chapter 13 chemistry foundation 2014
TRANSCRIPT
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CHAPTER 13 16)ACID BASE
EQUILIBRIA
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Contents
Definition of Acid and Base : Arrhenius
Brnsted-Lowry , conjugate acid-base pairs
Lewis
Strength of Acid and Base : Strong acid weak acid
Strong base weak base
Ionization of Acid and Base
Concepts of pH, pOH, pKa, pKb
Dissociation Constant : Ka, Kb, Kw
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Learning Outcomes
Able to differentiate and calculate acid
dissociation constants for weak and strongacids (applicable to bases)
Calculate pH and pOH
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Whenever an acid dissociates (ionizes) in water,
solvent molecules participate in the reaction
HA (g or aq) + H2O(l) A-(aq) + H3O
+(aq)
The H3O+is
called
hydronium ion
The terms hydrogen
ion = proton = H+ are
used interchangeably
Arrhenius Definition
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Strong acids: dissociate/ionize completely intoions in water.
Eg:HNO3(aq)+ H2O(l) H3O+(aq) + NO3
-(aq)
Weak acids: dissociate/ionize very slightly intoions in water.
Eg: HCN(aq) + H2O (l) H3O+(aq) + CN-(aq)
This classification correlates with classification ofelectrolytes: strong electrolytes dissociate/ionize completely, and
weak electrolytes dissociate/ionize partially.
Arrhenius Definition
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Brnsted-Lowry Acid-Base Definition
Acid
pro ton donor.Any species that donates an H+
such as HCl, HNO3, H3PO4
All Arrhenius acids areBrnsted-
Lowry acids
Base
proto n acceptor.
Any species that accepts an H+,
must contain lone pair of electrons to
bind the H+, such as NH3, CO32-, F- andOH-
Brnsted-Lowry bases are not
Arrhenius bases, but all Arrhenius bases
contain Brnsted-Lowry base OH-. -
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HCl (g) + H2O (l) H3O+(aq) + Cl-(aq)
HCl acts as a Brnsted-Lowry acid ( it donates a
proton to H2O)
H2O acts as a Brnsted-Lowry base ( it accepts a
proton from HCl)
Brnsted-Lowry Acid-Base Definition
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NH3(aq) + H2O (l) NH4+ (aq) +OH-
(aq) H2O acts as an acid, it donates the H
+
NH3
acts as a base, it accepts H+.
Thus, H2O is ampotheric: it acts as a base in
one case and as an acid in the other.
Brnsted-Lowry Acid-Base Definition
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NH3(aq) + H2O (l) NH4+ (aq) +OH-
(aq)
Brnsted-Lowry Acid-Base Definition
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Example
Determine acid and base in the following reactions:
1. HCO3-(aq) + HF (aq) H2CO3(aq) + F
-(aq)
2. HCO3-(aq) + OH-(aq) CO3
2-(aq) + H2O (aq)
3. SO32-(aq) + NH4
+(aq) HSO3-(aq) + NH3(aq)
4. HSO3-
(aq) + NH3(aq)
SO32-
(aq) + NH4+
(aq)
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Answer
Acid Base
1. HF (aq) HCO3-
(aq)2. HCO3
-(aq) OH-(aq)
3. NH4+(aq) SO3
2-(aq)
4. HSO3-(aq) NH3(aq)
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Conjugate Acid-Base Pairs
Consider the reaction of acid HA:
HA (aq) + H2O(l)
A
-
(aq) + H3O
+
(aq) Forward reaction: HA is the acid, H2O is the base
Reverse reaction: H3O+is the acid, A-is the base
HA and A-: differ in the presence or absence of aproton (base)
a conjugate acid-base pair.
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Add H+
In any acid-base reaction:
HNO2(aq)+ H2O (l) NO2- (aq) + H3O
+ (aq)
Acid Base Conjugate
base
Conjugate
acid
Remove H+
Conjugate Acid-Base Pairs
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Acids such as HCl , HNO3 , or H2SO4ionize in water to form a
hydrated proton and a species called the con jugate baseof
the acid;HCl (aq) H+(aq) + Cl-(aq)
AcidConjugate base
of HCl
Conjugate base of HCl results from the loss of one proton.
Conjugate Acid-Base Pairs
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Bases such as NH3react with water to form a hydrated
hydroxide ion and a species called the conjugate acidof
base.NH3(aq) + H2O(l) NH4
+(aq) + OH-(aq)
Base Conjugate acid
of NH3
Conjugate acid of NH3results from the gain of one proton.
Conjugate Acid-Base Pairs
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In NH3and H2O
NH3
(aq) + H2
O (l) NH4
+(aq) + OH-(aq)Base Acid Conjugate
acid
Conjugate
Base
Add H+
Remove H+
NH3 NH4+
base Conjugate acid
Add H+H2O OH
-
Acid Conjugate Base
Remove H+Note:
Conjugate Acid-Base Pairs
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a) What is the conjugate base of the following acids:HClO4; HCO3
-
- remove one proton ( H+) from the formula.
ClO4-; CO32-
b) What is the conjugate acid of the following bases:
CN-; H2O ; HCO3-
- add one proton to the formulaHCN ; H3O
+; H2CO3
Note: Hydrogen Carbonate ion, HCO3-is amphoteric: a
substance that can act as an acid or as a base.
Example
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Example
Determine the conjugate acid-base pairs in the
following:
1. HCO3-(aq) + HF (aq)H2CO3(aq) + F
-(aq)
2. HCO3-(aq) + OH-(aq)CO3
2-(aq) + H2O (aq)
3. SO32-(aq) + NH4+(aq)HSO3-(aq) + NH3(aq)
4. HSO3-(aq) + NH3(aq)SO3
2-(aq) + NH4+(aq)
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Answer
Acid Base Conj. acid Conj. Base
HF (aq) HCO3-(aq) H2CO3(aq) F-(aq)
HCO3-(aq) OH-(aq) H2O (l) CO3
2-(aq)
NH4+(aq) SO3
2-(aq) HSO3-(aq) NH3(aq)
HSO3-
(aq) NH3(aq) NH4+
(aq) SO32-
(aq)
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Other example: HSO3-
Acid
HSO3-(aq) + H2O(l) SO3
2-(aq) + H3O+
Remove H+
Add H+
Base
HSO3-(aq) + H2O(l) H2SO3(aq) + OH
- (aq)
Add H+
Remove H+
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Conjugate pairs in acid-base reactions
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Lewis Acid-Base Definition
OH-, H2O , an amine - are electron-pair donors
A base in the Brnsted-Lowry ( a proton acceptor)also base in the Lewis (an electron pair donor )
In the Lewis theory: a base can donate its electronpair to species other than H+
Acid Electron pair acceptor
Base Electron pair donor
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Lewis Acids with Electron-Deficient Atoms:
Eg: Reaction between NH3and BF3.
BF3has a vacant orbital in its valence shell- acts as
an electron pair acceptor (a Lewis acid) toward
NH3.NH3donates the electron pair.
Lewis Acid-Base Definition
N
H
H
H
B
F
F
F N
H
H
H B
F
F
F
Lewis base Lewis acid
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Lewis acid - molecules have an incomplete octet ofelectrons.
Eg. Cations Fe3+interacts with cyanide ions;
Fe3+ion has vacant orbitals - accept the electronpairs donated by the CN-ions.
Lewis Acid-Base Definition
Fe3+
+ C N6 C NFe 63-
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Note: Lewis acids and bases do not need to containprotons.
Therefore, the Lewis definition is the most generaldefinition of acids and bases.
Lewis acids generally have an incomplete octet (eg.BF3).
Transition metal ions are generally Lewis acids.
Lewis acids must have a vacant orbital (into which theelectron pairs can be donated).
Lewis Acid-Base Definition
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Metal Cation as Lewis Acid
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Definitions Comparison
Theory
Arrhenius
Brnsted-Lowry
Lewis
Acid Base
Forms H3O+ions in
water
Forms OH-ions in
water
Proton donorHCl + H2O H3O
++ Cl-Proton acceptor
NH3(aq) + H2O
NH4 + OH-
Electron pair acceptor
B
F
F
F
Electron pair donor
N
H
H
H
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Strength of Acid and Base:Objectives
Strength of acid-base
Strong acid - strong base / weak acid - weakbase
Strength in conjugate acid base determinethe direction of acid-base reaction
Water auto-ionization
Concept of pH
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Variation in Acid Strength
Strong acids: Ionize completely into ions in water.
Eg: HNO3(aq)+ H2O(l) H3O+(aq) + NO3
-(aq)
Weak acids: ionize very slightly into ions in water.
Eg: HCN(aq) + H2O (l) H3O+(aq) + CN-(aq)
This classification correlates with classification ofelectrolytes:
strong electrolytes dissociate/ionize completely
weak electrolytes dissociate/ionize partially.
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Strong Acids
Hydrohalic acidsHCl,
HBr,
HI
Oxoacids
(number of O exceeds no. of
ionisable H atoms by two or more)
HNO3H2SO4
HClO4HClO3
HCl (aq) + H2O (l) H3O+(aq) + Cl-(aq)
HNO3(aq) + H2O (l) H3O+(aq) + NO3
-(aq)
Strong acid will fully dissociate/ionize in water
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Weak acids
Hydrohalic acids HF
Acids in which H is not bonded
to O or to halogen
HCN
H2S
Oxoacids
(in which no. of O equals or exceed
by one the no. of ionizable H)
HClO,
HNO2
H3PO4
Organic acids
(general formula RCOOH)
CH3COOH
C6H5COOH
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Example
HF (aq)+ H2O (l)H3O+(aq)+ F
-(aq)
CH3COOH (aq)+ H2O (l)H3O+
(aq)+ CH3COO-
(aq) HNO2 (aq)+ H2O (l)H3O
+(aq)+ NO2-(aq)
Weak acid will partially ionize in water
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Strong Base
Soluble
compoundscontaining O2-
or OH-
The cations are
usually those ofthe most active
metalsM2O or MOH;
Where M= Group IA
metal
(Li, Na, K, Rb, Cs)
MO or M(OH)2;
Where M= Group IIA
metal
(Be, Mg, Ca, Sr, Ba)
Ionic hydrides
and nitrides
H-(aq) + H2O (l) H2(g) + OH-(aq)
N3-(aq) + 3 H2O (l) NH3(aq) + 3OH-(aq)
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Weak Bases
Compounds with
an electron-rich
nitrogen are weak
bases(none are
Arrhenius bases).
The common
structural
feature is an
N atom thathas a lone
pair
Ammonia: NH3
Amines
such asCH3CH2: NH2C5H5N:
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Relative Strengths of Acid-Base
Some acids are better proton donors than others.
The stronger the acid, the weaker its conjugate base.
Some bases are better proton acceptors.
The stronger the base, the weaker its conjugate acid
The strength of an acid tells the strength of its conjugatebase.
Strong acids: completely transfer their protons 100%dissociation Their conjugate bases have a negligibletendency to be protonated.
(E.g: HCl + H2O H3O++ Cl-)
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Consider the proton transfer in this:
HA(aq) + H2O (l) H
3O+(aq) + A-(aq)
If HA is a stronger acid than H3O+;
HA will transfer its proton to H2O ; equilibrium lies to
the right
If H3O+is a stronger acid than HA (if HA is a weak
acid);
equilibrium lies to the left
In every acid-base reaction, the position of theequilibrium favors transfer of the proton from thestronger acid to the stronger base to form the
weaker acid and the weaker base.
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Example: CN-reacts with water as follows:
CN- (aq) + H2O(l) HCN (aq) + OH-(aq)
A Cl-ion does not react with water to form HCl.
Which acid is stronger, HCl (aq) or HCN (aq)? Why?
HCl and HCN are the conjugated acids of the bases Cl-and
CN-.The information provided tells us that Cl- is a weaker
base than CN-.CN-accepts a proton from H2O to form HCN.
Principle: The weaker a base, the stronger its conjugated
acid, can conclude that HCl is a stronger acid than HCN
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Example
Predict the product of the following acid-base reactionsand also predict whether the equilibrium lies to the left
or to the right of the equation:1. Cl-(aq) + H3O
+(aq)
2. H2PO4-(aq) + H2O (l)
3. HS-
(aq) + HC2H3O2(aq)
4. HCO3- (aq) + OH (aq)
5. NH4+ (aq) + H2PO4
-(aq)
HCl (aq) + H2O(aq) Lies to the left
HPO42-(aq) +H3O
+(aq)Lies to the
leftH
2S (aq) + C
2H
3O
2
-
Lies to the
right
CO32-(aq) + H2O (l) Lies to the
rightNH3(aq) + H3PO4(aq)
Lies to the left
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Autoionization of water
Water ionises into H+and OH-(v. small extent).This process is called the autoionizationof water.
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The Ion-Product Constant for Water, Kw
We can write an equilibrium constant expression for the
autoionization of water: 2H2O(l) H3O+(aq) + OH-
(aq)
Because H2O(l) is a pure liquid, the expression can be
simplified:
22
eq
OH
OHOHK
3
weq KOHOHKOH 32
2
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The Ion-Product Constant for Water, Kw
Kw is called the ion-product constant.
At 25C the ion-product of water is always:
In neutral solution: [H3O+] = [OH-]
In acidic solution : [H3O
+
] exceeds [OH-
] In basic solution : [OH-] exceeds [H3O
+]
Note: The ion -product constant Kw is unaffected bywhether a solution is acidic or basic.
OHOHKw 3
14101
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Calculate the concentration of H3O+(aq) in a
solution in which [OH-] is 2.0 x 10-9M at25C.
M10x5.010x2.010x1.0
OH
10x1.0H
10x1.0OHH
69
1414
3
14
3
O
O
Example
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The pH scale
Express H+in terms of pH.
For [H+] = 1.0 x 10-3M.
pH = -log (1.0 x 10-3) = -(- 3.00 ) = 3.00
pH = -log [H+] = -log [H3O+]
Basic
[H+
] (M)
1.0x10-7[OH-] (M)
>1.0x10-7
=1.0x10-7
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Acidic Neutral Basic
[H3O+]
1 10-7
[OH-]
1 10-7
Neutral
[H3O+]1
[OH-]1 10-14
Acidic
1 10-1
1 10-2
1 10-31 10-4
1 10-5
1 10-6
1 10-13
1 10-12
1 10-111 10-10
1 10-9
1 10-8
[OH-][H3O+]1 10-6
1 10-5
1 10-4
1 10-3
1 10-13
1 10-8
1 10-9
1 10-10
1 10-111 10-12
1 10-14
Basic
1 10-2
1 10-1
1
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Example
Given [OH-] = 2.0 x 10-3, calculate [H+] (or [H3O+])
and pH at 25C.
pH = -log ( 5.0 X 10-12 )
= 11.3
MOH
KH W
12
3
14
100.5100.2
100.1
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The concentration of OH-can be expressed as pOH
Example:
A solution has a pH of 5.6 . Calculate the hydrogenion concentration.
pH = -log [H+], log [H+] = -5.6
[H+
] = 2.5 10-6
M
Other p scale
pOH = -log [OH-]
pH + pOH = pKw= -log Kw= 14.00
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Calculate [H+] for a solution with pOH of 4.75
pOH is defined aslog [OH-
] The pH and the pOH are related: pH + pOH = 14.00
pH = 14.00 - pOH = 14 - 4.75 = 9.25
pH = 9.25
log [H+] = -pH = -9.25
[H+] = 5.6 x 10-10 M.
Example
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Acid-base indicator
pH paper
pH meter
Methods for measuring pH
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Strong Acids
Strong acids are strong electrolytes and they ionizecompletely in solution.
HNO3(aq) + H2O (l) H3O+(aq) + NO3-(aq)
Example
What is the pH of a 0.04 M solution HClO4?
HClO4 is completely ionized: [H+] = 0.04 M
Hence, pH= -log (0.040) = 1.40
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Strong Bases
Most ionic hydroxides are strong bases (eg. NaOH,KOH and Ca(OH)2.)
Strong bases are strong electrolytes and dissociatecompletely in solution:
NaOH (aq)
Na
+
(aq) + OH
-
(aq)
The pOH of a strong base is given by the initialmolarity of the base.
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Example
What is the pH of a 0.011 M solution of Ca(OH)2?
Ca(OH)2 is a strong base.
Ca(OH)2 Ca2+ + 2OH-
0.011M 0.011M 2 0.011M
pOH = -log (0.022) = 1.66pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 1.66 = 12.34
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Acid Dissociation Constant, Ka
HA (g or aq) + H2O(l) A-(aq) + H3O
+(aq)
Stronger acidlarger Ka Weaker acidSmaller Kalower % HA ionizes
HAorHA
AH
KAOH
K eqeq3
HA
AOH
K 3
a
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Ka values for weak acids
Acid Ka % HA dissociated
1 M HClO2 1.12 x10-2 10 %
1 M CH3COOH 1.8 x10-5 0.42 %
1 M HCN 6.2 x10-10 0.0025 %
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Solving Problems Involving Weak-AcidEquilibria
1. Given equilibrium concentrations, find Ka.
2. Given Ka and other info, find other equilibriumconcentration.
HA (aq) + H2O(l) H3O+ (aq) + A- (aq)
Ionization is incomplete and some significantamount of undissociated acid remains at equilibrium.
[HA]
][A][HKa
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Problem-Solving Approach : ProblemsInvolving Weak-Acid Equilibria
1. Write the balanced equation and Kaexpression
2. Define x as unknown concentration that changesduring the reaction.
3. Construct a reaction table that incorporates theunknown.
4. Make assumptions that simplify the calculations.
5. Substitute the values into the Ka expression and solveforx
6. Check that the assumptions are justified.
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HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
Iffis the formal concentration of acid ,xbe theequilibrium concentration of H3O
+. Then,
HA (aq) + H2O (l) H3O+(aq) + A- (aq)
f - x x x
xf
x
[HA]
][A]O[HK
2
3a
Problem-Solving Approach : ProblemsInvolving Weak-Acid Equilibria
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The equation can be solved using quadratic eqn.
If f>>x, then
x2
Kafx=
Ifxis 5% of the initial concentration, we can use
the assumption.
Ifxis 5% of the initial concentration, it maybe
best to solve the quadratic equation.
fKa
Problem-Solving Approach : ProblemsInvolving Weak-Acid Equilibria
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Percent ionisation is another method to assess acid
strength.
% ionisation
The higher the percent ionisation, the stronger theacid.
x100[HA]
]O[H
initial
eqm3
Problem-Solving Approach : ProblemsInvolving Weak-Acid Equilibria
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A student prepared 0.10 M solution of formic acid HCHO2.
The pH= 2.38 at 25C.
1. Calculate Ka
2. Percentage of ionisation.
Example
MH
H
HpH
HCHO
CHOHK
a
3
2
2
102.4
38.2log
38.2log
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Calculate Ka
HCHO2(aq) H+(aq) + CHO2
-(aq)
Example
MHHHpH
HCHO
CHOHKa
3
2
2
102.4,38.2log,38.2log
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AnswerHCHO2(aq) H
+
(aq) + CHO2-
(aq)
Initial
Change
Eqm
(0.10 4.2 10-3)M 0.10 M
-4.2 10-3M
0.10 M 0 0
+4.2 10-3M +4.2 10-3M
4.2 10-3M 4.2 10-3M(0.10 - 4.2 10-3)M
4
33
108.110.0
102.4102.4
a
K
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The percentionisation of a weakacid decreases as its
concentrationincreases. 1.0
2.0
3.0
4.0
5.0
0.05 0.10 0.15
Per
centionised
Acid concentration (M)
%2.4%10010.0
102.4
%100%
3
2
initial
mequilibriu
HCHO
Hionisation
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Calculate the percentage of HF (Ka= 6.8 x 10-4)
molecules ionised in:
a. 0.10M HF solution
b. 0.010 M HF solution
Example
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Acids have more than one ionizable H atom
Example: Sulfurous acid, H2SO3.
H2SO3 (aq) H+(aq) + HSO3
-(aq) Ka1= 1.7 x 10-2
HSO3-
(aq) H+
(aq) + SO32-
(aq) Ka2= 6.4 x 10-8
Ka2 is much smaller thanKa1; easier to remove the firstproton from polyprotic acid than the second
Ka1is much larger thanKa2; can estimate the pH byconsidering onlyKa1.
Polyprotic Acid
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Example
CO2dissolved in water at 25C and 0.1 atm to form H2CO3 withconc. 0.0037 M.
What is the pH ?
CO2(aq) + H2O(l) H2CO3(aq)
[H2CO3] = 0.0037 M
H2CO3is a polyprotic acid:
H2CO3(aq) H+(aq) + HCO3
-(aq) ka1= 4.3 10-7
HCO3-(aq) H+ (aq) + CO3
2-(aq) ka2= 4.710-11
ANSWER = 4.40
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Weak Bases
There is an equilibrium between the base and theresulting ions
Weak base + H2O(l)conjugate acid + OH-(aq)Eg: NH3(aq) + H2O NH4
+(aq) + OH-(aq)
The base-dissociation constant , Kb, is defined as
The larger the Kb the stronger the base
][NH]][OH[NHK
3
4b
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Example
Calculate the concentration of OH- in a 0.15 Msolution of NH3. (Kb= 1.8 x 10
-5)
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
0.15 -x x x
x= [NH4+] =[OH-] = 1.6 x10-3 M
5
3
4b 10x1.8x0.15
(x)(x)
][NH
]][OH[NHK
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Example
A solution is made by adding Sodium Hypochlorite,NaClO, to water (2.0 L soln). The pH of solution is10.50. How many moles of NaClO were added?
Note:
NaClO is an ionic compound, strong electrolyte.
NaClO
Na+
+ ClO-
ClO-(aq) + H2OHClO(aq) + OH-(aq)
Kb= 3.3 x 10-7.
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Unit in molarity
pOH + pH = 14
pOH = - log [OH-]
pH = - log [ H+]
Given: pH = 10.50 ; base. Whats the [OH-]
Knowing the [OH-] then we know the Molarity of ClO-.
Molarity of NaClO mole ( in the same volume: 2.0 L )
NaClO Na+ + ClO-
1 mole 1 mole 1 mole
Answer
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ClO-(aq) + H2O(l) HClO(aq) + OH-(aq)
Let say we start with 1.0 M NaClO : conc. of ClO-is 1.0 M. Then,
ClO-(aq) + H2O(l) HClO(aq) + OH-(aq)
1 - x x x
x; concentration of OH-
* If we dont know the ClO-
. Let say ClO-
: zBut we know the [OH-] x . Then,
z - x x x
Answer
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Given pH = 10.50
from : pH + pOH = 14.0pOH = 14.0 - 10.50 = 3.50
pOH = - log [OH-]
[OH-] = 10 -3.50
= 3.2 x 10-4M
x = 3.2 x 10-4M
Answer
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ClO-(aq) H2O(l) HClO(aq) OH-(aq)
Initial z - 0 0
Change -3.2 10-4 - 3.2 10-4 3.2 10-4
final z - 3.2 10-4 - 3.2 10-4 3.2 10-4
Tabulate in equilibrium table:
ClO-(aq) + H2O(l) HClO(aq) + OH-(aq)
M0.31z
10x3.310x3.2z
)10x(3.2
][ClO][HClO][OHK
7
4
24
b
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Z = [ClO-] = 0.31 M
NaClO = 0.31 M
Mol of NaClO = 0.31 (mol/ L) x (2 L)= 0.62 mol
2L
Mol0.31
Volume(L)
MolMMolarity,
Answer
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Relationship between Kaand Kb
We need to quantify the relationship between
strength of acid and conjugate base.
NH4+(aq)NH3(aq) + H
+(aq)
NH3(aq) +H2O(l)NH4+(aq) + OH-(aq)
][NH
]][OH[NHK
][NH]][H[NHK
3
4b
4
3a
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NH4+(aq) NH3(aq) + H
+(aq)
NH3(aq) +H2O(l) NH4+(aq) + OH-(aq)
H2O(l) H+(aq) + OH-(aq)
When two reactions are added to give a third, the equilibrium
constant for the third reaction is the product of the equilibrium
constant for the two added reactions:
reaction 1+ reaction 2 = reaction 3
K1x K2 = K3
Relationship between Kaand Kb
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For a conjugate acid -base pair
wba
w
3
4
4
3ba
KKKK]][OH[H
][NH
]][OH[NH
][NH
]][H[NHKK
Therefore , the larger the Kathe smaller the Kb. That is, the
stronger the acid, the weaker the conjugate base.
Taking negative logarithms:
pKa = -log (Ka) and pKb= -log(Kb)
then pKa+ pKb= pKw
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END of CHAPTER 3