Transcript Analytical Chemistry I Lecture Note 산

Version 2012 Updated on 042312 Copyright © All rights reserved
Dong-Sun Lee, Prof., Ph.D. Chemistry, Seoul Women’s University
Chapter 12
Principles of Neutralization
Titration
Standard solutions for Acid-Base Titrations
Analyte Titrant vs Standard solution
N V = N’V’
The standard reagents used in acid-base titrations are always strong acids or
strong bases, most commonly HCl, HClO4, H2SO4, NaOH, KOH. Weak acids
and bases are never used as standard reagents because they react incompletely
with analyte.
N= unknown
V= measure
N’= known
V’= known
Primary standards for standardizing
Acids
Potassium acid phthalate
Sulfamic acid ( H2NSO3H)
HCl
Potassium hydrogen iodate
Bases
TRIS(hydroxymethylaminomethane)
Sodium carbonate
Borax (= sodium tetraborate )
HgO

General goal and procedure of acid-base titration 
1) Standardization
2) Determination
3) Titration curve
Plots of pH versus
volume of titrant
Acid or base
standard solution
unknown normality(N’)
unknown volume(V’)
14
12
10
pH
8
6
4
Acid or base
standard solution
known normality(N’)
unknown volume(V’)
2
0
0
5
10
15
20
25
Volume of NaOH added (ml)
4) Interpret titration curve ,
Primary standard solution
known normality(N)
known volume(V)
Analyte Sample
solution
unknown Ns
known Vs
understand what is
happening during titration
Finding the end point
1) Indicator
2) Potentiometry : Titration curve ( pH or mV vs Va )
1st derivative titration curve
2nd derivative titration curve
Gran plot
3) Conductometry
4) Spectrometry
Indicator
An acid-base indicator is itself an acid or base whose different protonated species
have different colors.
Ex. Thymol blue
H2In 
HIn–
Red
Yellow
H2In


Transition
range
pKa11
In2 –
Blue
H+ + HIn–
pKa1 = 1.7
pH = pKa1 + log [HIn– ]/[H2In]
if [HIn– ]/[H2In]  1:10
HIn–
pH = 0.70
Red color
[HIn– ]/[H2In] = 1:1
pH = pKa1 = 1.7
Orange color
[HIn– ]/[H2In]  10:1
pH = 2.70
Yellow color

H+ + In2 –
pKa2 = 8.9
The approximate pH transition range of most acid-type indicator is roughly pKa 1
Choosing an indicator
14.00
Phenolphthalein
12.00
acid form :
10.00
colorless
PP 8.0~9.0
8.00
transition
range(pH):
pH
6.00
8.0~9.0
4.00
Phenolphthalein
2.00
MR
MO
4.8~6.0
3.1~4.4
base form :
pink
0.00
0
5
10
15
20
25
30
35
Volume of 0.1000M HCl (ml)
The calculated pH titration curve for the
titration of 10.00ml of 0.1000M Na2CO3
with 0.1000M HCl.
Calculated titration curve for the reaction of 100 mL of 0.0100M base (pKb = 5.00)
with 0.0500 M HCl.
Indicator color
as a function of
pH (pKa=5.0)
Experiments. Standardization of 0.1000N NaOH
5
.
0
0
1
4
.
0
0
6
.
0
0
1
2
.
0
0
4
.
0
0
4
.
0
0
pH
8
.
0
0
6
.
0
0
 pH/ Va)/ Va
pH/ Va
1
0
.
0
0
3
.
0
0
2
.
0
0
2
.
0
0
0
.
0
0
2
.
0
0
4
.
0
0
1
.
0
0
4
.
0
0
2
.
0
0
0
.
0
0
0 5 1
0 1
5 2
0 2
5 3
0
V
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0
9
2
7
0
N
N
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0
.
0
6
8
6
0
N
K
H
P
2
5
.
0
0
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s
0
.
0
9
2
7
0
N
N
a
O
H
0
.
0
0
0 5 1
0 1
5 2
0 2
5 3
0
6
.
0
0
0 5 1
0 1
5 2
0 2
5 3
0
V
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0
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2
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0
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0
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0
6
8
6
0
N
K
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2
5
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0
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0
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7
0
N
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N
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2
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.
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0
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0
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2
7
0
N
N
a
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H
Primary standard KHP
Titration of
End point=18.50ml
204.22g/1000ml=1.0000N
25.00ml of
0.06860N×25.00ml= x N×18.50ml
0.35g/25ml = xN
KHP with
x=0.09270N
x = 0.06860N
NaOH
Calculation of concentration (Finding of NaOH concentration)
HOOCC6H4COOK (mw 204.44)  NaOH (mw 40.01)  1 Eq wt.
204.22 g  1 N × 1000 mL
20.422 mg  0.1 N × 1 mL
a g  x N’ × Ve mL
 x N’ = (a g × 1000 mL) / (204.22 g × Ve mL)
Experiments. Determination of acetic acid in venigar
14.00
4.00
5.00
3.00
12.00
4.00
2.00
8.00
(pH/Va)/Va
pH/Va
pH
10.00
3.00
2.00
6.00
1.00
0.00
-1.00
-2.00
1.00
4.00
-3.00
2.00
0
10
20
30
40
50
60
Volume of 0.0927N NaOH(ml)
Fig. The 1st derivative experimental titration curve.
0.1094N Venigar 25.00ml vs 0.0927N NaOH
-4.00
0.00
0
5
10
15
20
25
0
30
20
30
40
50
60
Volume of 0.0927N NaOH(ml)
Volume of 0.09270N NaOH (ml)
Fig. The 1st derivative experimental titration curve.
0.06860N PHP 25.00ml vs 0.0927N NaOH
10
Fig. The 2nd derivative experimental titration curve.
0.1094N Venigar 25.00ml vs 0.0927N NaOH
Standardization of NaOH
Titation of
End point = 29.50ml
0.09270N
25.00ml of Venigar
x ´ 25ml = 0.09270N ´ 29.50ml
with NaOH
Venigar : x = 0.1094N
Gran plot
HA = H+ + A–
Ka = [H+] fH+ [A–]fA– / [HA]fHA
HA vs NaOH
[A–] = (moles of OH– delivered) / (total volume)
= VbFb / (Vb + Va)
[HA] = (original moles of HA – moles of OH –) / (total volume)
= (VaFa – VbFb ) / (Vb + Va)
[H+] fH+ Vb = (fHA / fA– ) Ka{(VaFa – VbFb )/ Fb}
10–pH Vb = (fHA / fA– ) Ka(Ve – Vb)
10–pH Vb
Ka = [H+] f H+ VbFb fA– / (VaFa – VbFb ) fHA
Slope =–(fHA / fA– ) Ka
Ve
= –(fHA / fA– ) Ka Vb + (fHA / fA– ) KaVe
Vb(l)
Gran plot for the first equivalence point. This plot gives an estimate of Ve that
differs from that in Figure 12.7 by 0.2μL (88.4 versus 88.2μL). The last 10-20% of
volume prior to Ve is normally used form a Gran plot.
Conductometric end-point detection
Titration of strong acid with strong base
A
Neutralization :
14
HCl + NaOH  NaCl + H2O
B
12
Ve mL×0.1000M = 50.00 mL × 0.02000 M
10
Ve = 10.00 mL
pH
C
8
6
0.1000M
HCl
D
4
2
0
0
2
4
6
8
10
12
14
16
18
20
V a (mL)
0.02000M
NaOH
50.00 mL
Calculated titration curve for the reaction of
50.00 mL of 0.02000 M NaOH with 0.1000 M HCl.
A. Before the beginning of titration : Va = 0.00 mL.
NaOH  Na+ + OH–,
[OH–] = 0.02000 M
0.02000 M
[H+] = 1.00×10–14/ 2.000×10–2
pH = 12.30
B. Before the equivalence point : 0 < Va < Ve
remaining NaOH
Initial NaOH amount =F × Vi
added HCl amount = used NaOH amount = F× Va
Remaining NaOH amount =
Va
Ex.
Ve
[OH–] = {(Ve – Va )/ Ve}FNaOH {Vi /(Vi + Va)}
Va = 3.00 mL
[OH–] = {(10.00 – 3.00)/10.00}(0.02000){50.00 /(50.00 +3.00)} = 0.0132 M
[H+] = 1.00×10–14/ 0.0132 = 7.58×10–13
pH = 12.12
C. At the equivalence point : Va = Ve
+
–
H2O = H + OH ,
[H+] [OH–] = 1.00×10–14
[H+] = 1.00×10–7
D. After the equivalence point :
pH = 7.00
Va > Ve
excess HCl
NaOH
added HCl amount = FHCl× Va
initial NaOH amount = used HCl = F × Vi
Vi
Excess HCl amount = FHCl(Va–Vi)
Ve
Va
= FHCl (Va–Ve)
[H+] = FHCl {(Va– Ve) /(Vi + Va)}
Ex.
Va = 10.50 mL
[H+] = (0.1000) {10.50 –10.00) /(50.00 + 110.50)} = 8.26×10–4
pH = 3.08
Titration curves for NaOH with HCl.
A.
50.0 mL of 0.0500 M NaOH with 0.1000 M HCl.
B.
50.00 mL of 0.00500 M NaOH with 0.0100 M HCl.
Titration curves for HCl with NaOH.
A. 50.0 mL of 0.0500 M HCl
with 0.1000 M NaOH.
B.
50.00 mL of 0.000500 M HCl
with 0.00100 M NaOH.
pH
The effects of titrant and
analyte concentrations
on neutralization titration
curves.
14.00
13.00
12.00
11.00
10.00
9.00
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
0.00
B
A
0
5
10
15
20
25
30
35
Va of NaOH (mL)
Changes in pH during the totration of
strong acid with strong base.
A: 50 mL of 0.0500 M HCl vs 0.1000 M NaOH
B: 50 mL of 0.000500 M HCl vs 0.001000 M NaOH
For the titration with
diluted concentrations,
the change in pH in
equivalence point region
is markedly less than
concentrated solution
Weak acid titrated with strong base

1) Titration reaction :
CH3COOH + NaOH  CH3COO– Na+ + H2O
14.00
strong + weak  complete reaction
12.00
K = 1/Kb = 1.76
×109
A
10.00
pH
B
8.00
6.00
2) Calculation of equivalence point :
4.00
2.00
[CH3COOH] Vi mL = [NaOH] Ve mL
0.1000 M×5.00 mL = 0.1000 M × Ve mL
Ve= 5.00 mL
3) Titration curve :
0.00
0
1
2
3
4
5
6
7
8
Va of NaOH(mL)
Changes in pH during the titration of a
weak acid with a strong base.
A: 5.00 mL of 0.1000M HOAC vs 0.1000
M NaOH
B: 5.00 ml of 0.001000M HOAC vs
0.001000M NaOH
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
with 0.00100 M NaOH.
A. Before adding any titrant : Va = 0.00

HAC = H+ + AC–
[H+] = KaF =  1.75 ×10–5 ×0.1000
14.00
= 1.323 × 10–3
12.00
pH= 2.88
10.00
pH
8.00
0.1000M
NaOH
6.00
4.00
A
2.00
0.00
0
0.1000M
HAC
25.00ml
1 2
3
4
5
6
7
Va of 0.1000 M NaOH (mL)
Titration curve.
0.1000 M HAC vs 0.1000 M NaOH.

B. Before equivalence point : 0 < Va < Ve
HAC + NaOH  H2O + Na+AC–
[H+] = Ka[HAC] / [AC–]
14.00
initial HAC amount = F×Ve
– added NaOH amount
10.00
= used amount HAC= F×Va
8.00
Remaining HAC amount =F(Vi–Va)
6.00
[HAC] = {(F×Ve) – (F×Va)}/(Vi +Va)
4.00
[AC–] = (F×Va) / (Vi +Va)
2.00
pH = pKa + log [AC–] /[HAC]
0.00
at Va = Ve/2
B
12.00
pH= pKa = 4.76
Vi
Ve/2
0
1
2
3
4
5
6
7
Va of 0.1000 M NaOH (mL)
Titration curve.
0.1000 M HAC vs 0.1000 M NaOH.
Va
Ve

C. Equivalence point Va = Ve
14.00
HAC + NaOH  H2O + Na+AC–
12.00
10.00
C
8.00
initial
1
1
final
0
0
6.00
4.00
AC– = HAC + OH–
2.00
[OH–] = KbF’ =  KwF’ /Ka
0.00
0
1
2
3
4
5
6
7
Va of 0.1000 M NaOH (mL)
Titration curve.
0.1000 M HAC vs 0.1000 M NaOH.
F’=( F×Vi) / (Vi +Va)
1

D. After equivalence point Va> Ve
14.00
added NaOH amount = F×Va
D
12.00
initial HAC amount
10.00
= used NaOH amount = F×Ve
8.00
Excess NaOH amount = F×(Va –Ve)
6.00
4.00
2.00
[OH–] = FNaOH ×(Va– Vi) / (Vi +Va)
0.00
0
1
2
3
4
5
6
7
Va of 0.1000 M NaOH (mL)
Titration curve.
Vi
0.1000 M HAC vs 0.1000 M NaOH.
Va
Ve
The effect of acid strength (dissociation constant) on titration curves. Each curve
represents the titration of 50.00 mL of 0.1000 M acid with 0.1000 M base.
Weak base titrated with strong acid
1) Titration reaction :
NH3 + HCl  NH4+Cl–
14.00
2) Equivalence point :
A
12.00
NH3
HCl
0.1100 M×20.00 mL = 0.1000 M×Ve mL
Ve = 22.00 mL
10.00
8.00
6.00
3) Titration curve :
4.00
A. Initial
2.00
Va = 0
NH3 + H2O = NH4
+
+
OH–
0.00
0
10
20
30
40
0.1000M HCl volume (mL)
Kb = Kw / Ka = [NH4+][OH– ]/[NH3]
1.79×10–5 = [OH– ]2 / 0.1100–[OH– ]
[OH– ]=  KbF =1.4×10–3
pH = 11.15
The calculated titration curve for
the titration of 20.00 mL of 0.1100 M
ammonia with 0.1000 M HCl.
50
B. Before the equivalence point 0<Va< Ve
Major constituents: NH3 + NH4+Cl–
[OH]= Kb [NH3]/ [NH4
1
4
.
0
0
+]
1
2
.
0
0
initial NH3 amount = F×Ve
1
0
.
0
0
– added HCl amount
pH
8
.
0
0
= used amount NH3 = F×Va
6
.
0
0
Remaining NH3 amount =F(Vi–Va)
4
.
0
0
[NH3] = {(F×Ve) – (F×Va)}/(Vi +Va)
2
.
0
0
[NH4+] = (F×Va) / (Vi +Va)
0
.
0
0
0
pH = pKb + log [NH4+] / [NH3]
at Va = Ve/2
pOH= pKb = 4.74
pH = 9.26
Vi
Va
B
Ve/2
Ve
1
0
2
0
3
0
4
0
5
0
0
.
1
0
0
0
M
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1
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0
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C
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.
C. Equivalence point Va = Ve
1
4
.
0
0
NH3 + HCl  NH4+Cl–
1
1
final
0
0
1
0
.
0
0
1
C
8
.
0
0
pH
initial
1
2
.
0
0
6
.
0
0
NH4+ = NH3 + H+
4
.
0
0
[H+] = KaF’
2
.
0
0
F’=( F×Vi) / (Vi +Va)
0
.
0
0
0
1
0
2
0
3
0
4
0
5
0
0
.
1
0
0
0
M
H
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0
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1
1
0
0
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m
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a
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0
.
1
0
0
0
M
H
C
l
.
D. After equivalence point Va> Ve
added HCl amount = F×Va
1
4
.
0
0
initial NH3 amount
1
2
.
0
0
1
0
.
0
0
= used HCl amount = F×Ve
8
.
0
0
pH
Excess HCl amount = F×(Va –Ve)
D
6
.
0
0
4
.
0
0
[H+] = FHCl ×(Va– Ve) / (Vi +Va)
2
.
0
0
0
.
0
0
0
Vi
1
0
2
0
3
0
4
0
5
0
0
.
1
0
0
0
M
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C
lv
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Va
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0
0
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0
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1
1
0
0
M
a
m
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0
.
1
0
0
0
M
H
C
l
.
The effect of base strength (dissociation constant) on titration curves. Each curve
represents the titration of 50.00 ml of 0.1000 M base with 0.1000 M HCl.
Titration of Weak Base with Strong Acid
Comparison of Weak Acid/ Base with Strong Base/Acid
Weak Acid with Strong Base
Weak Base with Strong Base
Titration reaction
HA + OH- → H2O + A-
B + H2O → BH+ + OH-
Initial
[H+] = KaF
[OH-] =  KbF =1.4×10–3
Before the equivalence
point (0<V a<V e)
pH = pKa + log [A–] /[HA]
pH = pKb + log[NH4+]/[NH3]
Equivalence point
[OH–] = KbF’ =  KwF’ /Ka
[H+] = KaF’
F’=( F×Vi) / (Vi +Va)
F’=( F×Vi) / (Vi +Va)
After equivalence point
(Va>Ve)
[OH-] = FNaOH
(Va– Ve)
{ (V + V ) }
i
a
[H+
]
=
FHCl
{
(Va– Ve)
(Vi + Va)
}
Determining the pK values for amino acids
Amino acids contain both an acidic and a basic group.
alanine
All naturally occurring amino acids are left-handed (L) form.
The amine group behaves as a base, while the carboxyl group acts as an acid.
Aspartic acid
Curves for the titration of 20.00ml of 0.1000M alanine with 0.1000 M NaOH and 0.1000M HCl.
Note that the zwitterion is present before any acid or base has been added. Adding acid protonates
the carboxylate group with a pKa of 2.35. Adding base reacts with the protonated amine group with
a pKa of 9.89.
Plots of relative amounts of acetic acid and acetate ion during a titration.
Acid base titration in non-aqueous media
HOOCCHNH2 + HClO4
R

CH3COOH
HOOCCHNH3+ClO4–
R
(solvent)
0.010M CH3COOK
in acetic acid
Amino acid
in acetic
acid +
0.020M HClO4
in acetic acid
HClO4 + CH3COOK  CH3COOH + K+ClO4–
CH3COOH(solvent)
Titrants used in non-aqueous titrimetry
Acidic titrants
Perchloric acid
p- Toluenesulfonic acid
2,4-Dinitrobenzenesulfonic acid
Basic titrants
Tetrabutylammonium hydroxide
Sodium acetate
Potassium methoxide
Sodium aminoethoxide
Selected solvents for non-aqueous titration
Solvent
Amphiprotic
Aprotic or basic
Autoprotolysis constant
Dielectric
(pKHS)
constant
Glacial acetic acid
14.45
6.1
Ethylenediamine
15.3
12.9
Methanol
16.7
32.6
Dimethylformamide
Benzene
36.7
2.3
Methyl isobutylketone
13.1
Pyridine
12.3
Dioxane
2.2
n-Hexane
1.9
Acid and base strengths that are not distinguished in aqueous solution may be
distinguishable in non-aqueous solvents.
Ex. Perchloric acid is a stronger acid than hydrochloric acid in acetic acid
solvent,
neither acid is completely dissociated.
HClO4 + CH3COOH = ClO4–
strong acid
strong base
+ CH3COOH2+
weak base
K = 1.3×10–5
weak acid
HCl + CH3COOH = Cl– + CH3COOH2+
K = 5.8×10–8
Differentiate acidity or basicity of different acids or bases
differentiating solvent for acids …… acetic acid, isobutyl ketone
differentiating sovent for bases
…… ammonia, pyridine
Acid
Name
Ho
H2SO4 (100%)
Sulfuric acid
–11.93
H2SO4 SO3
Fuming sulfuric acid
–14.14
H SO3F
Fluorosulfuric acid
–15.07
H SO3F+10%SbF5
Super acid
H SO3F+7%SbF53SO3
Hammett acidity function, Ho, for aqueous solutions of acids.
–18.94
–19.35
Titration of a mixture of acids with tetrabutylammonium hydroxide in methyl isobutyl
ketone solvent shows that the order of acid strength is HClO4 > HCl > 2-hydroxybenzoic
acid > acetic acid > hydroxybenzene. Measurements were made with a glass electrode and
a platimum reference electrode. The ordinate is proportional to pH, with increasing pH as
the potential becomes more positive.
Summary
Acid-Base titration
Titration curve
Indicator
Detection of end-point
Potentiometry
Gran plot
Conductometry
Non-aqueous titration
Hammett acidity function, Ho