNO3
acid
Perchloric
HClO4
acid
Sulfuric acid
H2SO4
Weak
Hydrofluoric
acid
Acetic
acid
Formic
acid
Sulfurous
acid
Carbonic
acid
Phosphoric
acid
HF
HC2H3O2
HCHO2
H2SO3
H2CO3
H3PO4
Practice—Determine the [H3O+] in the
Following Solutions, Continued:
•
0.5 M HI
 Since HI is a strong acid, [H3O+] = [HI] = 0.5 M.
•
0.1 M HCHO2
 Since HCHO2 is a weak acid, [H3O+] << [HCHO2].
 Therefore, [H3O+] << 0.1 M.
Tro's Introductory Chemistry, Chapter
14
76
Strong Bases
• The stronger the base, the more
willing it is to accept H.
NaOH  Na+ + OH-
 Use water as the standard acid.
• Strong bases, practically all
molecules are dissociated into
OH– or accept Hs.
 Strong electrolyte.
 Multi-OH bases completely
dissociated.
• [HO–] = [strong base] x (# OH).
Tro's Introductory Chemistry, Chapter
14
77
Strong Bases, Continued
Lithium hydroxide
LiOH
Sodium hydroxide
NaOH
Potassium hydroxide
KOH
Calcium hydroxide
Ca(OH)2
Strontium hydroxide
Sr(OH)2
Barium hydroxide
Ca(OH)2
Tro's Introductory Chemistry, Chapter
14
78
Weak Bases
• In weak bases, only a small
fraction of molecules accept
Hs.
NH3 + H2O  NH4+ + OH-
 Weak electrolyte.
 Most of the weak base molecules
do not take H from water.
 Much less than 1% ionization in
water.
• [HO–] << [strong base].
Tro's Introductory Chemistry, Chapter
14
79
Weak Bases, Continued
Ammonia
Pyridine
NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
C5H5N(aq) + H2O(l)  C5H5NH+(aq) + OH−(aq)
Methyl amine CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH−(aq)
Ethyl amine
C2H5NH2(aq) + H2O(l)  C2H5NH3+(aq) + OH−(aq)
Bicarbonate
HCO3−(aq) + H2O(l)  H2CO3 (aq) + OH−(aq)
Tro's Introductory Chemistry, Chapter
14
80
Example 14.6—Determine the [OH−] in the
Following Solutions:
•
2.25 M KOH

•
Since KOH is a strong base, [OH−] = [KOH] x 1 = 2.25 M.
0.35 M CH3NH2
 Since CH3NH2 is a weak base, [OH−] << [CH3NH2].
 Therefore, [OH−] << 0.35 M.
•
0.025 M Sr(OH)2
 Since Sr(OH)2 is a strong base, [OH−] = [Sr(OH)2] x 2 = 0.050 M.
Tro's Introductory Chemistry, Chapter
14
81
Practice—Determine the [OH−] in the
Following Solutions:
Weak
Strong
•
•
0.05 M Ba(OH)2
0.01 M C5H5N
Lithium
hydroxide
Sodium
hydroxide
Potassium
hydroxide
Calcium
hydroxide
Strontium
hydroxide
Barium
hydroxide
LiOH
Ammonia
NH3
NaOH
Pyridine
C5H5N
KOH
Methyl
amine
CH3NH2
Ca(OH)2 Ethyl amine C2H5NH2
Sr(OH)2 Bicarbonate
Ba(OH)2
H2CO3
Practice—Determine the [OH−] in the
Following Solutions:
•
0.05 M Ba(OH)2
 Ba(OH)2 is a strong base.
 [OH−] = [Ba(OH)2] x 2 = 0.1 M.
•
0.01 M C5H5N
 C5H5N is a weak base, [OH−] << [C5H5N].
 Therefore, [OH−] << 0.01 M.
Tro's Introductory Chemistry, Chapter
14
83
Autoionization of Water
• Water is actually an extremely weak electrolyte.
 Therefore, there must be a few ions present.
• About 1 out of every 10 million water molecules form
ions through a process called autoionization.
H2O  H+ + OH–
H2O + H2O  H3O+ + OH–
• All aqueous solutions contain both H3O+ and OH–.
 The concentration of H3O+ and OH– are equal in water.
 [H3O+] = [OH–] = 1 x 10-7M at 25 °C in pure water.
Tro's Introductory Chemistry, Chapter
14
84
Ion Product of Water
• The product of the H3O+ and OH–
concentrations is always the same number.
• The number is called the ion product of
water and has the symbol Kw.
• [H3O+] x [OH–] = 1 x 10-14 = Kw.
• As [H3O+] increases, the [OH–] must
decrease so the product stays constant.
Inversely proportional.
Tro's Introductory Chemistry, Chapter
14
85
Acidic and Basic Solutions
• Neutral solutions have equal [H3O+] and [OH–].
[H3O+] = [OH–] = 1 x 10-7
• Acidic solutions have a larger [H3O+] than [OH–].
[H3O+] > 1 x 10-7; [OH–] < 1 x 10-7
• Basic solutions have a larger [OH–] than [H3O+].
[H3O+] < 1 x 10-7; [OH–] > 1 x 10-7
Tro's Introductory Chemistry, Chapter
14
86
Example—Determine the [H3O+] for a 0.00020 M
Ba(OH)2 and Determine Whether the Solution Is
Acidic, Basic, or Neutral.
Ba(OH)2 = Ba2+ + 2 OH– therefore:
[OH–] = 2 x 0.00020 = 0.00040 = 4.0 x 10−4 M

 

H
O
K w  3  OH
14 

1

10
H3O     4.0 104 
OH
Kw
[H3O+] = 2.5 x 10-11 M.
Since [H3O+] < 1 x 10−7, the solution is basic.
Tro's Introductory Chemistry, Chapter
14
87
Practice—Determine the [H3O+]
Concentration and Whether the Solution Is
Acidic, Basic, or Neutral for the Following:
• [OH–] = 0.000250 M
• [OH–] = 3.50 x 10-8 M
• Ca(OH)2 = 0.20 M
Tro's Introductory Chemistry, Chapter
14
88
Practice—Determine the [H3O+] Concentration
and Whether the Solution Is Acidic, Basic, or
Neutral for the Following, Continued:
• [OH–] = 0.000250 M
[H3
O+]
-14
1
x
10
-11
=
=
4.00
x
10
2.50 x 10-4
[H3O+] < [OH-1], therefore, base.
• [OH–] = 3.50 x 10-8 M
[H3
O+]
-14
1
x
10
-7
=
=
2.86
x
10
3.50 x 10-8
• Ca(OH)2 = 0.20 M
[H3
O+]
[H3O+] > [OH-1], therefore, acid.
[OH-1] = 2 x 0.20 = 0.40 M
-14
1
x
10
-14
=
=
2.5
x
10
4.0 x 10-1
[H3O+] < [OH-1], therefore,
base.
Tro's Introductory Chemistry,
Chapter
89
14
Complete the Table
[H+] vs. [OH-]
[H+] 100 10-1
+
H
OH-
10-3
10-5
+
H
OH-
10-7
10-9
+
H
10-11
H+
OH OH
10-13 10-14
H+
OH
[OH-]
Tro's Introductory Chemistry, Chapter
14
90
Complete the Table
+
[H ] vs. [OH ]
[H+] 100 10-1
+
H
OH-
Acid
10-3
10-5
+
H
OH-
[OH-]10-14 10-13 10-11
10-9
10-7
10-9
Base
10-11
H+
+
H
10-13 10-14
H+
OH OH OH
10-7
10-5
10-3
10-1 100
Even though it may look like it, neither H+ nor OH- will ever be 0.
The sizes of the H+ and OH- are not to scale
because the divisions are powers of 10 rather than units.
Tro's Introductory Chemistry, Chapter
14
91
pH
• The acidity/basicity of a solution is often
expressed as pH.
• pH = ─log[H3O+], [H3O+] = 10−pH
 Exponent on 10 with a positive sign.
 pHwater = −log[10-7] = 7.
 Need to know the [H+] concentration to find pH.
• pH < 7 is acidic; pH > 7 is basic; pH = 7 is neutral.
Tro's Introductory Chemistry, Chapter
14
92
pH, Continued
• The lower the pH, the more acidic the solution; the
higher the pH, the more basic the solution.
1 pH unit corresponds to a factor of 10 difference
in acidity.
• Normal range is 0 to 14.
pH 0 is [H+] = 1 M, pH 14 is [OH–] = 1 M.
pH can be negative (very acidic) or larger than 14
(very alkaline).
93
pH of Common Substances
Substance
pH
1.0 M HCl
0.0
0.1 M HCl
1.0
Stomach acid
1.0 to 3.0
Lemons
2.2 to 2.4
Soft drinks
2.0 to 4.0
Plums
2.8 to 3.0
Apples
2.9 to 3.3
Cherries
3.2 to 4.0
Unpolluted rainwater
5.6
Human blood
7.3 to 7.4
Egg whites
7.6 to 8.0
Milk of magnesia (saturated Mg(OH)2)
10.5
Household ammonia
10.5 to 11.5
1.0 M NaOH
14
94
Example—Calculate the pH of a 0.0010 M
Ba(OH)2 Solution and Determine if It Is
Acidic, Basic, or Neutral.
Ba(OH)2 = Ba2+ + 2 OH− therefore,
[OH-] = 2 x 0.0010 = 0.0020 = 2.0 x 10-3 M.
[H3O+] =
1 x 10-14
= 5.0 x 10-12M
-3
2.0 x 10
pH = −log [H3O+] = −log (5.0 x 10-12)
pH = 11.3
pH > 7 therefore, basic.
Tro's Introductory Chemistry, Chapter
14
95
Practice—Calculate the pH of the
Following Strong Acid or Base
Solutions.
• 0.0020 M HCl
• 0.0050 M Ca(OH)2
• 0.25 M HNO3
Tro's Introductory Chemistry, Chapter
14
96
Practice—Calculate the pH of the
Following Strong Acid or Base
Solutions, Continued.
• 0.0020 M HCl therefore, [H3O+] = 0.0020 M.
pH = −log (2.0 x 10-3) = 2.70
• 0.0050 M Ca(OH)2 therefore, [OH–] = 0.010 M.
[H3O+]
−14
1
x
10
−12
=
=
1.0
x
10
1 x 10−2
pH = −log (1.0 x 10−12) = 12.00
• 0.25 M HNO3 therefore, [H3O+] = 0.25 M.
pH = −log (2.5 x 10−1) = 0.60
Tro's Introductory Chemistry, Chapter
14
97
Complete the Table:
pH
pH
[H+] 100 10-1
+
H
OH-
10-3
10-5
+
H
OH-
[OH-]10-14 10-13 10-11
10-9
10-7
+
H
10-9
10-11
H+
OH
OH
10-7
10-5
Tro's Introductory Chemistry, Chapter
14
10-3
10-13 10-14
H+
OH
10-1 100
98
Complete the Table:
pH, Continued
pH
0
1
[H+] 100 10-1
+
H
OH-
Acid
3
10-3
5
7
9
10-5
10-7
10-9
+
H
OH-
[OH-]10-14 10-13 10-11
10-9
+
H
13
10-11
OH
10-5
Tro's Introductory Chemistry, Chapter
14
14
10-13 10-14
H+
H+
OH
10-7
Base
11
OH
10-3
10-1 100
99
Example—Calculate the Concentration
of [H3O+] for a Solution with pH 3.7.
[H3O+] = 10-pH
[H3O+] = 10-3.7
means 0.0001 < [H+1] < 0.001.
[H3O+] = 2 x 10-4 M = 0.0002 M.
Tro's Introductory Chemistry, Chapter
14
100
Practice—Determine the [H3O+] for
Each of the Following:
• pH = 2.7
• pH = 12
• pH = 0.60
Tro's Introductory Chemistry, Chapter
14
101
Practice—Determine the [H3O+] for
Each of the Following, Continued:
• pH = 2.7
[H3O+] = 10−2.7 = 2 x 10−3 M = 0.002 M
• pH = 12
[H3O+] = 10−12 = 1 x 10-12 M
• pH = 0.60
[H3O+] = 10−0.60 = 0.25 M
Tro's Introductory Chemistry, Chapter
14
102
pOH
• The acidity/basicity of a solution may also be
expressed as pOH.
• pH = ─log[OH−], [OH−] = 10−pOH
Exponent on 10 with a positive sign.
pOHwater = −log[10−7] = 7.
Need to know the [OH−] concentration to find
pOH.
• pOH < 7 is acidic; pOH > 7 is basic, pOH = 7
is neutral.
Tro's Introductory Chemistry, Chapter
14
103
pOH, Continued
• The lower the pOH, the more basic the solution;
the higher the pOH, the more acidic the
solution.
1 pOH unit corresponds to a factor of 10
difference in basicity.
• Normal range is 0 to 14.
pOH 0 is [OH−] = 1 M; pOH 14 is [H3O+] = 1 M.
pOH can be negative (very basic) or larger than
14 (very acidic).
• pH + pOH = 14.00.
Tro's Introductory Chemistry, Chapter
14
104
Complete the Table:
pOH
[H+] 100 10-1
+
H
OH-
10-3
10-5
+
H
OH-
[OH-]10-14 10-13 10-11
10-9
10-7
+
H
10-9
10-11
H+
OH
OH
10-7
10-5
10-3
10-13 10-14
H+
OH
10-1 100
pOH
Tro's Introductory Chemistry, Chapter
14
105
Complete the Table:
pOH, Continued
[H+] 100 10-1
+
H
OH-
Acid
10-3
10-5
+
H
OH-
[OH-]10-14 10-13 10-11
pOH 14
13
11
10-7
+
H
10-9
Base
10-11
10-13 10-14
H+
H+
OH
OH
OH
10-9
10-7
10-5
10-3
9
7
5
3
Tro's Introductory Chemistry, Chapter
14
10-1 100
1
0
106
Example—Calculate the pH of a 0.0010 M
Ba(OH)2 Solution and Determine if It Is
Acidic, Basic, or Neutral.
Ba(OH)2 = Ba+2 + 2 OH− therefore,
[OH−] = 2 x 0.0010 = 0.0020 = 2.0 x 10−3 M.
pOH = -log [OH−] = -log (2.0 x 10−3)
pOH = 2.70
pH = 14.00 - pOH = 14.00 - 2.70
pH = 11.30
pH > 7 therefore, basic.
Tro's Introductory Chemistry, Chapter
14
107
Practice—Calculate the pOH and pH of the
Following Strong Acid or Base Solutions.
• 0.0020 M KOH
• 0.0050 M Ca(OH)2
• 0.25 M HNO3
Tro's Introductory Chemistry, Chapter
14
108
Practice—Calculate the pOH and pH of the
Following Strong Acid or Base Solutions,
Continued.
• 0.0020 M KOH therefore, [OH–] = 0.0020 M.
pOH = −log (2.0 x 10-3) = 2.70
pH = 14.00 – 2.70 = 11.30
• 0.0050 M Ca(OH)2 therefore, [OH–] = 0.010 M.
pOH = −log (1.0 x 10-2) = 2.00
pH = 14.00 – 2.00 = 12.00
• 0.25 M HNO3 therefore, [H3O+] = 0.25 M.
pH = −log (2.5 x 10-1) = 0.60
pOH = 14.00 – 0.60 = 13.40
Tro's Introductory Chemistry, Chapter
14
109
Buffers
• Buffers are solutions that resist changing pH
when small amounts of acid or base are
added.
• They resist changing pH by neutralizing
added acid or base.
• Buffers are made by mixing together a weak
acid and its conjugate base.
Or weak base and its conjugate acid.
Tro's Introductory Chemistry, Chapter
14
110
How Buffers Work
• The weak acid present in the buffer mixture
can neutralize added base.
• The conjugate base present in the buffer
mixture can neutralize added acid.
• The net result is little to no change in the
solution pH.
Tro's Introductory Chemistry, Chapter
14
111
How Buffers Work, Continued
H2O
New
HA
HA
HA

A−−
+
H3 O+
Added
H3O+
Tro's Introductory Chemistry, Chapter
14
112
How Buffers Work, Continued
H2O
New
A−
HA
HA

A−−
+
H3 O+
Added
HO−
Tro's Introductory Chemistry, Chapter
14
113
A Buffer Made from Acetic Acid
and Sodium Acetate
• A buffer solution with a pH of 4.75 can be made by
mixing equal volumes of 1 M HC2H3O2 and 1 M
NaC2H3O2.
• Adding 10 mL of 0.1 M HCl to 1 L of this solution
will give a solution with a pH of 4.75.
 Adding 10 mL of 0.1 M HCl to 1 L of distilled water will
give a solution with pH of 3.0.
• Adding 10 mL of 0.1 M NaOH to 1 L of this solution
will give a solution with a pH of 4.75.
 Adding 10 mL of 0.1 M NaOH to 1 L of distilled water will
give a solution with pH of 11.0.
Tro's Introductory Chemistry, Chapter
14
114
Acetic Acid/Acetate Buffer
Tro's Introductory Chemistry, Chapter
14
115
Nonmetal Oxides Are Acidic
• Nonmetal oxides react with water to form
acids.
• Causes acid rain.
CO2 (g) + H2O(l) → H2CO3(aq)
2 SO2(g) + O2(g) + 2 H2O(l) → 2 H2SO4(aq)
4 NO2(g) + O2(g) + 2 H2O(l) → 4 HNO3(aq)
Tro's Introductory Chemistry, Chapter
14
116
What Is Acid Rain?
• Natural rain water has a pH of 5.6.
Naturally slightly acidic due mainly to CO2.
• Rain water with a pH lower than 5.6 is
called acid rain.
• Acid rain is linked to damage in ecosystems
and structures.
Tro's Introductory Chemistry, Chapter
14
117
What Causes Acid Rain?
• Many natural and pollutant gases dissolved in the air
are nonmetal oxides.
 CO2, SO2, NO2.
• Nonmetal oxides are acidic.
CO2 + H2O  H2CO3
2 SO2 + O2 + 2 H2O  2 H2SO4
• Processes that produce nonmetal oxide gases as waste
increase the acidity of the rain.
 Natural—volcanoes and some bacterial action.
 Man-made—combustion of fuel.
• Weather patterns may cause rain to be acidic in regions
other than where the nonmetal oxide is produced.
Tro's Introductory Chemistry, Chapter
14
118
pH of Rain in Different Regions
Sources of SO2 from Utilities
Damage from Acid Rain
• Acids react with metals and materials that contain
carbonates.
• Acid rain damages bridges, cars, and other metallic
structures.
• Acid rain damages buildings and other structures made
of limestone or cement.
• Acidifies lakes affecting aquatic life.
• Dissolves and leaches more minerals from soil.
 Making it difficult for trees.
Tro's Introductory Chemistry, Chapter
14
121
Damage from Acid Rain
circa 1995
circa 1935
Tro's Introductory Chemistry, Chapter
14
122
pH of Rain in Different Regions
Tro's Introductory Chemistry, Chapter
14
123
Sources of SO2 from Utilities
124
Acid Rain Legislation
• 1990 Clean Air Act attacks acid rain.
Forces utilities to reduce SO2.
• The result is acid rain in the northeast is
stabilized and begins to be reduced.
Tro's Introductory Chemistry, Chapter
14
125