ARGENTOMETRIC TITRATION || B.TECH FIRST YEAR CHEMISTRY PRACTICAL
ARGENTOMETRIC TITRATION
Title: Argentometric titration
Object:Estimation of cl ion in a given sample water by Argentometric method.
Theory:
Chloride ion (cl-) in a solution can be determined by titrating with a standard solution of AgNO3. Titrating method based upon AgNO3 is called Argentometric method .Addition of AgNO3 solution in a solutio. solution containing cl- ion causes precipitation of Agcl since solubility of Agcl is small and titration carry out under neutral condition in presence of K2cr2O7 indicator.
When AgNO3 is added to cl- ion gradually removed as Agcl precipitates by reaction-
Cl-(aq) + Ag+(aq) = Agcl(s) (white ppt)
The first permenant appearance of reddish precipitate of AgCrO4 in indicates thebend point of titration .
As concentration of cl- ion is large and that of CrO4²- ion is small,addition of Ag+ ion first produce Agcl(s) precipitate and Ag2CrO4(s) precipitate is not obtained .prexipitation of AgCrO4 stir only when concentration of Ag+ ion becomes large enough to exceed the solubility product (ksp) of Ag2CrO4(ksp=2.0×10^-12).At equivalent point in the titration of cl- ion byAgNO3 solution .All cl-ion are precipitate as Agcl(s) and there is an equilibrium.
Agcl=Ag+(aq) + cl- such that C Ag+ =C cl-
Now ,ksp= C Ag+×C Ag ²+ at the equivalent point.
This at equivalence point C Ag+=√ksp = 1.0×10^-5 mol/lit.It is possible to calculate concentration of CrO4²- ion required for precipitation of Ag2CrO4 at equivalence point in cl- ion titration from ksp value of AgCrO4.
AgCrO4(s) =2 Ag +(aq)+ CrO4²-
Ksp (Ag2CrO4)=(C Ag+)²×C CrO4²-
C CrO4²- = ksp/C Ag²+= (2×10^-12)/(1×10^-5)=0.02(M)
0.02(m) concentration of CrO4 ²- (indicator) will give red precipitate of AgCrO4 and correct end point detection is not possible .
Usually ,Lower concentration of K2CrO4 solution (0.005 to 0.001m)
Is used as indicator .An excess AgNo3 solution beyond equivalence point is required to be added for appearance of red precipitate of Ag2CrO 4 .This introduce error in the titration data.The error involved may be corrected by performing a blank titration.
Mohr titrationis to be carried out at a ph between 7.0 to 10.0.In acidic medium
Ag²+(aq)+ OH- (aq) = AgOH
For correct titration ,we need only Agcl(s) formation reaction till equivalence point and reaction of Ag+ ion with OH- ion leads to huge error.
OBSERVATION TABLE
Calculation:
Volume of water sample taken for titration in step 1-50 ml
Volume of standard AgNO3 solution required for titration =(v1-v2)ml=2.95ml
Therefore,strength of cl- ion in water sample ×volume sample= strength of AgNO3 solution (S1) ×volume of AgNO3 solution
Hence,strength of cl- ion in water sample=[(v1-v2)×s1]/50=0.0059
Amount of cl- ion in water sample in gm/lit=Normality of cl- ion ×Equivalent wt of cl- ion
[{(V1-V2) ×S2}/50]×35.5
Hence,the cl- ion concentration is =[{(V1-V2)×S2}/50]×35.5gm/lit
=0.2094gm/lit
CONCLUSION:
Hence,the concentration of cl- ion in water sample=0.2094gm/lit
Agcl=Ag+(aq) + cl- such that C Ag+ =C cl-
Now ,ksp= C Ag+×C Ag ²+ at the equivalent point.
This at equivalence point C Ag+=√ksp = 1.0×10^-5 mol/lit.It is possible to calculate concentration of CrO4²- ion required for precipitation of Ag2CrO4 at equivalence point in cl- ion titration from ksp value of AgCrO4.
AgCrO4(s) =2 Ag +(aq)+ CrO4²-
Ksp (Ag2CrO4)=(C Ag+)²×C CrO4²-
C CrO4²- = ksp/C Ag²+= (2×10^-12)/(1×10^-5)=0.02(M)
0.02(m) concentration of CrO4 ²- (indicator) will give red precipitate of AgCrO4 and correct end point detection is not possible .
Usually ,Lower concentration of K2CrO4 solution (0.005 to 0.001m)
Is used as indicator .An excess AgNo3 solution beyond equivalence point is required to be added for appearance of red precipitate of Ag2CrO 4 .This introduce error in the titration data.The error involved may be corrected by performing a blank titration.
Mohr titrationis to be carried out at a ph between 7.0 to 10.0.In acidic medium
Ag²+(aq)+ OH- (aq) = AgOH
For correct titration ,we need only Agcl(s) formation reaction till equivalence point and reaction of Ag+ ion with OH- ion leads to huge error.
OBSERVATION TABLE
Calculation:
Volume of water sample taken for titration in step 1-50 ml
Volume of standard AgNO3 solution required for titration =(v1-v2)ml=2.95ml
Therefore,strength of cl- ion in water sample ×volume sample= strength of AgNO3 solution (S1) ×volume of AgNO3 solution
Hence,strength of cl- ion in water sample=[(v1-v2)×s1]/50=0.0059
Amount of cl- ion in water sample in gm/lit=Normality of cl- ion ×Equivalent wt of cl- ion
[{(V1-V2) ×S2}/50]×35.5
Hence,the cl- ion concentration is =[{(V1-V2)×S2}/50]×35.5gm/lit
=0.2094gm/lit
CONCLUSION:
Hence,the concentration of cl- ion in water sample=0.2094gm/lit
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