Transcript Chapter 4 The Basic Approach to Chemical Equilibrium

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Chapter 9
Aqueous Solutions and Chemical Equilibria
1) Non electrolytes: Are substances that dissolve in water but do
not produce any ions and do not conduct an electric current.
2) Electrolytes:
Form ions when dissolved in water or other
solvents and produce solutions that conduct electricity.
1) Strong Electrolytes: Ionize essentially
completely. Strong conductor of electricity.
2) Weak Electrolytes: Ionize only partially.
Poorer conductor than strong electrolyte.
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Acids and Bases: An acid is a proton donor and a
base is a proton acceptor (Bronsted-Lowry
concept).
Conjugate Acids and Bases: A conjugate base is
the species formed when an acid loses a proton.
NH3 + H2O  NH4+ + OHbase1 acid2 conjugate acid1 conjugate base2
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Amphiprotic Solvents: A solvent that can act
either as an acid or as a base depending on the
solute. Water is the classic example.
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Three general types of solvents :
1) Protic solvent : amphiprotic solvents
-
2H2O  H 3O+ + OH–
possess both acidic and basic properties
undergoes self-ionization (=autoprotolysis)
ex. Water, alcohols, acetic acid, ammonia
2C2H5OH  C2H5OH2+ + C2H5O–
2HOAc  H2OAc+ + OAc–
ethylenediamine
2NH3  NH4+ + NH2–
2) Aprotic solvents
-
have no appreciable acidic or basic character
-
do not undergo autoprotolysis
-
ex. Benzene, carbon tetrachloride, pentane
3) Basic solvents
-
have basic properties but essentially no acidic
tendencies
-
do not undergo autoprotolysis
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ex. ethers, esters, pyridine and amines
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Strong acid: Reacts with water so completely that
no undissociated solute molecules remain.
Weak acid: Reacts incompletely with water to give
solution that contain significant amounts of both the
parent acid and its conjugate base.
Strong base: Completely dissociated in water
solution
Weak base: Incomplete dissociation in water
solution
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Relative Strength
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Strengths of Acids and Bases
In a leveling solvent, several acids are completely dissociated and show the
same strength.
In a differentiating solvent, various acids dissociate to different
degrees and have different strengths.
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• The tendency of a solvent to accept or
protons determine the strength of a solute
base dissolved in it.
H2O + HClO4
→
H 3O +
+
H2O + HCl
→
H 3O +
+
donate
acid or
ClO4Cl-
CH3COOH + HClO4

CH3COOH2+ + ClO4-
CH3COOH + HCl

CH3COOH2+ + Cl-
base1
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acid2
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acid1
base2
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Chemical Equilibrium & Equilibrium Constants
wW + xX
yY + zZ
Where, the capital letters represent the formulas of participating
chemical species and the lowercase letters are the small whole
numbers (# of moles) required to balance the equation. The
equilibrium-constant expression is
y
z
[Y] [Z]
w
x
K =
[W] [X]
Where, the bracketed terms are molar concentration if the
species is a dissolved solute or partial pressure (atm) if the
species is a gas. If one of the species is a pure liquid, a pure
solid, or the solvent in excess (dilute soln.), no term for this
species appear in the equilibrium-constant expression.
Equilibrium
constant http:\\asadipour.kmu.ac.ir...36
K is a temperature
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The Ion-Product Constant for Water
Aqueous solutions contain small concentrations of
hydronium and hydroxide ions as a consequence of the
dissociation reaction.
2H2O
H3O+ + OH

Equilibrium constants,
[H3O ] [OH ]
K =
[H2O]
2
The concentration of water in dilute aqueous solution is
enormous when compared with the concentration of
hydrogen and hydroxide ions. [H2O] can be taken as
constant,
K[H2O]2 = Kw = [H3O+][OH-] where the new constant is the
ion-product constant for water. At 25oC, Kw 1.00 x 10-14
-logKw = - log[H3O]+ - log[OH-]
pKwhttp:\\asadipour.kmu.ac.ir...36
= pH + pOH =slides14.00
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Temperature dependence of pH
H+ + OH-  H2O +Q
Body temperature = 37 oC
Blood pH = 7.35 ~ 7.45
[HCl] in gastric juice = 0.1 ~ 0.02M
if [H+] = 0.02M
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pH = 1.7
pOHhttp:\\asadipour.kmu.ac.ir...36
= 13.6 –1.7 = slides
11.9 at 37 oC
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Solubility-Products
Ba(IO3)2(s)
Ba2+(aq) + 2IO3-(aq)
2

[Ba ] [IO3 ]
K =
[Ba(IO3)2(S)]
2
K[Ba(IO3)2(s)] = Ksp = [Ba2+][IO3-]2
Where, the new constant, Ksp, is called the solubility-product.
The position of this equilibrium is independent of the amount of
Ba(IO3)2 as long as some solid is present.
Common Ion Affect
The common-ion effect is responsible for the reduction in
solubility of an ionic precipitate when a soluble compound
combining one of the ions of the precipitate is added to the
solution
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with the precipitate.
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Dissociation Constants for Acids and Bases
When a weak acid or a weak base is dissolved in water, partial
dissociation occurs,


[H3O ] [NO2 ]
HNO2 + H2O
H3O+ + NO2- , Ka =
[HNO2]
+
OH-


4
[NH ] [OH ]
Kb =
[NH3]
NH3 + H2O
NH4 +
,
Where Ka and Kb are acid dissociation constant for nitrous acid
and base dissociation constant for ammonia respectively.
Dissociation Constants for Conjugate Acid/Base Pairs:

NH3 + H2O
NH4+ + OH- ,
[NH4 ] [OH ]
Kb =
[NH3]
NH4+ + H2O
NH3 + H3O+ ,
[NH3] [H3O ]
Ka. Kb = [H3O+ ][OH- ] = Kw
Ka =

[NH
4 ]
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K w = K a. K b

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Calculation of pH : strong acids and bases
1.
Strong acid or base
Cacid or base >> 10–6 M
2.
Both
10–6 < Cacid or base < 10–8 M
3.
Water
Cacid or base << 10–8 M
Calculated pH as a function of the
concentration of a strong acid or
strong base dissolved in water.
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Buffer solutions
A buffered solution is one that resists change of pH
on adding acids or bases, or on diluting it with solvent.
The buffer consists of mixture of an acid and its conjugated base.
HA  A– + H+
Ka = [A–][H+] / [HA]
[H+] = Ka [HA] / [A–]
– log [H+] = – log Ka [HA] / [A–] = – log Ka – log [HA] / [A–] = – log Ka + log [A–]/[HA]
pH = pKa+ log [A–] / [HA]
pKa is constant  [A–]/[HA] 
100:1 or 1:100
pKa 2
10:1 or 1:10
pKa 1
1:1
pKa 0
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pH
Henderson-Hasselbalch
equation
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preparation
• 1) HA + A• 2) HA + OH• 3) A- + H+
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B + HB+
B + H+
HB+ + OH-
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Ex. Tris buffer
NH2
N+H3
C
HOH2C
HOH2C

CH2OH
HOH2C
HOH2C
Tris hydrochloride(FW=157.597) :
BH+Cl–
C
+
H+
CH2OH
 BH+ + Cl–
BH+ = B + H+
Find the pH of a solution prepared by dissolving 12.43g of tris plus 4.67g of tris
hydrochloride in 1.00 L of water. pKa = 8.075
Tris = B :
121.136 g/L = 1.00M
Tris hydrochloride = BH+ : 157.597 g/L = 1.00 M
12.43 g/L = x M
x = 0.1026 M
4.67 g/L = y M
y = 0.0296 M
pH = pKa+ log [B] / [BH+] = 8.075 + log (0.1026 / 0.0296) = 8.61
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Effect of dilution of a buffer solution
Ex. 1) Calculate the pH of a buffer prepared by mixing 0.250 mol of acetic acid with
0.100 mol of sodium acetate and diluting to 1.00 L.
pKa = – log 1.75×10–5 = 4.75
2) Calculate the pH when 10.0 ml of this buffer is diluted to 250 mL with water.
HOAC  H+ + OAC–
Ka = [OAC–][H+] / [HOAC]
[H+] = Ka [HOAC] / [OAC–]
pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.100M)/(0.250M) = 4.36
HOAC 0.250 M×10.0 ml = x M×250 ml
NaOAC 0.100 M ×10.0 ml = x M×250 ml
x = 0.0100 M
x = 0.00400 M
pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.00400M)/(0.0100M) = 4.36
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Fig 9-4, p.255
Effect of addition of acids or bases to buffer.
Buffer
HA  A- + H+
NaOH + HA  A-
HCl + A-  HA
pH of 100 ml of buffer consisting of 0.20M HA and 0.10M NaA
pH = pKa+ log [NaA] / [HA] = pKa+ log (0.10/0.20) = pKa+ 0.30
If 0.01 mole of strong base is added
HA =0.10M.
A- =0.2M
pH = pKa+ log [NaA] / [HA] = pKa+ log (0.20/0.10) = pKa– 0.30
pH = 0.60
Not very large
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Effect of addition of acids or bases to buffer.
Ex. 1) Calculate the pH of a buffer prepared by mixing 0.250 mol of acetic acid with
0.100 mol of sodium acetate and diluting to 1.00 L.
pKa = – log 1.75×10–5 = 4.75
2) Calculate the pH when 10.0 ml of 0.1M NaOH is added.
HOAC  H+ + OAC–
Ka = [OAC–][H+] / [HOAC]
[H+] = Ka [HOAC] / [OAC–]
pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.100M)/(0.250M) = 4.36
HOAC N1V1-N2V2 = 1000 ×0.250 -10 ×0.1=0.240 mole
NaOAC N1V1+N2V2 = 1000 ×0.100 +10 ×0.1=0.110 mole
pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.110)/(0.240) = 4.41
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Buffer capacity
The relative ability of a buffer solution to resist pH change upon addition of an
acid or a base.
 = dCb / dpH = – dCa / dpH
A buffer is most effective in resisting changes in pH when pH= pKa
(i.e., when [HA] = [A–]).
Choose a buffer for an experiment whose pKa is as close as possible to the
desired pH.
The useful pH range of a buffer is usually considered to be pKa  1 pH unit.
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Buffer Capacity
• [Buffer]=[Acid]+[Base]
• [Acid]
↑ & [Base]↑ Capacity ↑
10 20 8


10
9 30
[ Base]
• In equimolar buffersis
is important
[ Acid ]
[ Base]
10
12
15
5
 1 Capacity ↑



•
[ Acid ]
10 8
5 15
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Buffer capacity as a function of
the logarithm of the ratio
CNaA/CHA. The maximum buffer
capacity occurs when the
concentration of acid and
conjugated base are equal; that is,
when 0 = 1 = 0.5.
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Buffer capacity
Ex. ) Calculate the pH and capacity of a buffer prepared by mixing 0.250 mol of acetic
acid with 0.100 mol of sodium acetate and diluting to 1.00 L.
pKa = – log 1.75×10–5 = 4.75
pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.100M)/(0.250M) = 4.36
[OAC–] is less than [HOAC] and reduce sooner,by adding H+ ,
so it is determinant of capacity.
4.36 -1= 3.36= 4.75 + log ( 0.100M) - X/(0.250M) +X
-1.39 = log ( 0.100M) - X/(0.250M) +X
0.041=0.1-X/0.250+X
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X=0.064M
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(a) Cb versus pH for a solution containing 0.100F HF with pKa =5.00.
(b) Buffer capacity versus pH for the same system reaches a maximum
when pH = pKa. The lower curve is the derivative of the upper curve.
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The composition of buffer solutions as a function of pH:  value
CT = CHOAC + CNaOAC
0 : fraction of the total concentration of acid that is undissociated
0 = [HOAC] / CT
1 : the fraction dissociated
1 = [OAC–] / CT
Ka = [OAC–][H3O+] / [HOAC]
[OAC–] = Ka [HOAC] /[H3O+]
CT = CHOAC + CNaOAC = [HOAC] + [OAC–] = [HOAC] + Ka [HOAC] /[H3O+]
={[HOAC][H3O+]+[HOAC] Ka)]} / [H3O+]=[HOAC](Ka + [H3O+])/ [H3O+]
0 = [HOAC] / CT = [H3O+] / ([H3O+]+ Ka)
1 = [OAC–] / CT =
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Ka / ([H3O+]+ Ka)
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Variation in  of HOAC with pH
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The fraction of dissociation of weak electrolyte increases as the electrolyte is
diluted. Stronger acid is more dissociated than the weaker acid at all concentrations.
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1 value : Fraction of dissociation of a weak acid
5
B
e
n
z
o
i
c
a
c
i
d
K
=
6
.
2
6
x
1
0
a
3
S
a
l
i
c
y
l
i
c
a
c
i
d
K
=
1
.
0
7
x
1
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a
5
p
H
y
d
r
o
x
y
b
e
n
z
o
i
c
a
c
i
d
K
=
2
.
6
3
x
1
0
a
Ex. 0.0500 M benzoic acid
1
.
0
Ka = [A–][H+] / [HA] = (x)(x) / (F –x)

 x = [H+] = [A–] = 1.77 × 10–3
.
8
.
6
= 0.0354
= 3.54 %
Fractionfdisociaton(
  = x / F = 1.77 × 10–3 / 0.05
.
4
.
2
0
.
0
0
1
2
3
4
5
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a
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i
d
The fraction of dissociation of weak acids
increases as the acid is diluted.
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1 value: Fraction of association of a weak base
Ex. 0.0372 M cocaine solution
Kb = 2.6× 10–6
x = [BH+] = [OH–] = KbF = 3.10 × 10–4
 1 = x / F = 3.10 × 10–4 /0.0372
= 0.0083
= 0.83 %
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The End!!!!!!!
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