8. volumetric4(precipitation1)

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1st March 2011 1DrSabiha/CHM421/Dec 10-Apr11

TITRIMETRIC ANALYSIS

Topic 4PRECIPITATION TITRATION


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Fajans Method The end-point technique based on

adsorption indicator .

Limitation The samples should not be acidic.

Application in the determination ofneutral to slightly acidic samples only.


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At end point (Fajans Method) :

AgCl precipitates will adsorb excess Ag+ and that this adsorbed Ag+ will contributeexcess positive charges on surface ofprecipitate. The excess positive charges are

neutralized by a negatively-charged ionspresent in solution usually a dye (D)(indicator) in the above titration.

if AgD is coloured, the white precipitatewill turn colour just past the end point.


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The best indicator is FLUORESCEIN (weak acid)Indicator ionising slightly in water to form

flouresceinate :

HFn H+ + Fn- yellow green C20H12O5

at the end point :Ag+ + Fn- AgFn (red precipitate)

Silver Fluoresceinate (AgFn) is a red precipitate.RED or

PINK COLOUR is seen AT THE END-POINT.

Prior to the end-point, the solution is yellow-greendue to

the Fn- ions.


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Volhard Method (indirect method)

The samples can be acidic or alkaline.Volhard method is used to determine cation Ag+ as well

as anion Br- , I- and Cl- . A halide ion is added withstandard AgNO3 in excess and then the extra Ag+ isback titrated with standard KSCN with Fe3+ ion asthe indicator.

PrincipleX- + Ag+ AgX ( white ppt )

(excess)The extra Ag+ ion is back titrated with KSCN.


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In back titration :Ag+ + SCN- AgSCN ( white ppt )

The indicator used is Fe 3+ ion which will form a complex withSCN- at the end point.

The end point of the titration is shown by the colour of thesoluble complex [Fe(SCN)] 2+ which is red in colour.

End-point: Fe3+ + SCN- [Fe(SCN)]2+

a red soluble complex


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Chloride Determination by Volhard Method. More difficult than Mohr Method because AgCl ismore soluble than AgSCN precipitate in extreme pH

conditions.Therefore, at the end point the following reaction occursin the back titration.

AgCl(ppt) + SCN- SCN(ppt) + Cl-

To avoid this difficulty:1. Add a very high concentration of indicator (0.2 M) ofFe3+ so that [Fe(SCN)] 2+ is formed before a significantSCN- is added (and has the chance to react with AgCl).2. Remove AgCl precipitate by filtration.3. Add a few mL of nitrobenzene to coat the surface ofthe AgCl precipitate to prevent it from reacting withSCN-.


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Calculation examples :Example 125.0 ml of silver nitrate solution was precipitatedwith excess sodium chloride solution. 0.4335 g ofsilver chloride was formed. If 25.40 ml of the samesilver nitrate solution required 17.30 ml potassiumthiocyanate solution to reach the end-point.Calculate the concentration of silver nitrate andpotassium thiocyanate.


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Example 27.5124 g of sample X which contain 45.5% Br - is

dissolved in a 250 mL volumetric flask. 25.0 mLof the above solution is being titrated with0.1235 M AgNO3 solution.

1. Suggest what method can be used in the abovetitration? Name a suitable indicator that can be used todetect the end point. Calculate the volume of silver nitrate used atend-point.


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1. Mohr method

2. Chromate, CrO 42-

3. Br - + Ag+ = AgBr

g Br- = 7.5124 x 0.455 = 3.4181 g

Mol Br- = 0.0428 mol.Molarity Br - solution = 0.1711 MMmol Br- = 0.1711 x 25.0 =4.2775 mmol

= mmol of AgNO3Volume of AgNO3 = 34.64 mL#


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4th January 2011 DrSabiha/CHM421/Dec 10-Apr11 12

The arsenic in a 1.035g sample of a pesticide was converted to H3 AsO4 bysuitable treatment. The excess acid was then neutralised and 30.0 mL of0.0650 M AgNO3 was added to precipitate the arsenic quantitatively as Ag

3 AsO

4. The excess Ag+ in the filtrates, and washings from the precipitate

were titrated with 9.50 mL of 0.1000 M KSCN.

Ag+(aq) + SCN-(aq) AgSCN(s)

Calculate the percentage of As2O3 in the sample.


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4th January 2011 DrSabiha/CHM421/Dec 10-Apr11 13

H3 AsO 4 + 3AgNO 3 Ag 3 AsO 4 + 3HNO 3 Initial mol Ag + = 40x10 -3Lx0.085M

= 3.4x10 -3 mol

Excess mol Ag+

=11.27x10-3

Lx0.1M= 1.127x10 -3 molReacted mol Ag + = 3.4x10 -3 mol - 1.127x10 -3 mol

= 2.273x10 -3 molFrom reaction above, H 3 AsO 4 =2.273x10 -3 mol

3

= 7.577x10 -4 mol All arsenic in H 3 AsO 4 originated from As 2O 3 Mol As 2O 3=7.577x10 -4 molx2

= 1.515x10 -3 molMass of As 2O 3 in the sample = 1.515x10 -3 molx197.8432

= 0.2998g

Percent As 2O 3 in the sample = 0.2998 x 1001.223

= 24.51%

8. volumetric4(precipitation1)

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