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2. Volumetric Precipitation Titration
Precipitation titrations are volumetric methods based on the formation of a
slightly soluble precipitate. The substance is determined simply by converting it
into an insoluble form of known composition by adding a standard solution of the
titrant.
Solubility product
The factor governing the completeness of a precipitation reaction is the solubility
of the precipitate formed.
Let us consider what happens when the sparingly soluble salt AB comes in contact
with water:
AB →A+ + B−
(1)
+
−
AB (solid) → A + B (solution)
(2)
The equilibrium constant for this reaction is:
K
[A ] [B- ]
[AB]
For slightly soluble compounds, this equilibrium expression becomes.
Ksp = [A+] [B-]
The solubility product constant expression for a slightly soluble compound is
“the product of the concentrations of the constituent ions, each raised to the
power that corresponds to the number of ions in one formula unit.”
Let us consider in the same way the saturated solution of the sparingly soluble salt
Xm Yn which dissociates into m cation, Xn+ and n anions, Ym−. The equilibrium for this
saturated solution can be represented by the equation:
[Xn+]m [Ym-]n = SXm Yn
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1. Calculation of solubility product from
solubility
1. If the solubility of AgCl is 0.0015 g/l what is the
solubility product. (M.wt =143)
 Molar concentration of saturated solution =
0.0015/143= l.05xl0−5 g.mol/L
[AgCl] = [Ag+] = [Cl−]
SpAgC1 = [Ag+][Cl−] = (l.05xl0−5) (l.05xl0−5) = l.05xl0−10
SpAB = S2
(S= molar solubility)
2. Calculate the solubility product of Pb3(PO4)2
(solubility= 0.00014 g/1). (M.wt= 811.7 )
2. Calculation of solubility
from solubility product.
Calculate the solubility of
silver sulphide in pure
water.
SPAg2S = 6x10-50
Ag2S ⇋ 2 Ag+ + S2Ag2S = [Ag+]2[S2-] = 6x10-50
1 mole of Ag2S ⇋ 2 Ag+ + S2A mole
⇋2A + A
2
SPAg2S = [2A] [A]
6x10-50 = 4 A3
3
6
𝐴 = √4 𝑥10−50
Molar concentration =0.00014/811.7 = 1.4xlO−7 g.mol
If we consider we have A mol/1.
Pb3(P04)2 → 3Pb2+ + 2P043−
A mole
3A
2A
3
2
Pb3(P04)2=[3A] [2 A]
= [3(1.7xl0-7)]3 [2(1.7xl0-7)]2 = 1.5x 10−32
Fractional Precipitation
When a precipitating agent such as Ag+ is added to a solution containing two anions
e.g. chloride and iodide both of which form slightly soluble salt with Ag+. The
question which arises is: Which salt will precipitate first?
The solubility product of AgCl and Agl is l.lxl0−10 and 1.7x10−16 respectively.
It is the initial concentration of iodide is 0.1 M, upon addition of silver ion
precipitation of silver iodide commences when the molar concentration of silver ion
SpAgI 10 16

[Ag ] 

 10 15
is.
1
[I ]
10
And for chloride if the initial concentration is 0.1M silver chloride will begin to
precipitate when the molar concentration of silver ion is:
SpAgCl 10 10
[Ag  ] 

 10 9
1
[Cl ]
10
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So, when silver ion concentration reaches 10−15 iodide will be precipitated while
chloride will not until silver ion concentration is raised and by 10−9. Such rise in
silver ion concentration will only reach when practically all iodide is precipitated as
silver iodide. In fact, both Agl and AgCl will be precipitated simultaneously when:
SpAgI SpAgCl
[Cl- ] SpAgCl 10 10
[Ag  ] 

..........
....

 10 6
16
[I ]
[Cl ]
[I ] SpAgI 10
i.e. when the ratio of I− : Cl−= 1:106
This type of successive precipitation using the same precipitating agent is
known by fractional precipitation.
Factors Affecting Solubility of the Precipitate
1- Common ion effect on solubility (C.I.):
The common ion causes decrease of the solubility. If we have a saturated
solution of sparingly soluble salt e.g. AgCl . if Ag+ are added (from AgNO3) excess Ag+
will disturb the equilibrium and this ions will combine with Cl− to form precipitate AgCl
till equilibrium is again reached where [Ag+] [Cl−] = SpAgCl i.e. the solubility decreases. A
similar effect occurs if excess Cl− is added to the solution.
Example: Calculate the solubilities of AgCl in 0.001 M, 0.01M and 0.1MKCl.
In 0.001 MKCl [Cl−] = 10−3
in 0.01 M KC1 solution [Cl−] = 10−2
in 0.1 KC1 solution
10
+
−2
In a saturated solution of AgCl 1.1 xlO- =[Ag ][10 ]
the solubility will be
+
−
8
SpAgCl= [Ag ] [Cl ]
[Ag]1010−9
1.1 x l0−10 = [Ag+][10−3]
molar solubility of AgCl = 10−8
In pure water 1.05x
−7
[Ag+] 10
10−5
molar solubility of AgCl = 10−7
2- Increased solubility by complex formation: The solubility can be increased by
including an ion which forms a complex with one of ion components of the
precipitate, e.g. addition of CN- to AgCl ppt. will dissolve the ppt. due to the
formation of complex ion [Ag(CN)2] −.
Also silver ion forms complex with ammonia.
Ag+ + 2NH3
→
[Ag(NH3)2]+
3- Effect of temperature on solubility: Increase of temperature mostly increases
the solubility of precipitate.
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4- Diverse ion effect: The presence of diverse salts will generally increase the
solubility of precipitates, for example BaSO4 in presence of NaNO3 (diverse ion).
5- Effect of acids: The solubility of sparingly soluble sulphates (e.g. those of
barium, strontium and lead) increased in acids.
6- Effect of solvent: The solubility of most inorganic compounds is decreased by
the addition of organic solvents such as methanol, ethanol and acetone. For example,
the addition of about 20 vol% ethanol renders the solubility of lead sulphate
practically negligible.
Determination of end points in precipitation reaction
Many methods are utilized in determining end points in these reactions, but only
the most important will be mentioned here.
A- Formation of a coloured
Used for determination
Indicator
Medium
Titrant
precipitate (Mohr method):
Chloride and bromide (can’t used for CN- and I-?)
chromate ion CrO42- (brick red silver chromate at e.p)
neutral or faintly alkaline…??
AgNO3...... Argentometric titration
Interferences and limitations of Mohr method:
1. Only in neutral or faintly alkaline solution with pH values from about 6 to 10.
In acid solution
In acid solution, the CrO42− concentration is
greatly decreased
2H+ + 2 CrO42− → 2 HCrO4− → Cr2O72− + H2O
and dichromate is formed whose silver salt is
soluble. Therefore, no indicator precipitate
forms.

In alkaline medium
Silver will precipitate as
its oxide:
2 Ag+ + 2 OH− → 2 AgOH
→ Ag2O + H2O
Therefore, the halide solution should be neutralized before titration if
necessary, by adding NaHCO3 or dilute HNO3, as the case may be.
2. Cations which give insoluble chromate e.g. barium ions. They must be
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absent or removed before the titration.
3. The reverse titration of silver ion with chloride ion using chromate as
indicator is not feasible, the flocculated Ag2CrO4 formed initially, reacts
slowly with chloride especially near the end point.
4. Titration of iodide; and of thiocyanate is not successful because silver iodide
and silver thiocyanate adsorb chromate ions so strongly that a false and
indistinct end point is obtained.
B- Formation of a soluble coloured compound (Volhard method):
Used for determination
Indicator
Medium
Titrant
chlorides, bromides, and iodides indirect method
ferric nitrate or of ferric ammonium alum
Nitric acid solution
standard potassium or ammonium thiocyanate
Excess of standard silver nitrate solution is added, and the excess is back-titrated with
standard thiocyanate solution.
Ag+ + SCN− → AgSCN
Fe+++ + SCN− → [FeSCN]++ ( reddish color)
For the chloride estimation, we have the following two equilibria during the titration of
excess of silver ions:
Ag+ + Cl− → AgCl
Ag+ + SCN− →AgSCN
The two sparingly soluble salts will be in equilibrium with the solution hence:
SAgCl
[Cl  ]
1.2x10 10


 170
[SCN - ] SAgSCN 7.1x10 13
When the excess of silver has reacted, the thiocyanate may react with the
silver chloride, since silver thiocyanate is the less soluble until the ratio [Cl−] / [SCN−] in
the solution is 170.
AgCl + SCN− →AgSCN +Cl−
It is therefore absolutely necessary to prevent the reaction between the
thiocyanate, and the silver chloride. This may be affected in several ways:
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a) The silver chloride is filtered off before back titration.
b) After the addition of the silver nitrate, the suspension is boiled for about 3
minutes, cooled and then titrated immediately,
c) An immiscible liquid is added to "coat" the silver chloride particles and thereby
protect them from interaction with the thiocyanate. The most successful liquid
is nitrobenzene.
C- Use of adsorption indications (Fajan method):
Used for determination
Indicator
(termed adsorption
indicators)
chlorides, bromides, and iodides and thicyanate
either acid dyes, such as fluorescein, eosin, rose bengal,
dichloro-fluorescein, and di-iodo-diemthyl-fluorescein; or
basic dyes, such as rhodamine (6G), which are applied as the
halogen salts.
Titrant
AgNO3...... Argentometric titration
At the equivalence point the indicator is adsorbed by the
precipitate, and leads to a substance of different colour,



Before the e.p.: When a chloride solution is titrated
with a solution of silver nitrate, the precipitated silver
chloride adsorbs chloride ions; forming “primarily
adsorbed layer”, and it will hold by secondary
adsorption oppositely charged ions present in solution
(Fig. 2a).
at the equivalence point silver ions are present in
excess; these will now be primarily adsorbed, and
nitrate ions will be held by secondary adsorption (Fig.
2b).
If fluorescein is also present in the solution, the
negative fluorescein ion, which is much more strongly
adsorbed than the nitrate ion, is immediately adsorbed
with the formation of a pink complex of silver and a
modified fluoresceinate ion in the surface with the first
trace of excess of silver ions (Fig. 2c).
N.B. The indicator ion must be of opposite charge to
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the ion of the precipitating agent.
For chlorides, bromides, and iodides
and thicyanate
Fluorescein
,
Dichlorofluorescein
This is a very weak acid
The optimum pH range is between 7 and
10.
For chlorides, bromides, and iodides
and thicyanate
Stronger acid and may be utilised in
slightly acid solution of pH greater
than 4.4.
bromides, and iodides and thicyanate
(NOT FOR Cl-)
Stronger acid and can used down to a
pH of 1.2. The colour change is
sharpest in an acetic acid
Eosin
d- Turbidity method:

or
Liebig's method for cyanides. When a solution of silver nitrate is added to a
solution containing cyanide ions (e.g. alkali cyanide) a white precipitate is first
formed, but on stirring it re-dissolves owing to the formation of soluble stable
complex cyanide,
AgNO3 + 2 KCN → KAg(CN)2 + KNO3
Ag+ + 2 CN− —— [Ag(CN)2] −
When the above reaction is complete, further addition of silver nitrate solution
yields the insoluble silver argento cyanide.
Ag+ + [Ag(CN)2] −→ Ag[Ag(CN)2]
The end point of the reaction is therefore indicated by the formation of
permanent turbidity.
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In Denige's modification of Liebig method, iodide ion is used as indicator and
aqueous ammonia is introduced to solubilise the silver cyanide:
Ag[Ag(CN)2] + 4NH3 —— 2 [Ag(NH3)2]+ + 2CN−
The iodide ion and ammonia solution are added before the titration, the formation
of silver iodide (as turbidity) will indicate the end point.
[Ag(NH3)2]+ + I− → Agl + 2NH3

Application of Precipitimetry
1- Determination of zinc ions by precipitation with potassium ferrocyanide:
Zinc ions in neutral or acid medium react with potassium ferrocyanide solution
to form a very sparingly soluble potassium zinc ferrocyanide (at 60°-80°C)
3Zn++
+ 2K4Fe(CN)6
→
K2Zn3[Fe(CN)6]2
External indicators as uranyl nitrate and ammonium molybdate.
Titrant: ferrocyanide solution
Colour at e.p.: light brown colour
Or internal indicator method using diphenylamine. Diphenylamine is a redox
indicator; it changes its colour according to the ratio of the concentration of the
oxidised form (ferricyanide) to the reduced form (ferrocyanide) of redox system.
The colour change will be from blue violet (oxidised form) to colourless (reduced
form of indicator).
2- Determination of barium and of sulphates:
Weakly acidified barium solution (with HCl) gives a red precipitate of barium
rhodizonate with a freshly prepared solution of the sodium salt of rhodizonic acid.
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The solubility of the red compound is much greater than that of barium sulphate,
hence upon the addition of a sulphate solution to a solution of barium salt, the barium
will be completely precipitated when the red colour changes to colourless.
3- Determination of mixtures:
a) Chloride-iodide mixture:
By direct titration using Fajan’s method fluorscein indicator for total halide then
the iodide is determined using eosin.
b) Chloride cyanide, bromide cyanide or iodide cyanide mixtures:
Such mixtures are analysed using a combination of Liebig's and Volhard's
method. Liebig's procedure determines the cyanide only while the total is determined
by Volhard's
2CN− + Ag+ →
[Ag(CN)2] − Liebig's
[Ag(CN)2] − + Ag+ ——
Ag-Ag(CN)2 (Volhards)
So the volume of standard silver nitrate equivalent to cyanide by the Liebig's
method is ½ the volume of Volhards.
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