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© 2008 Brooks/Cole 1
Chapter 16: Acids and BasesChapter 16: Acids and Bases
Chemistry: The Molecular ScienceChemistry: The Molecular ScienceMoore, Stanitski and Jurs
© 2008 Brooks/Cole 2
Arrhenius Definition
Arrhenius: any substance which ionizes in water to produce…
Better version of the Arrhenius definition:
Acid:Acid: hydronium ions (H3O+) in water are acidic.
BaseBase: hydroxide ions (OH-) in water are basic.
• protons (H+) is an acid.acid. • hydroxide ions (OH-) is a base.base.
© 2008 Brooks/Cole 3
Acid-Base Reactions
A strong acid is an acid that ionizes completely in water (product favored).
)aq(OH)aq(NO)aq(HNOOH
333
2
)aq(OH)aq(Cl )aq(HClO2H
3
There are six strong acids:
HCl - hydrochloric acid HBr – hydrobromic acid
HI – hydroiodic acid
H2SO4 - sulfuric acidHNO3 – nitric acid
HClO4 – perchloric acid
© 2008 Brooks/Cole 4
Acid-Base Reactions
Strong Acids and Bases
A strong base is a base that is present entirely as ions (product favored).
The hydroxides of Group IA, IIA (except Be and Mg hydroxides) are strong bases.
)aq(OH)aq(Na)s(NaOH O
2H
So, wSo, why is NH3 (aq) basic?
© 2008 Brooks/Cole 5
Brønsted-Lowry Concept
An alternative definition:
NH3(g) + H2O(l) NH4+(aq) + OH-(aq)
indirectly producedBase:H+ acceptor
Acid:H+ donor
acidacid = proton (H+) donor
basebase = proton (H+) acceptor
Works for non-aqueous solutions and explains why NH3 is basic:
© 2008 Brooks/Cole 6
WeakWeak acids and bases do not fully ionize (reactant favored).
Brønsted-Lowry ConceptStrongStrong acids and bases almost completely ionize (product favored).
HNO3(aq) + H2O(l) H3O+(aq) + NO3-
(aq)
Note: the products are a new acid and base pair.LeChatelier – a more dilute solution (more water), will ionize more.
HF(aq) + H2O(l) H3O+(aq) + F-(aq)
Base:H+ acceptor
Acid:H+ donor
Base:H+ acceptor
Acid:H+ donor
100%
3%
© 2008 Brooks/Cole 7
Water’s Role as Acid or Base
Water acts as a base when an acid dissolves in water:
HBr(aq) + H2O(l) H3O+(aq) + Br-(aq)acid base acid base
Water is amphiproticamphiprotic - it can donate or accept a proton (act as acid or base).
But water acts as an acid for some bases:
H2O(l) + NH3(aq) NH4+(aq) + OH-(aq)
acid base acid base
© 2008 Brooks/Cole 8
Practice
• Complete the equations: (acid or base?, mass/charge balance?, single or double arrow?)
• HClO4 + H2O
• CN- + H2O
• H2S + H2O
• NO2- + H2O
• Ca(OH)2 + H2O
ClO4- + H3O+
HCN + HO-
HS- + H3O+
HNO2 + HO-
Ca+2 + 2OH-
© 2008 Brooks/Cole 9
Conjugate Acid-Base Pairs
Molecules or ions related by the loss/gain of oneone H+.
Conjugate Acid Conjugate Base
H3O+ H2O
CH3COOH CH3COO-
NH4+ NH3
H2SO4 HSO4-
HSO4- SO4
2-
HCl Cl-
donate H+
accept H+
NH4+ and NH2
- are notnot conjugate(conversion requires 2 H+)
© 2008 Brooks/Cole 10
Identify the base conjugate to HC2H3O2(aq) and the acid conjugate to HCO3
-(aq)
HCO3- + H2O OH-
+
HC2H3O2 + H2O H3O+ +
Conjugate Acid
Base:H+ acceptor
ConjugateBase
Acid:H+ donor
Conjugate Acid-Base Pairs
Show that HCO3- is amphoteric.
C2H3O2 -
H2CO3
© 2008 Brooks/Cole 11
Practice: Conjugate Acid-Base Pairs
Write the equation for the acid or base in water and identify the acid-base pairs.
HBr
HSO3- (as an acid)
HSO3- (as a base)
PH4+
© 2008 Brooks/Cole 12
Relative Strength of Acids & Bases
Strong acids are better H+ donors than weak acids.
Strong bases are better H+ acceptors than weak bases. Strong acids have weak conjugate bases. Weak acids have strong conjugate bases.
Strong acid + H2O H3O+ + weak conjugate baseFully ionized, reverse reaction essentially does not occur.
The conjugate base is weak.
Weakly ionized, reverse reaction readily occurs.
The conjugate base is strong.
Weak acid + H2O H3O+ + strong conjugate base
© 2008 Brooks/Cole 13
Conjugate acid Conjugate baseH2SO4 HSO4
-
HBr Br-
HCl Cl-
HNO3 NO3-
H3O+ H2O
H2SO3 HSO3-
HSO4- SO4
2-
H3PO4 H2PO4-
HF F-
CH3COOH CH3COO-
H2S HS-
H2PO4- HPO4
2-
NH4+ NH3
HCO3- CO3
2-
H2O OH-
OH- O2-
H2 H-
CH4 CH3-
Acid
streng
th in
crea
sing
Ba
se s
tre
ngth
incr
ea
sin
g
stong acids
strong baseextremelyweak acids
extremelyweak bases
Relative Strength of Acids & Bases
© 2008 Brooks/Cole 14
HF is a stronger acid than NH4+.
(NH3 is a stronger base than F-)
HF has greater tendency to ionize than NH4
+. (NH3 more readily accepts H+ than F-)
Reactant FavoredReactant Favored
NH4+ + F- NH3 + HF
ProblemProblemIs the following aqueous reaction product or reactant favored?
Acid
streng
th in
crea
sing
Ba
se s
tre
ngth
incr
ea
sin
g
Conj acid. Conj. base
H2SO4 HSO4-
HBr Br-
HCl Cl-
HF F-
NH4+ NH3
OH- O2-
H2 H-
CH4 CH3-
Relative Strength of Acids & Bases
© 2008 Brooks/Cole 15
Carboxylic Acids
H C
H
H
C
H
H
C
H
H
C
O
O H
nonacidic hydrogens
butanoic acid
acidic hydrogen
CH3 C
O
acetic acidOH acidic hydrogen CH3 C
O
O-acetate ion
C3H7COOH
CH3COOH
© 2008 Brooks/Cole 16
Amines (bases)
CH3NH2 (CH3)2NH (CH3)3N
proton acceptors
© 2008 Brooks/Cole 17
Autoionization of Water
Pure water conducts a very small electrical current.
AutoionizationAutoionization occurs:
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
Kw = ionization constant for water
Heavily reactant favored.
Kw = [ H3O+ ] [ OH- ]
Kw = 1.0 x 10-14 (at 25°C)
Base AcidAcid Base
© 2008 Brooks/Cole 18
Ionization Constant for WaterKw is a special equilibrium constant for the autoionization of water. It is T-
dependent.
T = 25°C (77°F) is usually used as the standard T.
T (°C) Kw
10 0.29 x 10-14
15 0.45 x 10-14
20 0.68 x 10-14
2525 1.01 x 101.01 x 10-14-14
30 1.47 x 10-14
50 5.48 x 10-14
Kw increases with temperature. Is it endo or exothermic?
© 2008 Brooks/Cole 19
Autoionization of Water
Hydronium and hydroxide ions are produced in equal numbers in pure water,
Thus, the concentrations of H3O+ and OH- in pure water are both 1.0 x 10-7 M.
C 25at )x)(x(100.1 o14
714 100.1100.1x
Kw = [ H3O+ ] [ OH- ]
2H2O(l) H3O+(aq) + OH-(aq)
© 2008 Brooks/Cole 20
Autoionization of WaterH3O+ and OH- are present in allall aqueous solutions.
Neutral solution.Neutral solution.
Pure water (@ 25°C): [H3O+] = 10-7 M = [OH-]
Acidic solutionAcidic solution
If acid is added to water:• [ H3O+ ] is increased, disturbing the equilibrium:
2 H2O H3O+ + OH-
• LeChatelier: equilibrium shifts to the left. [OH-]
• Equilibrium is reestablished: [ H3O+ ] > 10-7 M > [OH-]
• Product of ion concentrations is the same
[ H3O+ ][OH-] = Kw = 1.0 x 10-14
© 2008 Brooks/Cole 21
Autoionization of WaterBasic solution.Basic solution.
If base is added to water.• LeChatelier: equilibrium shifts to the left, [H3O+]
• New equilibrium: [H3O+] < 10-7 M < [OH-]
• [ H3O+ ][OH-] = Kw = 1.0 x 10-14 at 25°C
ExampleExample
Calculate the hydronium and hydroxide ion concentrations at 25°C in a 6.0 M aqueous sodium hydroxide solution.
Kw = [H3O+][OH-] = 1.0 x 10-14
NaOH (aq) is strong (100% ionized) so [OH-] = 6.0 M.
[H3O+](6.0) = 1.0 x 10-14
[H3O+] = 1.7 x 10-15 M [ OH- ] = 6.0 M
2 H2O H3O+ + OH-
© 2008 Brooks/Cole 22
Autoionization of Water
In a neutral solution, the concentrations of H3O+ (aq) and OH-(aq) are equal.
In an acidic solution, the concentration of H3O+ (aq) is greater than that of OH-(aq).
In a basic solution, the concentration of OH-(aq) is greater than that of H3O+ (aq).
The relative concentrations of H3O+ and OH- indicate acidic, neutral or basic character of a solution:
© 2008 Brooks/Cole 23
At 25 oC, we observe the following conditions.
Acid or Base
In an acidic solution, [H3O+ ] > 1.0 x 10-7 M.
In a neutral solution, [H3O+ ] = [OH-] = 1.0 x 10-7 M.
In a basic solution, [H3O+ ] < 1.0 x 10-7 M.
© 2008 Brooks/Cole 24
The pH scale
Can describe acidity in terms of [H3O+], but a logarithmic scale is more convenient:
pH = −logpH = −log1010[H[H33OO++]]
At 25°C a neutral aqueous solution has:
pH = −log10[1.0 x 10-7] = −(−7.00) = 7.00
acidic solutions: [H3O+]>1.0 x 10-7, pH < 7.00
basic solutions: [H3O+]<1.0 x 10-7, pH > 7.00
© 2008 Brooks/Cole 25
The pH of a Solution
For a solution in which the hydronium-ion concentration is 1.0 x 10-3, the pH is:
Note that the number of decimal places in the pH equals the number of significant
figures in the hydronium-ion concentration.
00.3)100.1log( 3 pH
© 2008 Brooks/Cole 26
Example – pH given [H3O+] ?
A sample of orange juice has a hydronium-ion concentration of 2.9 x 10-4 M. What is the pH?
]OHlog[pH 3
)109.2log(pH 4
54.3pH
© 2008 Brooks/Cole 27
Example - [H3O+] given pH?
]OHlog[pH 3
The pH of human arterial blood is 7.40 (acidic or basic?). What is the hydronium-ion concentration? (Guesss….> or < 1.0x10-7M?)
M100.410]OH[ 840.73
]OHlog[pH 3
]OHlog[pH 31010
pH3 10]OH[
10logx = x
© 2008 Brooks/Cole 28
The pOH scale
pOH = −logpOH = −log1010[OH[OH--]]
A neutral solution (25°C) has:
pOH = −log10[1.0 x 10-7] = −(−7.00) = 7.00
Since Kw = [ H3O+ ][ OH- ] = 1.0 x 10-14
−log(KW)= −log[H3O+] + (−log[OH-]) = −log(1.0 x 10-14)
pKw = pH + pOH = 14.00
© 2008 Brooks/Cole 29
Example
An ammonia solution has a hydroxide-ion concentration of 1.9 x 10-3 M. What is the pH of the solution?
we first calculate the pOH:
72.2)109.1log(pOH 3
Then the pH is:
28.1172.200.14pH
© 2008 Brooks/Cole 30
pH Calculations
Solution A: pOH = −log[ OH-] = 3.37
pH + pOH = pKw = 14.00
pH = 10.63
Solution B: pH = −log[ H3O+] = 8.12
A has higher pH, B is more acidic
Given two aqueous solutions (25°C).
Solution A: [OH-] = 4.3 x 10-4 M,
Solution B: [H3O+] = 7.5 x 10-9 M.
Which has the higher pH? Which is more acidic?
© 2008 Brooks/Cole 31
Practice - WSs
[H+] Kw =1x10-14 = [H+] [OH-] [OH-]
pH 14 = pH + pOH pOH
pH= -log [H+][H+] = 10-pH
pOH =-log [OH-][OH-] = 10-pOH
© 2008 Brooks/Cole 32
pH of Aqueous Solutions
Stomach Fluids
Lemon JuiceVinegar
WineTomatoesBlack Coffee
MilkPure waterBlood, seawaterSodium bicarbonateBorax solutionMilk of MagnesiaDetergentsAqueous ammonia
1 M NaOH
[H[H33OO++] [OH] [OH--]]
100 10-14
10-1 10-13
10-2 10-12
10-3 10-11
10-4 10-10
10-5 10-9
10-6 10-8
10-7 10-7
10-8 10-6
10-9 10-5
10-10 10-4
10-11 10-3
10-12 10-2
10-13 10-1
10-14 1
ExampleExample
Battery Acid
pHpH
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Bleach
© 2008 Brooks/Cole 33
Measuring pH
H3O+ concentrations can be measured with an:
• Electronic pH meterpH meter: fast and accurate preferred method.
• Acid-base IndicatorIndicator: substance that changes color within a narrow pH range may have multiple color change (e.g. bromthymol blue) one “color” may be colorless (e.g. phenolphthalein) cheap and convenient.
© 2008 Brooks/Cole 34
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
The acid ionization constantacid ionization constant is used to report the degree of ionization:
[A-][H3O+][HA]
Ka = (Water omitted, as usual)
Strong acids have large Ka values
Weak acid have small Ka values
Ionization Constants of Acids and Bases
When an acid ionizes in water:
© 2008 Brooks/Cole 35
Base Ionization Constants
For a base in water:
The base ionization constantbase ionization constant, Kb, is:
Kb =[BH+][OH-]
[B]
A-(aq) + H2O(l) HA(aq) + OH-(aq)
Kb =
[HA][OH-][A-]
B(aq) + H2O(l) BH+(aq) + OH-(aq)
If the base is an anion:
© 2008 Brooks/Cole 36
Larger Ka = stronger acid Larger Kb = stronger base
Ionization Constants
© 2008 Brooks/Cole 37
Larger Ka = stronger acid Larger Kb = stronger base
Ionization Constants
© 2008 Brooks/Cole 38
Acid and Base Strength
]HA[
]A][OH[Ka
3
[B]
]][OH[HBK b
)aq(OH)aq(HB )l(OH)aq(B 2
)aq(A)aq(OH )l(OH)aq(HA 32
Small value of Ka indicates that the numerator is smaller than the denominator. Equilibrium favors reactants or only a small amount dissociates – a weak acid.
Small value of Kb indicates that the numerator is smaller than the denominator. Equilibrium favors reactants or only a small amount dissociates – a weak base.
© 2008 Brooks/Cole 39
Write the ionization equation and ionization constant expression for the following acids and bases:
a.) HF
b.) HBrO
c.)NH3
d.) CH3NH2
Example
© 2008 Brooks/Cole 40
Ka Values for Polyprotic Acids
Some acids can donate more than one H+
Formula Name Acidic H’s
H2S Hydrosulfuric Acid 2
H3PO4 Phosphoric Acid 3
H2CO3 Carbonic Acid 2
HOOC-COOH Oxalic acid 2
C3H5(COOH)3 Citric acid 3
Each H+ ionization has a different Ka.• The 1st proton is easiest to remove• The 2nd is harder, etc.
© 2008 Brooks/Cole 41
Ka Values for Polyprotic Acids
Phosphoric acid (H3PO4) is weak. It has three acidic protons:
H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4- (aq)
Ka = 7.5 x 10-3
H2PO4-(aq) + H2O(l) H3O+(aq) + HPO4
2- (aq)
Ka = 6.2 x 10-8
HPO42-(aq) + H2O(l) H3O+(aq) + PO4
3- (aq)
Ka = 3.6 x 10-13
Ka
decreasingH
+ is harder to rem
ove
© 2008 Brooks/Cole 42
Practice – Relative Strength (Ka and Kb)
The following reactions all have Keq>1.
NO2- + HF HNO2 + F-
CH3COO- + HF CH3COOH + F-
HNO2 + CH3COO HNO2- + CH3COOH
Arrange the substances based on their relative base strength.
F-
CH3COOHHFHNO2
-
CH3COO-
HNO2
1 – strongest base2 – intermediate base3 – weakest base4 – not a B-L base
344214
© 2008 Brooks/Cole 43
Molecular Structure and Acid Strength
What makes a strong acid?• A weak H–A bond, so H+ can be easily removed!
smaller bond energy
larger acid strength
HX Bond Energy (kJ) Ka
HF weak acid 566 7 x 10-4
HCl 431 1 x 107
HBr 366 1 x 108
HI 299 1 x 1010
stro
ng a
cids
Consider the binary acidsbinary acids HF, HCl, HBr, HI.
H-A, bond energy decreases down a group.Acid strength increases.
© 2008 Brooks/Cole 44
Molecular Structure and Acid Strength
What makes a strong acid?• A weak H–A bond, so H+ can be easily removed!
Consider the binary acidsbinary acids SiH4, PH3, H2S, HCl.
The higher electronegativity, across a period, makes the bond more polar and hence more ionic – a stronger acid.
© 2008 Brooks/Cole 45
Oxoacids hydrogen, oxygen and one other element
H O-E higher the electronegativity on E, stronger the acid as
this weakens the bond between the O and H
Molecular Structure and Acid Strength
HOCl HOBr HOI
Electronegativity: Cl = 3.0, Br = 2.8, I = 2.5
Ka: 3.5 × 10-8 2.5 × 10-9 2.3 × 10-11
© 2008 Brooks/Cole 46
The acid strength increases as the number of oxygen atoms bonded to E increases. Strong oxoacid has at least two oxygen atoms per acidic H atoms.
Molecular Structure and Acid Strength
HClO HClO2 HClO3 *HClO4
Ka: 3.5 × 10-8 1.1 × 10-2 ≈ 103 ≈ 108
© 2008 Brooks/Cole 47
Problem Solving Using Ka and Kb
Example: Example: KKaa from pH from pH
Lactic acid is monoprotic. The pH of a 0.100 M solution was 2.43 at 25°C. Determine Ka for this acid.
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Using the pH information:- log [H3O+] = 2.43 [H3O+] = 10-2.43 = 0.0037 M
These hydronium ions are produced by:
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
and from the autoionization of water:
© 2008 Brooks/Cole 48
Ka = = 1.4 x 10-4 (0.0037)(0.0037)(0.100 – 0.0037)
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[ ]initial 0.100 1.0 x 10-7 0 0 (water autoionization)
change -0.0037 +0.0037 +0.0037
[ ]equil 0.100 – 0.0037 0.0037 0.0037
Ka =[H3O+][A-]
[HA]
Problem Solving Using Ka and Kb
© 2008 Brooks/Cole 49
Problem Solving Using Ka and Kb
Example: pH from Example: pH from KKaa
Determine the pH of a 0.100 M propanoic acid solution at 25°C. Ka = 1.4 x 10-5. What % of acid is ionized?
C2H5COOH + H2O(l) H3O+(aq) + C2H5COO-(aq)
Ka = = 1.4 x 10-5[H3O+][C2H5COO-] [C2H5COOH]
© 2008 Brooks/Cole 50
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[ ]initial 0.100 0* 0
change -x +x +x
[ ]equil 0.100 – x x* x
*Ignore the water contribution
1.4 x 10-5=x2
(0.100 – x)
[H3O+][A-][HA]
Ka = 1.4 x 10-5 =
Problem Solving Using Ka and Kb
Determine the pH of 0.001 M propanoic acid soln. at 25°C. Ka = 1.4 x10-5. What % of acid is ionized?
© 2008 Brooks/Cole 51
Problem Solving Using Ka and Kb
Because Ka is small, assume x < [HA]
Ka = 1.4 x 10-5 = ≈
x2
(0.100 – x)x2
0.100
x = (1.4 x 10-5 )(0.100) = 0.00118 M
(0.00118 << 0.100 M; the approximation is good)
pH = -log(0.00118) = 2.93
%-ionized = x100 = 1.18 %0.001180.100
<5% rulecan assume[HA]eq~[HA]0
© 2008 Brooks/Cole 52
What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.
Example 3 – calculating pH from known Kb
)aq(OH)aq(NHHC )l(OH)aq(NHC 55255
Initial 0.20 0 0Change -x +x +x
Equilibrium 0.20-x x x
© 2008 Brooks/Cole 53
Example 3 – calculating pH from known Kb
What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.
)x20.0(
x104.1
29
x+x0 00.20Initial
x0.20-xEquilibrium+x-xChange
x+x0 00.20Initial
x0.20-xEquilibrium+x-xChange
b55
55 K]NHC[
]OH][NHHC[
)aq(OH)aq(NHHC )l(OH)aq(NHC 55255
© 2008 Brooks/Cole 54
Example 3 – calculating pH from known Kb
What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.
)20.0(104.1
29 x
Solving for x we get
)104.1()20.0(x 92
][107.1)104.1()20.0( 59 OHx
© 2008 Brooks/Cole 55
Example 3 – calculating pH from known Kb
What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.
Solving for pOH
77.4)107.1log(]OHlog[pOH 5
23.977.400.14pH
© 2008 Brooks/Cole 56
Practice
Ka from pHPara-aminobenzoic acid (PABA), HC7H6NO2, is used in some sunscreen agents. A solution is made by dissolving 0.263 mol of PABA in enough water to make 750.0 mL of solution. The solution’s pH = 2.59. What is the Ka for PABA? (1.9x10-5)
pH from Ka Barbituric acid (Ka=1.1x10-4) is used in the manufacture of some sedatives. What is the pH for a 0.673 M solution of barbituric acid? (2.07)
© 2008 Brooks/Cole 57
Relationship between Ka and Kb values
For an acid-base conjugate pair: HA and A-
[HA][OH-][A-]Ka x Kb =
[H3O+][A-][HA]
= [H3O+][OH-] = Kw
)aq(CN)aq(OH )l(OH)aq(HCN 32
)aq(OH)aq(HCN )l(OH)aq(CN 2
Ka
Kb
)aq(OH)aq(OH )l(OH2 32
When two reactions are added, their equilibrium constants
are multiplied (chapter 14).wba KKK
© 2008 Brooks/Cole 58
Relationship between Ka and Kb values
Phenol, C6H5OH, is a weak acid, Ka = 1.3 x 10-10 at 25°C. Calculate Kb for the phenolate ion C6H5O-
Ka x Kb = 1.0 x 10-14
Kb = = 7.7 x 10-51.0 x 10-14
1.3 x 10-10
© 2008 Brooks/Cole 59
Acid Base Neutralization Reactions
SaltSalt = ionic compound formed in acid + base reaction
Strong acid + strong baseStrong acid + strong base form neutral salts.
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)
Na+(aq) + Cl-(aq) + H2O(l)
HX(aq) + MOH(aq) MX(aq) + H2O(l)acid base salt
Net ionic equation: H+(aq) + OH-(aq) H2O(l)
H3O+(aq) + OH-(aq) 2 H2O(l)
© 2008 Brooks/Cole 60
H3O+(aq) + OH-(aq) H2O(l) + H2O(l)
Keq = 1/Kw = 1x1014
strongbase
strongacid
weakbase
weakacid
• Na+ and Cl- are spectator ions.• Na+ is the v. weak conjugate acid of a v. strong base. • Cl- is the v. weak conjugate base of a v. strong acid.• Final solution has pH = 7.
Salts of Strong Bases and Strong Acids
H3O+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)
Na+(aq) + Cl-(aq) + 2 H2O(l)
100%
© 2008 Brooks/Cole 61
Salts of Strong Bases and Weak Acids
Weak acid + strong baseWeak acid + strong base form basic salts.
CH3COOH(aq) + NaOH(aq)
CH3COONa(aq) + H2O(l)
Net ionic: CH3COOH(aq) + OH-(aq)
CH3COO- + H2O(l)
CH3COOH(aq) + Na+(aq) + OH-(aq)
Na+(aq) + CH3COO-(aq) + H2O
100%
final solution
© 2008 Brooks/Cole 62
• acetate is a weak base (much stronger than water)• solution is basic (pH > 7.0).
CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-
(aq)
Calculate pH of solution from Kb of CH3COO-.
A hydrolysishydrolysis reaction – water is broken apart.
CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O(l)
baseacid base acid
Salts of Strong Bases and Weak Acids100%
© 2008 Brooks/Cole 63
pH of a Salt Solution
For a weak acid + strong base, pH depends on Kb
Larger Kb = stronger base
ExampleExample
What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb (CO3
2-) = 2.1 x 10-4.
(or equal volumes of 1.5M H2CO3 and 3.0 M NaOH are mixed, what is the pH of the mixture?)
© 2008 Brooks/Cole 64
Na+ is the conjugate acid of NaOH (strong base). It remains 100% ionized Na+ has no effect on pH.
CO32- is the conjugate base of HCO3
- (weak acid)
Most CO32- ions undergo hydrolysis
Generates HCO3- and OH-
Changes the pHCO3
2-(aq) + H2O(l) HCO3-(aq) + OH-(aq)
What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb = 2.1 x 10-4
pH of a Salt Solution
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CO32-(aq) + H2O(l) HCO3
-(aq) + OH-(aq)
[ ]initial 1.50 0 0change - x +x +x[ ]equil 1.50 - x x x
Kb = 2.1 x 10-4 = = ≈[HCO3-][OH-]
[CO32-]
x2
(1.50 – x)x2
1.50
x = 1.77 x 10-2 pOH = −log(1.77 x 10-2) = 1.75
pH = 14.00 - 1.75 = 12.25
What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb = 2.1 x 10-4
pH of a Salt Solution
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Salts of Weak Bases and Strong Acids
NH3(aq) + HCl(aq) NH4+(aq) + Cl-(aq)
The products of the neutralization reaction form an acidic solution:
NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)
NH3(aq) + H3O+(aq) + Cl-(aq)
NH4+(aq) + Cl-(aq) +
H2O(l)
Net ionic: NH3(aq) + H3O+(aq) NH4+(aq) + H2O(l)
The final pH depends on the Ka , larger Ka = more acidic
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pH of a Salt Solution
Find the pH of a 0.132 M NH4Br. Is this solution acidic, basic or neutral? Ka(NH4
+) = 5.6 x 10-10.
• The solution contains NH4+ ions and Br- ions.
• NH4+ is the conjugate acid of NH3 (a weak base).
It will partially ionize in solution giving NH3 and H3O+
Forms an acidic solution.
• Br- is the conjugate base of HBr (a strong acid) Br- ions stay fully ionized Does not change the pH.
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NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)
[ ]initial 0.132 0 0change - x +x +x[ ]equil 0.132 - x x x
x = 8.57 x 10-6 pH = −log(8.57 x 10-6) = 5.07
Ka = 5.6 x 10-10 = = ≈[NH3][H3O+][NH4
+]
x2
(0.132 – x)x2
0.132
pH of a Salt SolutionpH of a 0.132 M aq. solution of NH4Br? Is this acidic, basic or neutral? Ka(NH4
+) = 5.6 x 10-10.
Acidic!
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Salts of Weak Bases and Weak Acids
The most difficult. Consider qualitative results. e.g.
F-(aq) + H2O(l) HF(aq) + OH-(aq)
NH3 (aq) + HF(aq) NH4+ (aq) + F- (aq)
NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)
Ka(NH4+) = 5.6 x 10-10 ; Kb(F-) = 1.4 x 10-11
Ka (weak acid) > Kb (weak base)
The ammonium reaction is more favorable.
The acid wins! The solution will be acidic!
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Acids, Bases and Salts
Strong acidStrong acid + strong basestrong base → salt solution, pH = 7pH = 7
Strong acidStrong acid + weak baseweak base → salt solution, pH < 7pH < 7
Weak acidWeak acid + strong basestrong base → salt solution, pH > 7pH > 7
Weak acidWeak acid + weak baseweak base → salt solution, pH = ?pH = ?
Need K’s “strongest wins”
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Neutral Ions in Water (Table 16.5)
Anions Cl- NO3-
Br- ClO4-
I-
Cations Li+ Ca+2
Na+ Sr+2
K+ Ba+2
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What is the pH of a 0.10 M NaCN solution at
25 oC? The Kb for CN- is 2.5 x 10-5.
Example
Sodium cyanide gives Na+ ions and CN- ions in solution.
Only the CN- ion hydrolyzes.
)aq(OH)aq(HCN )l(OH)aq(CN 2
pH =11.2
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What is the pH of a 0.15 M NH4NO3 solution at
25 oC? The Ka for NH4+ is 5.6 x 10-10.
Example
NH4NO3 gives NH4+ ions and NO3
- ions in solution.
Only the NH4+ ion hydrolyzes.
)aq(OH)aq(NH )l(OH)aq(NH 3324
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The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.
The Common Ion Effect
(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232
Consider a solution of acetic acid (HC2H3O2),
in which you have the following equilibrium.
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The Common Ion Effect
The equilibrium composition would shift to the left
(LeChatelier Principle) and the degree of ionization of the acetic acid would decrease.
(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232
If we were to add NaC2H3O2 to this solution, it would provide C2H3O2
- ions which are present on the right side of the equilibrium.
This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect.
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What is the pH of a solution that is 0.025M in formic acid?
The Ka for formic acid is 1.7 x 10-4.
What is the pH of the same solution after 0.018 M in sodium formate, NaCH2O is added?
Example
(aq)OCH(aq)OH 23 )l(OH)aq(OHCH 22
Initial
Change
Equilibrium
0.025 0
-x +x +x
0.025-x x 0.018+x
From NaCH2O
0.018
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Initial 0.025 0 0.018Change -x +x +x
Equilibrium 0.025-x x 0.018+x
4107.1)025.0()018.0( x
Example
4
2
23a 101.7
x)(0.025x)x(0.018
O][HCH]O][CHO[H
K
(aq)OCH(aq)OH 23 )l(OH)aq(OHCH 22
x [H3O+]= 2.4 × 10-4
pH = -log(H3O+) = -log(2.4 x 10-4) = 3.62