Chemistry 221
Chapter 13: Precipitation Titrimetry
Precipitation titrimetry, which is one of the oldest analytical techniques, is based on
reaction that yield ionic compounds of limited solubility. The number of precipitating
agents that can be used is limited because of the slow rate of formation of most
precipitates. The most widely used precipitation reagent is silver nitrate, and
precipitation titrimetry based on silver nitrate are known as argentometric methods.
Titration curves for precipitation reactions are very similar to those involving strong
acids and strong bases. Consequently, they can be derived in the exact same way. One
need only substitute the solubility product of the precipitate for the ion-product constant
for the water. The following diagram illustrates such a titration curve.
Clearly, one can see the resemblance of this curve to that of an acid/base titration shown
below.
Both have a sharp inflection point. Furthermore, it is evident that at pre-equivalence and
post-equivalence points, the p-function does not chance very much. At the equivalence
point in contrast, a small volume of titrant corresponds to a large p-function change.
Figure 3 below shows the effect of product solubility on the sharpness of the endpoint.
The change in the p-function at the equivalence point becomes greater as the solubility
product becomes smaller. In other words  p-function increases as the reaction between
the analyte and the silver nitrate becomes more complete.
The endpoint produced by a chemical indicator usually consists of a color change.
Once again, we see the relation between argentometric titrations and acid/base titrations,
as the requirements for an indicator for the titrations are analogous. These requirements
are namely: (1) the color change should occur over a limited range in the p-function of the
reagent or the analyte, and (2) the color change should take place within the steep portion
of the titration curve for the analyte. Three indicators that have widespread use in
argentometric titrations are the Chromate ion, Flourescein and Iron(III) ion.
The chromate ion is employed as an indicator in the Mohr method. Sodium chromate
can serve as an indicator for the argentometric titrations of chlorine, bromine and cyanide
ions, by reacting with silver to form silver chromate (Ag2CrO4). The brick red Ag2CrO4
precipitates at the equivalence point region and thus is a useful indicator. The Mohr
method has an inherent systematic error that is correctable. At reagent concentrations
lower than .1M, the chromate ion imparts an intense yellow color to the solution that
hinders the identification of the red silver chromate. As a consequence, excess silver
nitrate is required before precipitation can be detected and as a result, the volume is
skewed. The correction that is often made to account for this error is making a blank
titration of a chloride free suspension of calcium carbonate. This volume can then be
used to adjust the experimental results. The Mohr method will follow these reactions:
Ag+ + Cl-
AgCl (s) white
2Ag+ + CrO4-2
Ag2CrO4 (s) red
The second popular indicator is Flourescein, an adsorption indicator used in the Fajans
method. An absorption indicator is an organic compound that tends to be absorbed onto
the surface of the solid in a precipitation titration. In the ideal scenario, the absorption
occurs near the equivalence point and results in a color change. Flourescein is a widely
used absorption indicator.
HO
OH
O
C OOH
In aqueous solution partial dissociation into H3O and the negatively charged Flourescein
ion. The Flourescein forms a bright red silver salt. These types of indicators give a rapid,
accurate and reliable titration endpoint.
The third indicator used is Iron(III) ion, used in the Volhard method. In this method,
silver ions are titrated with a standard solution of thiocyanate ion, where Fe3+ serves as
the indicator imparting a red color to the solution. This titration must be carried out in an
acidic solution lest the Fe(III) precipitates out as a hydrated oxide. The reaction that
accompany the Volhard method are:
Ag+ + ClSCN- + Ag+
Fe3+ + SCN-
AgCl (s) white
AgSCN (s) white
Fe(SCN)2+ red
The most important application of the Volhard method is for determing the presence of
halide ions. An excess of silver nitrate is added to the sample and back-titrated with a
standard thiocyanate solution.
Chapter 13 Questions
13-11 Titration of a .485 g sample by the Mohr method required 36.8 ml of standard
.1060M AgNO3 solution. Calculate the percentage of Cl in the sample.
13-12 A .1064 g sample of a pesticide was decomposed by the action of sodium biphenyl
in toluene. The liberated Cl was extracted with water and titrated with 23.28 ml of
.03337 M AgNO3, using an adsorption indicator. Express the results of this analysis in
terms of percent aldrin C12H8Cl6
13-13 A 100 ml sample of brackish water was made ammoniacal and the sulfide it
contained titrated with 8.47 ml of .01310 M AgNO3 . The net reaction is
2Ag+ + S-
Ag2S
Calculate the parts per million of H2S in the water.
13-14 A 2000 L water sample was evaporated to a small volume and treated with an
excess of sodium tetraphenylboron. The precipitated KB(C6H5)4 was filtered and then
redissolved in acetone. The analysis was completed by a Mohr titration, with 37.90 ml of
.03981 M AgNO3 being used. The net reaction is
KB(C6H5)4 + Ag+
AgB(C6H5)4 + K
Express the results of this analysis in terms of ppm of K.
13-21 A carbonate fusion was needed to free the Bi from .6423g sample containing the
mineral eulytite (2Bi2O3-3SiO2). The fused mass was dissolved in dilute acid, followed
which the Bi3+ was titrated with 27.36 ml of .03369 M NaH2PO4. The reaction is
Bi3+ + H2PO4-
BiPO4 + 2H+
Calculate the percentage purity of eulytite 1112 g/mol in the sample.
13-22 The theobromine (C7H8N4O2) in a 2.95 g sample of ground cocoa beans was
converted to the sparingly soluble silver salt C7H8N4O2Ag by warming in an ammonical
solution containing 25ml of .0100M AgNO3. After reaction was complete, all solids were
removed by filtration. Calculate the percentage of theobromine 180.1 g/mol in the
sample if the combined filtrate and washings required a back-titration with 7.69 ml of
.0108M KSCN.
13-23 A 20 tablet sample of soluble saccharin was treated with 20ml of .08181 M
AgNO3. After removal of the solid, titration of the filtrate and washings require 2.81 ml
of .04124 M KSCN. Calculate the average number of mg of saccharin 205.17 g/mol in
each tablet.
13-26 A 5.00nl aqueous suspension of elemental selenium was treated with 25 ml of
ammoniacal .0360 M AgNO3. The reaction is
6Ag(NH3)2+ + 3Se + 3H2O
2Ag2Se + Ag2SeO3 + 6NH4-
After this reaction was complete, nitric acid was added to dissolve the Ag2SeO3. The Ag+
from the dissolved Ag2SeO3 and the excess reagent required 16.74 ml of .01370 M
KSCN in a Volhard titration. How many milligrams of Se were contained in each ml of
sample.
13-27 A 2.4414 g sample containing KCl, K2SO4 and inert materials was dissolved in
sufficient water to give 250 ml of solution. A Mohr titration of a 50.00 ml aliquot
required 41.36 ml of .05818 M AgNO3. A second 50.00 ml aliquot was treated with
40.00 ml of .1083 M NaB(C6H5)4. The reaction is:
NaB(C6H5)4 + K+
KB(C6H5)4 + Na+
The solid was filtered, redissolved in acetone, and titrated with 49.98 ml of AgNO3
solution. Calculate the percentages of KCl and K2SO4 in the sample
13-28 A 1.998 g sample containing Cl- and ClO4- was dissolved in sufficient water to
give 250.0ml of solution. A 50.00 ml aliquot required 13.97 ml of .08551 M AgNO3 to
titrate the Cl-. A second 50.00 ml aliquot was treated with V2(SO4)3 to reduce ClO4- to
Cl- :
ClO4- + 4V2(SO4)3 + H2O
Cl- + 12SO4- + 8VO2+ + 8H+
Titration of the reduced sample required 40.12 ml of the AgNO3 solution. Calculate the
percentage of Cl- and ClO4- in the sample.
Chapter 13 Answers to Questions
13-11
36.8ml*.1060 * 1 / 1*.03543 / 1
* 100% =
.485g
28.5% Cl
13-12
23.28ml*.03337 * 1 / 1 * 1 / 6
* 100% =
.1064
44.41%
13-13
8.47ml*.01310 * 1 / 2*.03408 / 1
* 106ppm =
100mlH 2O * 1gH 2O
18.9ppm
13-14
37.90ml*.03981 * 1 / 1 * 39.098
=
2.000
29.50mg
13-21
27.36ml*.03369 * 1 / 1 * 1 / 4 * 1112
. g / mM
* 100% =
.64223
39.90%
13-22 AgNO3 consumed: 25 * .0100 - 7.69*.0108 = .1169
.1669 * 1 / 1*.1801
* 100% =
2.95
13-23 AgNO3 consumed: 20*.08181 - 2.81*.04124 = 1.52
1.02%
152
. * 1 / 1*.20517 * 10^ 3mg / g
=
20
15.60 mg/Tablet
13-26 AgNO3 consumed: 25*.0360 - 16.74*.01370 = .6707
.6707 * 2 / 4 * 3 / 2 * 78.96
=
5.00ml
13-27 mM KCl:
mM K+:
7.94mg/ml
41.36 * .05818 * 1/1 = 2.406
49.98 * .05818 *1/1 = 2.908
2.908-2.406 *1/1 = .2508 mM K2SO4
2.406*.07455
* 100% 
2.4414 * 50 / 250
36.73% KCl
.2508*.17426
* 100% 
2.4414 * 50 / 250
8.951% K2SO4
13-28 mM Cl = 13.97 * .08551 *1/1 = 1.195mM
mMCl + mM ClO4 = 40.12*.08551*1/1*1/1 = 3.431mM
mM ClO4 = 3.331 - 1.195 = 2.236 mM ClO4
1195
. *.03543
* 100% 
11998
.
* 50 / 250
10.59% Cl-
2.236*.09945
* 100% 
11998
.
* 50 / 250
55.65% ClO4-