Post on 28-Dec-2015
SILABUS
Metode titrimetri; titrasi netralisasi; analisis volumetri untuk reaksi pembentukan endapan dan senyawa kompleks (argentometri); titrasi pembentukan senyawa kompleks; analisis volumetri secara oksidasi-reduksi (redoksimetri); Analisis gravimetri
BUKU REFERENSI• David Harvey, Modern Analytical Chemistry, Mc Graw-Hill• Skoog, D. A. and West, D.M., Fundamentals of Analytical Chemsitry, Holt-Saunders International edition.• Vogel’s Textbook of Quantitative Chemical Analysis, Longman and Scientific, Essex•Christian, G.D., Analytical Chemsitry, 5th edition, John Wiley and Sons, Inc., New York.
Most common types of titrations :
• neutralization titrations
• oxidation-reduction titrations
• complex formation titration
• precipitation reactions
In a titration, increments of titrant are added to the analyte until their reaction is complete.
From the quantity of titrant required, the quantity of analyte that was present can be calculated.
TITRATIONS IN PRACTICE
Accurately add of specific volume of sample solution to a conical flask using a pipette
Known: volume of sample
Unknown: concentration of analyte in sample
1
Slowly add standard solution from a burette to the sample solution
Known: concentration of the titrant
2
Add until just enough titrant is added to react with all the analyte
The end point is signalled by some physical change or detected by an instrument
Known: volume of the titrant
Note the volume of titrant used
3
If we have:
HA + BOH BA + H2O
Then from the balanced equation we know:
1 mol HA reacts with 1 mol BOH
analyte titrant
We also know:
CBOH, VBOH and VHA and2
22
1
11
n
vc
n
vc
1
vc
1
vc BOHBOHHAHA HA
BOHBOHHA v
vcc
STANDARD SOLUTIONS
Standard solution:
Reagent of known concentration
Primary standard:
highly purified compound that serves as a reference material in a titration.
Determine concentration by dissolving an accurately weighed amount in a suitable solvent of known volume.
primary standard1. High purity 100.0%2. Stability toward air3. Absence of hydrate water4. Available at moderate cost5. Soluble6. Large F.W.
Secondary standard:
compound that does not have a high purity
Determine concentration by standardisation.
Titrate standard using another standard.
Standard solutions should:
• Be stable
• React rapidly with the analyte
• React completely with the analyte
• React selectively with the analyte
EQUIVALENCE
POINTThe amount of added titrant is the exact amount necessary for stoichiometric reaction with the analyte in the sample.
An estimate of the equivalence point that is observed by some physical change associated with conditions of the equivalence point.
Aim to get the difference between the equivalence point and the end point as small as possible.
Titration error: Et = Veq – Vep
Estimated with a blank titration
END POINTVS
Instruments can also be used to detect end points.
Respond to certain properties of the solution that change in a characteristic way.
E.g.: voltmeters, ammeters, ohmmeters, colorimeters, temperature recorders, refractometers etc.
Add excess titrant and then determine the excess amount by back titration with a second titrant.
BACK TITRATION
Used when:
• end point of back titration is clearer than end point of direct titration
• an excess of the first titrant is required to complete reaction with the analyte
If we have:
HA + BOH BA + H2O
Then from the balanced equation we know:1 mol HA reacts with 1 mol BOH
analyte titrant
If I add excess titrant and then react the excess with a second titrant as follows:
HX + BOH BX + H2Otitrant 2 excess
Then from the balanced equation we know:
1 mol HX reacts with 1 mol BOH
We also know:
CBOH, CHX and VHX and 2
22
1
11
n
vc
n
vc
1
vc
1
vc HXHXBOHBOH (excess)
BOH
HXHXBOH c
vcv (excess)
We also know vBOH(total)
vBOH(reacted) = vBOH(total) – vBOH(excess)
1
vc
1
vc (reacted)BOHBOHHAHA
HA
BOHBOHHA v
vcc
(reacted)
From our initial titration:
HA + BOH BA + H2O analyte titrant
we then know:
CBOH, VBOH(reacted) and VHA
and we want to find CHA!
In summary:
HA + BOH BA + H2Oanalyte titrant reacted
VHA CBOH
HX + BOH BX + H2Otitrant 2 titrant excess
VHX
CHX CBOH
BOH
HXHXBOH c
vcv (excess)
vBOH(reacted) = vBOH(total) – vBOH(excess)
2
22
1
11
n
vc
n
vc
CHA?
HA
BOHBOHHA v
vcc
(reacted)
Example:
50.00 ml of HCl was titrated with 0.02 M Ba(OH)2. The end point was reached (using bromocresol green as indicator after 20 ml Ba(OH)2 was added.
What is the concentration of the HCl?
2
22
1
11
n
vc
n
vc
Example:
A 0.8 g sample of iron ore is dissolve in acid. The iron is reduced to Fe2+ and titrated with 0.02 M KMnO4. 40 ml of titrant was added to reach the end point. Calculate the % Fe in the sample.
MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O
Example:
The CO in a 20 L sample of gas was converted to CO2
by passing the gas over iodine pentoxide heated to 150oC:
I2O5(s) + 5CO(g) 5CO2(g) + I2(g)
The iodine distilled at this temperature was collected in an absorber containing 8 mL of 0.01 M Na2S2O3:
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq)
The excess Na2S2O3 was back titrated with 2 mL of 0.001 M I2 solution.
Calculate the mg CO per liter of sample.
I2O5(s) + 5CO(g) 5CO2(g) + I2(g)
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq)
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq)
nreacted =
AD
DE
DE
XC
ES
S
REACTED
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq)
I2O5(s) + 5CO(g) 5CO2(g) + I2(g)
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq)
I2O5(s) + 5CO(g) 5CO2(g) + I2(g)