Download Chapter3 Titrimetric analysis

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Chapter3
Titrimetric analysis
3.1 General Principles
3.2 Acid-base Titration
3.3 Complexometric Titration
3.4 Oxidation-reduction Titration
3.5 Precipitation Titration
3.1 General principles
（mov）
3.1.1 Definitions
1. Titrimetric Analysis
In a titration, increments of the standard
solution—the titrant—is added to the analyte
by means of a buret until their reaction is
complete.
The volume of the titrant that completely reacts
with the analyte is measured.
From the required volume and the concentration
of the titrant, the amount of the analyte can be
calculated.
This analytical techniques is called titrimetric
analysis or volumetric analysis.
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2.Standard Solution
is a solution of exactly known concentration
5C2O42- + 2MnO4- + 16H+→10CO2 + 2Mn2+ +8H2O
analyte
3.Equivalence point (Stoichiometric point)
The point at which the theoretical amount of titrant has
been added is called the equivalence point of the
titration. (the point at which the titrant added just
completely reacts with the analyte.)
titrant
4.Indicator
An indicator is a compound with a physical property
(usually color) that changes abruptly near the
equivalence point
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5.End point
Requirements for Reactions Used in Titrimetric Analysis
is the point at which the color of the indicator changes or
the electrochemical or physical properties of the solution
change.
1. The reaction must proceed according to a definite chemical equation
6.Titration error
2. The reaction must proceed to virtual completion at the equivalence point.
the difference between the end point and the
equivalence point.
KSp low) .
without side reaction, which means that the reaction follow the definite
stoichiometric relationship.
The equilibrium constant of the reaction should be very large（K’MY, high;
The above two will provide a basis for stoichiometric calculation.
3. Some means must be available for determining when the equivalence
point is reached, such as an indicator or some instrumental methods to tell
the end point.
4. The reaction should be rapid.
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第三章 滴定分析法
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Standard Solutions : primary standard , standardization
„
1. The material should be of known composition and highly
pure; 99.9% pure, or better.
Only a few chemical reagents can meet these
requirements like K2Cr2O7; Na2C2O4; H2C2O4 ·2H2O
Na2B4O7 · 10H2O; CaCO3; NaCl; Na2CO3 .
2. It should undergo a rapid and stoichiometric chemical reaction
with the solution being standardized.
„
3. The material should stay stable . H2O, CO2, O2
„
4. It should have a high equivalent weight in order to minimize
the relative error caused by weighing.
K2Cr2O7, Na2C2O4, H2C2O4·2H2O, Na2B4O7·10H2O,
CaCO3, NaCl
More commonly, a solution with approximately
desired concentration is prepared, then it is
standardized by titrating a weighed portion of a
primary standard. (standardization)
For example, solutions of hydrochloric acid and
sodium hydroxide are prepared like this, and it is
customary to standardize one of them as a
secondary standard to obtain the concentration of
the other solution.
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3.1.2 Titrimetric Reactions
(1).Acid-base
Hundreds of compounds, both organic and
inorganic, can be determined by a titration based on
their acidic or basic properties. The end point of an
acid-base titration is usually detected by adding a
small amount of an indicator.
常用基准物质的应用
标定的溶液
基准物质
HCl标准溶液
无水Na2CO3， Na2B4O7.10H2O
NaOH标准溶液
KHC8H4O4， H2C2O4.2H2O
EDTA标准溶液
Zn，CaCO3
KMnO4标液
H2C2O4.2H2O，NaC2O4，As2O3
碘标液（I3-溶液）
As2O3
Na2S2O3标准溶液
K2Cr2O7，KIO3
AgNO3标液
常用基准物质的应用
(2). Complex Formation
M + Y == MY
The commonly used complexing agent in
complexometric
titrations
is
EDTA
(ethylenediaminetetraacetic acid) ,which reacts with
a number of metal ions, forming 1:1 stable
complexes.
NaCl，KCl
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3.1.3 Titrimetric Procedures and Calculations
(3). Precipitation
Ag+ + X- = AgX↓
X ----- chloride, bromide, iodide or thiocyanide
(SCN-).
visual-indicator methods or potentiometric titration
1. Titrimetric Procedures
manual titration, automatic titration, continuous analysis
2. Calculations in Titrimetric Analysis
“ Mole and Molarity (Amount concentration of substance)
c
(4). Oxidation-reduction
Several elements that have more than one
oxidation state can he determined by titration
with a standard oxidizing or reducing agent.
MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
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第三章 滴定分析法
=
n=
m
M
c=
n
V
the unit for the concentration is mol⋅L-1 or mmol ⋅L-1.
the elemental unit must be pointed out.
e.g. atom, molecule, ion, electron etc.
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“ Titer ----- the mass of analyte that is chemically equivalent
to 1 mL of the titrant. It is usually expressed as Tanalyte/titrant.
The unit is g·mL-1.
T
Example :
Fe / KMnO 4
A H2SO4 liquid, its density is 1.84 g/ml, and
the content of H2SO4 is 95.0%, calculate the
molarity of the H2SO4.
Solution: n of H2SO4 per liter:
n=m/M=1.84*1000*0.950/98.08=17.9 mol
c=n/v=17.9 mol·L-1
aAanalyte+bBtitrant=cC+dD
c
B
=
b
a
M
T
A / B
× 1 0
3
TA/B =
A
a
c B M A × 10 − 3
b
c -- the molar concentration of the titrant, mol·L-1;
a -- the coefficient of analyte, b is the coefficient of the titrant,
M is the molar mass of the analyte, g·mol-1.
It is convenient to obtain the mass of the analyte from
titer according to the following relation: m = TV
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“Weight Percent (mass fraction of substance) ---The
percentage of a component in a mixture or solution is usually
expressed as a weight percent (wt%)
m
wt % x = x × 100%
G
a
cVM x
wt % x = b
× 100%
G
a and b are the numbers of moles of analyte and titrant,
c is the concentration of the titrant,
V is the volume of the titrant required,
Mx is the molar mass of the analyte.
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Stoichiometric Relationships in Titrimetry
（1）Direct titration（one reaction）
reaction formula between titrant and analyte：
aA+bB=cC+dD
if it is at the stoichiometric point, a moles of A just completely
react with b moles of B, where A is the analyte and B is the
titrant:
thus：
thus nA / nB = a / b ， nA = (a / b) · nB
(c · V )A = (a / b )·（c · V )B
or：(c · V )A = (a / b)·(W / M)B
In acid-base titrations，the concentration of standard
solutions are around 0.1 mol/L , the volumes of titrant used are
20—30mL, thus the mass of the analyte needed can be
estimated.
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Solution:
Example
The reaction in titration is :
A 0.3000g portion of pure oxalic acid(H2C2O4
2H2O) can be neutralized by 22.59 mL of KOH
solution.(a) Calculate the concentration of KOH
solution; (b) Suppose that 20.00 mL of H2SO4
solution is titrated with the KOH solution requiring
25.34 mL, Calculate the concentration of H2SO4
solution.
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第三章 滴定分析法
-
H2C2O4 + 2OH = C2O42- + 2H2O
Thus: n(KOH) = 2 n(H2C2O4 ·2H2O)
M(H2C2O4 ·2H2O) = 126.1 g/mol
n(H2C2O4 ·2H2O)
= m(H2C2O4 ·2H2O) / M(H2C2O4 ·2H2O)
= 0.3000/126.1 = 2.379 × 10-3 mol
Then: n (KOH) = 2n(H2C2O4 2H2O)
= 4.758 × 10-3 mol
c (KOH) = n / V
= 0.2106 mol/L
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Example1:
(2) Diluting a solution
In diluting a solution, the amount of the solute keeps
constant, only the volume and concentration changes.
So: C1V1=C2V2
(3) Indirect titration（two or more reactions）
The stoichiometric relationship between the reactions
should be found out from the overall reactions.
For example, in standardization of Na2S2O3 solution by
using KBrO3 as a primary standard under acidic
condition. (two steps)
In standardization of Na2S2O3 solution by using KBrO3 as a
primary standard under acidic condition(two steps) .
（can not be direct titrated for complex products)
(1) KBrO3 reacts with an excess KI to release I2 in a acidic solution:
BrO3 + 6 I + 6H+ = 3 I2 + Br + 3H2O
n(BrO3 ) = 1 / 3 n(I2)
(2) The released I2 is titrated with Na2S2O3 (sodium thiosulfate )
solution
2 S2O32- + I2 = 2 I + S4O622n(I2 ) = 1/2 n(S2O3 )
(3) the relationship between n of BrO3- and S2O32n(BrO3 ) = 1/3 n(I2 ) = 1/6 n(S2O32-)
(cV)S2O32- = 6(W/M)BrO3
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Back titration：
Example2
Determination of Ca2+ by KMnO4 (several steps)
Ca2+
CaC2O4 ↓ dissolve C2O42-
KMnO4
2CO2
2 MnO4- + 5 C2O42- + 16 H+ = 2 Mn2+ + 10 CO2 + 8 H2O
A known excess of one standard reagent is added to the
analyte. Then a second standard reagent is used to titrate
the excess of the first reagent.
e.g. aluminum (complexometric titration)
Reaction1：
Al3+ + Y4-（accurate、excessive） = AlY
reaction2： Zn2+ + Y4- = ZnY
n (Ca2+) = n (C2O42- ) = 5 / 2 n (MnO4- )
(W / M ) (Ca2+) = 5/2 (cV) (MnO4- )
(nEDTA)total - (nEDTA )excessive = (nAl)
(nEDTA)excessive = (a / b)nZn
；a / b =1
(cV )EDTAtotal- (cV )Zn = (cV )Al
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3.2 Acid-base Titration
Displacement titration：
e.g. AlY
AlF62- + Y4-
3.2.1 Acid-base Equilibria
ZnY
1. Acid-base Theory
Among the theories for describing acid-base properties of
substances, Arrhenius theory and Brönsted theory are widely
adopted for aqueous systems.
According to Brönsted theory, an acid is a substance that can
give up protons. A base is any compound or ion that can
accept protons.
By definition, a strong acid or base is completely dissociated
in aqueous solution.
HCl == H+ + ClKOH == K+ + OH-
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第三章 滴定分析法
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A weak acid is one that is only partially dissociated in water.
HA + H2O == H3O+ + ADissociation constants of various acids or bases (Ka or Kb)
quantitatively describe the strengths of their acidity or basicity.
For polyprotic acids, such as H3A, the corresponding
relations between Ka and Kb should be:
Ka1
H3A = H++H2AKb3
The acid dissociation constant is:
Ka =
+
−
[ H ][ A ]
[ HA]
Weak bases, A-, react with water by abstracting a proton from
water:
A- + H2O == HA + OH[ HA][OH − ]
The base dissociation constant is:
Kb =
−
[A ]
KaKb=[H+][OH-]=
Kw
Ka2
H2A=H++HA2Kb2
Ka3
HA2-=H++A3Kb1
=10-14.
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pH = −lg [H+]
2. Calculation of pH
Ka1×Kb3=Ka2 × Kb2=Ka3 × Kb1=Kw
For example
write the PBE for the acid HA solution
Proton Balance Equation (PBE):
a) numbers of H+ obtained by conjugate base ==
numbers of H+ lost by conjugate acid
In the solution：the species that are present
in sufficient amount are HA, H2O
b) selecting zero levels ------ that are present in
sufficient amount and take part in the proton
transferring reactions
H2O + H2O == H3O+ + OH-
c) judging the species that lose protons and obtain
protons
PBE: [H+] == [A-] + [OH-]
not containing HA, H2O
d)writing the PBE equation（not containing the zero
levels)
HA + H2O == H3O+ + A-
Write the PBE for Na2S solution?
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Solutions of Strong Acids and Bases :
For a strong acid (HB), the following proton transferring
reactions occurs:
HB ==
H+
+B-
H2O == H + +OH PBE is:
[H + ] =[B-]+[OH-]
Solutions of Weak Acids and Bases :
For a monoprotic weak acid HA, PBE is：[H +]=[A -]+[OH -]
Since Ka=[H +] [A -]/ [HA] ; [OH -]=Kw/ [H+] , substituting the
above equations into the proton balance equation：
[ H + ] = K a [ HA] + K w ---exact equation
If cka≥20 Kw, Kw is negligible, thus
Suppose [B-] =C, [H + ] is 10-7 mol/L from water at 25°C
[ H + ] ≈ K a [ HA] = K a (C − [ H + ] )
If C>10-6 mol/L, H+ produced from autoprotolysis of water
can be neglected. So: [H + ] =[B -]=C
If C< 10-6 mol/L, [H + ] =C+Kw/ [H + ] , solving the quadratic
equation,
[H + ]=
C + C 2 + 4K w
2
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第三章 滴定分析法
-------exact equation
− K a + K a + 4cK a
2
2
[H + ] =
----approximate equation
If the concentration of the weak acid is not too low, that is
c/Ka ≥ 500,
[ H + ] = CK a
----simplified
----simplified equation
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Buffers : the buffer consists of a mixture of an acid and
its conjugate base
For HA + A- solution,
[ H + ][ A − ]
Ka =
[ HA]
− log[ H + ] = − log K a + log
[ A− ]
[ HA]
Distribution coefficient：The
The fraction of each
δ i = ci / c
species at equilibrium, expressed as δ
The distribution of the acidic and basic forms as a function of pH
----- distribution graph
Role：(1)understanding the acid-base titration deeply
Henderson-Hasselbalch equation:
pH = pK a + log
3. Calculation of Present Species
[ A− ]
[ HA]
([H+]=Ca/Cb Ka)
(2) judging whether the polyprotic acid (base) can be
titrated step by step
The titration is a process that the fraction of each
species and the pH change continuously.
(prepared from the weak base
A- and its conjugate acid HA)
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(1)For monoprotic acid, such as HA： C = [HA] + [A-]
C= [ HA](1 +
δ~pH：Distribution graph
（moving ）
Ka
)
[H + ]
the fraction of HA isδHA；the fraction of A- isδAδ HA =
[ HA]
1
[H + ]
=
=
+
Ka
[
H
] + Ka
C
1+ +
[H ]
δA =
−
Ka
[ A− ]
=
C
K a + [H + ]
Discuss：
1.δHA + δA-= 1
Fractions for various HA species (δ ) as a function of
pH is ：
2.When pH = pKa ；δHA = δA-= 0.5
3.When pH < pKa ；HA is predominate
4.When pH > pKa ；A- is predominate
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（moving）
(2)For diprotic acid,H2A
δH
2A
=
[ H + ]2
[ H + ] 2 + [ H + ]K a1 + K a1 K a 2
δA =
2−
δ HA =
−
[ H + ]K a1
[ H + ]2 + [ H + ]K a1 + K a1 K a 2
K a1 K a 2
[ H + ]2 + [ H + ]K a1 + K a1 K a 2
a.when pH < pKa1，H2A
predominate
b. When pKa1< pH <pKa2,HApredominate
c. when pH>pKa2，
predominate
pKa1=1.23；pKa2=4.19
d. When pH=2.75时,δHA- is the
biggest
δHA- =0.938;δH2A =0.028;δA2- =0.034
(H2C2O4)
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第三章 滴定分析法
A2 -
(3) For triprotic acid (H3PO4)
（moving）
four existence：
existence：H3PO4 ；H2PO4-；HPO42-；PO43-；
δ：
δ3
δ2
δ1
δ0
δ3 =[H+]3 /{ [H+]3+[H+]2 Ka1+[H+]Ka1 ×Ka2+Ka1×Ka2 ×Ka3 }
δ2 =[H+]2 Ka1/{[H+]3+[H+]2 Ka1+[H+]Ka1 ×Ka2+Ka1×Ka2 ×Ka3 }
δ1 =[H+]Ka1×Ka2/{[H+]3+[H+]2 Ka1+[H+]Ka1 ×Ka2+Ka1×Ka2 ×Ka3 }
δ0=Ka1×Ka2 ×Ka3/{[H+]3+[H+]2Ka1+[H+]Ka1 ×Ka2+Ka1 ×Ka2 ×Ka3 }
Discuss：
pKa1=2.12；pKa2=7.20； pKa3=12.36
（1）the difference between pKa is
high, coexistence is little；
（2）when pH=4.7，
δ2 =0.994 δ3 =δ1 = 0.003
（3）when pH=9.8， δ1=0.994
δ0 =δ2 = 0.003
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4. Activity and Activity Coefficient
The relationship between activity, a, and the concentration,
c is:
a=γc,
is:
where γ is the activity coefficient.
In 1923. P. Debye and E. Huckel derived an equation: the
activity coefficient of B(rB) (at 25°
25°C )is:
where ZB is the charge on the ionic species taken without
regard to sign; I is the ionic strength of the solution.
The ionic strength of the solution is defined by:
1
2
∑
Functions:
(1) 确定滴定终点时，消耗的滴定剂体积；
(2) 判断滴定突跃大小；
lgrB= -0.509ZB2I1/2,
I =
3.2.2 Titration Curve
C iZ i
(3) 确定滴定终点与化学计量点之差。
(4) 选择指示剂；
2
滴定曲线的计算。
i=1
where Ci is the concentration and Zi is the charge of each
ionic species in the solution.
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3.2.2 Titration Curves(pH~volume of the titrant added)
(moving）
Strong Acid Versus Strong Base
Supposing that 20.00 ml 0.1000 mol/L HCl is titrated with
0.1000 mol/L NaOH
(1）Before the titration，volume of NaOH added is 0.00 ml：
0.1000 mol/L HCl, pH=1
( 2）when 18.00 ml of NaOH solution is added：
[H+]= 0.1000 × (20.00-18.00)/(20.00+18.00)=5.26×10-3 mol/L
pH=2.28
(3) when 19.98 ml of NaOH solution is added (about half
drop to equivalence point):　[H+]=c ×VHCl/V=0.1000×(20.00-19.98)/(20.00+19.98)
=5.0 × 10-5 mol/L
pH=4.3
(4)At the stoichiometric point： 20.00mL
[H+]=10-7 mol/L , pH=7
(5) when 20.02 ml of NaOH solution is added：, (about
half drop pass)
[OH-]=n(NaOH)/V=(0.1000 × 0.02)/(20.00+20.02)
=5.0 × 10-5 mol/L
pOH=4.3 , pH=14-4.3=9.7
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Weak Acid Versus Strong Base
Discuss:
1) In a titration, the calculation of pH
should be careful because the volume is
changing with the titrant added.
2）when the titrant added from 0 to 18
mL, the change of pH is
2.28-1=1.28；
In the vicinity of the stoichiometric point,
(0.05 mL, 1 drop), a break is occurred,
change of pH: 9.70-4.30=5.4
3）end point is not definitely the same as
the stoichiometric point ， but the
difference
between
them
is
less
than±
than±0.02mL, the relative error is less
than±
than±0.1%.
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第三章 滴定分析法
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0.1000mol/L NaOH titrates 20.00 mL 0.1000mol/L HAc
1)before the titration, the solution is 0.1000mol/L HAc
solution.
H + = ca Ka = 0.1000 × 10 −4.74 = 10 −2.87 ; pH = 2.87
2）As 19.98 mL of NaOH is added, we have a buffer
solution of HAc and Ac;
ca=0.02×0.1000/(20.00+19.98)=5.00×10-5 mol/L
cb=19.98×0.1000/(20.00+19.98)=5.00×10-2 mol/L
[H+]=Ka×ca/cb=10-4.74[5.00×10-5/(5.00×10-2)]=1.82×10-8
pH=7.74
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3）at the stoichiometric point,
the solution of NaAc（weak base）：
cb =20.00×0.1000/(20.00+20.00) =5.00×10-2 mol/L
pKb=14- pKa=14-4.74 = 9.26
[OH-]= (cb × Kb)0.5 = (5.00×10-2 ×10-9.26)0.5 =5.24×10-6 mol/L
pOH = 5.28 ;
pH = 14-5.28 = 8.72
4）after the stoichiometric point,
volume of titrant added: 20.02 mL
[OH-]=(0.1000×0.02)/(20.00+20.02)
=5.0×10-5 mol/L
pOH=4.3 , pH=14-4.3=9.7
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The influence of concentration on the break
Discuss ：
1)before stoichiometric point, the curve for a strong acidstrong base titration is much flatter than for the weak acid；
2) due to partial ionization, the pH of the starting solution of
the weak acid is higher, the magnitude of the break is
smaller
3) methyl orange can not be used as an
indicator in the titration of weak acid.
4) with the decrease of Ka，the magnitude
of break is smaller，Ka at 10-9, the break
disappear;
5) for direct titration of weak acid：
cKa≥10-8
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3.2.3 AcidAcid-Base Indicators
The magnitude of break
increase with the increment of
concentration
Weak Base Versus Strong Acid
Similarly to weak acid versus strong base
1．The principle of indicator
2．How to choose and how to use；
3．Ranges for the indicator to change color。
condition for direct titration:cKb≥10-8.
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（1）principles of acid-base indicator
AcidAcid-base indicator：weak organic acids or bases which
dissociated and ionic forms show different colors.
phenolphthalein：diprotic acid，
acid，colorless. pH range of
changing color：
color： 8－10，
10，colorless to red,
red, used in titrating of
acid with base.
moving
base.
Methyl orange：base，
base，used in titration of base with acid.
pH range of changing color：
4.4，
，red to yellow.
color： 3.1-3.1--4.4
yellow. moving
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第三章 滴定分析法
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Discussion：
Assume that the indicator is a weak acid, designated
by HIn，the equilibrium in solution is
HIn == H+ + In[ H + ][ In − ]
(red)
(blue
Ka= [ HIn]
red)
(blue))
From the above equation, we obtain pH = pKa −lg[HI−n]
[In ]
It is obviously that：
the color of solution is decided by the ratio of:
[In-] / [HIn]
[In ] show the shade of the base color
[HIn] show the shade of the acid color
1．KHIn is known，
known，the color changes with [H+] .
when [In-] / [HIn
[HIn]] = 1 middle color
[HIn]
pH = pKa −lg −
= 1/10 acid color
[In ]
= 10/1 base color
so the pH range for indicator of color change ：pKHIn ± 1
2. pH range for indicator of color change is decided by KHIn,
3. Selection of indicator: pKa of the indicator should be close
to the pH of the stoichiometric point.
4. The amount of indicator added should be kept minimal.
5．Mixed indicator：
indicator：
some indicators are mixed to improve sensitivity and to
narrow the transition range.
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3.2.4 Some Acid-base Methods
酸碱指示剂
1. Titration of Mixtures Containing Sodium Carbonate
NaOH ,Na2CO3 ,NaHCO3
（moving）
1) At the first end point (phenolphthalein), HCl consumed is V1
mL
The reactions are:
HCl + Na2CO3 == NaHCO3 + NaCl
HCl + NaOH == NaCl + H2O
2) At the second end point (methyl orange ), HCl consumed V2 mL
The reaction is:
HCl + NaHCO3 == NaCl + CO2↑ + H2O
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Discussion：
2. Boric acid
（1） V1>V2 : NaOH(V1-V2) , Na2CO3 (V2)
Boric acid (pKa=9.24) is too weak to be directly titrated
，however, it can react with polybasic alcohols forming
complex acid (pKa is about 6), pH of stoichiometric point
is about 9.
（2） V1=V2 : Na2CO3
（3） V1<V2 : Na2CO3 (V1), NaHCO3 (V2-V1)
（4） V1=0 : NaHCO3
（5） V2=0 : NaOH
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第三章 滴定分析法
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3．Nitrogen
（3）Kjeldahl method　（1）Distill method
Nigrogen in organic compounds (protein,
etc.)
Ammonia can be liberated by addition of excess NaOH
solution and is caught in excess standard acid, then the excess
acid is titrated：
treat the organic material with hot concentrated
sulfuric acid, the nitrogen converted to ammonium
sulfate. + distill method
NH４+ +OH- =NH3 (g) +H2O
NH３＋ HCl = NH４+＋ Cl－
（2）Formaldehyde method (甲醛法)
６ＨＣＨＯ＋４ NH４+ =(ＣＨ２)６Ｎ４Ｈ+ ＋３Ｈ+＋６Ｈ２Ｏ
2 SO4 ,CuSO4
Cm H n N ⎯H⎯
⎯ ⎯⎯→ CO2 + H 2O + NH 4 +
hexamethylene-tetramine六亚甲基四胺
choose phenolphthalein as indicator
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4. SiO2
3.3 Complexometric Titration
accuracy
Gravimetric
method
Titrimetric
method
more
Time-consu
ming
more
3.3.1 Metal-chelate Complexes
Ligand: The molecules or ions that donate an unshared
electron pair to the metal ion to form a metal-ligand bond.
Such as ammonia and water.
less
less
Coordination number of a metal ion: the number of metalligand bonds it forms.
Formation constant, Kf: An equilibrium-constant expression
in the complexes formation. Also called stability constant, Ks.
KOH
6HF
3H2O
SiO2 ---→ K2SiO3 ---→ K2SiF6↓ ----→ 4HF
Dissociation constant, Kd: An equilibrium-constant expression
in the complexes dissociation. Simply the reciprocals of the Kf.
For example:
Cd2++4I-=CdI42-,
Kf=[CdI42-]/[Cd2+][I-]4 ,
Kd=1/Kf=[Cd2+][I-]4/[CdI42-]
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Chelating agent: An organic agent that has two or more
groups capable of complexing with a metal ion.
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3.3.2 EDTA
moving
EDTA
Ethylene diamine
tetraacetic acid
Chelate:
Chelate: the complex formed is called a chelate.
chelate.
Chelometric titration: a type of complexometric titration,
the titration with a chelating agent.
Complexing groups: the groups which contain a pair of
free electrons capable of complexing with a metal ion.
EDTA: 6 complexing group
five or sixsix-member ring is relative free of strain and is
Ni + Y = NiY
more stable and is often formed in chelates.
chelates.
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第三章 滴定分析法
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氨羧试剂的特点： 动画）
a.
1. Acid-base Properties
moving
配位能力强；氨氮和羧氧
两种配位原子；
b. 与金属离子能形成多个多
元环，配合物的稳定性高
；
Fractions of each EDTA species as a function of pH
1) pH >12,
c. 与大多数金属离子1∶1配
位， 计算方便；
Y4- is predominate;
2). Y4- is effective
species to chelate with
metal ions;
d.多元弱酸；EDTA可获得两
个质子，生成六元弱酸；
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2. EDTA Complexes
Table KMY for some metal - EDTA complexes
Side reactions and conditional stability constant
M + Y == MY
KMY：
a ．For alkaline metals
，lg KMY<3, the chelates
are less stable；
ｂ．For alkaline earth
metal, lgKMY=8-11；
ｃ．Transition metals, rare earth metal and Al3+, lgKMY=15-19
ｄ．Trivalent, quaternary metals and Hg2+, lgKMY>20.
配合物的稳定性受两方面的影响：金属离子自身性
质和外界条件。
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Effect of pH and αY（H）
Table lgαY（H）at different pH
αY（H） =[Y’]/[Y]
ａ．αY（H）decreases
with the increase of pH；
the ratio of total concentration of all forms of the
uncomplexed EDTA [Y’] to the concentration of the
reactive EDTA species [Y].
(αY（H）is the reciprocal of the fraction of Y4- (δ))
ｂ．αY（H）is bigger, the
effect of hydrogen ion is
higher；
c. commonlyαY（H） >1, [Y’]>[Y].
whenαY（H） =1, [Y’]=[Y]；
d．α Y（H） =1/δ
By means of α Y（H） , a conditional stability constant
may be calculated for metal-EDTA complex at any pH.
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第三章 滴定分析法
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11
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3. Conditional Formation Constant
reaction ： Mn+ + Y4- = MY
KMY =[MY]/（[Mn+][Y4-]）
αY(H) = [Y']/[Y4-]
so
[MY] /（
/（[M][Y'
[M][Y']）= KMY / αY（H） = K MY’
MY’
lgK'MY = lgKMY - lgα Y（H）
In some cases, other side reaction effects have to be
considered.
Example
Calculated the value of lgKZnY’ at pH=2.0 and pH=5.0
Solution：
• From Table of K：lgKZnY=16.5
• at pH=2.0 , lgαY（H）=13.5
• at pH=5.0 , lgαY（H）= 6.6
• lgK'MY = lgKMY - lgαY（H）
• so,
• at pH=2.0, lgK'ZnY =16.5-13.5=3.0 can not be titrated
• at pH=5.0时, lgK'ZnY=16.5-6.6=9.9 can be titrated
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Minimum pH for effective titration
The tolerable error is ± 0.1%, ∆pM for the visual end point
ÞpH effect on the titration ：
detection is ±0.2, so the criterion for effective titration of a
（1）the higher the pH is，the lower the α is, the
single metal is:
higher KMY‘
is beneficial to the complexometric
titratioin；
ÞK’MY = [MY]/（[M][Y’]）= c/(c ×
0.1% × c × 0.1%)=1/(c × 10-6)
（ 2 ） if pH is too high，metal ion may hydrolyze,
forming hydroxide.
lgcK’MY≥6
ßThere are minimum pH values at which K’ is high
lgα Y（H） ≤lgKMY + lgC - 6
enough and metal ion do not hydrolyze.
ßcalculation of minimum pH for a
given metal ion.
ßlgα Y（H） ~minimum pH for
different metal
àThe minimum pH value for a given metal ion
depends on the tolerable error and the required pM
change in the end point：
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Suitable pH range：
lgcK’MY≥6
1）pHmin
When [c]=10-2 mol/L 时, lgK’MY≥8
lgα Y（H） ≤lgKMY - lgK’MY =lgKMY -8
2)pHmax
the metal ion do not hydrolyze at this pH
1
pH = 14 − ( pK M (OH ) n + lg c M )
n
n——金属离子的价态
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第三章 滴定分析法
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3.3.3 EDTA Titration Curves
pM ~ mililiters of EDTA or percent titration.
The conditional stability constant is used in the calculation.
For example: 20 mL 0.01000 mol/L
with 0.01000 mol/L EDTA
Ca2+
solution is titrated
3)at the stoichiometric point:
[CaY]=0.01/2=0.005 mol/L ;[Ca2+]=[Y] ;KMY=1010.69
from the condition constant:
0.005/X2 = 1010.69 ；[Ca2+]=3.2×10-7 mol/L ；pCa=6.49
4)0.02mL of excess EDTA is added beyond the equivalence
point
1．At pH 12αY(H)=0；
=0；KMY’
MY’=KMY
1）before titration，the concentration of Ca2+ is：
[Ca2+]=0.01 mol/L ； pCa=-lg[Ca2+]=-lg0.01=2.00
2)as 19.98mLof EDTA is added,
[Ca2+]=0.01000×0.02/(20.00+19.98)=5×10-6 mol/L； pCa=5.3
[Y]=0.01000× 0.02/(20.00+20.02)=5×10-6 mol/L
pCa=7.69
2．At pH=9，lgα Y（H） =1.29,
lgKMY’=lgKMY-αY（H）=9.40
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Effect of pH and K’MY on the magnitude of break
1)．Effect of pH on the magnitude of break
The higher pH is, the
bigger the break is.
2)．Effect of K’MY on the
magnitude of break
The higher K ‘MY is, the
bigger the break is.
moving
3.3.4 Metal Ion Indicators
Indicators: complex agent，forms a complex of a
different color with the metal ion being titrated.
For example： before titration, EBT is added into Mg2+
solution（pH 8-10)，color of the solution is red：
EBT+ Mg2+ = Mg2+-EBT
(Blue)
(red )
at the stoichiometric point，EDTA break up Mg2+EBT the color of the solution changes to blue：
Mg2+-EBT+ EDTA = EBT + Mg2+- EDTA
(red)
(Blue)
the indicator should be used in suitable pH
H2InpH <6
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Requirement for metallochromic indicators
（1）In titration pH，there is color difference between the
complex and the dissociate indicators；
（2）Ks of the complex between the indicator and the metal
In-3
>12
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1). Indicator blocking : in the titration of a metal with
EDTA, the indicator can not be displaced by EDTA(the metal
does not freely dissociate from an indicator)--- the metal is
said to block the indicator.
e.g. EBT can be blocked by Fe3+、Al3+、Cu2+、Ni2+，
triethanolamine 。
should be suitable；
（moving）
a．not be too large,can be displaced by the EDTA
b．not be too small。Or EDTA will replace it before the
stoichiometric point.
C. the metal indicator complex should be 10~100 times less
stable than the metal-titrant complex.
2009-1-11
第三章 滴定分析法
HIn-2
8-11
2). Indicator rigidifying: in the titration of a metal with
EDTA, if the complex formed by a metal and indicator does
not dissolve in water, and produces precipitation or colloid(胶
体）， the metal is said to rigidify the indicator.
e.g. PAN can be rigidified at low temperature, organic solvent
or heating.
（moving）
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Commonly used metallochromic indicators
（1）Eriochrome black T (EBT) ：
•suitable pH value is 9-10,
•used in the titration of Zn2+、Mg2+、Cd2+、Pd2+.
•Color change from red to blue
•the indicator should be kept as dilute as possible and
still give a good color.
（2）CalconCalcon-carboxylic acid 钙指示剂
•at pH=7，
pH=7，purple；
purple；at pH=12pH=12-13：
13：blue；
blue；
•used in titration of Ca2+ at pH>12.5,
•color changes from red to blue。
blue。
（3）PAN
•pH=1.9~12.2,
pH=1.9~12.2,
•color change form red to yellow
•used in titration of Cu2+, etc, and rare earth metals
3.3.5 EDTA Titration Techniques
1. Direct Titration
------Determination of the hardness of water
Successive Titrations by pH Adjustment
Suppose that in a solution containing metal ions M and N,
both the ions can react with EDTA forming complexes,
The requirements that must be satisfied for the separate
titration is:
∆ lgcK≥6
If cm=10cn,
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------ determination of Al３＋
+ Y (excess) == AlY (heat)
Y + Zn2+ == ZnY
c N K NY '
⎞
⎟⎟ ≥ 6
⎠
∆lgK≥5
4. Indirect Titration -------determination of PO43 －
e.g.： PO4 3－ + Bi3+ (adequate) === BiPO４(s)
Bi3+(excess) + Y === BiY
5. Masking
1.Masking agent is often used to eliminate interference
in complexometric titration
3. Displacement Titration
-------determination of Ag＋测定（Ag＋ cannot be
titrated directly, why?）
2Ag+ + Ni(CN)42- (excess) → 2Ag(CN)2- + Ni2+
Ni2+ + Y ====
c M K MY '
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2. Back Titration
Al３＋
⎛c
= ∆ lg K + lg ⎜⎜ M
⎝ cN
TE ≤ ± 0 . 5 % ∆ pM = ± 0 . 2
lg
NiY
2. Masking agent can form more stable complexes with
the interferential ion than EDTA
for example in titration of Ca2+ and Mg2+ for
determining water hardness, the interference of Al3+
and Fe3+ may be eliminated by adding triethanolamine
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3.4.1 Redox Reactions
3.4 Oxidation-reduction Titration
Oxidation-reduction reactions are those in
which there is a net change in the
oxidation numbers of one or more
elements in the reaction substances. The
word redox is frequently applied to
oxidation-reduction reactions
1. Electrode potentials
Ox+ne=Red,
The potential of the electrode is given by Nernst equation:
O
ϕ ox/Red = ϕ ox/Red
+
ϕ° : standard potential of the electrode;
R: gas constant(8.314 J/K mol)
T: absolute temperature;
F: Faraday constant (96485 C/mol)
n: the number of the transferred electrons.
At 25°C,
2009-1-11
第三章 滴定分析法
RT
a
ln ox
nF aRed
O
ϕ ox/Red = ϕ ox/Red
+
0.059 aox
lg
n
aRed
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Two problems exist in applying above equation:
1)．a or γ is not known，a = γ c
2)．Side reaction of the reactant or product
should be considered，
(side reaction coefficient：αM=[M’]/[M]
[M’] total concentration,[M] effective
concentration)
O
ϕox/Red = ϕox/Red
+
RT γ oxαRedcox
RT cox
O'
ln
= ϕox/Red
+
ln
nF γ RedαoxcRed
nF cRed
so the conditional potential is introduced:
O'
O
= ϕ ox/Red
+
ϕox/Red
RT γ oxα Red
ln
nF γ Redα ox
O
ϕ ox/Red = ϕ ox/Red
+
O'
O
= ϕ ox/Red
+
ϕ ox/Red
1). The effect of concentration of oxidized species
and reduced species
2). Side reaction effect (Precipitation, complex)
3). Acidity effect when hydrogen or hydroxyl ions
are involved in redox reactions.
0.059 γ oxα Red
lg
n
γ Redα ox
then
O'
ϕ ox/Red = ϕ ox/Red
+
0.059 Cox
lg
n
CRed
When cox/cRed = 1, conditional potential=actual potential.
2009-1-11
Factors affect conditional potential
RT γ oxα Redcox
RT cox
O'
ln
= ϕ ox/Red
+
ln
nF γ Redα ox cRed
nF cRed
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For example, consider the following reaction
2Cu2+ + 4I- == 2CuI ↓ + I2
ϕθCu2+/Cu+=0.16V, ϕθI2/I-=0.54V
According to the related standard potentials, it seems that Cu+
would be oxidized by I2, but in fact, Cu2+ can oxidize I- to
virtual completion because I- reacts with Cu+ forming CuI
precipitate.
Ksp（CuI） = [Cu+][I-]=1.1×10-12
If the oxidized form（cOX）has side reaction,
potential decrease；if the reduced form（cRed） has
side reaction, potential increase.
ϕ Cu 2+ /Cu + = ϕ O 2+
Cu
/Cu +
+ 0.059 lg
[Cu 2 + ][I − ]
[Cu 2 + ]
= ϕ O 2+ + + 0.059 lg
Cu /Cu
K Sp[CuI ]
[Cu + ]
If [Cu2+]=[I-]=1.0 mol/L then:
2009-1-11
2. Equilibrium Constant
In a redox titration, it should be estimated that:
1).Whether the reaction can proceed completely, that is
whether the end-point error can satisfy the requirement;
2). the requirement for the potential to react completely.
For the redox reaction
n2AOx + n1BRed == n2ARed + n1BOx
K=
(C ARe d ) n2 (C BOx ) n1
(C AOx ) n2 (C BRe d ) n1
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第三章 滴定分析法
ϕ ' Cu
2+
/Cu +
= 0.87
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As equilibrium is reached, we have
ϕθ'A +
0.059 C AOx
0.059 C BOx
lg
= ϕθ'B +
lg
n1
C ARe d
n2
C BRe d
lg K ' = lg
(C ARe d ) n2 (C BOx ) n1
(C AOx ) n2 (C BRe d ) n1
=
(ϕθ ' A − ϕθ ' B ) ⋅ n1 ⋅ n2
0.059
K’ is the conditional constant, the larger the K’ is, the
more completely the reaction proceeds。
K’ is decided by the difference of the two conditional
potentials and n1 、n2
when n1=n2=1, to ensure that the reaction shall go to at least
99.9% completion, at the stoichiometric point,the following
requirements must be fulfilled:
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3.4.2 Rate of Redox Reactions
cARed / cAOx ≥ 103 ; cBOx / cBRed ≥ 103
then
∆ϕ = ϕ AO′ − ϕ BO′ ≥
0.059
0.059
lg(103n1103n2 ) =
6 ≈ 0.36V
n1n2
n1n2
∆ϕ = ϕ AO′ − ϕ BO′ ≥
0.059
0.059
lg(103n1103 n2 ) = 3(n1 + n2 )
n1n2
n1n2
so
The factors affect the rate of redox reactions：
1). Concentration of the reactants（increment of C
will increase the rate）；
2 ) . catalyst（change reaction process, reduce the
activation energy）；
3).temperature（normally, rate will be 2~3 times
faster when the temperature increase every 10
degree).
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For example：
example：titration of oxalic acid with
KMnO4
(1）the reaction of KMnO4 with C2O42- should be heated
to 75-85OC，to elevate the rate. But the temperature
should not be too high, or part of oxalic acid will
decompose.
( 2 ） Mn2+ produced in the reaction will catalyze the
reaction itself --self-catalyzed reaction
The role of Mn2+：catalyst；reduce the potential of the
Mn3+/Mn2+ couple so as to prevent the induced reaction （
conjugate reaction).
Induced reaction： 由于一种氧化还原反应的发生而促进另一种
氧化还原反应进行的现象，称为诱导作用，反应称为诱导反应或
共轭反应。
2MnO4- + 10Cl- + 16H+ == 2Mn2+ + 5Cl2 + 8H2O 极慢，(Fe2+)
3.4.3 Titration Curves
For example: Consider the titration of Fe2+
with standard Ce4+ (in 1mol⋅L-1 HClO4,
calomel electrode as the reference electrode)
The titration reaction:
Ce4+ + Fe2+ =
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2）at the stoichiometric point：
4+
/Ce 3+
the potential at stoichiometric point ϕe ：
0.059 cCe 4+
0.059 cFe3+
O′
+
lg
= ϕ Fe
+
lg
3+
/Fe 2+
n1
cCe 3+
n2
cFe 2+
O′
n1ϕ e = n1ϕ Ce
+ 0.059 lg
4+
/Ce 3+
O′
n2ϕ e = n2ϕ Fe
+ 0.059 lg
3+
/Fe 2+
ϕe =
cCe3+
cFe3+
reactant：c Ce 4+ , c Fe 2+ low and equal
product：c Ce 3+ , c Fe 3+ high and equal
n1ϕ1O′ + n2ϕ 2O′
n1 + n2
ϕe =(0.767 +1.70)/(1+1)=2.467/2=1.23 V
cFe2+
at equivalenc e point，
第三章 滴定分析法
(n1 + n2 )ϕ e = n1ϕ1O′ + n2ϕ 2O′
cCe 4+
O′
O′
( n1 + n2 )ϕ e = n1ϕ Ce
+ n2ϕ Fe
+ 0.059 lg
4+
3+
/Ce 3+
/Fe 2+
2009-1-11
Fe3+
1）before stoichiometric point(99.9% completion):
0.059 cFe3+
99.9
ϕ Fe3+ /Fe2+ = ϕ FeO′3+ /Fe2+ +
lg
= 0.767 + 0.059 lg
= 0.94 V
0 .1
n2
cFe2+
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O′
ϕ e = ϕ Ce
Ce3+ +
cCe 4+ cFe3+
cCe3+ cFe 2+
3)after the stoichiometric point，0.1% excess of Ce4+:
ϕ Ce
4+
/Ce3+
O′
= ϕ Ce
+
4+
/Ce 3+
0.059 cCe 4+
0 .1
lg
= 1.70 + lg
= 1.52 V
99.9
n1
cCe3+
the magnitude of the break is 1.52-0.94 V
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1.7
3.4.4 Redox indicators
potential (V)
1.4
1.1
0.8
0.5
0
50
100
150
200
titrant added (%)
the magnitude of the end point break in
redox titration is decided by the conditional
constant K’，K’is mainly decided by the
difference between the two conditional
potential.
1．Redox Indicators
the redox indicators are highly colored dyes
that are weak reducing or oxidizing agents that
can be oxidized or reduced;
The oxidized and reduced form of redox
indicators show different colors.
In the vicinity of the stoichiometric point,
the break of the potential transferred the
indicator from one form to the other, and the
color changes.
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For example: Diphenylaminesufonic
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Requirement for redox indicators:
1). The transition range must fall within the end point break
2). Redox indicator reaction must be rapid
3).the reaction must be reversible
O′
E = E In
Transition
range:
O′
E In ±
0.059
Oxidized form
+
lg
Reduced form
n
0.059
Oxidized form
1 10
=
；（
~ ）
n
Reduced form
10
1
When n=1, a 0.12V change of the break is required.
2．Starch Indicator
used for titration involving iodine. Starch forms a very
reversible complex with I2。
Very sensitive.
3. Self-indicator
the color change of the standard solution or the analyte
at the stoichiometric point to indicate the end point.
For example: KMnO4 --- H2C2O4
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3.4.5 Applications of Redox Titrations
2. Oxidation with Potassium Permanganate
1. Preadjustment
1)．standard solution（indirect）
boiling→ standing over night →filtering to remove MnO2
→standardizing with oxalic acid.
Preoxidiation
S2O82- + Ag+ ＝ SO42- + SO4- + Ag2+
2H2O2 ＝ O2 + 2H2O
Prereduction
Stannous chloride (SnCl2) , Chromous chloride, Zinc
Sn2+ + 2HgCl2 ＝ Sn4+ + Hg2Cl2↓ + 2Cl-
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第三章 滴定分析法
2MnＯ４－＋5C2O４2 －＋16H＋=2Mn2＋+10CO２↑＋8H２O
note:
① rate：self-catalyzed
②temperature：75～８5℃，not too high or too low
③acidity： 0.5～1.0mol/L H2SO4，to avoid Fe3＋ inducing
KMnO4 to oxidize Cl－, HCl can not be used as an acidic
medium
④end point：self-indicator (30s)
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5. methods Involving Iodine
２)．Application
for the determination of metal ion that can be
precipitated by C2O４２－,
Ca2＋+C2O４２－—→CaC2O４↓—→hot sulfuric acid—→
H２Ｃ２Ｏ４ —→titrate（KMnO4 standard solution）
Determination of formic acid
in strong basic medium:
MnO４－＋HCOO－＋３OH－= CO３－＋MnO42-＋２H2O
the excess MnO４－ is titrated by ferrous standard solution.
１)．iodimetry
redox analysis based on the oxidation nature of I2 and the
reduction nature of I－.
I3－+ 2e = 3I－， EＩ２／Ｉ-＝0.545 V
I2:weak oxidant，I－: moderate reductant，
direct iodimetry: I2 standard solution directly titrate a reductant
indirect iodimetry: I－ react with strong oxidant create I2，a
reductant (Na2S2O3) is used to titrate I2，, also called iodometry.
Reaction of iodometry：
I2 : S2O３2－ = 1:2
I2 ＋ 2S2O３2－= S４O６2－＋2Ｉ－
in neutral or weak acid medium
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碘量法: I2 ＋ 2S2O３2-= S４O６2-＋２I-
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I2 : S2O３2－ = 1:2
反应在中性或弱酸性中进行，pH过高，I2会发生岐化反应：
3Ｉ2＋6OH- = IO３－＋5I-＋3H2O
4I2 + S2O32- + OH- == 8I- + 2SO42- + 5H2O
I2 : S2O３2－ = 4:1
在强酸性溶液中，Na２Ｓ２Ｏ３会发生分解， I－容易被氧化。通
常pH<9。
碘法中的主要误差来源：①I2易挥发，②I－在酸性条件下容
易被空气所氧化。措施：加入过量KI，生成I3－络离子；氧
化析出的I2立即滴定；避免光照；控制溶液的酸度。
碘法中常用淀粉作为专属指示剂；硫代硫酸钠溶液为标
准溶液。
2. application
（1）iodometry for determination of Cu2＋
2 Cu2＋ + 4I－ = 2CuI↓ +I2
I2＋ 2S2O32－= S4O62－＋2I-
S2O32- + 2H+ == H2SO3 + S
n Cu
H2SO3 + I2 + H2O == SO42- + 2I- + 4H+
I2 : S2O３2－ = 1:1
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2＋
: n Na
=
2Ｓ2Ｏ3
1:1
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3.5 Precipitation Titration
（2）Karl Fisher method for determination of minim
water
principle：I2 oxidizes SO2 and consumes water
3.5.1 Precipitation Reactions
Argenometry:
Ag+ + Cl- → AgCl (s)
Ag+ + SCN- → AgSCN (s)
ＳＯ2 ＋Ｉ2 ＋２Ｈ2Ｏ == Ｈ2ＳＯ４＋２ＨＩ
The solubility product constant
total reaction：
K=
C５H５N·I2 ＋C５H５N·SO2 ＋C５H５N＋H2O ＋CH３OH
==２C５H５N·HI＋ C５H５NHOSO2 ·OCH３
Karl Fisher reagent is the mixture of
I2、SO2、C５H５N and CH３OH.
So
then
KspAgCl = [Ag+][Cl-] (γAg+=1,γCl-=1)
KSP’=[A’][B’]=KSPαAαB
第三章 滴定分析法
(αAgCl=1)
a AgCl
K = a Ag + aCl − = [ Ag + ]γ Ag + [Cl − ]γ Cl −
If side reaction exists,
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a Ag + a Cl −
[ A] =
[ A' ]
αA
, [ B] =
[ B ]'
αB
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3.5.2 Titration Curves
For example
:0.1mol.L-1 KCl
The factors influence the titration:
with
0.1mol.L-1 AgNO3
1)c :
when Ksp is constant,
the larger the c,the
larger the break of the
curve
1) At the beginning of the titration pCl=1
2) At the stoichiometric point
Ksp（AgCl） = [Ag+][Cl-]； [Ag+]=[Cl-]
[Cl-]= (Ksp（AgCl）)1/2=1.25×10-5 mol.L-1
So
pCl=4.90
2)Ksp:
3) after the stoichiometric point
[Cl-]=Ksp（AgCl）/ [Ag+]
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3. 5.3 End-point Detection
1. Mohr titration –using potassium chromate as
indicator
1) This method can be used to determine the chloride , and
the standard solution is AgNO3.
Reactions: Ag＋ ＋ Cl－ ＝ AgCl↓
At the stoichiometric point ：
［Ag＋］＝(ＫspAgCl )1/2＝1.25×10－５
ＣrO４２－(yellow)＋２Ag＋ = Ag２CrO４↓　（red）
So the concentration of the ＣrO４２－ is：
［ＣrO４２－］=ＫspAg２CrO４ /[Ag＋]２＝５.８× 10－2 mol/L
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2. Volhard titration—using Ferric Alum as indicator
The Volhard method is a procedure for titrating silver(Ⅰ)
with standard potassium thiocyanine.
Equations for the Volhard titration of silver are given
below:
Ag＋＋ SCN－ = AgSCN↓
(Titration reaction)
Fe3＋＋ SCN－ = FeSCN２＋
（End-point reaction)
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第三章 滴定分析法
when the c is constant,
the smaller the Ksp, the
larger the break of the
curve
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.2)
The Mohr titration should be performed at pH
6.5~10.5
If the solution is too acidic, ［CrO４２－］is smaller
2H+ + 2CrO42- == 2HCrO4- == Cr2O72- + H2O
If the solution is alkali(pH＞10) , it can produce AgOH
precipitate
3)The method can not be used for back-titration（返滴定）
method
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1)The titration with thiocyanate is carried out in acid
solution.
2)This method can be used to determine chloride and
other anions.
Reactions: Ag+ + X-1 ==AgX(s) + excess Ag+
Titration reaction
SCN－ + excess Ag+ == AgSCN↓
back-titration reaction
Fe3＋＋ SCN－ = FeSCN２＋
end-point reaction
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3. Fajans’ titration –using Adsorption indicator
In the adsorption indicator method for halide ions,
the end point reactions occurs on the surface of the
silver halide precipitate.
The adsorption indicators are some colored organic
compounds. Fluorescein（荧光黄） is a kind of
adsorption indicator.
It can be adsorbed on the surface of the colloidal
particles and change color.
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第三章 滴定分析法
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