Transcript Document

Volumetric Analysis
Volumetric analysis refers to any procedure in
which the volume of reagent needed to react
with the analyte is being measured.
In this chapter we will learn:
- the principles in volumetric procedure
- methods in volumetric analysis
One method in volumetric
analysis is titration
In titration:
- substance to be
analysed is known as the
analyte
- the solution added to the
analyte is known as the
titrant
- titrant is usually delivered
Stirrin
g bar
Titration
- the point when the quantity of added titrant is
the exact amount necessary for stoichiometric
reaction with the analyte is called the
equivalence point
Example :
5HO-C-C-OH + 2MnO4- + 6H+
10CO2 + 2Mn2+
+ 8H2O
If the unknown contains 5.0 mmol of oxalic acid,
the equivalence point is reached when 2.0 mmol
of MnO4- has been added. Hence
Equivalence point is the ideal theoretical result we
seek in a titration
What we actually measure is the end point, which
is marked by a sudden change in the physical
property of the solution
Methods for determining the end point are:
- detecting a sudden change in the voltage or
current between a pair of electrodes
- observing an indicator colour change.
An indicator is a compound which changes
colour abruptly near the equivalence point.
This change is caused by the disappearance
of the analyte or appearance of excess titrant
- monitoring the
absorption of
light
What is the difference between
equivalence point and end
point?
Titration error – by choosing a physical property
whose change is easily observed one can
minimise the titration error so that the end point
is very close to the equivalence point
Estimation of titration error – blank titration :
5HO-C-C-OH + 2MnO4- + 6H+
10CO2 + 2Mn2+
+ 8H2O
- solution containing no oxalic acid was titrated
with MnO4- to determined how much is needed to
form an observable purple colour
- this volume of MnO4- is then subtracted from the
volume observed in the analytical titration
The concentration of titrant used can be derived if :
- the titrant was prepared by dissolving a weighed
amount of pure reagent in a known volume of
solution
- such a reagent is known as a primary standard
because the reagent is pure enough (should be
 99.9% pure)to be weighed and used directly.
- reagents used as primary standards should not
decompose under ordinary storage or upon
heating
Standard Solution
- reagent should not contain hydrate water
- reagent should be available at moderate cost
- reagent should be soluble in the solution
What if a primary standard is not available ?
- titrate reagent (to be used as titrant in analysis)
against a weighed primary standard in order to
determine the concentration of titrant – process
is known as standardization
- after standardization, the reagent (to be used as
titrant in analysis) is known as the standard
solution
Example :
The calcuim content of urine can be determined
by:
- precipitating Ca2+ as calcium oxalate in basic
solution: Ca2+ + C2O42Ca(C2O4).H2O(s)
- wash the precipitate and dissolve it in acid to
- heat the dissolved oxalic acid to 60oC and
titrate with standardized KMnO4 until the purple
end point is observed
Standardization :
Suppose 0.3562 g of Na2C2O4 is
dissolved in a 250.0 ml volumetric
flask.
Concentration of the
oxalate solution is:
(0.3562g Na2C2O4)(134.00 g/mol Na2C2O4)
0.250 ml
= 0.01063M
Moles of C2O42- in 10.0ml = (0.01063)(0.010l)
= 1.063 x 10-4 mol
Since:
5NaO-C-C-ONa + 2MnO4- + 6H+
10CO2 +
2Mn2+ + 8H2O
2 mol MnO4- requires 5 mol oxalate
(2mol MnO4-)
moles of MnO4- =
(5mol C2O4
2-)
(mol C2O42-)
= 0.04253 mmol
molarity of
MnO4-
=
0.04253 mmol
48.36 ml
= 8.794 x 10-4 M
Analysis of Unknown :
Ca in a 5.0 ml urine sample was precipitated,
redissolved and required 16.17ml of standard
MnO4- solution. Find the concentration of Ca2+
in the urine.
In 16.17ml of standard MnO4- solution, there
are:
(0.01617l)(8.794 x 10-4 M)
= 1.422 x 10-4 mol MnO4-
1.422 x 10-4 mol MnO4- will react with:
mol C2O42-
(5mol C2O42-)
= (2mol MnO -)
4
(mol C2O42-)
= 0.03555 mmol
From :
Ca2+ + C2O42-
Ca(C2O4).H2O(s)
there is one oxalate ion for each Ca2+ in Ca(C2O4)
.H2O(s)
there must be 0.03555 mmol Ca2+ in 5.00ml
urine
[Ca2+] = (0.03555 mmol )/5ml
= 0.00711M
When titrant is added to the analyte until the
reaction is complete
direct titration
Some reactions require an excess of the titrant
for complete reaction with the analyte. In such
cases, the first titration will be followed by a
titration with a second standard reagent. The
second standard reagent is used to titrate the
excess of the first reagent
back titration
Back titration is also applicable if the end point of
the back titration is clearer than the end point of
the direct titration.
Titration Curve
A titration curve is a graph showing how the
concentration of one of the reactants varies as
titrant is added
Titration curve tells us how concentrations of
analyte and titrant vary during a titration. From it
we can:
- understand the chemistry that occurs during a
titration
- learn how experimental control can be exerted
to influence the quality of an analytical titration
- since concentration varies over many orders of
magnitude, the graph plots the p function { pX =
-log[X]} against the volume of X
Example :
Consider the titration of 25.00 ml of 0.1000M I- with
0.05000M Ag+, given:
AgI(s)
I- + Ag+
Ksp
=[Ag+][I-]
=8.3 x 10-17
This implies:
I- + Ag+
AgI(s)
K = 1/ Ksp =1.2 x 1016
Since the equilibrium constant for the titration
reaction is large, this means the equilibrium lies
far to the right.
What does this mean experimentally?
Each aliquot of Ag+ reacts completely with I- .
At the equivalence point, there will be a sudden
increase in the Ag+ concentration because all the
I- has been consumed and now unreacted Ag+
exists in the solution
What volume of Ag+ is needed to reach the
equivalence point ?
Let Ve denote the volume of Ag+ is needed to
reach the equivalence point
Since 1 mol of Ag+ reacts with 1 mol of I- , thus
(0.02500l)(0.1000 mol I- /l) = Ve(0.050000 mol I- /l)
Ve = 0.05000 l = 50.00 ml
Before the equivalence point :
Consider the point when 10.00ml Ag+ has been
added to the I- solution:
- solution has more moles of I- than Ag+
- moles of I- =original moles of I- - moles of Ag+
added
=(0.02500 l)(0.100 mol/l) –
(0.01000 l)(0.05000 mol/l)
=0.002000 mol ISince the volume of solution is now 35.00 ml,
[I- ]
0.002000 mol I0.03500 l
=
[Ag+ ]
=
Ksp
[I- ]
= 0.05714 M
= 1.4 x 10-15 M
pAg+ = -log[Ag+ ] = 14.84
At the equivalence point :
All the AgI precipitates and some may redissolve
to give equal concentrations of Ag+ and I- . Hence
let [Ag+] = [I- ] = x
Ksp = [Ag+][I- ] = x2 = 8.3 x 10-17
x = 9.1 x 10-9
pAg+ = -log [Ag+]
= -log(9.1 x 10-9) = 8.04
After the equivalence point :
When V > Ve, the concentration of Ag+ is
determined by the excess Ag+ added from the
buret
Ag+
Suppose that V
= 52.00ml
mol Ag+ =(0.00200l)(0.05000 mol Ag+/l)
= 0.000100 mol
Since the total volume = 77.00ml,
[Ag+] = (0.000100mol)(0.07700l)
= 1.30 x 10-3 M
pAg+ = -log (1.30 x 10-3 ) = 2.89
- equivalence point
is the steepest
point of the curve
- concentration of
reactant affects the
titration
- for stoichiometries
other than 1:1, the
equivalence point is
not at the centre of
the steepest
section of the curve
- in practice,
conditions are
chosen such that
the titration curves
are steep
Ksp also affects the steepness of the titration
curve
At the equivalence point, the titration curve is
steepest for the least soluble precipitate
The larger the equilibrium constant for any
titration reaction, the more pronounced will be the
change in concentration near the equivalence
Kjeldahl Nitrogen
Analysis
- developed in 1883
- one of the most accurate and widely used
method for determining nitrogen in
substances
Procedure :
- substance is decomposed and dissolved
(digested) in boiling H2SO4 to which K2SO4 has
been added. Selenium-coated boiling chips are
used to catalyze the reaction
organic C H N
boiling
H2SO4
NH4+ + CO2 + H2O
- after digestion, the solution containing NH4+ is
made basic
NH4+ + OHNH3 + H2O
- the liberated NH3 is distilled into a known
amount of HCl
NH3 + H+
NH4+
- Excess unreacted HCl is titrated with standard
NaOH to determine how much HCl has been
consumed
H+ + OHH2O