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Chapter Fifteen
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Slide 1 of 31
•1
Equilibrium In Solutions Of
Weak Acids And Weak Bases
weak acid:
HA + H2O H3O+ + A[H3O+][A-]
Ka =
[HA]
weak base:
B + H2O 
HB+ + OH-
[HB+][OH-]
Kb =
[B]
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•2
Some Acid-Base Equilibrium Calculations
• cHA≈[HA]
[H3O+][A-]
[H3O+][A-]
Ka = --------------------= ---------------cHA –[H3O+]
cHA
• cHA > [HA]
Analytical C> Equilibrium C
• - the calculations can be simplified.
• - When Macid/Ka or Mbase/Kb > 100,
• - When Ka or Kb<1×10-4 (In usual Conc.)
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•3
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•4
An Example
1.Determine the concentrations of H3O+, CH3COOH
and CH3COO-, and the pH of 1.00 M CH3COOH
solution. Ka = 1.8 x 10-5.
2. What is the pH of a solution that is 0.200 M in
methylamine, CH3NH2? Kb = 4.2 x 10-4.
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•5
Are Salts Neutral, Acidic or Basic?
Salts are ionic compounds formed in the
reaction between an acid and a base.
1. NaCl
Na+ is from NaOH , a strong base
Cl- is from HCl, a strong acid
H2O
NaCl (s) → Na+ (aq) + Cl- (aq)
Na+ and Cl- ions do not react with water.
The solution is neutral.
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•6
Are Salts Neutral, Acidic or Basic?
2. KCN
K+ is from KOH , a strong base
CN- is from HCN, a weak acid
H2O
KCN (s) → K+ (aq) + CN- (aq)
K+ ions do not react with water, but CN- ions do.
CN- + H2O 
HCN + OH-
hydrolysis
The OH- ions are produced, so the solution is
basic.
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•7
Are Salts Neutral, Acidic or Basic?
3. NH4Cl
NH4+ is from NH3 , a weak base
Cl- is from HCl, a strong acid
H2O
NH4Cl (s) → NH4+ (aq) + Cl- (aq)
Cl- ions do not react with water, but NH4+ ions do.
NH4+ + H2O 
H3O+ + NH3
hydrolysis
The H3O+ ions are produced, so the solution is
acdic.
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•8
Are Salts Neutral, Acidic or Basic?
3. NH4CN
NH4+ is from NH3 , a weak base
CN- is from HCN, a weak acid
H2O
NH4CN (s) → NH4+ (aq) + CN- (aq)
NH4+ + H2O 
H3O+ + NH3
Ka
hydrolysis

HCN + OH-
Kb
hydrolysis
CN- + H2O
(Ka>Kb ,Acidic)’’’(Ka< Kb,Basic)‘’’ (Ka= Kb,Nutral)
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•9
Ions As Acids And Bases
Certain and anion ions can cause an aqueous solution to
become acidic or basic due to hydrolysis.
•
•
•
•
Salts of strong acids and strong bases form neutral solutions.
Salts of weak acids and strong bases form basic solutions.
Salts of strong acids and weak bases form acidic solutions.
Salts of weak acids and weak bases form solutions that are
acidic in some cases, neutral or basic in others.
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•10
Strong Acids And Strong Bases
Strong acids:
HCl, HBr, HI, HNO3, H2SO4, HClO4
Strong bases:
Group IA and IIA hydroxides
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•11
An Example
Indicate whether the solutions
(a) Na2S and (b) KClO4 are acidic, basic or neutral.
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•12
The pH Of a Salt Solution
What is the pH of 0.1M NaCN solution?
What is the pH of 0.1M NH4Cl solution?
What is the pH of 0.1M NH4CN solution?
Ka of HCN=1.0×10-9.
Kb for NH3=1.0×10-5
Ka x Kb = Kw
so, Kb = Kw/Ka
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•13
Prentice-Hall ©2002
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Chapter Fifteen
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Slide 14 of 31
•14
Common Ion Effect Illustrated
•CH3COOH

CH3COO- + H+
blue-violet:
yellow:
pH < 3.0
pH > 4.6
((1.00 M CH3COOH))
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((1.00 M CH3COOH + 1.00 M CH3COONa))
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•15
The Common Ion Effect
Calculate the pH of 0.10 M CH3COOH solution.
Ka of CH3COOH=1.0×10-5
Calculate the pH of 0.10 M CH3COONa solution.
Calculate the pH of 0.10 M CH3COOH/
0.10 M CH3COONa solution.
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•16
Depicting Buffer Action
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•17
Buffer Solutions
• A buffer solution is a solution that changes
pH only slightly when small amounts of a
strong acid or a strong base are added.
• A buffer contains
CH3COOH  CH3COONH3  NH4+
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Acidic buffer
Alkalin buffer
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•18
How A Buffer Solution Works
• The acid component of the buffer can neutralize
small added amounts of OH-, and the basic
component can neutralize small added amounts
of H3O+.
• CH3COOH  CH3COO- + H+
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•19
•Ionization constant of an acid
H O  A 

Ka 
3

eqb
eqb
HA
eqb
•Taking log of the equation on both sides,
H O  A 

LogKa  Log
3

eqb
HA
eqb
eqb
(ab)
b

Since Log
can be written as Log (a )  Log  
c
c

 [A ] 

LogKa  Log H 3O   Log 

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Chapter Fifteen
 [ HA]  Slide 20 of 31
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•20
•Ionization constant of an acid
 [A ] 
LogK  Log H O   Log 



 [ HA] 
•Multiplying both sides of the equation by -1
a
3
 [A ] 
 LogKa   Log H 3O   Log 

 [ HA] 


but, LogKa  pKa and  Log H 3O    pH
 [A ] 
 [A ] 
pK a  pH  Log 

 pH  pK a  Log 
 [ HA] 
 [ HA] 
•Henderson-Hasselbach equation


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Slide 21 of 31
•21
Henderson-Hasselbalch Equation
For Buff Solutions
[conjugate base]
pH = pKa + log
[conjugate acid]
If [conjugate acid] = [conjugate base], pH = pKa
Requirement:
- [B] / [A] between 0.10 and 10
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•22
Buffer Capacity
• There is a limit to the capacity of a buffer solution
to neutralize added acid or base, and this limit is
reached before all of one of the buffer
components has been consumed.
• In general, the more concentrated the buffer
components in a solution, the more added acid
or base the solution can neutralize.
• As a rule, a buffer is most effective if the
concentrations of the buffer acid and its
conjugate base are equal. [B]=[A]
• [Buffer]=[Acid]+[Base]
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•23
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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•24
Calculations in Buffer Solutions
1) A buffer solution is 0. 1 M NH3 (pKb=5)and 1.0 M NH4Cl.
(a) What is the pH of this buffer?
 [ B] 
pH  pK a  Log 

 [ A] 
pKb  5  pK a  9
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0.1
pH  9  log
 9 1  8
1.0
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•25
Calculations in Buffer Solutions
1) A buffer solution is 0. 1 M NH3 (pKb=5)and 1.0 M NH4Cl.
(a) What is the pH of this buffer?
(b) If 5 mmol NaOH is added to 0.500 L of this solution,
what will be the pH?
500  0.1  5
55
pH  9  log
 9  log
 9  0.954  8.046
500  1  5
495
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•26
Calculations in Buffer Solutions
1) A buffer solution is 0. 1 M NH3 (pKb=5)and 1.0 M NH4Cl.
(a) What is the pH of this buffer?
(b) If 5 mmol NaOH is added to 0.500 L of this solution,
what will be the pH?
(c) If 5 mmol HCl is added to 0.500 L of this solution,
what will be the pH?
500  0.1  5
45
pH  9  log
 9  log
 9  1.050  7.950
500  1  5
505
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•27
Calculations in Buffer Solutions
1) A buffer solution is 0. 1 M NH3 (pKb=5)and 1.0 M NH4Cl.
(a) What is the pH of this buffer?
(b) If 5 mmol NaOH is added to 0.500 L of this solution, what
will be the pH?
(c) If 5 mmol HCl is added to 0.500 L of this solution, what
will be the pH?
(d) If 5 mmol NH4Cl is added to 0.500 L of this solution, what
will be the pH?
500  0.1
50
pH  9  log
 9  log
 9  1.004  7.996
500  1  5
505
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•28
Calculations in Buffer Solutions
2) What concentration of acetate ion in
500 ml of 0.500 M CH3COOH (pKa=5)
produces a buffer solution with pH = 4.00?
B
B
pH  pK a  log  4  5  log
A
A
B
B
 1  log 
 0.1  [ B]  0.05M
A 0.5
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Chapter Fifteen
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Slide 29 of 31
•29
Calculations in Buffer Solutions
2) What concentration of acetate ion in500 ml of 0.500 M
CH3COOH (pKa=5) produces a buffer solution with pH = 4.00?
•How many mg?
mg
500  0.05 
 mg  2050
82
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•30
Acid-Base Indicators
• An acid-base indicator is a weak acid having
one color and the conjugate base of the acid
having a different color. One of the “colors” may
be colorless.
HIn + H2O  H3O+ + In• Acid-base indicators are often used for
applications in which a precise pH reading isn’t
necessary.
• A common indicator used in chemistry
laboratories is Phenolphetalein.
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•31
Prentice-Hall ©2002
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Chapter Fifteen
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Slide 32 of 31
•32
Prentice-Hall ©2002
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Chapter Fifteen
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Slide 33 of 31
•33
Neutralization Reactions
• Neutralization is the reaction of an acid and a base.
• Titration is a common technique for conducting a
neutralization.
• At the equivalence point in a titration, the acid and base
have been brought together in exact stoichiometric
proportions.
• The point in the titration at which the indicator changes
color is called the end point.
• The indicator endpoint and the equivalence point for a
neutralization reaction can be best matched by plotting a
titration curve, a graph of pH versus volume of titrant.
• In a typical titration, 50 mL or less of titrant that is 1 M or
less is used.
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•34
Drawing titration Curve For
Strong Acid - Strong Base
HCl + NaOH → NaCl +H2O
Calculate the pH at the
some points
and draw the curve.
4 essential points.
1)initial point
2)equivalence point
3)before the equivalence point
4)beyond the equivalence point
Ml
pH
‫محیط تیترانت‬
0
15
19
19.5
19.9
20
20.1
20.5
21
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40
•35
Drawing titration Curve For
Strong Acid - Strong Base
HCl + NaOH → NaCl +H2O
4 questions.
1)What are the present compounds?
2)Which of them is effective on pH?
3)How much are the concentrations?
4)What is the relationship between their Conc. And pH?
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•36
Drawing titration Curve For
Strong Acid - Strong Base
HCl + NaOH → NaCl +H2O
Calculate the pH at the following points in the
titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH.
(a) initial pH. (Before the addition of any NaOH) .
Answer Q1. There are:HCl & H2O
Answer Q2. HCl
Answer Q3. [HCl]
Answer Q4. pH=-log[H+]
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•37
Drawing titration Curve For
Strong Acid - Strong Base
HCl + NaOH → NaCl +H2O
Calculate the pH at the following points in the
titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH.
b)equivalence point.
Answer Q1. There are:NaCl & H2O
Answer Q2. H2O
Answer Q3.
Answer Q4. pH=7
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•38
Drawing titration Curve For
Strong Acid - Strong Base
HCl + NaOH → NaCl +H2O
Calculate the pH at the following points in the
titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH.
c)before the equivalence point.
Answer Q1. There are:HCl,NaCl & H2O
Answer Q2. HCl
Answer Q3.
N1V 1  N 2V 2
N
HCl

V1  V 2
Answer Q4. [H+]=N pH=-log[H+]
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•39
Drawing titration Curve For
Strong Acid - Strong Base
HCl + NaOH → NaCl +H2O
Calculate the pH at the following points in the
titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH.
d)after the equivalence point.
Answer Q1. There are:NaOH,NaCl & H2O
Answer Q2. NaOH
N 2V 2  N1V 1
Answer Q3.

N OH
V1  V 2
Answer Q4. [OH-]=N pOH=-log[OH-] pH=14-pOH
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•40
Titration Curve For
Strong Acid - Strong Base
•pH is low at the beginning.
•pH changes slowly until just
before equivalence point.
•pH changes sharply around
equivalence point.
•pH = 7.0 at equivalence point.
•Further beyond equivalence
point, pH changes slowly.
•Any indicator whose color
changes in pH range of 4 – 10
can be used in titration.
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•41
Drawing titration Curve For
weak acid- Strong Base
CH3COOH + NaOH → CH3COO- + Na+ +H2O
Calculate the pH at the some points and draw the curve.
Ka=1×10-5
5 essential points.
1)initial point
2)equivalence point
3)beyond the initial point
4)before the equivalence point
5)beyond the equivalence point
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Chapter Fifteen
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Slide 42 of 31
•42
Drawing titration Curve For
weak acid- Strong Base
CH3COOH + NaOH → CH3COO- + Na+ +H2O
4 questions.
1)What are the present compounds?
2)Which of them is effective on pH?
3)How much are the concentrations?
4)What is the relationship between their Conc. And pH?
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Chapter Fifteen
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Slide 43 of 31
•43
Drawing titration Curve For
weak acid- Strong Base
CH3COOH + NaOH → CH3COO- + Na+ +H2O
Calculate the pH at the following points in the
titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M
NaOH.
(a) initial pH. (Before the addition of any NaOH) .
Answer Q1. There are: CH3COOH & H2O
Answer Q2. CH3OOH
Answer Q3. CH3OOH

+
[
H
]  Ka  C
Answer Q4. pH=-log[H ]
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Slide 44 of 31
•44
Drawing titration Curve For
weak acid- Strong Base
CH3COOH + NaOH → CH3COO- + Na+ +H2O
Calculate the pH at the following points in the
titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M
NaOH.
b)equivalence point.
Answer Q1. There are: CH3COO- , Na+ & H2O
Answer Q2. CH3COON1V 1  N 2V 2
Answer Q3.
N
V1  V 2

Answer Q4. pOH=-log[OH-]
Ka×Kb=Kw
b
[OH ]  K  C
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Slide 45 of 31
•45
Drawing titration Curve For
weak acid- Strong Base
CH3COOH + NaOH → CH3COO- + Na+ +H2O
Calculate the pH at the following points in the
titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M
NaOH.
c)beyond the initial point.
Answer Q1. There are: CH3COOH, CH3COO- ,Na+ & H2O
Answer Q2. CH3COOH, CH3COON1V 1  N 2V 2
Answer Q3.
N 2V 2


Na
b
V1 V 2
V1 V 2
[ B]
Answer Q4.
N
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pH  pK a  log
Chapter Fifteen
[ A]
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slides
Slide 46 of 31
•46
Drawing titration Curve For
weak acid- Strong Base
CH3COOH + NaOH → CH3COO- + Na+ +H2O
Calculate the pH at the following points in the
titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M
NaOH.
d)before the equivalence point.
Answer Q1. There are: CH3COOH, CH3COO- ,Na+ & H2O
Answer Q2. CH3COOH, CH3COON1V 1  N 2V 2
Answer Q3.
N 2V 2

Na

V1 V 2
b
N
Answer Q4.
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[ B]
pH  •http:\\asadipour.kmu.ac.ir........57
pK a Chapter
log Fifteen slides
[ A]
V1 V 2
Slide 47 of 31
•47
Drawing titration Curve For
weak acid- Strong Base
CH3COOH + NaOH → CH3COO- + Na+ +H2O
Calculate the pH at the following points in the
titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M
NaOH.
e)after the equivalence point.
Answer Q1. There are:NaOH, CH3COO- , Na+ & H2O
Answer Q2. NaOH
N 2V 2  N1V 1
Answer Q3.

N
OH
V1  V 2
Answer Q4. [OH-]=N pOH=-log[OH-] pH=14-pOH
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Slide 48 of 31
•48
Titration Curve For
Weak Acid - Strong Base
•The initial pH is higher because
weak acid is partially ionized.
•At the half-neutralization point,
pH = pKa.
•pH >7 at equivalence point
because the anion of the
weak acid hydrolyzes.
•The steep portion of titration
curve around equivalence point
has a smaller pH range.
•The choice of indicators for the
titration is more limited.
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•49
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Slide 50 of 31
•50
Application of Ka
•The Ka of nicotinic acid, HNic, is 1.4e-5. A solution containing
0.22 M HNic. What is its pH? What is the degree of ionization?
•Solution:
HNic = H+ + Nic–
0.22-x
x
x
•
x2
Ka = ———— = 1.4e-5
0.22 – x
(use approximation, small indeed)
x =  (0.22*1.4e-5) = 0.0018
•
pH = – log (0.0018) = 2.76
•Degree of ionization = 0.0018 / 0.22 = 0.0079 = 0.79%
•
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Slide 51 of 31
•51
Determine Ka and percent ionization
•Nicotinic acid, HNic, is a monoprotic acid. A solution
containing 0.012 M HNic, has a pH of 3.39. What is its
Ka? What is the percent of ionization?
•Solution:
HNic  H+ + Nic–
0.012-x x
x
•
x = [H+] = 10–3.39 = 4.1e-4
[HNic] = 0.012 – 0.00041 = 0.012
•
(4.1e-4)2
Ka = ————— = 1.4e-5
0.012
•Degree of ionization = 0.00041 / 0.012 = 0.034 = 3.4%
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Slide 52 of 31
•52
Using the quadratic formula
•The Ka of nicotinic acid, HNic, is 1.4e-5. A solution containing
0.00100 M HNic. What is its pH? What is the degree of ionization?
•Solution:
HNic = H+ + Nic–
0.001-x x
x
•
x2
Ka = —————— = 1.4e-5
0.00100 – x
x2 + 1.4e-5 x – 1.4e-8 = 0
–1.4e–5 +  (1.4e–5)2 + 4*1.4e-8
x = —————————————————— = 0.000111 M
2
pH = – log (0.000111) = 3.95
•
•Degree of ionization = 0.000111/ 0.001 = 0.111 = 11.1%
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•% ionization
Degree of or percent ionization
•The degree or percent of
ionization of a weak acid always
decreases as its concentration
increases, as shown from the table given
earlier.
•Deg.’f ioniz’n
0.220
0.012
0.001
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•Concentration of
acid
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0.8%
3.4 %
11.1 %
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Polyprotic acids
•Polyprotic acids such as sulfuric and carbonic acids have
more than one hydrogen to donate.
•
H2SO4 → H+ + HSO4–
completely ionized
•
HSO4–  H+ + SO42–
Ka1 very large
H2CO3  H+ + HCO3–
HCO3–  H+ + CO32–
Ka1 = 4.3e-7
Ka2 = 4.8e-11
•
•
Ascorbic acid (vitamin C)
is a diprotic acid, abundant
in citrus fruit.
Ka2 = 0.012
•Others:
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H2S, H2SO3, H3PO4, H2C2O4 (oxalic acid) …
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Species concentrations of diprotic acids
•Evaluate concentrations of species in a 0.10 M H2SO4
solution.
•Solution:
H2SO4 → H+ +
(0.1–0.1) 0.10
•
HSO4–  H+
+
0.10–y
0.10+y
•
(0.10+y) y
————— = 0.012
(0.10-y)
HSO4–
0.10
completely ionized
SO42–
Ka2 = 0.012
y Assume y = [SO42–]
•[SO42–] = y = 0.01M
[H+] = 0.10 + 0.01 = 0.11 M;
[HSO4–] = 0.10-0.01 = 0.09 M
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Species concentrations of weak diprotic acids
•Evaluate concentrations of species in a 0.10 M H2S solution.
•Solution:
H 2 S = H+ +
(0.10–x) x+y
HS–
x-y
•
HS–
x–y
•
(x+y) (x-y)
————— = 1.02e-7
(0.10-x)
= H+
x+y
+
S2–
y
Ka1 = 1.02e-7
Assume x = [HS–]
Ka2 = 1.0e-13
Assume y = [S2–]
(x+y) y
———— = 1.0e-13
(x-y)
•[H2S] = 0.10 – x = 0.10 M
[HS–] = [H+] = x  y = 1.0e–4 M;
[S2–] = y = 1.0e-13 M
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•0.1>> x >> y:
x+ y = x-y = x
x = 0.1*1.02e-7 =
1.00e-4
y = 1e-13
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