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•1
•Neutralization
•Reactions
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•2
• Standard Solutions: strong acids or strong
bases because they will react completely
– Acids: (HCl), (HClO4), (H2SO4)
– Bases: (NaOH), (KOH)
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•3
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•4
Thymol blue
(thymolsulphonepht
halein) is a
brownish-green or
reddish-brown
crystaline powder
that is used as an pH
indicator. It is
insoluble in water
but soluble in
alcohol and dilute
alkali solutions. It
transitions from red
to yellow at pH 1.2–
2.8 and from yellow
to blue from at pH
8.0–9.6.
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Bromophenol
blue (3',3",5',5"tetrabromophenol
sulfonphthalein) is
an acid-base
indicator whose
useful range as an
indicator lies
between pH 3.0
and 4.6. It
changes from
yellow at pH 3.0 to
purple at pH 4.6;
this reaction is
reversible.
Chloropheno
l red is an
indicator dye
that changes
color from
yellow to
violet in the
pH range 4.8
to 6.7. The
lamda max is
at 572 nm.
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A solution of
phenol red is used
as a pH indicator:
its color exhibits a
gradual transition
from yellow to red
over the pH range
6.6 to 8.0. Above
pH 8.1, phenol red
turns a bright pink
(fuchsia) color. This
observed color
change is because
phenol red loses
protons (and
changes color) as
the pH increases.
48slides
Bromothymol
Blue (also
known as
dibromothymol
sulfonephthalei
n, Bromthymol
Blue, and BTB)
is a chemical
indicator for
weak acids and
bases The pKa
for
bromothymol
blue is 7.10.
Slide 5 of 31
•5
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•6
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•7
Acid/Base Indicators
Many substances display colors that depend on the
pH of the solutions in which they are dissolved. An
acid/base indicator is a weak organic acid or a weak
organic base whose undissociated form differs in
color from its conjugate form. e.g., the behavior of
an acid-type indicator, HIn, is described by the
equilibrium
HIn + H2O
In- + H3O+
acid color
base color
The equilibrium for a base-type indicator, In, is
In + H2O
InH+ + OHbase
color
acid
color
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•8
…continued…
The equilibrium-constant expression for the
dissociation of an acid-type indicator takes the form
•+
•Rearranging leads to
 H 3O  In 
Ka 
 HIn
 HIn
 H 3O   Ka • In 
•+
The hydronium ion concentration determines the ratio
of the acid to the conjugate base form of the indicator
and thus determines the color developed by the
solution.
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•9
…continued…
The color imparted to a solution by a typical indicator
appears to the average observer to change rapidly only
within the limited concentration ratio of approximately 10
to 0.1
 HIn 10

 In  1
and its base color when
 HIn 1

 In  10
The color appears to be intermediate for ratios between
these two values. These ratios vary considerably from
indicator to indicator.
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•10
…continued…
 HIn
 H 3O   Ka
 In 
For the full acid color,
[H3O+] = 10Ka
and similarly for the full base color,
[H3O+] = 0.1Ka
To obtain the indicator pH range, we take the
negative logarithms of the two expression:
pH (acid color) = -log (10Ka) = pKa - 1
pH (basic color) = -log (0.1Ka) = pKa + 1
indicator pH range = pKa  1
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•11
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•12
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•13
Variables:
1)temperature,2)ionic strength of medium
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3)presence
organic solvents
4)presence
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•14
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•15
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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•16
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
Slide 17 of 31
•17
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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•18
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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•19
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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•20
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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•21
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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•22
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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•23
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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•24
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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•25
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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•26
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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•27
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
[ A]
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Slide 28 of 31
•28
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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pH  •http"\\asadipour.kmu.ac.ir........52
pK a Chapter
log Fifteen slides
[ A]
V1 V 2
Slide 29 of 31
•29
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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•30
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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•31
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•32
•
Titration curves for HCl with NaOH.
A.
50.0 ml of 0.0500 M HCl
•
B.
•
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with 0.1000M NaOH.
50.00 ml of 0.000500 M HCl
with 0.00100 M NaOH.
Slide 33 of 31
•33
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•
Titration curves for the titration of
acetic acid with NaOH.
•
A. 0.1000 M acetic acid
•
with 0.1000M NaOH.
•
B. 0.001000 M acetic acid
•
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34 of 31
with 0.00100Slide
M NaOH.
•34
•
Titration curves for HCl with NaOH.
A.
50.0 ml of 0.0500 M HCl
•
B.
•
with 0.1000M NaOH.
50.00 ml of 0.000500 M HCl
with 0.00100 M NaOH.
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•
Titration curves for the titration of
acetic acid with NaOH.
•
A. 0.1000 M acetic acid
•
with 0.1000M NaOH.
•
B. 0.001000 M acetic acid
•
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with 0.00100Slide
M NaOH.
•35
•General Shapes of Titration Curves
•Effect of pKa
•Effect of initial
•concentration
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•36
•Effect of Ka
•The effect of acid strength (dissociation constant) on titration curves. Each curve
represents
titration of 50.00 ml of
0.1000
M acid with 0.1000 M base.Slide 37 of 31
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•37
•Effect of Kb
•The effect of base strength (dissociation constant) on titration curves. Each curve
represents
titration of 50.00 ml of
0.1000
M base with 0.1000 M HCl.Slide 38 of 31
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•38
non-aqueous acid base titration
1)Solubility 2) acid or base strength
•Acid and base strengths that are not distinguished in aqueous solution
may be distinguishable in
non-aqueous solvents.
HClO4 > HCl in acetic acid solvent,
•
•
•
•
neither acid is completely dissociated.
HClO4 + CH3COOH  ClO4–
strong acid
strong base
+ CH3COOH2+
weak base
HCl + CH3COOH  Cl–
K = 1.3×10–5
weak acid
+ CH3COOH2+
K = 5.8×10–8
• Differentiate acidity or basicity of different acids or bases
•
differentiating solvent for acids
…… acetic acid
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……Fifteen
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Slide 39 of 31
•39
Reaction between weak acid and weak base
–Both the weak acid and weak base remain largely undissociated
and neutralization involves proton transfer from the weak acid to
the weak base. Consider acetic acid and ammonia:
•CH3COOH + NH3  CH3COO- + NH4+ is composed of
•CH3COOH + H2O  CH3COO- + H3O+ K1 = Ka = 1.8 x 10-5
•NH3 + H2O  NH4+ + OHK2 = Kb = 1.8 x 10-5
•H3O+ + OH-  2 H2O
K3 = 1/ Kw = 1 x 1014
Kn = Koverall = K1 x K2 x K3 = Kb Ka / Kw = 3.2
x 104
•Therefore, the Reaction Still Shifts significantly to the
right -- ionizing much of the component present in the
smaller amount
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•40
Titration problems
1. What volume of 0.10 mol/L NaOH is needed
to neutralize 25.0 mL of 0.15 mol/L H3PO4?
2. 25.0 mL of HCl(aq) was neutralized by 40.0
mL of 0.10 mol/L Ca(OH)2 solution. What
was the concentration of HCl?
3. A truck carrying sulfuric acid is in an accident.
A laboratory analyzes a sample of the spilled
acid and finds that 20 mL of acid is neutralized by 60 mL of 4.0 mol/L NaOH solution.
What is the concentration of the acid?
4. What volume of 1.50 mol/L H2S will neutralPrentice-Hall
Chapter Fifteen 32.0 g NaOH?
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ize a ©2002
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•41
Titration problems
1. (3)(0.15 M)(0.0250 L) = (1)(0.10 M)(VB)
VB= (3)(0.15 M)(0.0250 L) / (1)(0.10 M) = 0.11 L
2. (1)(MA)(0.0250 L) = (2)(0.10 M)(0.040 L)
MA= (2)(0.10 M)(0.040 L) / (1)(0.0250 L) = 0.32 M
3. Sulfuric acid = H2SO4
(2)(MA)(0.020 L) = (1)(4.0 mol/L)(0.060 L)
MA = (1)(4.0 M)(0.060 L) / (2)(0.020 L) = 6.0 M
4. mol NaOH = 32.0 g x 1 mol/40.00 g = 0.800
(2)(1.50 mol/L)(VA) = (1)(0.800 mol)
VPrentice-Hall
mol)
/
(2)(1.50
mol/L)
=
0.267
L
A= (1)(0.800
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•42
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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•43
Alpha Values
• Def.: the relative equilibrium concentration of the weak
acid/base and its conjugate base/acid
• (titrating HOAc with NaOH):
[HOAc]
0 
Ct
[ H3O  ]
0 
[H3O  ]  K a
[OAc- ]
1 
Ct
Ka
0 
[H3O  ]  K a
α0 + α1 = 1
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•44
•Plots of relative amounts of acetic acid and acetate ion during a titration.
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•45