Post on 12-Feb-2016
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INTRODUCTION TO INTRODUCTION TO TITRIMETRYTITRIMETRY
Most common types of titrations :
• acid-base titrations
• oxidation-reduction titrations
• complex formation
• 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 PRACTICETITRATIONS 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
nvc
nvc
1vc
1vc BOHBOHHAHA
HA
BOHBOHHA v
vcc
OR if we have:
H2A + 2BOH 2BA + 2H2O
Then from the balanced equation we know:
1 mol H2A reacts with 2 mol BOH
analyte titrant
We also know:
CBOH, VBOH and VHA and2
22
1
11
nvc
nvc
2vc
1vc BOHBOHHAHA
HA
BOHBOHHA 2v
vcc
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.
STANDARD SOLUTIONSSTANDARD SOLUTIONS
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
The 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 POINTEND POINTVSEQUIVALENCE EQUIVALENCE
POINTPOINT
Indicators used to observe the end point (at/near the equivalence point)
Thymol blue indicator
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 of unreacted titrant by back titration with a second titrant.
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
BACK TITRATIONBACK TITRATION
For example: If I add excess titrant …
H2A + 2BOH 2BA + 2H2O
Known: vanalyte, cBOH and vBOH(total)
analyte titrant reacted
… and then react the excess with a second titrant as follows:
HX + BOH BA + H2Otitrant 2 excess
Known: vHX, cHX and cBOH
Unknown: vBOH(reacted)
Step 1: From the second titration HX + BOH BA + H2O
titrant 2 excessKnown: vHX, cHX and cBOH
BOH
HXHXBOH c
vcv (excess)
Step 2: Calculate Unknown vBOH(reacted)Known: vBOH(total)
vBOH(reacted) = vBOH(total) – vBOH(excess)
Step 3: From the first titration H2A + 2BOH 2BA + 2H2O
analyte titrant reactedKnown: vanalyte, cBOH and vBOH(reacted)
AH
BOHBOHAH
22 2v
vcc (reacted)
Work backwards!
Example:
50.00 ml of HCl was titrated with 0.01963M Ba(OH)2. The end point was reached (using bromocresol green as indicator) after 29.71 ml Ba(OH)2 was added.
What is the concentration of the HCl?
Example:
A 0.8040 g sample of iron ore is dissolve in acid. The iron is reduced to Fe2+ and titrated with 0.02242 M KMnO4. 47.22 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 20.3 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 8.25 mL of 0.01101 M Na2S2O3:
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq)
The excess Na2S2O3 was back titrated with 2.16 mL of 0.00947 M I2 solution.
Calculate the mg CO per liter of sample.