dr. s. m. condren principles of volumetric analysis titration titrant analyte indicator equivalence...
TRANSCRIPT
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Dr. S. M. Condren
Principles ofVolumetric Analysis
titration
titrant
analyte
indicator
equivalence point vs. end point
titration error
blank titration
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Dr. S. M. Condren
Principles ofVolumetric Analysis
primary standard
1. High purity 100.02%
2. Stability toward air
3. Absence of hydrate water
4. Available at moderate cost
5. Soluble
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Dr. S. M. Condren
Principles ofVolumetric Analysis
standardization
standard solution
Methods for establishing concentration
direct method
standardization
secondary standard solution
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Dr. S. M. Condren
VolumetricProcedures and Calculations
relate the moles of titrant to the moles of analyte
# moles titrant = # moles analyte
#molestitrant=(V*M)titrant
=
#molesanalyte=(V*M)analyte
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Dr. S. M. Condren
EXAMPLE: What is the molarity of an HCl solution if it took 39.72 mL of the 0,09782 M NaOH solution to titrate 25.00 mL HCl solution?
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Dr. S. M. Condren
EXAMPLE: What is the molarity of an HCl solution if it took 39.72 mL of the above NaOH solution to titrate 25.00 mL HCl solution?at equivalence point:
# molesacid = # molesbases
Vacid x Macid = Vbase x Mbase
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Dr. S. M. Condren
EXAMPLE: What is the molarity of an HCl solution if it took 39.72 mL of the above NaOH solution to titrate 25.00 mL HCl solution?at equivalence point:
# mLacid x Macid = # mLbase x Mbase
# mLbase x MbaseMacid = ------------------------
# mLacid
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Dr. S. M. Condren
EXAMPLE: What is the molarity of an HCl solution if it took 39.72 mL of the above NaOH solution to titrate 25.00 mL HCl solution?at equivalence point:
# mLbase X MbaseMacid = ------------------------
# mLacid
(39.72 mL)(0.09782 M)= ------------------------------- = 0.1554 M
(25.00 mL)
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Dr. S. M. Condren
Acid-Base Indicators
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Dr. S. M. Condren
Figure 7.5 Spectrophotometric Titration
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Dr. S. M. Condren
Precipitation Titration Curve
p-function pX = - log10[X]
precipitation titration curve
four types of calculations
initial point
before equivalence point
equivalence point
after equivalence point
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Dr. S. M. Condren
Precipitation Titration Curve
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
titration curve => pAg vs. vol. AgNO3 added
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
titration curve => pAg vs. vol. AgNO3 added
initial point
after 0.0 mL of AgNO3 added
at the initial point of a titration of any type, only analyte is present, no titrant is present, therefore pAg can not be calculated.
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.
before equivalence point
pAg can be accurately calculated only after some AgBr has started to form. This may take a few mL of titrant
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.before equivalence point
after 5.0 mL of AgNO3 added
VNaBr*MNaBr - VAgNO3*MAgNO3MNaBr unreacted = -------------------------------------
VNaBr + VAgNO3
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.before equivalence point
after 5.0 mL of AgNO3 added
VNaBr*MNaBr - VAgNO3*MAgNO3MNaBr unreacted = -------------------------------------
VNaBr + VAgNO3
(50.00mL*0.00500M) -(5.00mL*0.01000M)MNaBr unreacted = -------------------------------------------------------
(50.00 + 5.00)mL
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.before equivalence point
after 5.0 mL of AgNO3 added
(50.00mL*0.00500M) -(5.00mL*0.01000M)
MNaBr unreacted = ------------------------------------------------------(50.00 + 5.00)mL
MNaBr unreacted = 3.64 X 10-3M
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.before equivalence point
after 5.0 mL of AgNO3 added
MNaBr unreacted = 3.64 X 10-3M
[Br-]total = [Br-]NaBr unreacted + [Br-]dissolved AgBr
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.before equivalence point
after 5.0 mL of AgNO3 added
MNaBr unreacted = 3.64 X 10-3M
[Br-]total = [Br-]NaBr unreacted + [Br-]dissolved AgBr
[Br-]total = 3.64 X 10-3M + [Ag+]dissolved AgBr
where [Br-]dissolved AgBr = [Ag+]dissolved AgBr
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.before equivalence point
after 5.0 mL of AgNO3 added
[Br-]total = 3.64 X 10-3M + [Ag+]dissolved AgBr
where [Br-]dissolved AgBr = [Ag+]dissolved AgBr
except very near the equivalence point,
3.64 X 10-3M >> [Ag+]dissolved AgBr
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.before equivalence point
after 5.0 mL of AgNO3 added
[Br-]total = 3.64 X 10-3M + [Ag+]dissolved AgBr
except very near the equivalence point,
3.64 X 10-3M >> [Ag+]dissolved AgBr
thus [Br-]total ~ 3.64 X 10-3M
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.before equivalence point
after 5.0 mL of AgNO3 added
[Br-]total ~ 3.64 X 10-3M
pBr = - log10(3.64 X 10-3) = 2.44
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.before equivalence point
after 5.0 mL of AgNO3 added
Ksp = [Ag+][Br-] = 5.2 X 10-13M2
- log Ksp = - log[Ag+] - log[Br-]
= - log(5.2 X 10-13)
pKsp = pAg + pBr = 12.28
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.before equivalence point
after 5.0 mL of AgNO3 added
Ksp = [Ag+][Br-] = 5.2 X 10-13M2
- log Ksp = - log[Ag+] - log[Br-]
= - log(5.2 X 10-13)
pKsp = pAg + pBr = 12.28
if pBr = 2.44
pAg = pKsp - pBr = 12.28 - 2.44 = 9.84
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.equivalence point
at 25.00 mL of AgNO3 added
At the equivalence point, [Ag+] = [Br-]
thus
Ksp = [Ag+][Br-] = 5.2 X 10-13M2
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.equivalence point
at 25.00 mL of AgNO3 added
becomes when [Ag+] = [Br-]
[Ag+]2 = 5.2 X 10-13M2
[Ag+] = 7.21 X 10-7M
pAg = 6.14
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.after equivalence point
After the equivalence point there is very little change in the amount of precipitate present (except very close to the equivalence point)
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.after equivalence point
thus, at 25.10 mL of AgNO3 added
VAgNO3*MAgNO3 - VNaBr*MNaBrMAgNO3 unreacted = -----------------------------------
VAgNO3 + VNaBr
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.after equivalence point
thus, at 25.10 mL of AgNO3 added
VAgNO3*MAgNO3 - VNaBr*MNaBrMAgNO3 unreacted = -------------------------
VAgNO3 + VNaBr
(25.10 mL * 0.01000 M) - (50.00 mL * 0.00500 M)MAgNO3 = --------------------------------------------------------------
unreacted (25.10 + 50.00)mL
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.after equivalence point
thus, at 25.10 mL of AgNO3 added
(25.10 mL * 0.01000 M) - (50.00 mL * 0.00500 M)MAgNO3 = --------------------------------------------------------------
unreacted (25.10 + 50.00)mL
MAgNO3 unreacted = 1.33 X 10-5M
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.after equivalence point
thus, at 25.10 mL of AgNO3 added
MAgNO3 unreacted = 1.33 X 10-5M
[Ag+]total = [Ag+]AgNO3 unreacted + [Ag+]dissolved
AgBr
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Dr. S. M. Condren
EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.01000 M AgNO3.after equivalence point
thus, at 25.10 mL of AgNO3 added
[Ag+]total = [Ag+]AgNO3 unreacted + [Ag+]dissolved
AgBr
[Ag+]total = 1.33 X 10-5M + [Ag+]dissolved AgBr
[Ag+]total ~ 1.33 X 10-5M
pAg = 4.88
0
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Dr. S. M. Condren
Vol of titrantpAg5.00 9.84
25.00 6.1425.10 4.88
Precipitation Titration
0
2
4
6
8
10
12
0.00 5.00 10.00 15.00 20.00 25.00 30.00
Vol of AgNO3 added
pA
g
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Dr. S. M. Condren
Vol of titrantpAg5.00 9.84
10.00 9.68 2.60 0.0025
15.00 9.47 2.81 0.001538
20.00 9.13 3.15 0.000714
21.00 9.03 3.25 0.000563
22.00 8.90 3.38 0.000417
23.00 8.72 3.56 0.000274
24.00 8.41 3.87 0.000135
24.50 8.11 4.17 6.71E-05
25.00 6.1425.10 4.8826.00 3.88 0.0001316
27.00 3.59 0.0002597
28.00 3.41 0.0003846
29.00 3.30 0.0005063
30.00 3.20 0.000625
35.00 2.93 0.0011765
40.00 2.78 0.0016667
45.00 2.68 0.0021053
Precipitation Titration
0
2
4
6
8
10
12
0.00 10.00 20.00 30.00 40.00 50.00
Vol of AgNO3 added
pA
g
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Dr. S. M. Condren
Figure 7.6 Precipitation titration Curves
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Dr. S. M. Condren
End-Point Detection
argentometric titrations
1. Mohr - titration of Cl- with AgNO3
- K2CrO4 gives brick red color to precipitate, Ag2CrO4
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Dr. S. M. Condren
End-Point Detection
argentometric titrations
2. Volhard - titration of Ag+ with Cl-
- KSCN in the presence of Fe+3 gives blood red color to precipitate, FeSCN+2
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Dr. S. M. Condren
End-Point Detection
argentometric titrations
3. Fajans - titration of Cl- with AgNO3
- absorption indicator, dye absorbs on to the surface of the particles of AgCl, attracted by excess Cl- absorbed on the surface of the precipitate
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Dr. S. M. Condren
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Dr. S. M. Condren
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Dr. S. M. Condren
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Dr. S. M. Condren
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Dr. S. M. Condren
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Dr. S. M. Condren