argentometry.pptx
TRANSCRIPT
argentometry is a type of titration involving the silver(I) ion. Typically, it is used to determine the amount of chloride present in a sample
According to the indicator used, three methods can be described:◦Mohr's method◦Fajans method◦Volhard Method
i) Guidance in precipitation titration calculation
Find Ve (volume of titrant at equivalence point)
Find y-axis values:- At beginning- Before Ve
- At Ve
- After Ve
Example: For the titration of 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+. The reaction is:
Ag+(aq) + Cl–(aq) AgCl(s) Ksp = 1.8×10–10
Find pAg and pCl of Ag+ solution added(a) 0 mL (b) 10.0 mL (c) 25.0 mL (d) 35.0 mL
(a) 0 mL Ag+ added (At beginning)[Ag+] = 0, pAg can not be calculated.[Cl–] = 0.0500, pCl = 1.30
(b) 10 mL Ag+ added (Before Ve)
14.8
60.1
102.71050.2
108.1
][][
1050.2 0.60
5.1][
0.60 0.10 0.50
50.1
1
1
1
100.0 0.10
1
0500.0 0.50
92
10
2
pAg
pCl
MCl
KAg
MmL
ClmmolCl
mLmLmLV
Clmmol
Agmmol
Clmmol
AgmL
AgmmolAgmL
ClmL
ClmmolClmL
ClmmoledprecipitatClmmoloriginalLeftClmmol
sp
total
(c) 25 mL Ag+ added (At Ve)
AgCl(s) Ag+(aq) + Cl–(aq) Ksp = 1.8×10–10
s = [Ag+]=[Cl–]Ksp = 1.8×10–10 = s2
[Ag+]=[Cl–]=1.35x10–5
pAg = 4.89 pCl = 4.89
(d) 35 mL Ag+ added (After Ve)
93.1
82.7
1053.11018.1
108.1
][][
1018.1 0.85
00.1][
0.85 0.35 0.50
00.1
1
100.0 0.25
1
100.0 0.35
82
10
2
pAg
pCl
MAg
KCl
MmL
AgmmolAg
mLmLmLV
Agmmol
AgmL
AgmmolAgmL
AgmL
AgmmolAgmL
AgofwithVAgmmoledprecipitatAgmmoloriginalLeftAgmmol
sp
total
e
ii) Construct a titration curveExample: Titration of 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+
pCl
pAg
iii) Diluting effect of the titration curves
25.00 mL 0.1000 M I– titrated with 0.05000 M Ag+
25.00 mL 0.01000 M I– titrated with 0.005000 M Ag+
25.00 mL 0.001000 M I– titrated with 0.0005000 M Ag+
The effect of Reaction Completeness on Titration Curves
Figure 13-5 Effect of reaction completeness on precipitation curves . For each curve , 50.00 mL of a 0.0500M solution of the anion was titrated with 0.1000 M AgNO3 . Note that smaller values of Ksp give much sharper breaks at the end point
Example: A 25.00 mL solution containing Br– and Cl– was titrated with 0.03333 M AgNO3. Ksp(AgBr)=5x10–13, Ksp(AgCl)=1.8x10–10. (a)Which analyte is precipitated first?(b)The first end point was observed at 15.55 mL.
Find the concentration of the first that precipitated (Br– or Cl–?).
(c) The second end point was observed at 42.23 mL. Find the concentration of the second that precipitated (Br– or Cl–?).
Solution:(a)Ag+
(aq) + Br–(aq) AgBr(s) K = 1/Ksp(AgBr) = 2x1012
Ag+(aq) + Cl–(aq) AgCl(s) K = 1/Ksp(AgCl) = 5.6x109
Ans: AgBr precipitated first
(b)
Br M 0.02073L 1
mL 1000
mL 25
1
Ag mol 1
Br mol 1
Ag L 1
Ag mol 0.3333
mLAg 1000
Ag L 1
1
Ag mL 15.55
(c)
Cl M 0.03557L 1
mL 1000
mL 25
1
Ag mol 1
Cl mol 1
Ag L 1
Ag mol 0.3333
mLAg 1000
Ag L 1
1
Ag mL 15.55)-(42.23
Volhard method: A colored complex (back titration). Analysing Cl– for example:
Step 1: Adding excess Ag+ into sampleAg+ + Cl– → AgCl(s) + left Ag+
Step 2: Removing AgCl(s) by filtration/washingStep 3: Adding Fe3+ into filtrate (i.e., the left Ag+)Step 4: Titrating the left Ag+ by SCN–:
Ag+ + SCN– → AgSCN(s)Step 5: End point determination by red colored
Fe(SCN)2+ complex. (when all Ag+ has been consumed, SCN– reacts with Fe3+)SCN– + Fe3+ → Fe(SCN)2+
(aq) Total mol Ag+ = (mol Ag+ consumed by Cl–) + (mol Ag+ consumed by SCN–)
Mohr Method: A colored precipitate formed by Ag+ with anion, other than analyte, once the Ve reached. Analysing Cl– and adding CrO4
2– for example:
Precipitating Cl–:Ag+ + Cl– → AgCl(s) Ksp = 1.8 x 10–10
End point determination by red colored precipitate, Ag2CrO4(s):2Ag+ + CrO4
2– → Ag2CrO4(s) Ksp = 1.2 x 10–12
Before Ve
(Cl– excess)Greenish yellow solution
AgCl(s)Cl–
1stlayer
After Ve (Ag+ excess)
AgCl(s)Ag+ In–
pink1stlayer
3) Fajans Method: An adsorbed/colored indicator. Titrating Cl– and adding dichlorofluoroscein for example: