Chapter 11. Titrations: Taking Advantage of Stoichiometric Reactions.
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Titrations: Taking Advantage of Stoichiometric Reactions
Volumetric titrimetry is used to measure the volume of a solution of known concentration that is needed to react completely with the analyte.
Gravimetric titrimetry is like volumetric titrimetry, but the mass is measured instead of the volume.
Coulometric titrimetry is where the reagent is a constant direct electrical current of known magnitude that consumes the analyte; the time required to complete the electrochemical reaction is measured.
The benefits of these methods are that they are rapid, accurate, convenient, and readily available.
Et is the titration error
Vep is the actual volume used to get to the end point
Veq is the theoretical value of reagent required to reach the end point
Equivalence Points and End Points
One can only estimate the equivalence point by observing a physical change associated with the condition of equivalence.
Compounds that meet or even approach these criteria are few, and only a few primary standards are available.
Standard solutions play a key role in titrimetric methods.
Desirable Properties of Standard Solutions:
Describe the preparation of a 5.0 L of 0.10 M Na2CO3 (105.99 g/mol)
from the primary standard solution.
Amount Na2CO3 = n Na2CO3 (mol) = Volume solution x c Na2CO3 (mol/ L)
= 5 L x 0.1 mol Na2CO3 = 0.5 mol Na2CO3
Mass Na2CO3 = mNa2CO3=0.5 mol Na2CO3 x 105.99 g Na2CO3 =53 g Na2CO3
The solution is prepared by dissolving 53 g of Na2CO3 in water and diluting to 5 L.
How would you prepare 50mL portions of standard solutions that are
0.005 M, 0.002 M, and 0.001 M in a standard 0.01 M Na+?
To solve this the relationship Vconcd x cconcd = Vdil x cdil
Vconcd = Vdil x cdil = 50mL x 0.005 mmol Na+ /mL = 25mL
cconcd 0.01mmol Na+ /mL
To produce 50mL of 0.005 M Na+, 25mL of the concentration solution
should be diluted to 50mL.
The following two examples show the two types of volumetric calculations.
The first involves computing the molarity of solutions that have been
standardized against either a primary standard or another standard solution.
The second example involves calculating the amount of analyte in a sample
from titration data.
A 50mL volume of HCl solution required 29.71mL of 0.01963 M Ba(OH)2 to reach an end point with bromocresol green indicator. Calculate the molarityof the HCl.
Stoichiometric ratio= 2 mmol HCl/ 1 mmol Ba(OH)2
Amount Ba(OH)2 = 29.71 mL Ba(OH)2 x 0.01963 mmol Ba(OH)2
Amount HCl = (29.71 x 0.01963) mmol Ba(OH)2 x 2 mmol HCl
1 mmol Ba(OH)2
C HCl = (29.71 x 0.01963 x 2) mmol HCL
= 0.023328 mmol HCl = 0.0233M
Titration of 0.2121 g of pure Na2C2O4 (134 g/mol) required 43.31 mL of KMnO4. What is the molarity of the KMnO4 solution?
Stoichiometric ratio = 2 mmol KmnO4/ 5 mmol Na2C2O4
Amount Na2C2O4= 0.2121 g Na2C2O4 x 1 mmol Na2C2O4
0.134 g na2C2O4
Amount KMnO4 = 0.2121 mmol Na2C2O4 x 2 mmol KMnO4
0.134 5 mmol Na2C2O4
C KMnO4 = ( 0.2121 x 2) mmol KMnO4
0.134 5 = 0.01462M
43.31 mL KMnO4
A 0.8040g sample of an iron ore is dissolved in acid. The iron is then reduced to Fe2+ and titrated with 47.22mL of 0.02242 M KMnO4 solution. Calculate the results of this analysis in terms of percent Fe (55.847 g/mol).
Stoichiometric ratio = 5 mmol Fe2+/ 1 mmol KMnO4
Amount KMnO4 = 47.22mL KMnO4 x 0.02242 mmol KMnO4
Amount Fe2+ = (47.22 x 0.02242) mmol KMnO4 x 5 mmol Fe2+
1 mmol KMnO4
Mass Fe2+ = (47.22 x 0.02242 x 5) mmol Fe2+ x 0.055847 g Fe 2+
% Fe2+ = (47.22 x 0.02242 x 5 x 0.055947) g Fe 2+ x 100% = 36.77%
0.8040 g sample