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Chapt. 18 – Acids and Bases
18.1 Acids and Bases: An Introduction
18.2 Strengths of Acids and Bases
18.3 Hydrogen Ions and pH
18.4 Neutralization
Section 18.1 Introduction to Acids and Bases
• Identify the physical and chemical properties of acids and bases.
• Classify solutions as acidic, basic, or neutral.
• Compare the Arrhenius, Brønsted-Lowry, and Lewis models of acids and bases.
Different models help describe the behavior of acids and bases.
Section 18.1 Introduction to Acids and Bases
Key Concepts
• The concentrations of hydrogen ions and hydroxide ions determine whether an aqueous solution is acidic, basic, or neutral.
• An Arrhenius acid must contain an ionizable hydrogen atom. An Arrhennius base must contain an ionizable hydroxide group.
• A Brønsted-Lowry acid is a hydrogen ion donor. A Brønsted-Lowry base is a hydrogen ion acceptor.
• A Lewis acid accepts an electron pair. A Lewis base donates an electron pair.
Acid/Base Properties
Acids have sharp taste or are sour• Carbonic acid (CO2 in water) - soda
• Phosphoric acid (colas)• Citric acid (lemons, citrus fruits)• Acetic acid (vinegar)
Bases taste bitter, feel slippery
Taste and feel are extremely dangerous ways to classify chemicals!
Industrial Chemicals - US
Acids & bases occupy 7 of top 20 spots in list of most widely produced industrial chemicals
These are commodity chemicals – relatively cheap
Sulfuric has many industrial uses
Typical ReactionsSingle replacement reaction: reaction of metals & acids to form metal salt and hydrogen
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
Double replacement reaction: generation of gas & water
NaHCO3(s) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g)
Base reacting with acid: neutralization
Practice
Reactions of acids
Problems 1(a-b), 2 page 635
Problems 64(a-b) page 672
Problems 1 – 2 page 989
Properties - Litmus
AcidBlue litmus Pink
BaseRed litmus Blue
Properties – Conductivity
Pure water essentially nonconductor
Any electrolyte (material that produces ions when dissolved in water) raises conductivity
Acids and bases are electrolytes – raise conductivity
Relation to Ions in SolutionAll aqueous solutions contain hydrogen ions (H+) and hydroxide ions (OH-)Acid – more H+ than OH-
Base – more OH- than H+
Neutral – equal concentrations
+
-
Water – Self Ionization
Water produces small but equal numbers of H+ and OH- (it is neutral)
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
H3O+ = hydronium ion (water molecule covalently bonded to H+)
Mostly OK to use H+ instead of H3O+
H2O(l) H+(aq) + OH-(aq)
Arrhenius Acid/Base Model
Acid: Contains hydrogen & ionizes to produce H+ in aqueous solution
HCl(g) H+(aq) + Cl-(aq)
Base: Contains OH- group & ionizes to produce OH- in aqueous solution
NaOH(s) Na+(aq) + OH-(aq)
Model limitation: NH3(g) when dissolved in water produces OH- (base)
Brønsted-Lowry Acid/Base Model
Broadens Arrhenius model
Acid: Hydrogen ion donor
Base: Hydrogen ion acceptor
Uses concept of conjugate acid-base pairs
Conjugate Acid-Base Pairs
Generalized acid dissociation (n=1)
HX(aq) + H2O(l) H3O+(aq) + X-(aq)
Forward reaction:• HX donates H+ to H2O acid
• H2O is base in this process
Acid Base
Reverse reaction:• H3O+ donates H+ to X conjugate acid
• X- is conjugate base in this process
Conjugate
Acid
Conjugate
Base
Brønsted-Lowry Acid/Base Model
HF(aq) + H2O(l) H3O+(aq) + F-(aq)
+++ -
Acid? HF
Base? H2O
Conjugate Acid? H3O+
Conjugate Base? F-
Brønsted-Lowry Acid/Base Model
HF(aq) + H2O(l) H3O+(aq) + F-(aq)
All acids and bases that fit Arrhenius definition also fit Brønsted-Lowry definition
Brønsted-Lowry Base
NH3(aq) + H2O(l) NH4+(aq)+OH-(aq)
NH3 not Arrenhius base
base conjugate acid
acid conjugatebase
Practice
Brønsted-Lowry Model
Problems 3(a-c), 4 page 640
Problem 9 page 643
Problems 19 (a-b) page 649
Problem 69, page 672
Problems 3 – 4 page 989
Amphoteric Substances
acid
base
HF(aq) + H2O(l) H3O+(aq) + F-(aq)
NH3(aq) + H2O(l) NH4+(aq)+OH-(aq)
H2O(l) acting as either an acid or a base
Compounds with this capability called amphoteric
Amphoteric Substances
Can act as either acid or base
Water most common example
Also common is bicarbonate ion, HCO3-
HCO3-(aq) + OH-(aq) CO3
2-(aq) + H2O(l)
acid base conjugate conjugate base acid
HCO3-(aq) + H3O+(aq) H2CO3(aq) + H2O(l)
base acid conjugate conjugate acid base
Lewis Acid/Base ModelMost general model of acids & bases
Same G.N. Lewis that gave us Lewis electron dot structures and shared electron pair model for covalent bond
Acid: electron pair acceptor ( / share)
Base: electron pair donor ( / share)
Includes all substances classified as Brønsted-Lowry acids & bases & many more; also reactions that do not necessarily occur in aqueous solution
Other Lewis Bases
:Cl:....
-H - N - H
H
..
(HCl) (NH4+)
Lewis Acid/Base Model
Formation of HF molecule
Fluoride ion donates electron pair; empty 1s orbital of H accepts pair (covalent bond formed)
Lewis Acid/Base Model
Reaction of gaseous boron trifluoride and gaseous ammonia
Electron deficient B accepts lone pair from ammonia (formation of coordinate covalent bond)
Lewis Acid/Base ModelGaseous sulfur trioxide reacts with solid magnesium oxide
SO3(g) + MgO(s) MgSO4(s)
Lewis acid/base part of reaction:
Will return to this reaction shortly
3 Acid-Base Models (Table 18.2)
Mono- and Polyprotic Acids
Monoprotic – capable of donating only one hydrogen ion
• HCl
Polyprotic – can donate more than one• H2SO4
Above examples simple - inorganic acids, all hydrogens ionize
More complicated for organic acids
Mono- and Polyprotic Acids - HnX
HnX acids that dissociate into nH+(aq) ions and Xn-(aq) ions
X=Cl hydrochloric acid n=1
X=NO3 nitric acid n=1
X=C2H3O2 acetic acid n=1
X=SO4 sulfuric acid n=2
X=CO3 carbonic acid n=2
X=PO4 phosphoric acid n=3
mon
opro
ticpo
lypr
otic
Step-Wise Ionization
Sulfuric acid
H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq)
HSO4-(aq) + H2O(l) H3O+(aq) + SO4
2-(aq)
Conjugate
BaseAcid
Step-Wise Ionization
Phosphoric acid
H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4-(aq)
H2PO4-(aq) + H2O(l) H3O+(aq) + HPO4
2-(aq)
HPO42-(aq) + H2O(l) H3O+(aq) + PO4
3-(aq)
Conjugate
BaseAcid
Amphoteric
H1
H2H3
Monoprotic AcidsAcetic acid: organic acid; HF: inorganic acid; each has one ionizable hydrogen atom (one that is involved in a polar bond)
Organic AcidsOrganic acids contain carboxylic acid group, COOH
R - C - O - H
=O
Formula often written as RCOOHR = (almost) anythingFor acetic acid, R = CH3 (methyl)CH3 hydrogens not ionizable – monoprotic even though has 4 H atoms
Ionizable hydrogen
R - C – O + H+
=O–
Organic Acids
Malonic acid (1,3-Propanedioic acid)
How many ionizable hydrogens?
?
2
Anhydrides
Oxides that become acids or bases when added to water
Nonmetal oxides (C, S, N) – acids• CO2(g) + H2O(l) H2CO3 (Carbonic Ac)
• N2O3(g) + H2O(l) 2HNO2 (Nitrous Ac)
• SO3(g) + H2O(l) H2SO4 (Sulfuric Ac)
Metal oxides – bases• CaO(s) + H2O(l) Ca2+(aq) + 2OH-(aq)
Anhydrides and Acid RainNatural rain water: pH 5.6
• Naturally slightly acidic due mainly to CO2
Acid rain: Rain water with pH < 5.6Due to pollutant gases dissolved in air (nonmetal oxides = acid anhydrides)
• SO2, SO3, NO2
Acid rain linked to damage in ecosystems and structures
Damage from Acid RainReacts with metals and materials that contain carbonates
Damages bridges, cars, and other metallic structures
Damages buildings and other structures made of limestone or cement
Acidifies lakes affecting aquatic life
Dissolves and leaches minerals from soil• Makes it difficult for trees to grow
Practice
Acids and Bases Intro – other topics
(neutrality, Arrenhius model, polyprotic acids, ionizable hydrogens, anhydrides)
Problems 5 – 8, 10 – 11 page 643
Problems 55 - 63, 64 (c-d) page 672
Problems 5 – 6 page 989
Chapt. 18 – Acids and Bases
18.1 Acids and Bases: An Introduction
18.2 Strengths of Acids and Bases
18.3 Hydrogen Ions and pH
18.4 Neutralization
Section 18.2 Strengths of Acids and Bases
• Relate the strength of an acid or base to its degree of ionization.
• Compare the strength of a weak acid with the strength of its conjugate base.
• Explain the relationship between the strengths of acids and bases and the values of their ionization constants.
In solution, strong acids and bases ionize completely, but weak acids and bases ionize only partially.
Section 18.2 Strengths of Acids and Bases
Key Concepts
• Strong acids and strong bases are completely ionized in a dilute aqueous solution. Weak acids and weak bases are partially ionized in a dilute aqueous solution.
• For weak acids and weak bases, the value of the acid or base ionization constant is a measure of the strength of the acid or base.
Acid Strength and Concentration
...are not the same thing
e.g. 1.0 M HCl and 1.0 M H2CO3
stronger acid
same concentrations!Can have a dilute solution of a strong acid and a concentrated solution of a weak acid
Acid Strength
0.10 M HCl conducts electricity better than 0.10 M HC2H3O2 (acetic acid)
Hyd
roch
loric
Ace
tic
Acid Strength
Strong acids dissociate completely
Limited number of them (see Table 18.3)
HCl hydrochloric
HBr hydrobromic
HI hydroiodic
3 binary halogen acids
except HF
HClO4 perchloric
HNO3 nitric
H2SO4 sulfuric
Relationship Between Strengths of Acids and Their Conjugate Bases
The stronger the acid, the weaker is the attraction of the ionizable H for the rest of the molecule
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 conj base
Weak acid: HF + H2O F– + H3O+ Strong conj base
Acid Strength & Brønsted-Lowry Model
Strong acid HA has weak conj. base A-
H2O(l) stronger base than A-
Reaction equilibrium lies far to right
+ A-
conjugate base of HA
HA + H2O(l)
acid base
H3O+(aq)
conjugateacid of B
Strong/Weak vs Concentrated/Dilute
Concentrated/Dilute• Number of acid/base molecules per
volume• 6 M HCl concentrated, 0.01 M HCl dilute
Strong/Weak• Measure of degree of ionization• HCl strong acid• CH3COOH (acetic) weak acid
Acid Strength & Brønsted-Lowry Model
Weak acid HA has strong conj. base A-
A- stronger base than H2O(l) Reaction equilibrium lies far to left
+ A-
conjugate base of HA
HA + H2O(l)
acid base
H3O+(aq)
conjugateacid of B
Acid Ionization ConstantsFocus on equilibrium for weak acidsHCN(aq) + H2O(l) H3O+(aq) + CN-(aq)
Write K following standard rules
Ka = [H3O+(aq)] [CN-(aq)] = 6.2x10-10 [HCN(aq)]
Ka acid ionization constant
Small number for weak acids – major species is undissociated form of acid
Table 18.4, page 647 lists values at 25C
Acid Ionization Constants - pKa
HCN(aq) + H2O(l) H3O+(aq) + CN-(aq)
Ka = [H3O+(aq)] [CN-(aq)] = 6.2x10-10 [HCN(aq)]
Ka acid ionization constant
Many texts will utilize quantity pKa
pKa = log(Ka)
For HCN, pKa = 9.21
[Note: SF after DP for log based on SF of number in front of power of ten]
Weak Acids & Ka Values
Strongest Acids
Weakest Acids
Acid Ka Value Conjugate Base
Weakest Bases
Strongest Bases
HF 6.3 x 10-4 F-
HNO2 5.6 x 10-4 NO2-
HCO2H 1.8 x 10-4 HCO2-
CH3CO2H 1.8 x 10-5 CH3CO2-
HOCl 4.0 x 10-8 OCl-
NH4+ 5.6 x 10-10 NH3
HCN 6.2 x 10-10 CN-
Practice
Strengths of Acids
Problems 12 – 14 page 647
Problems 65 – 68, 71 page 672
Problems 17, 18, 21 page 649
Problems 7 – 8 page 989
Dilute/concentrated vs weak/strong
Problem 74 page 672
Strong Bases
Alkali & some alkaline earth metal hydroxides are strong Arrhenius bases
M(OH)n(s) Mn+(aq) + n OH-(aq)
Dissociate completelyLimited number (see table 18.5)
NaOH KOH RbOH CsOH (alkali metals)Ca(OH)2, Sr(OH)2, Ba(OH)2 (alkaline
earths)
Strong BasesPicture complicated by low solubility of some
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)
Calcium hydroxide Ksp = 6.5x10-6
Base strength based on dissolved formCa(OH)2 strong base – dissolved form
[Ca(OH)2(aq)] completely dissociates
Weak Bases
Weak bases mostly Brønsted-Lowry bases
B(aq) + H2O(l) OH-(aq) + BH+(aq)
Most common weak base either ammonia or amine (organic compound with –NH2)
CH3NH2(aq) + H2O(l) OH-(aq) + CH3NH3+(aq)
Kb = [CH3NH3+(aq)] [OH-(aq)] = 4.3x10-10
[CH3NH2(aq)]
Kb= base ionization constant (see table 18.6)
Weak Base Equilibria
Ammonia NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)
Methyl amine
CH3NH2(aq) + H2O(l) CH3NH3+(aq) +
OH−(aq)
Ethyl amineC2H5NH2(aq) + H2O(l) C2H5NH3
+(aq) + OH−(aq)
BicarbonateHCO3
−(aq) + H2O(l) H2CO3 (aq) + OH−
(aq)
Practice
Strengths of Bases
Problems 15 (a-d), 16, 20 page 649
Problems 70, 72, 73 page 672
Problems 7 – 8 page 989
Chapt. 18 – Acids and Bases
18.1 Acids and Bases: An Introduction
18.2 Strengths of Acids and Bases
18.3 Hydrogen Ions and pH
18.4 Neutralization
Section 18.3 Hydrogen Ions and pH
• Explain pH and pOH.
• Relate pH and pOH to the ion product constant for water.
• Calculate the pH and pOH of aqueous solutions.
pH and pOH are logarithmic scales that express the concentrations of hydrogen ions and hydroxide ions in aqueous solutions.
Section 18.3 Hydrogen Ions and pH
Key Concepts
• The ion product constant for water, Kw, equals the product of the H+ ion concentration and the OH– ion concentration.
Kw = [H+][OH–]
• The pH of a solution is the negative log of the hydrogen ion concentration. The pOH is the negative log of the hydroxide ion concentration. pH plus pOH equals 14.
pH = –log [H+]
pOH = –log [OH–]
pH + pOH = 14.00• A neutral solution has a pH of 7.0 and a pOH of 7.0 because
the concentrations of hydrogen ions and hydroxide ions are equal.
Ion Product Constant for Water (Kw)
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
or
H2O(l) H+(aq) + OH-(aq)
Keq = [H3O+(aq)] [OH-(aq)]
= [H+(aq)] [OH-(aq)]
Keq given special symbol, Kw
Kw = 1.0 x 10-14 at 25oC
Ion product=Kw in any aqueous solution
Ion Product Constant for Water (Kw)
Kw = [H3O+(aq)] [OH-(aq)] = 1.0 x 10-14
In pure water only source of ions is water dissociation, so at 25oC (298 K)
[H3O+] = [OH-] = 1.0 x 10-7 M
H2O(l) H2O(l) H3O+(aq) OH-(aq)
+ ++ -
[H+] and [OH-] for General Case
H2O(l) H+(aq) + OH-(aq)
Kw = [H+(aq)] [OH-(aq)] = 1.0 x 10-14
H+ or OH- can come from source other than water (i.e., an acid or a base)
Le Châteliers principle equilibrium shifts to left, depressing either H+ or OH- concentrations
Kw must remain constant
Calculation of [H+] and [OH-]
Example problem 18.1, page 651
Kw = [H+(aq)] [OH-(aq)] = 1.0 x 10-14
Given [H+] = 1.0x10-5 M, [OH-] = ?
[OH-]= Kw / [H+]
[OH-]= 1.0 x 10-14 /1.0x10-5 = 1.0x10-9 M
Practice
Kw and Water Ionization Equilibrium
Problems 22(a-d), 23 page 651
Problems 37, 38 page 658
Problems 77 – 78, 80 page 673
Problems 11 – 12 page 989
pH
Defined as
pH = – log[H+]
Book says scale 0 to 14
Not strictly true! 10 M HCl pH = -1
10 M NaOH pH = 15
Remember log properties
pH 3 solution has 100x [H+] of pH 5
pH 0 - 7 acid pH 7 - 14 basic
pH – Common MaterialsSimilar to Fig 18.14, page 652
pH – Common Materials Similar to Fig 18.14, page 652
Calculating pH
Example problem 18.2, page 653
pH of neutral solution at 298 K?
Kw = 1.00x10-14 = [H+] [OH-]
Neutral solution, [H+] = [OH-]
[H+] = 1.00x10-7
pH = – log[H+] = 7.00
Practice
Calculate pH from [H+]
Problems 24(a-b), 25 page 653
Problems 42(a-b) page 658
Problems 82, 83 page 673
Problem 13 page 989
pOH
pOH = – log[OH-]
pOH 0 - 7 basic pH 7 - 14 acidic
pH + pOH = 14.00
Calculating pOH & pH
Sample problem 18.3, page 654
Ammonia cleaner [OH-] = 4.0x10-3 M
pOH, pH?
pOH = – log[OH-] = 2.40
pH + pOH = 14.00 pH = 11.60
Practice
Calculate pH or pOH from [H+] or [OH-]
Problems 27(a-d), 28(a-b), 29 page 654
Problems 41, 42(c-d), 43 page 658
Problems 80, 81 page 673
Problems 14 - 15 page 989
Calculating Ion Concs from pH[H+] = antilog(-pH)
If pH=5, [H+] = antilog(-5) = 10-5
Problem 18.4, page 655
Blood pH = 7.40 @ 298 K; [H+], [OH-] ?
[H+] = antilog(-7.40) = 4.0x10-8 M
[OH-] = Kw / [H+] = 1.0x10-14 / 4.0x10-8
[OH-] = 2.5 x10-7 M(book uses pH + pOH = 14.00 and antilog(-pOH)
Practice
Calculate [H+] or [OH-] from pH
Problems 30(a-d), 31 page 655
Problem 40 page 658
Problem 79 page 673 (also other topics)
Problems 16 - 17 page 989
pH of Strong Acids & Bases
Strong acids and bases ~100% ionizedHCl(aq) H+(aq) + Cl-(aq)
(Also HBr, HI, HClO4, HNO3, H2SO4) [H+] = Acid concentration
NaOH(aq) Na+(aq) + OH-(aq) [OH-] = Base concentration
Ca(OH)2(aq) Ca2+(aq) + 2OH-(aq) [OH-] = 2 x Base concentration
Using pH to Calculate Ka
HF(aq) H+(aq) + F-(aq) (weak acid)
pH of 0.100 M solution = 2.12; Ka?
Ka = [H+(aq)] [F-(aq)] / [HF(aq)]
[H+(aq)] = antilog(-2.12) = 7.6x10-3 M (2 SF)
If ignore contribution of H2O to H+(aq)
[F-(aq)] = [H+(aq)]
[HF(aq)] = 0.100 M – 7.6x10-3 M
Ka = 7.6x10-3 7.6x10-3 / 0.092 = 6.3x10-4
Using pH to Calculate Ka
Sample problem 18.5, page 657
pH 0.100 M formic acid = 2.38; Ka?
HCOOH(aq) H+(aq) + HCOO-(aq)
[H+] = antilog(-2.38) = 4.2x10-3 (2 SF)
[HCOO-(aq)] = [H+(aq)]
[HCOOH(aq)] = 0.100 M - 4.2x10-3 M
Ka = [H+(aq)] [HCOO-(aq)] / [HCOOH(aq)]
Ka = 4.2x10-3 4.2x10-3 / 0.096 = 1.8x10-4
Practice
Use pH or pOH to calcuate Ka
Problems 32(a-b), 33(a-c), 34 page 657
Problem 39 page 658
Problem 84* (avoid) page 673
Problems 20 - 21 page 989
* Relies on assumption that 1st ionization of chromic acid acts like strong acid (no hydrolysis of HCrO4
-)
Measuring pH
Indicators – quick, cheap but not precise
pH meter – can measure to 0.001 pH units under ideal conditions with proper calibration
Methods for Measuring pH
pH meterpH (indicator) paper
Chapt. 18 – Acids and Bases
18.1 Acids and Bases: An Introduction
18.2 Strengths of Acids and Bases
18.3 Hydrogen Ions and pH
18.4 Neutralization
Section 18.4 Neutralization
• Write chemical equations for neutralization reactions.
• Explain how neutralization reactions are used in acid-base titrations.
• Compare the properties of buffered and unbuffered solutions.
In a neutralization reaction, an acid reacts with a base to produce a salt and water.
Section 18.4 Neutralization
Key Concepts
• In a neutralization reaction, an acid and a base react to form a salt and water.
• The net ionic equation for the neutralization of a strong acid by a strong base is H+(aq) + OH–(aq) → H2O(l).
• Titration is the process in which an acid-base neutralization reaction is used to determine the concentration of a solution.
• Buffered solutions contain mixtures of molecules and ions that resist changes in pH.
Neutralization Reaction
Double replacement reaction of acid and base in aqueous solution to produce a salt and water
• Salt – ionic compound, cation from base and anion from acid
Mg(OH)2(aq) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
Base Acid Salt Water
Neutralization & Net Ionic Equation
Salt formed from neutralization of strong acid and strong base will be a strong electrolyte (fully ionized)HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
Complete ionic equation:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Na+(aq) + Cl-(aq) + H2O(l)
Net ionic: H+(aq) + OH-(aq) H2O(l)
Practice
Neutralization reactions
Problem 49 page 668
Problems 85, 90 page 673
Problems 22, 23 page 989
Acid-Base TitrationMethod for determining concentration of an unknown acid (base) solution by reacting it with a base (acid) of known concentration HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
Stoichiometry applies - if know concentration & volume of NaOH needed to neutralize & initial volume of HCl, can calculate initial concentration of HCl
TitrationTechnique that uses reaction stoichiometry to determine concentration of unknown solutionTitrant (known solution) added from buretIndicators – color-changing chemicals added to help determine when reaction complete (not required if pH measured)Endpoint of titration occurs when reaction complete
Acid-Base TitrationTitrant (solution in buret) is a baseAs base added to acid, H+
reacts with OH– to form H2O; still have excess acid present, so no color changeAt titration’s endpoint,just enough base has been added to neutralize allacid. At this point,indicator changes color
Acid-Base Titration Steps1. Measure volume of unkown.
Determine initial pH
2. Fill buret with titrating solution of known concentration (standard solution)
3. Slowly add standard solution while recording pH until neutralization reaction reaches stoichiometric point (equivalence point)
At equivalence
point, pH changes
rapidly with small
addition of NaOH
Acid-Base Titration - Indicators
Often convenient to use pH indicator to determine equivalence point (point at which number of moles of base added equals number of moles of acid originally present) [OHadd
-(aq) = H+start(aq) for 1:1 type system)
End point: pH at which indicator changes color
Must choose indicator so pH at end point and pH at equivalence point are ~ same
Acid-Base Titration Equivalence Point
Equivalence point for strong acid-strong base titration: pH 7
For weak acid-strong base, pH > 7
For strong acid-weak base, pH < 7
Different from 7 due to salt hydrolysis (topic discussed later)
Must choose pH indicator end point accordingly
Matching of equivalence
point and pH indicator end point – titration of strong acid with strong
base
Matching of equivalence
point and pH indicator end point – titration of weak acid with strong
base
Acid Base Indicators
are weak acids or bases themselves
Hin(aq) + H2O(l) H3O+(aq) + In-
(aq)
Color A Color B
Acid Base Indicators
Ka =[H3O+] [In-]
[Hin]
[In-][Hin]
=Ka
[H3O+]
i.e. ratio of colors changes with pH
Hin(aq) + H2O(l) H3O+(aq) + In-
(aq)
Indicators – See Fig 18.24, p 662
3 Very Common Indicators
Step 1
Known volume of
unknown in flask with indicator
Known concentration
of titrant in buret
Step 2
Titrant added
Pink color appears
momentarily but then
disappears upon mixing
Step 3
Just enough titrant added so that pink
color permanently
remains
Titration Calculation - Prob. 18.6 p 664
18.28 mL of 0.1000 M NaOH for 25.00 mL of formic acid (HCOOH)
1. Write balanced neutralization eqn.
HCOOH(aq)+ NaOH(aq) HCOONa(aq)+H2O(l)
2. Calculate moles titrant added
18.3 mL NaOH x 1 L NaOH = 0.0183 L NaOH 1000 mL NaOH
0.0183 L NaOH x 0.100 mol NaOH = 0.00183 mol L NaOH NaOH
Acid-Base Titration CalculationHCOOH(aq)+ NaOH(aq) HCOONa(aq)+H2O(l)
3. Use mole ratio: calculate moles unknown1.83x10-3 mol NaOH x 1 mol HCOOH
1 mol NaOH
= 1.83x10-3 mol HCOOH4. Calculate concentration of unknown
using moles and initial volumeMHCOOH = 1.83x10-3 mol HCOOH
0.0250 L HCOOH MHCOOH = 7.31x10-2 M
Practice
Acid-Base Titration
Problems 44 – 46 page 664
Problems 52, 54 page 668
Problems 86 – 89, 92 – 93 page 673
Problems 24 – 25 page 989
Salt HydrolysisPut salt in pure water
Unless salt product of strong acid/strong base reaction, pH won’t be neutral
Neutral Basic AcidicNaNO3 KF NH4Cl
Salt Hydrolysis
KF(s) K+(aq) + F-(aq)
KOH strong base, HF weak acid (see table 18.4, page 647)
F-(aq) + H2O(l) HF(aq) + OH-(aq)
F-(aq): Brønsted-Lowry base
Solution basic (OH- generated)
Salt Hydrolysis
NH4Cl(s) NH4+(aq) + Cl-(aq)
NH3 weak base, HCl strong acid
NH4+(aq) + H2O(l) NH3(aq) + H3O+ (aq)
NH4+(aq): Brønsted-Lowry acid
Solution acidic (H3O+ generated)
Practice
Salt Hydrolysis
Problems 47(a-d), 48 page 665
Problems 52, 54 page 668
Problems 91(a-b) page 673
Problems 26 – 27 page 990
Buffered Solutions
Buffer: Solution that resists change in pH when limited amounts of acid or base added
Add 0.01 mol HCl to solution• 1 L water – pH change: 7.0 to 2.0• Buffer – typical pH change 0.1 unit
Buffer Composition
Mixture of either:(a) Weak acid & salt of its conjugate
baseHF (weak acid) + KF (F- conj base)
(b) Weak base & salt of its conjugate acid
NH3 (weak base) + NH4Cl (NH4+ conj
acid)
Standard buffers mostly made using (a)Having both acid & base present makes buffer resist pH changes
How Buffers Work
Weak acid in buffer mixture can neutralize added base
Conjugate base in buffer mixture can neutralize added acid
Net result: little to no change in solution pH in either case
H2O
How Buffers Work
HA + H3O+A−A−
AddedH3O+
NewHA
HA
H2O
HAHA + H3O+
A−
AddedHO−
NewA−
A−
How Buffers Work
Equimolar Buffer Systems Table 18.7
HF/KF Buffer0.1 M HF (weak ac) + 0.1 M KF (F- conj base)
HF(aq) H+(aq) + F-(aq)F-(aq) from 2 sources:
• Dissociation of HF(aq)• Dissociation of KF(aq)
Add acid – equilibrium shifts to left so H+ consumed
Add base – H+ neutralized to H2O, equil. shifts to right to generate more H+
Either way, pH change minimized
Acetic Acid / Sodium Acetate Buffer
Mix equal volumes of 1 M HC2H3O2 and 1 M NaC2H3O2: buffer pH = 4.75
Add 10 mL of 0.1 M HCl to: • 1 L of buffer, pH = 4.75• 1 L of DI water, pH = 3.0
Add 10 mL of 0.1 M NaOH to:• 1 L of buffer, pH = 4.75• 1 L of DI water, pH = 11.0
Buffer Capacity
Amount of acid or base buffer can absorb without significant pH change0.1 M HF (weak ac) + 0.1 M KF (F- conj base)
HF(aq) H+(aq) + F-(aq)
Add more H+ than available F-, pH must drop
Remove more H+ than can be replaced by remaining HF, pH must rise
pH of Buffer
Easy to determine for equimolar buffers
Equimolar mix NaH2PO4 & Na2HPO4
H2PO4-(aq) H+(aq) + HPO4
2-(aq)
Acid conj. Base
Ka = [H+(aq)][HPO42-(aq)] / [H2PO4
-(aq)]
Since [HPO42-(aq)] = [H2PO4
-(aq)]
pH = -log[H+] = -log(Ka) = -log(6.2x10-8)
pH = 7.21 (see table 18.7, p 667 for others)
Practice
Buffers
Problems 51, 53 page 668
Problems 88, 102, 103 pages 673-4