Analytical chemistry
Analytical chemistry is the area of chemistry responsible for characterizing the composition
of matter, both qualitatively (what is present) and quantitatively (how much is present).
Analytical chemistry is the study of the separation, identification, and
quantification of the chemical components of natural and artificial materials.
Analytical chemistry is also focused on improvements in experimental design,
chemometrics, and the creation of new measurement tools to provide better chemical
information.
Analytical chemistry has applications in forensics, bioanalysis, clinical analysis,
environmental analysis, and materials analysis.
qualitative analysis
An analysis in which we
determine the identity of the
constituent species in a sample.
quantitative analysis
An analysis in which we determine how
much of a constituent species is present
in a sample.
characterization analysis
An analysis in which we evaluate a
sample’s chemical or physical
properties.
fundamental analysis
An analysis whose purpose is to improve
an analytical method’s capabilities.
Chemical Analysis : A process that provides chemical or physical
information about the constituents in the sample or the sample
itself.
Qualitative analysis
Quantitative analysis
gives an indication of the identity of determines the amount of one or
the chemical species in the sample.
more of these components
what is present?
how much is present?
Analytical methods can be separated into
classical and instrumental
Classical methods (also known as
wet chemistry methods)
use only chemicals for separations such as
precipitation, extraction and qualitative
analysis by color, odor, or melting point.
Quantitative analysis is achieved by
measurement of weight or volume.
Gravimetric
Volumetric
(Chemical
analysis
analysis
tests )
Instrumental methods
use an apparatus to measure physical
quantities of the analyte such as light
absorption,
fluorescence,
or
conductivity.
Spectroscopy
Electrochemistry
Chromatography
 a sample is a limited quantity of something
which is intended to be similar to and
represent a larger amount of that thing(s).
 The components of interest in the sample are
called analytes,
 the remainder of the sample is the matrix.
 Technique is a
chemical or physical
principle that can be used to analyze a
sample.
 A method is the application of a technique
for the determination of a specific analyte in
a specific matrix.
 A procedure is a set of written directions
detailing how to apply a method to a
particular sample.
Constituent : A component of a
sample; it may be further
classified as:
 A major constituent
% 1-100
 Minor constituent
% 0,01-1
 Trace constituent
˂ %0,01
Statistical Analysis of the Results
• Measurement : An experimental
determination of an analyte’s
chemical or physical properties.
• Measurement data give only an
estimate of the ‘true’ value.
•
•
•
•
Length :
Mass :
Time:
Temperature:
True Result
The 'correct' value for a measurement which remains unknown except
when a standard sample is being analysed. It can be estimated from the
results with varying degrees of precision depending on the experimental
method.
Mean
The mean, X, is the numerical average obtained by dividing the sum of the
individual measurements by the number of measurements. (The average
value of a set of data -X-).
Median
The median, Xmed, is the middle value when data are ordered from the
smallest to the largest value. Or the value for a set of ordered data, for which
half of the data is larger in value and half is smaller in value ( –Xmed).
Range (Spread )
The range, w, is the difference between the largest and smallest values in the
data set. It is a measure of precision.
Range = w = Xlargest – Xsmallest
N
∑ xi
i=1
X1 + X2 +X3 +......+ Xn
X = ------------- = ---------------------------------------N
N
Mean
Precision
is the closeness of data to other data that have been obtained in exactly the same
way. Sometimes called the variability, it can be represented statistically by the
standard deviation or relative standard deviation.
Accuracy
The closeness of an experimental measurement or result to the true or
accepted value. Accuracy is usually expressed in terms of error or absolute
error. (Bias)
Error
The difference between the true result and the measured value.. E = xi - xt
Absolute Error
defined as the actual difference between the true result and the
experimental value in the same units.
Relative Error
is the absolute error divided by the true value.
Uncertainty
The range of possible values for a measurement.
Er =
xi - xt
----------- . %100
Xt
(percentage relative error)
Er =
xi - xt
------------- . ‰1000
Xt
(parts per thousond)
Calculate the mean, median,
absolute and relative error for the
following data set
Fe concentration
19,4ppm
19,5ppm
19,6ppm
19,8ppm
20,1ppm
20,3ppm
Classification of Errors
On the basis of their origin, errors may usually be classified as determinate or
indeterminate.
Systematic
(determinate)
Error
Are the errors which can be avoided or whose mahnitude can be determined.
They are (in principle at least) measurable and for which a correction may be
made.
-
Random
(indeterminate)
Error
Gross Error
operational and personal error
instrumental and reagent error
errors of methods
Sampling Errors
They are due to causes over which the analyst has no control and which in
general are so intangible taht they are incapable of analysis.
They do not
have a definite measurable value.
Statical Evulation of the Random (indeterminate) Error
2
2
– (x-μ) /2 σ
y=
e
--------------------------------------
σ (2 П)^ 1/2
x herbir ölçümün değeri
μ bir çok ölçümün aritmetik ortalaması
x – μ ortalamadan sapma
σ standart sapma
y her bir x - μ değeri için bulunma frekansı
Deviation
Standard Deviation
is a measure of the precision of a population of data
Variance
(σ2 or s2) The square of the standard deviation. The variance
is a parameter describing in part either the actual probability
distribution of an observed population of numbers, or the
theoretical probability distribution of a sample of numbers.
Relative Standard
Deviation (RSD)
coefficient of
variation
is the percent relative standard deviation
Degree of Freedom
An independent variable. For N measurements it is equal
to N -1
Calculate the standard
deviation of the following
experimental results
Lead concentration in blood
0,752ppm
0,756ppm
0,752ppm
0,751ppm
0,760ppm
Spool =
∑ (xi – x1 )2 + ∑ (xi – x2 )2 + ...
( --------------------------------------------)^1/2
N1 + N2 + N3 ... – Nt
• Calculate the stardard deviation of the method.
Sample No
Number of measurements
Hg concentration ppm
1
3
1,80 1,58 1,64
2
4
0,96 0,98 1,02 1,10
3
2
3,13 3,35
4
6
2,06 1,93 2,12 2,16 1,89 1,95
5
4
0,57 0,58 0,64 0,49
6
5
2,35 2,44
2,70 2,48 2,44
7
4
1,11 1,15
1,22 1,04
What are the standard deviation, the relative
percent relative standard deviation for the folowing data
3.056 - 3.080 - 3.094 - 3.107- 3.112 - 3.174- 3.198
standard
deviation,
and
the
What are the standard deviation, the relative
percent relative standard deviation for the following data
3.056 - 3.080 - 3.094 - 3.107- 3.112 - 3.174- 3.198
standard
deviation,
and
the
1,76 (±0,03) + 1,89 (±0,02 ) - 0,59(±0,02) = ?
[ 1,76 (±0,03) x 1,89 (±0,02 ) ] / 0,59(±0,02) =?
The pH of a solution is defined as
pH = –log[H+]
where [H+] is the molar concentration of H+. If the pH
of a solution is 3.72 with an absolute uncertainty of
±0.03, what is the [H+] and its absolute uncertainty?
The pH of a solution is defined as
pH = –log[H+]
where [H+] is the molar concentration of H+.
If the pH of a solution is 3.72 with an absolute
uncertainty of ±0.03, what is the [H+] and its
absolute uncertainty?
The Nature and Origin of
Errors
On the basis of their origin, errors may
usually be classified as determinate or
indeterminate. The first are those
having a value which is (in principle at
least) measurable and for which a
correction may be made. The second
fluctuate in a random manner and do
not have a definite measurable value.
Indeterminate errors arise from the
unpredictable minor inaccuracies of
the individual manipulations in a
procedure. A degree of uncertainty is
introduced into the result which can
be assessed only by statistical tests.
The deviations of a number of
measurements from the mean of the
measurements should show a
symmetrical or Gaussian distribution
where μ is the mean and s is the standard deviation. The
width of the curve is determined by s, which is a useful
measure of the spread or precision of a set of results, and is
unique for that set of data. An interval of μ ± s will contain
68.3% of the statistical sample, whilst the intervals μ ± 2s
and μ ± 3s will contain 95.5% and 99.7% respectively.
Confidence Intervals for Populations
confidence interval:
Range of results around a mean value that could be
explained by random error.
there is a 68.26% probability that a member
selected at random from a normally distributed
population will have a value in the interval of
m ± 1s.
In general, we can write
Xi = m ± zs
where the factor z accounts for the desired level
of confidence. Values reported in this fashion are
called confidence intervals.
Confidence intervals also can be reported using the mean for a sample of size n, drawn from a population of known s.
The standard deviation for the mean value, s , which also is known as the standard error of the mean
What is the 95% confidence interval for the analgesic tablets if an analysis of
five tablets yields a mean of 245 mg of aspirin? the population standard
deviation for the amount of aspirin in a batch of analgesic tablets is known to
be 7 mg of aspirin.
What is the 95% confidence interval for the analgesic tablets if an analysis of
five tablets yields a mean of 245 mg of aspirin? the population standard
deviation for the amount of aspirin in a batch of analgesic tablets is known to
be 7 mg of aspirin.
Confidence Intervals for Samples
Once the reliability of a replicate set of measurements has
been established the mean of the set may be computed as
a measure of the true mean. Unless an infinite number of
measurements is made this true mean will always
remain unknown. However, the t-factor may be used to
calculate a confidence interval about the experimental
mean,
the confidence interval as a way to report the most
probable value for a population’s mean, m, when the
population’s standard deviation, s, is known.
Since s2 is an unbiased estimator of s2, the term z in
The confidence interval equation is replaced with the
variable t, where t is defined such that t  z at all
confidence levels.
t is a statistical factor derived from the normal error curve
s is the estimated standard deviation and N is the number of
results.
t is a statistical factor derived from the normal error curve
s is the estimated standard deviation and n is the number of
results.
What is the 95% confidence interval for the folowing data?
3.056 - 3.080 - 3.094 - 3.107- 3.112 - 3.174- 3.198
The accepted value for the chloride content of
a standard sample obtained from extensive
previous analysis is 54.20%.
Five analyses of the same sample are carried
out by a new instrumental procedure,
54.01, 54.24, 54.05, 54.27, 54.11%
being the results obtained. For the 95%
confidence interval, is the new method giving
results consistent with the accepted value?
The accepted value for the chloride content of
a standard sample obtained from extensive
previous analysis is 54.20%.
Five analyses of the same sample are carried
out by a new instrumental procedure,
54.01, 54.24, 54.05, 54.27, 54.11%
being the results obtained. For the 95%
confidence interval, is the new method giving
results consistent with the accepted value?
Dixon’s Q-test.
Statistical test for deciding if an outlier can be
removed from a set of data. The Q-test compares the difference between the suspected
outlier and its nearest numerical neighbor to the range of the entire data set.
outlier: Data point whose value is much larger or smaller than the remaining data.
Qexp =
differences between outlier and Nearest value to it
------------------------------------------------------------------------Range (spread)
Qexp is greater than Q(a, n), then the null hypothesis is rejected and the
outlier may be rejected. When Qexp is
less than or equal to Q(a, n) the suspected outlier must be retained.
Q (% 90 confidence
interval)
0,94
0,76
0,64
0,56
0,51
0,47
0,44
0,41
Number of measurements
3
4
5
6
7
8
9
10
The following masses, in grams, were recorded in an experiment to determine the average mass of
a U.S. penny.
3.067 3.049 3.039 2.514 3.048 3.079 3.094 3.109 3.102
Determine if the value of 2.514 g is an outlier at % 90 confidence level.
Comparing Two Sample Means
Tables below show results for two separate experiments to determine the mass of a circulating U.S.
penny. Determine whether there is a difference in the means of these analyses at a %95 confidence
level.
Tables below show results for two separate experiments to determine the mass of a circulating U.S.
penny. Determine whether there is a difference in the means of these analyses at a %95 confidence
level.
The %w/w Na2CO3 in soda ash can be determined by an acid–base titration.
The results obtained by two analysts are shown here. Determine whether the
difference in their mean values is significant at a %95 confidence level
Comparing Two Sample Variances
F-test Statistical test for comparing two variances to see if their
difference is too large to be explained by indeterminate error. The Ftest can be extended to the comparison of variances for two samples
If Fexp is greater than Fcri, the analysis is not under statistical control.
•
Tables below show results for two separate experiments to determine the mass of a circulating U.S.
penny. Determine whether there is a difference in the precisions of these analyses at %95
confidence level. the variances are s2A = 0.00259 and s2B= 0.00138.
A new method for Ni analysis in the stainless stel is being compared
with another method, is the new method is acceptable?
the results of the analysis of vitamin B are shown below,.
81,10 79,30 80,40 79,70
Calculate confidence interval for 95% confindence level
If the true value 80,00, is there any significant difference between
mean values at a %95 confidence level
Sistematik Belirli Hata
1- Sabit hatalar : analiz edilen numune miktarından bağımsızdır.
2- Orantılı hatalar : numunenin miktarı ile orantılı olarak azalır veya artar
F Testi (standart sapmaların karşılaştırılması)
•
İki farklı kişinin yaptığı analizlerin veya iki farklı metotla elde edilen sonuçların
standart sapmalarının karşılaştırılmasında kullanılır. Farklı sonuçların standart
sapmalarının karesi birbirine bölünür ve sonuç tablodaki F değeri ile karşılaştırılır.
Aqueous Solution Chemistry
 Precipitation Reactions
 Acids, Bases, and Neutralization Reactions
 Oxidation–Reduction (Redox) Reactions
 Complex formation reactions
Solutions
Homogeneous mixtures are called solutions.
Solvent
solute
is the component present in greatest
amount or determines the state of
solution exist
is a component present in a lesser
quantity and it is said to be dissolved in
the solvent.
Solvent is the component present in
greatest amount or determines the state of
solution exist.
A solute is a component present in a
lesser quantity and it is said to be
dissolved in the solvent.
Solubility
is the maximum amount of solute that dissolves in a given
quantity of solvent at a specific temperature to produce a
saturated solution
saturated
solution.
The solution in which no more solute can be dissolved is
said to be saturated solution.
Super saturated is a solution in which has higher amount of solute than
solution.
saturated solution
Unsaturated
solution
is a solution in which contain less amount of solute than
saturated solution
nonelectrolyte
Substances such as sucrose or ethyl alcohol, which do not produce ions
in aqueous solution, are called nonelectrolytes.
electrolytes
Substances such as NaCl or KBr, which dissolve in water to produce
conducting solutions of ions, are called electrolytes.
strong electrolytes,
weak electrolytes.
Compounds that dissociate to a large extent
(~100%) into ions when dissolved in water are
said to be strong electrolytes,
compounds that dissociate
to only a small extent are weak electrolytes.
Solution Concentrations
Mass of solute
Percent by mass = ------------------------ x 100
Mass of solution
Volume of solute
Percent by volume = ------------------------- x 100
Volume of solution
Mass of solute
Mass / volume percent = --------------------- x 100 (gr/100ml)
Volume of solution
Solution Concentrations
Molarity
Number of moles of solute
= ----------------------------------------------Number of liters of solution
Number of moles of solute
Molality = -------------------------------------------------Number of kilograms of solvent
Moles of component i
Mole fraction = --------------------------------------------------------total moles of all solution components
Parts per million (ppm),
mg solute
1 ppm = -------------------kg solution
Parts Per Billion(ppb)
mg solute
1 ppb = ----------------------kg solution
Mol
number
m
n = --------MA
Molarity
n
M= ------V
mol/L
m
d = ---------V
gr/mL
density
Diluting Concentrated Solutions
Minitial x Vinitial = mol number = Mfinal x Vfinal
Minitial x Vinitial = Mfinal x Vfinal
Dilution
The addition of solvent to solution is called dilution. it does not change the
amount of solute in a solution but changes the solution concentration..
Find the molarity of a solution that 23.4g of Na2SO4 was dissolved in water and diluted to
250.0ml
Na2SO4: 142
calculate the molarity of H2SO4 solution when we dilute 50.0 mL of a
solution of 2.00 M H2SO4 to a volume of 200.0 mL.
There is a 2000ml of 8M H2SO4 solution in the lab. A student needs 800ml of 2M
H2SO4 solution to carry out an experiment. In order to prepare this solution ahow many ml of 8 M H2SO4 solution should the student use? b- how many ml of
distilled water should be added?
Calulate the molarity of the solution prepared by mixing 400ml of 0.6M HNO 3
solution with 200ml water?
Q1
How can you prepare 500ml 0.10 M H2SO4 solution from 3.0M H2SO4?
Q2
How would you prepare 500.0 mL of 0.2500 M NaOH solution starting from a
concentration of 1.000 M?
What is the final concentration if 75.0 mL of a 3.50 M glucose solution is diluted
to a volume of 400.0 mL?
Q3
• question.
A solution has a density of 1.235g/ml and contains 90.0%
glycerol C3H8O3 and 10% H2O by mass. Determine
a- the molarity of C3H8O3
b- the molarity of H2O
c- mole fraction of C3H8O3
d- mole percent of H2O
e- the molality of H2O
Reactions in Aqueous Solution
Precipitation
reactions
Are the reactions, an insoluble solid is formed, the solid
product is called as a precipitate.
Ag+(aq) + Cl-(aq) → AgCl(s)
Solubility
is the maximum amount of solute that dissolves in a given
quantity of solvent at a specific temperature to produce a
saturated solution
saturated
solution.
The solution in which no more solute can be dissolved is
said to be saturated solution.
Super saturated is a solution in which has higher amount of solute than
solution.
saturated solution
Unsaturated
solution
is a solution in which contain less amount of solute than
saturated solution
• Solubility of most of solids increase with temperature.
• Solubility of gases decrease with increased temperature
• Solubility of a gas increases as the gas pressure is increased.
• Types of solvent and solute is important,
like dissolves like, generally nonpolar substances are soluble in nonpolar
substances, polar solvents dissolves ionic and polar substances.
solubility rules
Soluble salts
salts of 1A groups (Na, K, Li,) and NH4+ are soluble
all nitrates, acetates and perchlorates are soluble
NO3 (Nitrat), CHCOO- (asetat) ,ClO4 (perklorat)
all chlorides (halogens) are soluble except (AgCl, Hg2Cl2, PbCl2)
most sulfates (SO42+ ) are soluble except (Sr SO4, Ca SO4 , Ba SO4 , Pb SO4 )
Cl-, Br-, I- SO42Li+, Na+, K+, Rb+, Cs+ NH4+ NO3-
ClO4-
CH3CO2-
solubility rules
Slightly soluble
all OH- hydroxides are insoluble except those of 1A group (Na, K, Li,) and NH4+
all sulfides S2- are insoluble except those of 1A group (Na, K, Li,) and NH4+
all carbonates are insoluble except those of 1A group (Na, K, Li,) and NH4+
All PO43- are insoluble except those of 1A group (Na, K, Li,) and NH4+
CO32-, PO43- OH-, S2-
• What will happen if Na2CO3 and CaCl2 solutions are mixed ?
• What will happen if CuSO4 and NaNO3 solutions are mixed ?
Ionic compounds of alkali metals (like NaCl, KBr,
LiNO3 ..) generally dissolve completely in water.
solubility product
constant
But many ionic compounds are only slightly
soluble in water. We use the term insoluble for
such compound.
SrCrO4(s)  SrCrO4(aq)
SrCrO4(aq) 
Sr2+(aq)
+ CrO4
Ksp = [Sr2+] [CrO42-]
2-(aq)
Ksp is the solubility product constant and is equal
to the product of the concentration of the ions
involved in the equilibrium, each raised to the
power of its coefficient in the equilibrium
equation. Ksp has a fixed value for a given system
at a particular temperature.
Solubility equilibrium
Ksp
Solubility equilibrium
Ksp
Al(OH)3  Al3+ + OH-
1.3x10-33
ZnS  Zn2+ +S2-
1.6x10-24
BaCO3 Ba2+ +CO32-
5.1x10-9
MgCO3 Mg2+ +CO32-
3.5x10-8
PbCl2  Pb2+ + 2Cl-
1.6x10-5
AgCl  Ag+ + Cl-
1.8x10-10
question: write the solubility product constants
for
Ca3(PO4)2, Ag2CrO4, CaF2, PbSO4
question:
Solubility and Ksp
solid barium sulfate is shaken in
contact with pure water at 25°C for several days.
At the equilibrium saturated barium sulfate
Solubility
is
the
quantity
of
substance that dissolves in a given
solution contains 1.04x10-5M Ba2+, what is the Ksp
for BaSO4
quantity of water. It is often
expressed as grams of solid per
100g of water. Ksp and solubility
are
related
and
one
calculated from other.
can
be
question: Ksp for CaF2 is 3.9x10-11, what is the
solubility of CaF2 in water in grams per liter.
Common ion effect
The solubility of a slightly soluble ionic
compound is lowered when a second
soluble that furnishes a common ion is
added to solution.
question:
calculate the molar solubility of
Ag2SO4 in 1.0 M Na2SO4.
1.4x10-5
Ksp for Ag2SO4 is
Ksp and Precipitate
Qip is generally called ion product, it is the
formation
product of ion concentrations
raised to
appropriate powers.
Qip < Ksp no precipitate occur
SrCrO4(aq)  Sr2+(aq) + CrO42-(aq)
Qip = Ksp solution just saturated
Qip = [Sr2+] [CrO42-]
Qip > Ksp precipitation should occur
Selective prepicitation or fractional precipitation
One ion is precipitated while the other(s) remains in solution. For this purpose
there must be significant differences in their Ksp.
question: 0.10 M AgNO3 is adding slowly to a solution that has [CrO42-] =
0.010M and
[Br-]= 0.10M
( Ksp for Ag2CrO4 is 1.4x10-5, Ksp for
AgBr is 5.0 x10-13 )
show that AgBr (s) precipitates before than Ag2CrO4(s)
at the point where Ag2CrO4 begins to precipitate, what is the [Br-] remaining in
solution?
is separation of Br- and CrO42- by fractional precipitation feasible?
Activity Effects
activity
True thermodynamic constants use a species activity in place of its molar concentration (a).
activity coefficient
The number that when multiplied by a species’ concentration gives that species’ activity (g).
ionic strength
A quantitative method for reporting the ionic composition of a solution that takes into account the
greater effect of more highly charged ions (m).
Calculate the ionic strength of 0.10 M NaCl.
Repeat the calculation for a solution of 0.10 M Na2SO4.
Calculate the solubility of Pb(IO ) in a matrix
of 0.020 M Mg(NO ) .
3
3
2
2
Quantitatif Analysis
Titrimetry and Gravimetry
Gravimetry: is a method in which the signal is
a mass or change in mass to find the amount
(or concentration) of analyte in the sample. .
2 H2 (g)
2 molecules of H2
two moles of H2
4 grams of H2
reacts with
+
O2 (g)

2 H2O (s)
1molecules O2
one mole O2 and
reacts with
32 g O2
and
two moles H2O is produced.
36 g H2O
is produced.
Reaksiyon Stokiometrisi
2 H2 (g)
2 molekül H2
2 mol H2
4 gram H2
+
O2 (g)

2 H2O (s)
1 molekül O2 ile reaksiyona girer 2 molekül H2O oluşur
1 mol O2 ile reaksiyona girer ve
32 g O2
ile reaksiyona girer ve
2 mol H2O oluşur
36 g H2O oluşur
what mass of AgBr is formed when a solution containing 3.45 g of KBr
is mixed with a solution containing 7.28 g AgNO3?
KBr(aq) + AgNO3 (aq)
 AgBr(s) + K+ (aq) + NO3- (aq)
Gravimetry: is a method in which the signal is a mass or change in mass to find the
amount (or concentration) of analyte in the sample.
Types of Gravimetric Methods
precipitation
gravimetry
A gravimetric method in which the signal is the mass of a precipitate.
volatilization
gravimetry
A gravimetric method in which the loss of a volatile species gives rise to the
signal. In determining the moisture content of food, thermal energy vaporizes the H2O.
Fe3+ + OH- → Fe(OH)3 →Fe2O3
The amount of carbon in an organic compound may be determined by using the chemical
energy of combustion to convert C to CO2.
CaCO3 (k)→CaO (k) + CO2(g)
electrogravimetry
A gravimetric method in which the signal is the mass of an electrodeposit on
the cathode or anode in an electrochemical cell.
particulate
gravimetry
the analyte is determined following its removal from the sample matrix by
filtration or extraction. The determination of suspended solids is one example
of particulate gravimetry.
Quantitative Calculations In precipitation gravimetry
the relationship between the analyte and the precipitate is determined by the
stoichiometry of the relevant reactions.
Fe3+ + OH- → Fe(OH)3 →Fe2O3
Grams of analyte = k
 grams of precipitate
Gravimetric factor :
2
GF = ---- x
1
is a stochiometric ratio between the analyte and its precipitate
Formula weight of Fe
-------------------------------------molecular weight of Fe2O3
Mass of the Fe =
2x 56
the mass of the precipitate x -------------------------160
Quantitative Calculations In precipitation gravimetry
the relationship between the analyte and the precipitate is determined by the
stoichiometry of the relevant reactions.
% X (the percentage of analyte) =
Grams of analyte = k
 grams of precipitate
Gravimetric factor :
a
GF = ---- x
b
mass of the analyte X
--------------------------------------- x 100
sample mass
is a stochiometric ratio between the analyte and its precipitate
Formula weight of the analyte
---------------------------------------------------molecular weight of the final precipitate
Mass of X analyte =
the mass of the precipitate x GF
An ore containing magnetite, Fe3O4, was analyzed by dissolving a 1.5419 –g
sample in concentrated HCl, giving a mixture of Fe2+ and Fe3+ . After adding
HNO3 to oxidize any Fe2+ to Fe3+ the resulting solution was diluted with water
and the Fe3+ precipitated as Fe(OH)3 by adding NH3. After filtering and rinsing,
the residue was ignited, giving 0.8525 g of pure Fe2O3. Calculate the
%w/w Fe3O4, in the sample.
An impure sample of Na3PO3 weighing 0.1392 g was dissolved in 25 mL of
water. A solution containing 50 mL of 3% w/v mercury(II) chloride, 20 mL of
10% w/v sodium acetate and 5 mL of glacial acetic acid was then prepared.
The solution containing the phosphite was added dropwise to the second
solution, oxidizing PO3 3– to PO4 3– and precipitating Hg2Cl2. After digesting,
filtering, and rinsing, the precipitated Hg2Cl2 was found to weigh 0.4320 g.
Report the purity of the original sample as %w/w Na3PO3
Phosphorus has an oxidation state of +3 in PO3 3– and +5 in PO4 3–; thus,
oxidizing PO3 3– to PO4 3– requires two electrons.
The formation of Hg2Cl2 by reduction of HgCl2 requires 2 electrons as the
oxidation state of each mercury changes from +2 to +1.
Since the oxidation of PO3 3– and the formation of Hg2Cl2 both require two
electrons, we have Moles Na3PO3 = moles Hg2Cl2
An impure sample of Na3PO3 weighing 0.1392 g was dissolved in 25 mL of water. A
solution containing 50 mL of 3% w/v mercury(II) chloride, 20 mL of 10% w/v sodium
acetate and 5 mL of glacial acetic acid was then prepared. The solution containing the
phosphite was added dropwise to the second solution, oxidizing PO3 3– to PO4 3– and
precipitating Hg2Cl2. After digesting, filtering, and rinsing, the precipitated Hg2Cl2 was
found to weigh 0.4320 g. Report the purity of the original sample as %w/w Na3PO3
Moles Na3PO3 = moles Hg2Cl2
Soru: to find the %w/w of NaCl in a mixture, 0,9532 g sample is
disolved in pure water and precipitated with AgNO3. After drying
the AgCl precipitate ie weighed 0,7033g. What is the %w/w of NaCl
in a mixture? (NaCl: 58,44 AgCl: 143,32)
A typical gravimetric analysis procedure may be divided into five stages:
sample pretreatment; precipitation; filtration; drying and ignition; weighing.
1- sample pretreatment
2 - precipitation;
3- filtration;
5- drying and ignition;
6- weighing.
Çökelek ve çöktürücülerin özellikleri.
Precipitant A reagent that causes the precipitation of a soluble
species. Must be selevtive for one analyte.
Precipitate: is the product of a simple reaction between the analyte
and precipitant. The precipitate must be formed quantitatively and within a
reasonable time. Its solubility should be low enough for a quantitative
separation to be made. It must be readily filterable and, if possible, have a
known and stable stoichiometric composition when dried so that its weight can
be related to the amount of analyte present. Failing this, it must be possible to
convert the precipitate to a stoichiometric weighable form (usually by ignition).
The theoretical and experimental details of precipitation gravimetry
Precipitates form in two ways; by nucleation and by particle growth.
Nucleation :
A few ions, atoms or molecules come together to form
a stable solid.
Particle growth
is the addition of new preciptates onto existing
nucleus..
If nucleation predominates, a large number of very small particles results, if
particle growth predominates a smaller number of larger particles is obtained.
Solutions with a large, positive value
of relative supersaturation (RSS)
show high rates of nucleation,
producing a precipitate consisting of
numerous small particles.
When RSS is small, precipitation is
more likely to occur by particle growth
than by nucleation.( smaller number of
larger particles is obtained)
Controlling Particle Size
Following precipitation and digestion, the precipitate must be separated from the supernatant solution
and freed of any remaining impurities, including residual solvent. These tasks are accomplished by
filtering, rinsing,and drying the precipitate. The size of the precipitate’s particles determines the ease
and success of filtration. Smaller, colloidal particles are difficult to filter because they may readily pass
through the pores of the filtering device. Large, crystalline particles, however, are easily filtered.
A solute’s relative supersaturation, RSS, can be expressed as
Q is the solute’s actual concentration,
S is the solute’s expected concentration at equilibrium,
Q – S is a measure of the solute’s supersaturation when precipitation begins.3
A large, positive value of RSS indicates that a solution is highly supersaturated.
Such solutions are unstable and show high rates of nucleation, producing a
precipitate consisting of numerous small particles. When RSS is small,
precipitation is more likely to occur by particle growth than by
nucleation.
Colloidal solution: the particle size are in the range of 10-7-10-4 cm. colloidal solutions
are genereally not filterable.
Crystalline suspensions. Particle size are around 10-1 mm, they have the tendency to
precipitate.
A precipitate’s solubility usually increases at higher temperatures, and adjusting pH
may affect a precipitate’s solubility if it contains an acidic or basic anion.
Temperature and pH, therefore, are useful ways to increase the value of S. Conducting
the precipitation in a dilute solution of analyte, or adding the precipitant slowly and
with vigorous stirring are ways to decrease the value of Q.
When RSS is higher, colloidal solutions will form, when RSS is lower crystalline
solutions will form
Colloidal solution:
coagulation
The process of smaller particles of precipitate clumping together
to form larger particles.
Heating and string.
Peptization
İs a process by which a coagulated colloid returns to smaller
particles. It could be happen when preciptate is being
washed. HNO3, HCl, NH4Cl gibi elektrolitler eklenir.
A 101.3-mg sample of an organic compound known to contain Cl is burned in pure O2 and the
combustion gases collected in absorbent tubes. The tube used to trap CO2 increases in mass by
167.6 mg, and the tube for trapping H2O shows a 13.7-mg increase. A second sample of 121.8 mg
is treated with concentrated HNO3 producing Cl2, which subsequently reacts with Ag+, forming
262.7 mg of AgCl. Determine the compound’s composition, as well as its empirical formula.
Çökeltinin Saflığı
Normal olarak çözeltide kalması istenen bileşiklerin çökeltiye geçmesi safsızlıklara,
ürün kütlesinin yanlış tartılmasına dolayısıyla analizde hataya neden olur.
Yüzey Adsorpsiyonu normal koşullarda çözünen bileşiğin çöken teneciklerin
yüzeyinde tutunmasıdır. Koloidal çökeleklerde görülür. Yıkama,
Yeniden çötürme ile azaltılabilir.
Karışık kristal
oluşumu
kristalin yapısında bulunan bir iyonun yerine ortamda bulunan
başka bir iyonun yer değiştirerek çökmesidir.
Hapsetme ve
Mekanik
sürüklenme
kristaller arasındaki boşluklarda bir bileşiğin kalmasıdır. Yavaş
çötürme ve bağıl aşırı doygunluk kontrol edilerek azaltılır.
Titration
•
Titration is a procedure for determining the concentration of a solution
by allowing a carefully measured volume to react with a standard solution
of another substance, whose concentration is known. By finding the volume
of the standard solution that reacts with the measured volume of the first
solution, the concentration of the first solution can be calculated.
• equivalent point
The point that all reactants are consumed, Stoichiometric mol numbers of both
reactants are equal.
•
an indicator a compound that change its color around equivalent point. such
as phenolphthalein, is colorless in acidic solution but turns pink in basic
solution.
Calculations in the volumetric titrimetry
Mol number
m
n = --------MA
Molarity
n
M= ------V
mol/L
Calculations in the volumetric titrimetry
aA + bB
↔ cC + dD
a mol A react with b mol B
When A is a standard (titrant) the volume of A used in the titration and its molarity
can be used to calculate the mol number of B
A 25.0 mL sample of vinegar (dilute acetic acid, ) is titrated and
found to react with 94.7 mL of 0.200 M NaOH. What is the molarity
of the acetic acid solution?
Soru: antiasit ilaç tabletlerinin anabileşeni CaCO3 tür. 0.542 gr olarak tartılan bir
tablet HCl ile titre edildiğinde, reaksiyonun tamamlanması için 38.5 ml 0.200M
HCl harcandığına göre tablet içindeki CaCO3 yüzdesi nedir?
CaCO3
+ HCl → Ca2+ + CO2 + Cl- + H2O
soru: 0,2879g sodyum okzalatı (Na2C2O4) asidik çözeltide aşağıdaki
redoks tepkimesine göre titre etmek için 25,12mL KMnO4 çözeltisi
harcandığına göre KMnO4 ün molaritesi nedir?
C2O42- + MnO4- → Mn2+ + CO2
Titrimetric Analysis
konsantrasyonu bilinen bir çözeltinin analit ile reaksiyona giren miktarının ölçümüne
dayanan kantitatif analiz metotlarıdır. Standart çözelti reaksiyon tamamlanıncaya dek
ortama yavaş yavaş eklenir.
Volumetry
Standart reaktifin
hacminin ölçülmesi
temeline dayanan bir
titrimetrik metottur
- İndirgenme-Yükseltgenme
reaksiyonları
- Asit Baz reaksiyonları
- Kompleks oluşum
reaksiyonları
- Çöktürme titrasyonları
Gravimetric Titrimetry
Coulometric titrimetri
Standart
reaktifin
kütlesinin
ölçülmesine dayanan titrimetrik
metottur
Analit ile bir reaksiyonun
tamamlanması için gerekli
kulon
cinsinden
yük
miktarının
ölçüldüğü
titrimetrik metottur.
Titrant: The reagent added to a solution containing the analyte and whose volume is
the signal.
titration: is a process in which a standard reagent is added to a solution of an analyte until the
reaction between the analyte and reagent is completed.
equivalence point The point in a titration where stoichiometrically equivalent
amounts of analyte and titrant react.
Indicator A colored compound whose change in color signals the end point of a
titration.
end point The point in a titration where we stop adding titrant.
titration error The determinate error in a titration due to the difference between the
end point and the equivalence point.
Veq:
Venepoint:
Acids and Bases
Acids and Bases
Acids;
Have a sour taste,
dissolve metals such as zinc
and carbonate minerals
change color of litmus to red
Acetic acid in vinegar,
Citric acid in lemons
Bases
Have a bitter taste,
Have a slippery feel
change color of litmus to blue,
React with dissolved metal to
form prepiciate
Household cleaning products
Acid-Base Definitions
Arrhenius Acid-Base
Definition
Brønsted-Lowry Acid-Base
Definition
Lewis acid-base
Definition
Arrhenius Acid-Base Definition (1884)
An acid is a substance that contains hydrogen and dissociates to
produce Hydrogen ion : H+
HCl(aq)  H+(aq) + Cl-(aq)
A base is a substance that contains the hydroxyl group and
dissociates to produce Hydroxide ion : OH –
NaOH (aq)  Na+ (aq) + OH -(aq)
Neutralization is the reaction of an H+ ion from the acid and
the OH - ion from the base to form water, H2O
H+(aq) + OH-(aq)
<=> H2O(l)
Brønsted-Lowry Acid-Base Definition (1923)
An acid is a species having a tendency to donate an H+ ion.
HCl
+

H 2O
Cl–
+
H3O+
A base is a species having tendency to accept an H+ ion.
NH3 +
H 2O

NH4+
+
The Conjugate Pairs in Some Acid-Base Reactions
Conjugate Pair
Acid
+
Base
Base
+
Conjugate Pair
Acid
OH-
In the Brønsted-Lowry perspective, one species donates a proton and another
species accepts it: an acid-base reaction is a proton transfer process.
question : show the conjugated acid base pairs in the reactions
below
HF
+
H2O

F–
+
H3O+
H2PO4–
NH4+
+
OH–
+ CO32–


HPO42–
NH3
+
+
H2O
HCO3–
The Lewis acid-base definition :
• A base is any species that donates an electron pair
• An acid is any species that accepts an electron pair.
Strengths of Acids
• Strong Acids; An acid that completely ionized in water, is
called as a strong acid
• HCl(aq) + H2O  H3O+(aq) + Cl-(aq)
• Weak acid is an acid that partly ionized in water.
• CH3COOH (aq) + H2O CH3COO-(aq) + H3O+(aq)
Percent Ionization
HA + H2O ↔ H3O+ + A-
[H3O+] from HA
Degree of ionization =
[HA] originally
[H3O+] from HA
Percent ionization =
[HA] originally
x 100%
Strong acids.
1. The hydrohalic acids HCl, HBr, and HI
2. Oxoacids in which the number of O atoms exceeds the number of ionizable H
atoms by two or more, such as HNO3, H2SO4, HClO4
Weak acids.
1. The hydrohalic acid HF
2. Those acids in which H is bounded to O or to halogen, such as HCN and H2S
3. Oxoacids in which the number of O atoms equals or exceeds by one the number
of ionizable H atoms, such as HClO, HNO2, and H3PO4
4. Organic acids (general formula RCOOH), such as CH3COOH and C6H5COOH
Strong bases. A base that completely ionized in water, is called as a strong base.
Soluble compounds containing O2- or OH- ions are strong bases.
NaOH (aq)  Na+ (aq) + OH -(aq)
1) M2O or MOH, where M= Group 1A(1) metals (Li, Na, K, Rb, Cs)
2) MO or M(OH)2, where M = Group 2A(2) metals (Ca, Sr, Ba) [MgO and Mg(OH)2
are only slightly soluble, but the soluble portion dissociates completely.]
A weak base is a base that partly ionized in water. Many compounds with an
electron-rich nitrogen are weak bases (none are Arrhenius bases). The common
structural feature is an N atom that has a lone electron pair in its Lewis structure
NH3(g) +H20(l)  NH4+ (aq) + OH -(aq)
1) Ammonia (:NH3)
2) Amines (general formula RNH2, R2NH, R3N), such as CH3CH2NH2, (CH3)2NH,
(C3H7)3N, and C5H5N
The Meaning of Ka, the Acid Dissociation Constant
Ka is the equilibrium constant of the ionization reaction of a weak acid.
Kb represents the ionization constant of a base.
Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25°C)
The pH Scale
•
•
The pH is defined as the negative logarithm in base 10, of the hydronium
ion concentration
• pH = - log[H3O+]
The pOH is defined as the negative logarithm in base 10, of the hydroxyl
ion concentration
• pOH = - log[OH-]
pH of an acidic solution < 7.00
pH of a neutral solution = 7.00
pH of a basic solution
> 7.00
Calculating [H3O+], pH, [OH-], and pOH
Strong acid (or
base)
[H+] = acid
concentration
Weak acid (or
base)
Strong and weak acid
(or base)
calculate [H+] from
Ka
[H+] is generally equal to
strong acid (effect of weak
acid neglect unless strong
acid is too dilute)
question : What is the pH of a solution that contains 10-2
M hydronium ion ?
question : Calculate the [H3O+], pH, [OH-], and pOH of
0.0024 M hydrochloric acid solutions at 25°C.
• question : The weak acid hypochlorous acid is formed
in bleach solutions. What is the [H3O+] of a 0.125 M
HClO solution?
Ka = 3.5 x 10^-8
Buffer solution
• is a solution whose pH changes only very slightly
upon the addition of small of either an acid or a base.
• Buffer solutions contain a weak acid and its conjugate
base ( its salt) or a weak base and its conjugate acid
( its salt).
• question : calculate the the pH of a buffer solution containing
0.246 M NH3 and 0.0954 M NH4Cl (for NH3 Kb = 1.74x
10-5)
• NH3 +H2O
 NH4+ + OH-
Hidrolysis
Salts of the weak acids (or weak bases) react with water
to form the weak acid (or weak base).
Salt
+ Water ↔ weak acid + OH-
equilibrium constant of this reaction is Kh=Kw/Ka
the pH of the solution can be calculated from Kh.
question: calculate the ph of the 0.01 M sodium acetat solution.
For acetic acid, Ka= 1.8x10-5
Polyprotic Acids
Phosphoric acid:
A triprotic acid.
H3PO4 + H2O ↔ H3O+ + H2PO4-
Ka = 7.1x10-3
H2PO4- + H2O ↔ H3O+ + HPO42-
Ka = 6.3x10-8
HPO42- + H2O ↔ H3O+ + PO43-
Ka = 4.2x10-13
Ionization Constants of Some Polyprotic Acids
Phosphoric Acid
• Ka1 >> Ka2
• All H3O+ is formed in the first ionization step.
• H2PO4- essentially does not ionize further.
• Assume [H2PO4-] = [H3O+].
• [HPO42-] ≈ Ka2 regardless of solution
molarity.
Calculating Ion Concentrations in a Polyprotic Acid Solution.
For a 3.0 M H3PO4 solution, calculate:
(a) [H3O+];
]
(b) [H2PO4-];
[HPO42-]
(d) [PO43-
H3PO4 + H2O ↔ H2PO4- + H3O+
Initial conc.
3.0 M
0
0
Changes
-x M
+x M
+x M
Eqlbrm conc.
(3.0-x) M
xM
xM
Acid Base titrations
A titration in which the reaction between the analyte and titrant is
an acid–base reaction.
Standard
reactants:
acid–base titrations were conducted using H2SO4, HCl, and
HNO3 as acidic titrants, and K2CO3 and Na2CO3 as basic
titrants.
Indicator :
A colored compound whose change in color signals the end point of a
titration.
End points were determined using visual indicators such as litmus,
which is red in acidic solutions and blue in basic solutions,
Acidimetry:
Determination of the Base content of a sample by titrating strong acid
solution. Diluted solutions of HCl, H2SO4, HClO4 are used as a titrant. Since
HNO3 gives some extra reactions, it can not be used for acidimetry.
Alkalimetry:
Determination of the acid contents of a sample by titrating with a strong base
solutions like NaOH, KOH ve Ba(OH)2
Visual Indicators for Acid-base Titrations
an acid–base indicator are the weak organic acids. the weak acids
and bases derivatives of organic dyes can serve as a useful means
for determining the end point of a titration. Because such
compounds have at least one conjugate acid–base species that is
highly colored,
Indicator A colored compound whose change in color signals the
end point of a titration.
an acid–base indicator changes color with in changing the pH of the
solution.
In general terms a visual indicator is a compound which changes from one colour
to another as its chemical form changes with its chemical environment
If the indicator is present in an environment where a
titration reaction generates or consumes the X species,
the indicator will change with the concentration of X in
the solution and the colour of the solution will be
determined by the ratio [InB]/[InA]. As a general guide,
the eye will register a complete change from one colour
to the other when this ratio changes from 10:1 to 1:10.
The pH at which an acid–base indicator changes color is determined by
its acid dissociation constant. For an indicator that is a monoprotic weak
acid, HIn, the following dissociation reaction occurs
HIn(aq) + H2O(l)↔H3O+(aq) + In–(aq)
For example, the acid-base indicator methyl orange has a pKIn of 3.7 and will thus
change colour over the pH range 2.7–4.7.
The ultimate sharpness of the end point will further depend upon the rate at which pX
is changing at the end point of the titration.
Acid-base Titration Curves
titration curve: A graph showing the progress of a titration as a function of the
volume of titrant added.
A titration curve provides us with a visual picture of how a property, such as pH, changes as
we add titrant
Titrating Strong Acids and Strong Bases
Titrating a Strong Titrating a
StrongAcid with a
Base with a
Strong Base
Strong Acid
Titrating a Weak
Titrating a Weak
Acid with a Strong Base with a Strong
Base
Acid
Titration Curves
Titrating a Strong
Acid with a Strong
Base
starts with a low pH, then pH changes less till equivalent
point then there will be sudden change (increase) at the
equivalent point.
Titrating a Strong
Base with a Strong
Acid
The curve starts at the basic region, till equivalent point
pH decrease less but at the equivalent point sudden
change (decrease) will occur.
Titration Curves
Titrating a Weak Acid
with a Strong Base
Titrating a Weak Base
with a Strong Acid
the titration curve when titrating a diprotic weak acid, H2A,
with a strong base.
pH Calculations in the Titrating of a Strong Acid with a
Strong Base
Before
point
equivalence
At the equivalence
point
After
point
equivalence
question: 50.0ml 0.050 M HCl solutrion is being titrated with 0.100
M NaOH, calculate the pH when
0 ml base added,
10 ml base added, 20ml base added,
25ml base added,
30 ml base added, and 40 base added,
Draw the titration curve.
pH Calculations in the Titrating of a weak Acid with a Strong Base
Buffer solution
A solution containing a conjugate weak acid/weak base pair
that is resistant to a change in pH when a strong acid or
strong base is added.
Buffer capacity
is the mol number of a stronge acid or base needed to add for
changing the pH 1,0L solution in 1 unit
The pH of a buffer changes within the range of pH = pKa ± 1
Question: calculate the the pH of a solution containing 0.246 M NH3 and 0.0954
M NH4Cl
(for NH3 Kb = 1.74x 10-5)
NH3 +H2O
 NH4+ + OH-
Titrating of a weak Acid with a Strong Base
Hidrolysis
Salts of the weak acids (or weak bases) react with water
to form the weak acid (or weak base).
Salt
+ Water ↔ weak acid + OH-
equilibrium constant of this reaction is Kh=Kw/Ka
the pH of the solution can be calculated from Kh.
Question : calculate the pH of a 0.01 M sodium acetat solution.
Ka= 1.8x10-5
pH Calculations in the Titrating of a weak Acid with a Strong
Base
.
Adding 0ml base
Calculate the pH from Ka of the weak acid
Before
the Calculet the pH of the buffer solution
Equivalence point
At the
point
Equivalence Calculate pH of the salt of the weak acid (Hydrolysis)
After the Equivalence Calculate the pH from the concentraion of the base (not
point
consumed in the reaction)
Soru: 50ml 0.10 M acetic acid is being titrated with 0.10 M NaOH. calculate
the pH when.
0ml base added ,
10 ml base added
45 ml base added
50 ml baz ilavesi
51 base added 55 ml base added
Ka= 1.8x10-5
Titrasyonda İndikatörden gelen Hatalar
Belirli hatalar
Rastgele hatalar
indikatörün renk değişim pH renk
değişimi
izlenirken
aralığının titrasyonun eşdeğerlik yanılgılarından kaynaklanır.
noktasındaki pH ının faklı olması
sonucu olur. Uygun indikatör seçimi,
tanık deney düzeltmesi ile azaltılır.
göz
complexometric titrations
A complexometric titration is based on the essentially stoichiometric
reaction of a complexing agent with another species to form a complex
species that is only slightly dissociated and that is soluble in the titration
medium.
Complex
The product of the complexation reaction is called a metal–ligand
complex.
Ligant
A Lewis base that binds with a metal ion
Chelat
bir metalin bir ligantla bağlanarak oluşturduğu halkalı kompleks tir
formation
constant
The equilibrium constant for a reaction in which a metal and
a ligand bind to form a metal–ligand complex (Kf).
EDTA
The most widely use ligands is ethylenediaminetetraacetic
acid, EDTA, forms strong 1:1 complexes with many metal
ions.
Dissosiation constants of EDTA:
K1= 1.02x10-2 K2= 2.14x10-3 K3= 6.92x10-7 K4= 5.50x10-11
EDTA çözeltilerinde α değerlerinin hesaplanması
EDTA Hidrojen iyonlarını kaybederek
şu türlere iyonlaşır
Y4- nin çözeltideki
mol kesri
[Y4- ]+[HY3-]+[H2Y2-]+[H3Y-]+[H4Y]
K1= 1.02x10-2 K2= 2.14x10-3
K3= 6.92x10-7 K4= 5.50x10-11
[Y4-]
α4= -------------CEDTA
CEDTA= [Y4- ]+[HY3-]+[H2Y2-]+[H3Y-]+[H4Y]
Buradan
K1K2K3K4
K
sabitleri α4= ------------------------------------------------------------------------kullanılarak
[H+]4 + K1[H+]3-+ K1K2[H+]2- + K1K2K3[H+] + K1K2K3K4
K1K2K3K4
α4 = -----------------------------CEDTA
olarak bulunur
EDTA Metal İyonları Kompleksleri
Mn+
+ Y4-
↔ MYn-4
K’MY (durum sabiti)
[MYn-4 ]
KMY = -----------------[Mn+] [Y4-]
[MYn-4 ]
K’MY = ----------------- = α4 KMY
[Mn+] CEDTA
Calculate the Concentration of Ni2+ in a soltion which
prepared as mixing
50.0ml 0.0300 M Ni2+ and
50.0mlml 0.0500 M EDTA at pH= 3
At pH=3 , α4= 2.5x10-11 KNiY=4.2x1018
EDTA titrasyon eğrileri
Verilen pH değeri için durum sabiti değeri bulunur hesaplamalarda
kullanılır.
Eşdeğerlik
ortamda metal iyonu fazlası vardır ve
noktası öncesinde harcanmadan
kalan
metal
iyonu
konsantrasyonu hesaplanır.
Eşdeğerlik
noktasında
ortamda sadece kompleks vardır ve bir miktar
ayrışır, ayrışan metal iyonu konsantrasyonu
durum sabitinden hesaplanır.
Eşdeğerlik
noktası
sonrasında
EDTA fazlası vardır ve artan EDTA
konsantrasyonu durum sabitinde yerine
konarak
metal
iyonu
konsantrasyonu
hesaplanır.
50.0ml 0.0500 M Mg2+ nın 0.0500 M EDTA ile
titrasyonunda pH=10 na tamponlanıyor,
0ml, 5.0ml, 50.0ml ve 51.0 ml EDTA ilaveleri için
Mg2+ konsantrasyonunu hesaplayınız ve titrasyon
eğrisini çiziniz.
PH=10 için α4= 3.5x10-1 Kol= 6.2 x 108
EDTA titrasyonları için İndikatörler
Genel olarak metal iyon indikatörleri kullanılır. Bunlar metal
iyonlarına bağlandığında renk değiştiren organik boyalardır.
MetalIn + EDTA → MetalEDTA + In
Renk 1
renksiz
renksiz
renk 2
Chelometric titrations may be classified according to their manner of performance: direct
titrations, back titrations, substitution titrations, or indirect methods.
direct
titrations
back titrations
substitution
titrations
indirect methods
0,300 g metal wire is being dissolved in nitric acid and diluted to 100,0ml with pure
water. 25.00ml sample is taken from this solution and added 0.052 M EDTA and
then neutralized with NaOH. After that the pH of the solution is buffered to pH=5.5.
and excess EDTA is back titrated with 17.61ml 0.02299M Zn2+. Calculate the Ni2+
percent in the metal wire. (Ni: 58,71g/mol EDTA:372,0g/mol)
Application of Equilibrium Calculations to Complex Systems
when aqueous solutions contain several species that interact with one
another and water that yield two or more simultaneous equilibria.
PbS ↔ Pb2+ + S2S2- + H2O ↔ HS- + OHHS- + H2O ↔ H2S + OH2 H2O ↔ H3O+ + OH-
Steps for solving problems involving several equilibria
1.
2.
3.
4.
5.
6.
7.
8.
9.
Write all chemical equations and balance them
Write all equilibrium constant wxpressions.
State (undrline) which quantity (unknown) you want to find
(solve).
Write mass balance expression for the system.
Write the charge balance equation (IF possible)
Compare the number of equations and the number of
unknowns.
Make suitable approximations to simplify the algebra.
Solve the algebraic equations, find the unknown defined at
step 3.
Check the validity of approximations.
mass balance equation: (which is simply a statement of the conservation of
matter.)
An equation stating that matter is conserved, and that the total amount
of a species added to a solution must equal the sum of the amount of
each of its possible forms present in solution.
charge balance equation: (is a statement of solution
electroneutrality.) An equation stating that the total concentration of positive
charge in a solution must equal the total concentration of negative charge.
Total positive charge from cations = total negative charge from anions.
Write the mass balance equation for 1 liter of 0.250mol H3PO4
aqueous
solution.
Write charge balanse equation for a solution which contains
H+, OH-, K+, H2PO4-, HPO42- and PO43-
Calculate the Hg22+ cocentration in a saturated Hg2Cl2 solution.
Ksp= 1.2x10-18 Kw=1x10-14
Calculate the molar solubility of Mg(OH)2 in water.
Ksp Mg(OH)2 = 7.1x10-12 Kw=1x10-14
Calculate the molar solubility of Fe(OH)3 in water.
Ksp Fe(OH)3 = 2.0x10-39 Kw=1x10-14
Calculate the solubility of calcium oxalate in a solution which
the Hydronium ion cocentration is 1.0 x 10-4
For CaC2O4 Ksp= 2.3 x 10-9
for H2C2O4 Ka1= 5.36 x 10-2, Ka2= 5.42 x 10-8
Kw=1x10-14
Calculate the solubility of AgBr in 0.1 M NH3 solution.
For AgBr Ksp= 5.2 x 10-13
for Ag(NH3)+ Kformation oluşum= 2 x 103
for Ag(NH3)2 Kformation= 6.9 x 103
for NH3 Kb= 1.76 x 10-5
Soru: 0.1 M Pb2+ ve 0.1 M Fe2+ içeren bir çözeltiye H2S
eklenerek
Pb2+ ve Fe2+ iyonları birbirinden ayrılması için
gerekli koşulları belirleyiniz. Kçç(PbS)=3x10-28,
Kçç(FeS)=8x10-19 H2S için K1=9.6x10-8 K2=1.3x10-14
Calculate the solubilty of HgS in water by using all equations
which occur.
for HgS Ksp = 5 x 10-54
Kw= 1.0 x 10-14,
for H2S Ka1= 9.6x10-8 Ka2= 1.3x10-14