1. The document describes the argentometric titration method to determine the chloride ion concentration in water. Silver nitrate (AgNO3) is used as the titrant and reacts with chloride ions to form insoluble silver chloride precipitate. 2. The procedure involves standardizing the silver nitrate solution against a standard sodium chloride solution. Then, an unknown water sample is titrated with the standardized silver nitrate solution. 3. The chloride ion concentration in the water sample is calculated based on the volume of silver nitrate used and its known normality. The amount of chloride ion present per liter of sample water is reported as the result.

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Determination of chloride ion in water by argentometric method
Dr. Mausumi Adhya
HOD and Associate Professor
Supreme Knowledge Foundation, West Bengal, India

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Theory
Argentometric titration is the most important precipitation process in titrimetric analysis. The
amount of chloride ion present in water is estimated by silver nitrate (AgNO3) using potassium chromate
(K2CrO4) as indicator. The reactions occurred in this titration are
Ag++ Cl-= AgCl (white precipitate)
Ag++ CrO4
2-= Ag2CrO4 (light brown precipitate)
When the ionic product exceeds the solubility product, the precipitation takes place. The
solubility product of AgCl is lower than Ag2CrO4. Hence as long as Cl- is available the less soluble AgCl
is precipitated out. After removing of all Cl- ions, a slight excess of AgNO3 produces insoluble Ag2CrO4.
AgNO3 is a secondary standard and hence primary standard NaCl is used to find out the exact strength of
AgNO3.
Materials
1. Apparatus: Burette with burette stand, pipette, conical flask, beaker, volumetric flask, reagent bottle,
measuring cylinder, dropper
2. Chemicals: NaCl, AgNO3, K2CrO4, Sample water
Procedure
1. Preparation of reagents
(I) Preparation of 250 ml N/50 NaCl
Gram equivalent weight of NaCl = Molecular weight of NaCl/1= (23+35.45)/1= 58.45 gm
To prepare 250 ml N/50 NaCl gm equivalent required=[58.45×250]/[1000×50]=0.292 g
0.292 gm of NaCl is accurately weight out into a 250 ml volumetric flask and the solid is dissolved in
distilled water. The water is added upto the mark. Finally the solution is made uniform by shaking.
(II) Preparation of 1000 ml N/50 AgNO3
Gram equivalent weight of AgNO3 = Molecular weight of AgNO3/1= 107.87+14+16×3=169.87 gm
To prepare 1000 ml N/50 AgNO3 gm equivalent required=169.87/50= 3.4 g
About 3.4 gm AgNO3 is taken in 1000 ml volumetric flask and dissolved in distilled water. The water is
added upto the mark. Finally the solution is made uniform by shaking. The solution is then transferred
into an amber-coloured glass bottle.
(III) 5% aqueous solution of potassium chromate solution
About 5 gm K2CrO4 is dissolved in 100 ml distilled water.
2. Standardization of AgNO3 by standard NaCl

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10 ml standard NaCl solution is pipetted out in a 100 ml conical flask. Few drops of chromate indicator
are added to it. The mixture is titrated keeping the flask against white background with AgNO3 by
constant stirring. Red colour precipitate is formed for each drop addition of AgNO3 and disappeared.
AgNO3 solution is added until a light brown colour persists in the solution. Experiment is repeated three
times.
3. Estimation of chloride ion in sample water
25 ml sample water is pipetted out in a 100 ml conical flask. Few drops of chromate indicator are added
to it. The mixture is titrated with standard AgNO3 solution running from burette with constant stirring
until pale brown colour persists in the solution. Experiment is repeated three times.
Results
Table 1: Standardization of AgNO3 by standard NaCl
Number of
observation
Volume of
NaCl (ml)
Burette reading Mean volume
of AgNO3
(ml)
Strength of
NaCl ( N)
Strength of
AgNO3 (N)
Initial Final Actual
1
10 X N/50 S
2
3
Strength of AgNO3 = (10×N/50)/X =S (N)
Table 2: Estimation of chloride ion in sample water
Number of
observation
Volume of
sample water
(ml)
Burette reading Mean volume
of AgNO3
(ml)
Strength of
AgNO3 (N)
Amount of
chloride in
g/L
Initial Final Actual
1
25 Y S
2
3
Calculation
AgNO3 + Cl- = AgCl + NO3
-
Here 1 equivalent of AgNO3 reacts with 1 equivalent of Cl-.
Therefore,
1000 ml 1(N) AgNO3 = 35.46 g of chloride
Y ml S (N) AgNO3=]35.46×Y×S]/1000 g chloride
Hence 25 ml sample water contains [35.46 × Y × S]/1000 g chloride
Therefore, amount of chloride ion present in 1 L sample water = [35. 46× Y × S]/25 g

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Conclusion
The amount of chloride ion is present in sample water sample is…………………….. g/L
Precautions
1. Solutions must be prepared by distilled water.
2. Silver nitrate solutions should be prepared by wearing glavs.
3. Silver nitrate solutions should be kept in dark into an amber-coloured glass bottle to avoid the
decomposition of silver nitrate.
Viva Voce
1. Why the titration is called argentometry titration?
Answer. In this titration the amount of chloride ion is estimated by precipitation with standard AgNO3
solution. Hence the titration is called argentrometry titration.
2. What is the principle behind silver chloride first precipitated out and then silver chromate?
3. Why silver nitrate is used as a secondary standard solution?
Answer. AgNO3 is a secondary standard as expose in light it decomposes to silver, which appear black.
2AgNO3= 2Ag+2NO2+O2
4. Why silver nitrate solution should be kept in amber-coloured bottle?
Answer. Silver nitrate solutions should be kept in dark into an amber-coloured glass bottle to avoid the
decomposition of Silver nitrate.
5. Write all reactions occurred in estimation of chloride ion in water.
Amount of chloride ion present in water
= [Volume of AgNO3 required for titration of sample water × strength of AgNO3 ×35.46/ volume of sample water
taken] g/L