PRECIPITATOMETRIC(PRECIPITATION)TITRATION by nooryudhi

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									PRECIPITATOMETRIC
  (PRECIPITATION)
     TITRATION
       An application method of
  Inorganic Pharmaceutical Analysis


          Lecturer : Dr. Tutus Gusdinar
       Pharmacochemistry Research Group
              School of Pharmacy
        INSTITUT TEKNOLOGI BANDUNG
Precipitatometric Titration
• Compared to acid-base or reduction-oxidation
  titrations the precipitatometric titration has no much
  more methods
• Difficult to select a suitable indicator
• Difficult to obtain an accurate precipitate composition,
          p    p                       y
  the coprecipitation effect is oftenly occured.
  Solubility = a saturated concentration of analyte
  (crysstaline or solid form) dissolved in solvent at a
  defined temperature.
       BaSO4( )
           4(p)                 Ba2+ + SO42-
  Solubility Product Constant : Ksp = [Ba2+] [SO42-]
  (at equilibrium state)
                                       Solubility product
Saturated solution is occured when a
continuous addition of a substance is
performed into its solvent until dissolution
            terminated,
precess is terminated or it can be
achieved by increasing ionic concentration
 ntil                 formation.
until the precipitate formation

Solubility is influenced by temperature,
          properties,
solvent properties and other ions
existed in a solution.
         y
Solubility
  s


                     Saturated




              (concentration of ion)
             X(      t ti     fi )
     Factors influencing the solubility

1.   Temperature
2.   Solvent properties
3.   Common ions
4.
4    Ionic activity
5.   pH
6.   Hydrolysis
7.     eta yd o yde
     Metal hydroxyde
8.   Complex compound formation
         1. Temperature effect

Most of inorganic salts increase its
                         temperature.
solubility by increasing temperature Its
better to use hot/warm solution for filtering
  d       hi       i it t Exception :
and washing precipitate. E       ti
precipitates which are slighty soluble in
hot/warm solution (e.g. Hg2Cl2,
  g
MgNH4PO4) could be filtered after
previously stored in refrigerator.
                 2. Solvent effect
Most of inorganic salts dissolve in water but not in organic
solvent. Water molecule has a higher dipole moment and
could be attracted by cations or anions to form hydrate
ions Like hydrogen ion forming a hydrated ion H3O+, free
ions.                                           ion,H
energy released by ion–solvent interaction could increase
attractive ionic force to precipitate more solid lattice.
Crysstaline ions have no attractive force into organic
                       y
solvents, its solubility is smaller than those in water. In
chemical analysis, solubility difference could be used as
basic of separation of many compounds. Example : mixed
  f C (NO        d Sr(NO          be         t d in l
of Ca(NO3)2 and S (NO3)2 can b separated i solvent        t
mixture containing alcohol and eter, which yields a soluble
Ca(NO3)2 and an insoluble Sr(NO3)2.
              3.
              3 Common ion effect
Precipitate dissolves more easily in water than in solution
   t i i
containing common i          For        l in AgCl, l bilit
                      ions. F example, i A Cl solubility
product of [Ag+][Cl-] < its constant of solubility product (Ksp
AgCl = 1x10-10) in pure water, where [Ag+]=[Cl-] = 1x10-5 M;
                          water
when AgNO3 is added upon [Ag+] = 1x10-4 M, the [Cl-]
decreases into 1x10-6 M and reaction shifts to the right side
                        M,
as : Ag+ + Cl-                AgCl
                               p   p
There is salt addition to the precipitate while concentration
of Cl- decreases.
This technique of common ion addition is oftenly used for :
    1) completion of precipitation process
    2) precipitate washing with a solution containing
       common ion effect
If there is exessed common ions in a solution, the
precipitate solubility will be greater than estimated Ksp,
then common ion addition must be limited up to 10%.
Example : Calculate molar concentration of CaF2 dissolved
    in a) water; b) CaCl2 0,01 M; c) NaF 0,01 M. Ksp CaF2 =
    4x10 11 Hydrolysis is
    4 10-11. H d l i i neglected.
                                l t d
         CaF2(p)               Ca2+ + 2 F-
                    C 2+
a) Solubility s = [Ca2 ], then [F-] = 2s
  )S
    [Ca2+][F-] = Ksp
    s . (2s)2 = 4x10-11, then s = 2,1 x 10-4 M
b) [Ca2+] = (0,01+s) ; [F-] = 2s; then s = 3,2 x 10-5 M
c) [Ca2+] = s ; [F-] = (0,01+s); then s = 4 x 10-7 M
Common ions decrease precipitate solubility, [F-] effect is
greater than [Ca2+] effect.
   Solubility of Ag-halide in Na-halide at 18 oC
Solubility of AgX, M
     1

   10-2

                            AgCl
   10-4
                    g
                   AgBr
   10-6
            AgI
      8
   10-8


   10-10


            10-8          10-6     10-4     10-2       1

                                          Concentration of NaX, M
                   4. Common i effect
                   4 C       ion ff t

Most of precipitated compounds increase its solubility in a
solution containing substance not reacted to precipitate
ions. This phenomena is called as ionic activity effect or
                                   effect      example
diverse ion effect or neutral salt effect. For example, solubility
of AgCl and BaSO4 in KNO3 solution.
         [KNO3] (M)                 5
                          [AgCl]x10-5 M                  5
                                              [BaSO4]x10-5 M
        0,000 (water)         1,00                 1,00
           0,001
           0 001              1,04
                              1 04                 1,21
                                                   1 21
                                          Δ=12%           Δ=70%
           0,005              1,08                 1,48
           0,010
           0 010              1,12
                              1 12                 1,70
                                                   1 70
Molarity is an ionic activity occured in a high diluted solution,
more concentrated solution decreases faster the activity
coefficient (f), caused by greater different charge of
attractive ionic force. The ionic effectivity (in equlibrium state)
decreases as well, and addition of precipitate is required to
recover ionic activity.
        aAg+ . aCl- = Kosp (Ksp at a defined state of ionic activity)
        fAg+[Ag+] . fCl- [Cl-] = Kosp
        [Ag+][Cl-] = Kosp / fAg+ . fCl- = Ksp
The lower activity coefficient of both ion yields the greater
molar concentration of product. Increased BaSO4 solubility is
greater than solubility of AgCl, or ionic activity coefficient of
divalent ion is smaller than those of univalent ion.
In highly diluted solution the f = 1 and the Ksp = Kosp.
       Relative augmentation of the solubility
        of AgCl and BaSO4 in KNO3 solution
s/so

 1,7

 16
 1,6
                                             BaSO4
 1,5

 1,4

 13
 1,3

1,2                                           AgCl

1,1

10
1,0
        0,001          0,005          0,01       KNO3, M
Example : Calculate molar solubility of BaSO4 in KNO3
0,01 M solution using activity coefficient calculated by Debye-
Huckel equation. Solution of KNO3 (1:1) has ionic strength
equal to its molarity of 0,01M. Read from the table : fBa2+ =
0,667 ; fSO42- = 0,659.

Ksp = 1,00 x 10-10 / 0,667 x 0,659 = 2,27 x 10-10 = s2
then s = 1,51 x 10-5 M.

For it’s performed in neglected very low ionic solubility, the
activity effect is not a serious problem in the chemical analysis.
Precipitation process in high ionic concentration is quiet rare.
Hence, ionic activity influence gives no significant error.
                       5. pH effect
Solubility of weak acid salt depends on pH of the solution.
For example : oxalic, sulfide, hydroxyde, carbonate,phosphate.
Proton reacts with the anion to form weak acid, and increases
salt solubility.
a) Monovalent salt :         MA(p)                    M+ + A-
                             HA + H2O                 H3O+ + A-
Analytical concentration Ca = [A-] + [HA] = [A-]{[H3O+]+Ka}/Ka
Fraction of A- : [A-]/Ca = Ka / {[H3O+]+Ka = α1
                 [ ]            {[    ]
                 [A-] = α1 .Ca
Substituted to the Ksp = [ +][A-] = [M+]. α1.Ca
                       p [M ][      [ ]
Ksp/α1 = Keff = [M+].Ca
Keff = Effective equilibrium constant, varied on the pH
       for pH depends to α1.
b) Divalent salt :
      MA2                    M2+ + 2 A2-
H2A + 2 H2O                  2 H3O+ + A2-


    Kef = Ksp/α12 = [M2+] Ca2

[A2-] = α2 . Ca

α2 = Ka1.Ka2 / { [H3O+]+[H3O+]Ka1+Ka1Ka2 }


    Kef = Ksp/α2 = [M2+] . Ca
Molar concentration of iron species in a ferric hydroxyde
    solution as function of pH in room temperature


log C
                          Fe3+         [Fe3+][OH-]3 = Ksp = 2 x 10-39
   -1
                                       [FeOH2+][H+]/[Fe3+] = 9 x 10-4
             FeOH2+                    [Fe(OH)2+][H+]2/[Fe3+] = 5 x 10-7
    -2                                 [Fe2(OH)24+][H+]2/[Fe3+]2 = 1,1 x 10-3
    -3

    -4
               Fe(OH)2+
    -5

    -6

                          Fe2(OH)24+
    -7

         0            1          2            3             4           5
                                                                            pH
      Solubility of HgS at 20 oC as function of pH in a
        solution containing total sulfide H2S + HS-
log [Hg]total, log [Hg(HS)2], dst                                 log [H2S],log [HS-]

                    H2S                                 HS-
                                                                             -2
 -6



                                     Hg total                                -3
 -7


                                                                              4
                                                                             -4

 -8


                                                   Hg(HS)2                   -5
          Hg.HS2-                HgS22-
  9
 -9
      4           5          6            7        8          9         10
                                              pH
                  6 yd o ys s effect
                  6. Hydrolysis e ect
A weak acid salt dissolved in water changes pH of the solution.
     MA                           M+ + A-
     A- + H2O                     HA + OH-

     y
A very weak acid HA has lower Ka and an insoluble MA has
lower Ksp. At a lower [A-] hydrolysis reaction is completed.
   p              p
Depend to the Ksp it should show two extreeme conditions :
a) A very low solubility of precipitate where pH is not changed
   by hydrolysis reaction.
b) A high solubility of precipitate where OH- ion produced from
   water molecule is neglected.
             7. Metal hydroxyde effect
As occured by hydrolysis effect, when a metal hydroxyde
              water,                    changed.
dissolved in water the pH will be not changed
   M(OH)2                     M2+ + 2 OH-
   OH- + H2O                  H3O+ + OH-
       [M2+][OH-]2 = Ksp
       [H3O+][OH-] = KwK
       Charge balance : 2 [M2+] + [H3O+] = [OH-]
Molar l bilit        be l l td from th
M l solubility can b calculatd f                       ti
                                        these 3 equations.
When M(OH)2 dissolved then [OH-] increases, this anion will
 hif        dissociation reaction to the l f (H2O is f
shift water di    i i         i       h left              d)
                                                  i formed) :
   M(OH)2 (p)                 M2+ + 2 OH-
   2H2O                     H3O+ + OH-
Depend to solubility of OH- it should show two extreeme
conditions :
a) A very low solubility of precipitate where pH is not
   changed by the reaction.
   [H3O+] = [OH-] = 1,0 x 10-7
                     10
              Ksp = [M2+][OH-]2
              s = Ksp / (1,0 x 10-7)2
                        (1 0
b) A high solubility of precipitate increases [OH-], but
   [H3O+] is very low (neglected).
   Charge balance of these equation is either 2[M2+] =
   [OH-] or [OH-] = 2s
   [         [
              Ksp = [M2+][OH-]2 = s (2s)2
                         3
                    s=       Ksp/4
   8. Complex compound formation effect
Slightly soluble salt solubility is influenced by a compound
                                  cation.
forming complex to the metal cation Complexing ion could be
an anion or a neutral molecule which is common or diverse to
the precipitate; e.g. hydrolysis effect of complexing ion of OH-.
                 eg
Example : NH3 is used for separing Ag from Hg.
Ag+ + NH3                      Ag(NH3)+      K1 = 2,3 x 103
                                                  23
Ag(NH3)+ + NH3                 Ag(NH3) 2 + K2 = 6,0 x 103

Non-complexed silver fraction (β2) can be calculated as follow:
β2 = 1 / { 1 + K1[NH3] + K1K2[NH3]2 } = [Ag+] / CAg
Ksp = [Ag+][Cl-] = β2 CAg [Cl-]
Ksp/β2 = Kef = CAg [Cl-]
Example : Calculate molar solubility of AgCl in the solution
          0,01
  of NH3 0 01 M (as a final concentration of free ammonia
  solution). Ksp AgCl = 1,0 x 10-10. Stability constant K1 =
  2,3 x 103 and K2 = 6,0 x 103.

β2 = 1 / {1 + 2,3 + 103(10-2) + 1,4 x 107 (10-2)2 = 7,1 x 10-4
         {     ,         (        ,       (          ,
Keff = 1,0 x 10-10 / 7,1 x 10-4 = 1,4 x 10-7
s = CAg = [Cl-]
           [
s2 = 1,4 x 10-7 , and s = 3,4 x 10-4 M

In existed precipitating ions, most of precipitate could form
                  compound                step,
soluble complex compound. In the first step the solubility
decreases into the minimum caused by common ion effect,
but then it increases after formating complex compound in
enough quantity.
AgCl forms soluble complex with Ag+ and Cl- :
      AgCl + Cl-                AgCl2-
      AgCl2- + Cl-              AgCl32-
      A Cl + Ag+
      AgCl     A                A 2Cl+
                                Ag

     Curve          sol bilit    solution
     C r e of AgCl solubility in sol tion of NaCl and AgNO3
 (AgCl is more soluble in AgNO3 0,1 M and NaCl 1 M than in water)

        -1         -2   -3        -4         -4         -3   -2     -1

        log[Cl-]                        -3                        log[Ag+]

                                        -4

                                         5
                                        -5

                                        -6

                             log[Ag+]   -7        log[Cl-]
                 Titration Curve
  In a titration of 50 ml of NaCl 0.10 M solution with a solution
               0.10 M,
  of AgNO3 0 10 M calculate chloride ion concentrations
  during the titration and plot a titration curve of pCl vs ml of
  AgNO3. Ksp AgCl = 10 x 10-10.

Condition before titration : [Cl-] = 0.10 M or pCl = 1.00
                             [                 p

                          g
After addition of 10 ml AgNO3 :
                            Ag+    +    Cl-         AgCl (p)
               initial       1.00 mmol   5.00 mmol
               change       -1.0 mmol   -1.0 mmol
               equilibrium -             4.0 mmol
   [Cl-] = 4.00 mmol / 60.0 ml = 0.067 M or pCl = 1.17
After addition of 49.9 ml AgNO3 :
                             Ag+      +   Cl-          AgCl (p)
               initial         4.99 mmol 5.00 mmol
                 h
               change          4.99     l 4.99
                             - 4 99 mmol - 4 99 mmol l
               equilibrium -               0.01 mmol
                                          4
   [Cl ] = 0.01 mmol / 99 9 ml = 1 0 x 10-4 M or pCl = 4 00
   [Cl-] 0 01          l 99.9 l     1.0           Cl    4.00

At h Equivalent P i :
A the E i l     Point
                             Ag+     +     Cl-       AgCl (p)
              initial          5.00 mmol 5.00 mmol
              change         - 5.00 mmol - 5.00 mmol
              equilibrium -                    -
  [Ag+] = [Cl-] then [Ag+][Cl-] = Ksp = 1.0 x 10-10
              [Cl-] = 1.0 x 10-5 or pCl = 5.00
                  60.0
After addition of 60 0 ml AgNO3 :
                              Ag+    + Cl-          AgCl (p)
              initial         6 00 mmol 5 00 mmol
                              6.00        5.00
              change        - 5.00 mmol - 5.00 mmol
              equilibrium     1.00
                              1 00 mmol        -
   [Ag+] = 1.00 mmol / 110 ml = 9.1 x 10-3 M
                      2.04           10.00 2.04 7.96
              pAg = 2 04 or pCl = 10 00 – 2 04 = 7 96



Generally for halide salts :
              A + + X-
              Ag                         A X( )
                                         AgX (p)
Equilibrium constant : K = 1 / [Ag+][X-] = 1 / Ksp
Smaller th Ksp, hi h the K of a tit ti reaction.
S ll the K         higher th      f titration      ti
    ARGENTOMETRIC
    TITRATION CURVES
                                                                 AgI


   13                                                            AgBr

   11
             Ksp AgCl = 1 x 10-10                                AgCl
pX 10
 X                            12
             KspAgBr = 2 x 10-12
             KspAgI = 1 x 10-16
    8

    6

    4

    2
        10       20       30        40   50   60     70     80
                                                   ml of AgNO3
             Titration Feasibility
As in an acid-base titration, the feasible K value of precipitation
reaction could be calculated from its ionic concentrations.
Example :
   A solution of 50 ml of NaX 0.10 M is titrated with 50 ml of
   A NO3 0 10 M C l l           h       d Ksp f AgX h
   AgNO 0.10 M. Calculate the K and K of A X when
   addition of 49.95 ml of the titrant could achieve a completed
                        reaction,
   stoichiometric ionic reaction where the pX changed in 2 00  2.00
   units at the addition of 2 drops (0,10 ml) of titrant solution.
                                           salt,
Consider NaX as a complete dissolved salt the reaction is :
                       Ag+ + X-                AgX (p)
     1/Ksp.
K = 1/Ksp An addition of one drop of titrant performed before
the equivalent point consumes 4.995 mmol of Ag+, and
dissolved Ag+ needed to reach the equivalent point is :
(50 x 0.10) – 4.995 mmol = 0.005 mmol.
[X-] = 0.005 mmol / 99.95 ml = 5 x 10-5 M or pX = 4,30
a) If ΔpX = 2 00 then pX = 6 30 or [X-] = 5 x 10-7 M
             2,00            6.30                 7

The titration consumed 50.00 ml of titrant means one drop
          f il   ion
excess of silver i :
   [Ag+] = 0.05 x 0.10 mmol / 100.05 ml = 5 x 10-5 M.
                      5         7
   K = 1 / {(5 x 10-5)(5 x 10-7)} = 4 x 1010
   Ksp = 1 / (4 x 1010) = 2,5 x 10-11 M

b) If we consider ΔpX = 1.00 the K = 4 x 109 will be obtained.

  In the titration of X- with Ag+ ions, the ΔpX value at the equivalent
                                                    concentrations.
         point depends on the analyte and titrant concentrations
      These phenomena are equal to those in acid-base titration.
        Before equivalent point : smaller the [X-], higher the pX.
       At the equivelent point : smaller the [X-], smaller the ΔpX.
                                 ,
At a smaller titrant concentation, the lower
  curve branchement after the equivelent
  point as well as the lower of ΔpX at the
  point,
  equivalent point.

To obtain a good feasible titration of Cl-
 (analyte) with Ag+ (titrant) solutions, it is
 required both concentrations of analyte
 and titrant smaller than 0.10 M.
       Methods in
precipitatometric titration
p    p
    Argentometric method
    Mercurimetric method
    Kolthoff titration
Argentometric titration
Argentometry is the most usefull method of
precipitatometric titration, as it caused of very low
                                                 salts,i.e.
solubility product of halide (or pseudohalide) salts i e

Ksp AgCl = 1,82 . 10-10         Ksp AgCN = 2,2 . 10-16
Ksp AgCNS = 1,1 . 10-12         Ksp AgI = 8,3 . 10-17
Ksp AgBr = 5,0 . 10-13

There are 3 techniques of end point d
Th             h i       f d i determination  i i
• method of Mohr (indicator : chromic potassium)
• method of Volhard (indicator : ferric salt)
• method of Fajans (indicator : fluorosceince)
             ARGENTOMETRY               – MOHR
Mohr titration is used for determination of halide or
pseudohalide in a solution Chromate ion (CrO42-) is
                    solution.
added to serve as indicator. At the end point the
chromate ion is combined with silver ion to form the
sparingly soluble, red, silver chromate, Ag2CrO4.
         Ksp Ag2CrO4 = 1,2 . 10-12 mol3.L-3
                          12                L
         Ksp AgCl       = 1,82 . 10-10 mol2.L-2
 [ Please consider the stoichiometric unit of these ionic reactions ]


Although the solubility product constant (Ksp) of
AgCrO4 is close to the Ksp of silver (pseudo)halida,
  g                                   (       )
these silver salts have different solubility.
Mohr titration has to be performed at a neutral or
weak basic solution of pH 7-9 (or pH 6-10).

In a lower pH (acid) solution the chromate-
dichromate equilibrium decreases the sensitivity of
[CrO42-] then inhibits formation of Ag2CrO4
       ],
precipitate.
2 CrO42- + 2 H+                Cr2O72- + H2O
Chromate                     Dichromate

In a higher pH (basic) solution, Ag2O will be formed
as a black precipitate.
        Ag+ + Cl-              AgCl (p)
     Ag+ + CrO42-           Ag2CrO4 (p) red

   Solubility of Ag2CrO4 > Solubility of AgCl
      (8 4 x 10-5 M)
      (8.4                  (1 35 x 10-5 M)
                            (1.35


If Ag+ solution is added to a Cl- solution
containing of small quantity of CrO42-, then AgCl
will firstly precipitated, while Ag2CrO4 has not
yet, and [Ag+] increases progressively until
solubility product of the ions reach the value of
Ksp Ag2CrO4 (2,0 x 10-12) to form red precipitate.
   th     i l t     i t    A     Cl 5 00
At the equivalent point : pAg = pCl = 5.00

          [Ag ][CrO 2- 2 00
          [A +][C O42 ] = 2.00 x 10-1212

Then [CrO42-] = 2.00x10-12 / (1.0x10-5)2 = 0.02 M

This chromate concentration is too high as
                  f C O 2- disturb i
yellow colour of CrO42 di t b visual
  ll       l                             l
observation of Ag2CrO4 (red). In laboratory work
               ll t 0 005 /d 0.01 for
we use usually at 0.005 s/d 0 01 M f
minimizing titration error. And it could be
        t db       i di t bl k tit ti         b
corrected by an indicator blank titration, or by
AgNO3 solution standized to a pure chloride salt
(equal condition to the sample titration) .
(     l      diti   t th         l tit ti )
Mohr titration is limited at pH 6-10 (or pH 7-9).
Ionic     ti i b i         l ti    ill b
I i reaction in basic solution will be :
Ag+ + OH-            2AgOH             Ag2O + H2O
                        2-
In acid sol tion [CrO42 ] will decrease until onl a
        solution,            ill          ntil only
few of ionized HCrO4-, as the reaction will be
followed with the dichromate formation :
2H+ + CrO42-         2HCrO4-          Cr2O72- + H2O
     (chromate)                   (dichromate)
 the [CrO42-]
If            is too low ( < 0,005 M) then the
reaction needs excessive addition of [Ag+] to
precipitate Ag2CrO4; this might be a titration
error.
error
Dichromate ion, Cr2O72-, can not be used as
indicator as Ag2Cr2O7 precipitate is easily
dissolved in this ionic solution.
Mohr method can be used for determining Br-g
and CN- in a weak basic solution, but it is not
feasible for I- and CNS- as of the salt
precipitate adsorption.

Ag+ can not be titrated with Cl- solution using
indicator of CrO42-, because Ag2CrO4 will be
                               g
early obtained and slowly dissolves near the
equivalent point. This shlould be overcome by
  q         p                                    y
performing a back titration technique : to the
solution of Ag+ we add standardized excessive
              g
Cl- solution, then the excess of Cl- is titrated
with standardized Ag+ using indicator of CrO42-
                     g      g
.
          ARGENTOMETRY             - VOLHARD
Volhard titration is an indirect (back titration)
technique which is used if reaction is too slow or if
there is no appropriate indicator selected for
determining the equivalent point.
Titration principle :
Silver solution is added excessively to a (pseudo)halide
Br- + Ag+                             AgBr (precipitate)
        excess
After reaction has completed, the precipitate is filtered, then
the filtrate is titrated with a standardized thiocyanate solution.
Ag+ +      SCN-                          AgSCN (solution)
Fe(III) indicator reacts with thiocyanate ion to form a
red colour solution :
  Fe3+ + SCN-                          [Fe(SCN)]2+
The       ti        i     id      diti      in basic
Th reaction requires acid condition, as i b i
  solution the ferric ions form Fe(OH)3 precipitate.
                                   39
             Ksp Fe(OH)3 = 2.10-39 mol3L-3  3

             ( [ 3+] = 10-2 M is usually used)
               [Fe                       y     )

     p                                               p           q
Example : A solution of KBr is titrated with Volhard procedure requires
  addition of 100 ml of excessed AgNO3 0,095 M, then titrated with 18,3
  ml of KSCN 0,100 M using a Fe3+ indicator. Calculate Br- concentration in
              solution.
  the initial solution
Volhard method is very common for ion
determination based on Ag+ and Cl- reactions, as of
                          solubility.
very low precipitate salt solubility Acid condition is
required to avoid hydrolysis of Fe3+ indicator. In a
                                           (Hg2+
neutral solution, some of colour cations (H 2 ,
       l l i              f l         i
Co2+, Ni2+, Cu2+) will precipitate and disturb the
stoichiometric reaction.
Volhard method can be used as a direct titration of
Ag+ with CNS- as well as back titration of Cl-, Br- and
I- d t   i ti     These B - and I- can not be disturb
   determination. Th     Br     d          t b di t b
by the CNS- as solubility of AgBr is equal to those of
AgCNS, while solubility of AgI < solbility of AgCNS.
Titration error of Cl- determination would happen if
AgCl reacts with CNS- :
   AgCl(s) + CNS-               AgCNS + Cl-
Solubility of AgCNS < solubility of AgCl, then the
reaction in above will shift to the left side of the
reaction, and the result of Cl- analysis will be
lower.
lower But this could be prevented by filtering AgCl
precipitate or nitrobenzen (poison !) addition
before titration with CNS-. Nitrobenzen becomes a
lipid layer between precipitate and CNS- solution.
              ARGENTOMETRY –            FAJANS
Fajans titration use adsorption indicators, i.e. organic
compounds which is adsorbed into colloidal
precipitate surface during the titration processes.
Example : Fluoresence in form of its fluorescenate (yellowish
  green) anion react with Ag+ to form an intensive red
  precipitate which is adsorbed to AgCl precipitate surface
  caused by ionic pair interaction.
  Fi t step of titration tit i
  First t     f tit ti titrasi     Fi l step of titration : A +I d-
                                   Final t    f tit ti      Ag Ind

        Cl
        Cl-                               Ag+ Ind-
                    Cl-         Cl-
                                         Ag+     Ag+ Ind-
        Cl-               Cl-                  Ag+
                                Cl-
              Cl
              Cl-                                   Ag+ Ind-
                                       AgCl     Ag+
      AgCl                Cl-
               Cl-                                    Ag+ Ind-
Adsorption of colour organic compound on a
precipitate surface could induce intramolecular
           shift,                       colour.
electronic shift then change solution colour
This phenomenon is usually used for end titration
detection of silver salt precipitation.
d      i    f il      l      i i i
Before Eq.Pt. :
      (AgCl).Cl-       M+
      Primary layer    Secondary layer   Excess of Cl-


       q
After Eq.Pt. :
      (AgCl).Ag+       X-
      Primary layer
            y y        Secondary layer
                               y y       Excess of Ag+
                                                    g
A precipitate tends to adsorp easily ions formating
slighty soluble salts with ions of the precipitate lattice.
Then, Ag+ or Cl- could be more easily adsorb by AgCl
precipitate than by Na+ or NO3-. These anions in the
p    p             y
solution will be attracted by the precipitate to form an
                  layer.
ionic secondary layer

Fluorescein is a weak organic acid, f
Fl           i i        k     i              fluoresceinate i
                                  id forms a fl          i t ion
which is not adsorbed by colloidal AgCl precipitate during
excess of Cl-. B t in th solution with more excess of Ag+ th
           f      But i the l ti    ith                f A the
fluoresceinate anion will be adsorbed to form a Ag+ shielding
layer, f ll
l                d by h     the l ti      l     to i k
       followed b change th solution colour t pink.
The factors that must be consider in choosing
an adsorption indicator :
                       point, don t
1) At the equivalent point don’t let the AgCl
  precipitate grow too fast to form a big coagulant,
  for this could sharply decrease adsorptivity of
  the precipitate surface to indicator molecules.
                 happened
  But if it was happened, we can overcome by
  addition of dextrin molecules (as a protective
                     solution,
  colloid) into the solution so that increase
  particles dispersion, the colour change will be
  reversible and after equivalent point we can
  perform back titration using a standardized
             (Cl ) solution.
  chloride (Cl-) solution
2) Adsorption of indicator should be happen near
   and more fastly at the equivalent point. A bad
   indicator performance will result a too strong
   adsorption process resulting which can
   substitute ion adsorbed before the equivalent
   point.
3) The pH must be weel controlled for maintaining
   indication ionic concentration asam, in basic or
   acidic solution. For example, fluorescein (Ka =
   10-7) in a solution having pH > 7 will release
   small quantity of fluoreseinate ion, hence we
   can not observe colour change of the indicator.
   Fluorescein is used feasibly in a solution of pH
   7-10, difluorescein (Ka=10-4) is at pH 4-10.
4) It is recommended to choose ionic form of
  indicator with ionic charge in opposite of titrant
  ion.
  ion Adsorption of indicator can not happen
  before an excess of titrant ion.
  I a titration of A + solution with Cl- methyl violet
  In tit ti      f Ag    l ti    ith       th l i l t
  (as chloride salt of an organic base) could be
        d      indicator f h i
  used as an i di t of choice.
  Before an excess of Cl- in the solution that
  resulting a negative charge on colloid layer,
  cation should not be adsorbed. For this
  condition, we can use dichlorofluorescein, but it
  must be added just close to the equivalent point.
                     Adsorption Indicators

INDICATOR              ANALYTE         TITRANT   REACTION
                                                 CONDITION

Diklorofluorescein     Cl-             Ag+       pH = 4
Fluorescein            Cl-             Ag+       pH = 7 – 8
Eosin                  Br-, I-, SCN-   Ag+       pH = 2
Thorin                 SO42-           Ba2+      pH = 1,5 – 3,5
Bromcresol green       SCN-            Ag+       pH = 4 – 5
Methyl violet          Ag+             Cl-       acid solution
Rhodamin 6G            Ag+             Br-       HNO3 upto 0,3 M
Orthochrome T
O th h                 Pb2+            CrO 2-
                                       C O42         t l 0 02      l
                                                 neutral 0,02 M soln
Bromphenol blue        Hg22+           Cl-       solution of 0,1 M
                  List of Precipitation Titrations

ANALYTE                 TITRANT        INDICATOR               METHOD

Cl-, Br-                AgNO3          K2CrO4                  Mohr

Cl-,Br-,I-,SCN-
    Br I SCN            AgNO3          Adsorption              Fajans

Br-,I-,SCN-,AsO43-      AgNO3 + KSCN    Fe(III)                Volhard
                                                               (not filtered)
Cl-,CN-,CO32-,S2-       AgNO3 + KSCN    Fe(III)                Volhard
C2O42-,CrO42-                                                   (filtered)

F-                      Th(IV)         Alizarin                Fajans
SO42-                   BaCl2          Tetrahydroxyquinoline   Fajans
PO43-                   PbAc2          Dibromofluorescein      Fajans
C O42
CrO 2-                  PbAc
                        PbA 2          Fluorescein
                                       Fl       i              Fajans
                                                               F j
Ag+                     KSCN           Fe(III)                 Volhard
Zn2+                    K4Fe(CN)6      Diphenylamine           Fajans
Hg22+                   NaCl           Bromphenol blue         Fajans
         MERCURIMETRIC TITRATION
  Hg2+ + 2 Cl-              HgCl2 (as for other halides)

When halide ions is titrated with mercuric nitrate solution,
[Hg2+] is not found at the equivalent point caused of
precipitation of HgCl2 during the titration process.
After equivalent point, [Hg2+] increases, react with indicator
to form a Hg-indicator complex, e.g. Nitropruside form
white precipitate, acid solution of diphenylcarbazide or
diphenylcarbazon in forms intensive violet colour solution.
   Mercurimetric titration requires a blanc titration:
   0,17                  0,1                        0,05 N.
   0 17 ml of Hg(NO3)2 0 1 N for 50 ml of HgCl2 0 05 N
This blanc titration volume varies as [HgCl2]EquivalentPoint
For the excess of [Hg2+] reacts with HgCl2 :
        HgCl2 + Hg2+                2 HgCl+
                      KOLTHOFF TITRATION
Determination of Zn2+ (as titrant) which is precipitated with standard solution of
K-Ferrocyanide
         2 K4Fe(CN)6 + 3 Zn2+
               ( )                                      K2Zn3[Fe(CN)6]2 + 6 K+
                                                             [ ( )
         potassium ferro (II) cyanide          potassium zink ferro (II) cyanide

                                                                    uranylnitrate,
Titration end point is detected by using external indicator such as uranylnitrate
ammonium molybdate, FeCl3, etc, which needs a special technical skill. So it is
better to use internal indicators such as diphenylamine, diphenylbenzidine,
                sulfonate, etc.
diphenylamine sulfonate etc
A redox reaction of Fe2+      Fe3+ has rduction potential (at 30 oC) as follow :
         E = Eo + 0,060 log [Fe(CN)63-] / [Fe(CN)64-]

Acidic solution of ferro-ferric cyanide has much lower reduction potential
compared to those required to oxydize the indicator, forms intensive coloured of
oxidized form. When Zn2+ is added to this solution then a Zn-ferrocyanide will be
formed, followed with increasing reduction potential for Fe(CN)64- removed from
the solution. After Fe(CN)64- completely reacted, a sharp increase of reduction
                       ( )         p     y       ,      p
potential which is followed by appearing blue colour of oxidized form of indicator,
caused by excess of Zn2+.in the solution.
The END

								
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