analytic chemistry
TRANSCRIPT
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Solutions
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Solution Compositions
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Units of Concentration
*Formalityidentical to molarity that is used for
solutions of ionic salts that do not exist as
molecule in solid or in solution.
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Units of Concentration
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Units of Concentration
for salts: net charge of an ion
for redox reactants: no. of electrons lost or gain
Normality = fx Molarity
f:
for acids: no. of replaceable H+
for bases: no. of replaceable OH-
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Units of Concentration
Mole fraction
Mole percent
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Units of Concentration
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Units of Concentration
1 ppm = 1 mg/L1 ppb = 1 g/L
1 ppt = 1 ng/L
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Example 1
What is the molarity of water at 40C?
Ans. 55.5 M
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Example 2
The concentration of glucose in normal
spinal fluid is 75 mg/100 g. What is
the molal concentration?
Ans. 4.2 x 10-3M
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Example 3
A 34.00%-by-mass solution of H3PO4in
water has a density of 1.209 g/cm3at
200
C. What is the molarity andmolality of this solution?
Ans: 4.19 M, 5.26 m
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Example 4
The hardness of water (hardness count) is
usually expressed as parts per million
(by mass of CaCO3), which is equivalent
to milligrams of CaCO3per liter of water.
What is the molar concentration of Ca2+
ions in a water sample with hardness
count of 175?
Ans. 1.75 x 10-3M.
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Acids and Bases
Arrhenius Theory
Bronsted-Lowry Theory
Lewis Theory
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Acids and Bases
Arrhenius Theory of Acid and Bases:
An acid produces H+and
a base produces OH-
in aqueous solutions.
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Acids and Bases
Brnsted-Lowry Theory of Acid and Bases:
An acid is a proton donor, and
a base is a proton acceptor.
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Acids and Bases
Brnsted-Lowry Theory of Acid and Bases:
*For every acid, there is a conjugate base and for every
base, there is a conjugate acid.
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Acids and Bases
Brnsted-Lowry Theory of Acid and Bases:
Role of Solvent:
- Makes to act the substance to behave as an acid or base.
Amphotericrefer to a substance that can act as acid or base.
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Acids and Bases
Brnsted-Lowry Theory of Acid and Bases:
Classification of Solvents:
Solvent Example
1. Protophilic (proton seeking) H2O, NH32. Protegenic (proton generating) H2O,HOAc
3. Amphiprotic (both 1 & 2) H2O, EtOH
4. *Aprotic C6H6, CCl4
*independent of proton seeking and generating.
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Acids and Bases
Lewis Theory of Acids and Bases
A Lewis acidis a species (an atom, ion or
molecule) that is an electron-pair acceptor.
A Lewis baseis a species that is an electron-
pair donor.
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Acids and Bases
Lewis Theory of Acids and Bases
Ligandsmolecules or ions that behave as
Lewis bases.
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Acids and Bases
Lewis Theory of Acids and Bases
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Strengths of Acids and Bases
STRONG WEAK
Acids Bases Acids Bases
HCl LiOH HF NH3
HI NaOH HCN Organic basesHBr KOH H3PO4 Amines
HClO4 RbOH H3BO3
HClO3 CsOH H2CO3
HNO3
Mg(OH)2
H2
SO3
HBrO3 Ca(OH)2 Carboxylic acids
*H2SO4 Ba(OH)2 Polyprotic acids
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Strengths of Acids and Bases
*HClO4, HBr, H2SO4, HCl and HNO3are of
same strengths in water; but in glacial acetic
acid, their strengths are:
HClO4 > HBr > H2SO4> HCl > HNO3
This phenomenon is known as levelling effect
of water.
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p-Functions
p-Functions are a method of expressing
concentrations, especially very large or very small
values.
pHcommonly known p-Function
- defined as negative logarithm of hydronium ionconcentration.
pH = - log[H3O+]
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* pH can be less than 0 or greater than 14.
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pH Calculations
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pH Calculations
Other values of Kw...
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pH Calculations
For strong acids, wherein water contributes to
H3O+in solution;
H2O + H2O H3O+ + OH-
I: 55.5 M 55.5 M MSA 0
: - X -X + X + X
F: 55.5-X 55.5-X MSA+ X X
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pH Calculations
Kw = [H3O+] OH = 1 x 104
Kw = [MSA X] X = 1 x 104
X2 MSA X 1 x 104 = 0
Then,
pOH = -log [X]
pH = 14 + log [X]
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pH Calculations
For strong bases, wherein water contributes to
OH- in solution;
H2O + H2O H3O+ + OH-
I: 55.5 M 55.5 M 0 MSB
: - X -X + X + X
F: 55.5-X 55.5-X X MSB+ X
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pH Calculations
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Example 5
What is the pH of 1 x 10-8HNO3solution?
Ans. 6.98
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Example 6
What is the pH of pure water at 500C?
Ans. 6.63
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Example 7
How much water is to be added to 12 M HCl
in order to prepare 1600 mL solution of pH
= 1.50?
Ans. 1595.784 L
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Example 8
Calcium hydroxide solution has a
concentration of 0.05 M. Calculate its
pH.
Ans. 13
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Example 9
A solution is prepared by adding 125.0 mL of
0.025 M HNO3 to 150.0 mL of 0.020 M
HCl. Determine the concentration of pOH
of the resulting mixture.
Ans. 12.35
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pH Calculations
For weak acids:
I: MWB 55.5 M 0 0
: - X -X + X + X
F: MWBX 55.5-X X X
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pH Calculations
For weak acids:
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pH Calculations
For weak bases:
I: MWB 55.5 M 0 0
: - X -X + X + X
F: MWB- X 55.5-X X X
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pH Calculations
For weak bases:
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pH Calculations
For weak acids and
acidic salt:
X2
K x M K = 0
* K = Kafor Weak
acid
K = Kw/K
bfor
Acidic salt
pH = -log [X]
For weak bases andbasic salt:
X2 K x M K = 0
* K = Kafor Weak base
K = Kw/K
afor Basic
salt
pH = 14 + log [X]
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pH Calculations
Percent Ionization
equilibrium concentration of ionized acidinitial concentration of acid
X 100
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Example 10
Calculate the pH of a 0.010 M solution
of iodic acid (HIO3, Ka= 0.17).
Ans. 2.02
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Example 11
What mass of benzoic acid, HC7H5O2 is
needed to dissolve in 350.0 mL of
water to produce a solution having a
pH of 2.85? Ka= 6.3 x 10-5.
Ans. 1.4 g
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Example 12
Caproic acid, HC6H11O2, found in small
amounts in coconut and palm oils, is
used in making artificial flavors. A
saturated aqueous solution of the
acid contains 11 g/L and has pH =
2.94. Calculate the Kafor the acid.Ans. 1.4 x 10-5
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Example 13
What is the percent ionization of
propionic acid in a solution that is
0.45 M HC3H
5O
2? pKa = 4.89.
Ans. 0.53%
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Example 14
What is the % ionization in 0.10 M NH3?
Ans. 1.33 %
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Common Ion Effect
Common Ion Effect
Shift in equilibrium when one or more ions that
are part of the equilibrium are introduced from
an outside source. reduction in the dissociation of the weak
electrolyte
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Example 15
Calculate the [H3O+] in a 0.0045 M benzoic
acid (HC7H5O2) solution. Ka= 6.3 x 10-5.
Calculate the [H3O+] in a 0.0045 M benzoicacid a solution which contains 0.001 M
NaC7H5O2.
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Hydrolysis Reaction of Salts
Acidic Salt: NH4Cl
NH4++ H2O H3O
++ NH3
KH= KW/KNH3
Basic Salt: NaCN
CN- + H2O HO- + HCN
KH= KW/KHCN
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pH of Salts
Acidic Salt: pH = 7 log[Csalt/Kb]
when Csalt/KH>>> 1000
Basic Salt: pH = 7 + log[Csalt/Ka]
when Csalt/KH>>> 1000
pH of Salts
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Example 16
What is the pH of an aqueous solution thatis 0.089 M NaOCl? pKa= 7.54
Ans. 10.24
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Example 17
What weight (in grams) of NH4Cl is needed
to be dissolved in 200 mL of water to
provide a solution having a pH of 4.50?
Kb= 1.8 x 10-5
Ans. 19.2 g
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Buffered Solutions
A buffered solution is one that resists a change in
pHwhen either hydroxide ions or protons areadded.
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Buffered Solutions
Buffer Capacity
- refers to the amount of acid or base that abuffer can neutralize before its pH changes.
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Buffered Solutions
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Buffered Solutions
pH of a Buffered Solution
or
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Example 18
A buffered solution contains 0.25 M NH3and 0.40 M NH4Cl. Calculate the pH.
Ans. 9.05
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Example 19
How many mL of pure formic acid (s.g =
1.22) must be mixed to 325 mL of 0.0664
M NaOH solution to obtain a buffer
solution of pH 3.25? Ka= 1.7 x 10-4
Ans. 3.51 mL
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Example 20
What is the pH of the resulting solution
made by mixing 5 mL of 0.2178 M HCl
and 15 mL of 0.1156 M NH3
?
Kb= 1.8 x 10-5
Ans. 9.02
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Acid-Base Indicators
Acid-Base Indicatoris a substance whose color depends
on the pH of the solution to which it is added.
Two forms of acid-base indicators:
1. Weak acid (represented by HIn.
2. Conjugate base (represented by In-.
HIn + H2O H3O+ + In-
acid color base color
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Acid-Base Indicators
HIn + H2O H3O+ + In-
acid color base color
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Acid-Base Indicators
Name pH range pKa Color change
Thymolphthalein 1.70 1.2-2.8 R-Y
Methyl Orange 3.46 3.1-4.4 R-Y
Bromocresol Green 4.66 3.8-5.4 Y-B
Methyl Red 5.00 4.2-6.3 R-O
Bromothymol Blue 7.10 6.2-7.6 Y-B
M-cresol Purple 8.32 7.6-9.2 Y-Purple
Thymol Blue 8.96 8.0-9.6 Y-B
Phenolphthalein 9.00 8.3-10.0 C-Pink
Thymolphthalein 10.0 9.4-10.6 C-B
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Example 21
A particular indicator has a color red and acolor blue in its acid and base form
respectively. If this indicator has a Ka= 3x 10-5 , by how much must the pHchange in order to change the indicatorfrom 75% red to 75% blue?
Ans. 0.95
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Gravimetry
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Gravimetry
Gravimetric Methods:
- are methods that depend upon measuring the
mass (i.e., gravity).
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Gravimetry
Three types of gravimetric methods:
1) Volatilization Methods
2) Extraction Methods
3) Precipitation Methods
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Gravimetry
Three types of gravimetric methods:
Volatilization Method
The analyte are volatilized at suitable temperature.
The volatilized species are collected.
The collected samples are weighed directly or weighed by difference.
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Gravimetry
Three types of gravimetric methods:
Extraction Method
The analyte to be determined is extracted w/ appropriate
solvent.
The mass of the purified extract is related tothe amount of the analyte.
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Gravimetry
Three types of gravimetric methods:
Precipitation Method
A sample is dissolved in an appropriate solvent.
The analyte is precipitated by a reagent that yields a sparingly
soluble product.
The precipitate is converted to a product of known composition by heat treatment.
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Gravimetric Methods
Basic calculations:Percentage of analyte in a sample is calculated
using a Gravimetric factor GF.
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Gravimetric Methods
Basic calculations:
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Example 22
A sample containing NaBr and KBr onlyweighs 312.54 grams. The samplewas dissolved in water and treated
with excess AgNO3. The precipitateformed was found to weigh 532.55grams. Calculate % KBr in the
sample.Ans. 49%
23
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Example 23
A 0.1005 gram sample of an ioniccompound containing chloride ionsand an unknown metal is dissolved in
water and treated with an excess ofAgNO3. If 0.0445 g of AgClprecipitate forms, what is the % by
mass of Cl-
in the original compound?Ans. 10.95%
E l 24
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Example 24
What weight of Mn ore should be taken
so that the percentage of MnO2in the
ore would be twice the mass of
Mn3O4 precipitate obtained in
milligram?
Ans. 57.0 mg
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Titrimetric Methods
Tit i t i M th d
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Titrimetric Methods
Titrimetric methods are analytical procedures in
which the amount of an analyte is determined
from the amount of a standard reagent requiredto react completley with the analyte.
Tit i t i M th d
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Titrimetric Methods
Types:
1. Precipitimetry
2. Acidimetry/Alkalimetry3. Compleximetry
4. Reductimetry/Oxidimetry
Tit i t i M th d
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Titrimetric Methods
Standard Solution
- a reagent used to titrate the analyte.
It must:
1. Have a precisely known concentration.
2. Generally is added from a buret.
Tit i t i M th d
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Titrimetric Methods
Standard Solution
Tit i t i M th d
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Titrimetric Methods
Primary Standard
- a highly purifiedcompound that serves
as a reference materials.
Tit i t i M th d
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Titrimetric Methods
Primary Standard
It must have the ff. characteristics:
a. High purity
b. Stable in airc. Absence of hydrated water molecules
d. Moderate cost and easy availability
e. Solubility in the titration solutions
f. Large formula weight (molecular weight)
*Compounds that do not meet all these criteria are called secondary standards.
Tit i t i M th d
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Titrimetric Methods
Some commonly used Primary Standards:
For Bases:
Benzoic Acid, C6H5COOH (f = 1) Oxalic Acid, H2C2O42H2O (f = 2)
Potassium Biiodate, KH(IO3)2(f =1)
Potassium Hydrogen Phthalate (KHP),C6H4(COOH)(COOK) (f =1)
Sulfamic Acid (HSO3NH2)(f =1)
Tit i t i M th d
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Titrimetric Methods
Some commonly used Primary Standards:
For Acids:
Calcium Carbonate, CaCO3 (f = 2) Mercuric oxide, HgO (f = 2)
Sodium Carbonate, Na2CO3(f = 2)
Tris-hydroxymethylaminomethane (THAM),(CH2OH)3CNH2(f =1)
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Example 26
How many grams of KHP are neededto neutralize 167.33 mL of 0.99955 M
NaOH?
Ans. 34.157 grams
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Example 27
In standardizing a solution of NaOHagainst 1.431 gram of KHP, the
analyst uses 35.50 mL of the alkali
and has to run back with 6.12 ml of
acid (1mL = 4.1 mg NaOH). What is
the molarity of the NaOH solution?
Ans. 0.2151 M
A li ti f A id B Tit ti
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Applications of Acid-Base Titration
1. Determination of Organic Nitrogen-Kjeldahl Method
2. Double indicator method for mixtureof basesWarder Titration
3. Acid Number
4. Saponification Number
Kjeldahl Method (Determination of Organic
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j ( g
Nitrogen)
Step 1. Digestion The sample is oxidized in hot, concentrated
sulfuric acid, H2SO4and turns black. .
Step 2. Distillation
The oxidized solution is cooled and then
treated with NaOH to liberate ammoniagas:
NH4++ OH- NH3(g)+ H2O
Kjeldahl Method (Determination of Organic
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Nitrogen)
Step 3. Titration
Using an excess amount of HCl. . .
NH3+ HCl NH4Cl
The excess HCl is determined using a standard
NaOH solution
HCl + NaOH NaCl + H2O
Kjeldahl Method (Determination of Organic
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Ammonia distilled is collected in a boricacid solution. . .
NH3+ H3BO3 NH4++ H2BO3
-1
Titrate the H3BO3-NH3solution with
standard acid. . .
H2BO3-1+ H3O
+ H3BO3+ H2O
Nitrogen)
Kjeldahl Method (Determination of Organic
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Percentage Protein in the sample
%protein =%N * f
f = 5.70 (cereals)
= 6.25 (meat products)
= 6.38 (dairy products)
Nitrogen)
Example 28
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Example 28
A 758-mg sample of full cream milk wasanalyzed by the Kjeldahl method; 38.61
mL of 0.1078 M HCl were required to
titrate the liberated ammonia. Calculatethe % N in the sample.
a.12.04% b. 7.69% c. 15.59% d. 10.93%
Example 29
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Example 29
A 7.443-gram sample beef was analyzed for itsN content and the liberated NH3 was
collected in a 43.25 mL of 0.4330 M HCl and
a 15.00 mL back titration with 0.0250 MNaOH was required. Determine the % protein
in the sample using 6.25 as factor for meat
products.
a.12.44% b. 21.57% c. 32.54% d. 10.98%
Example 30
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Example 30
A 3000 mg sample of flour was taken through aKjeldahl method. Upon digestion, the
ammonia liberated was collected into 200 mL
of 0.995 M H3BO3 solution. If this solutionrequired 25.25 mL of 0.3315 M HCl for
titration to methyl red end point, what is the
percentage of protein in flour? Use 5.70 for
cereal products.
Ans. 22.27%
Example 31
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A 500 mg sample of each mixture wasanalyzed for its alkaline content using 0.1025
M HCl via double indicator method.
Example 31
Mixture V0->Ph V0->MR
1 4.27 10.18
2 0.01 6.19
3 5.12 10.24
4 6.37 6.38
5 5.63 9.04
Determine the weightcompositions of each
mixture.
Acid Number
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Acid Number
Acid number - mass (mg) of KOH that will neutralizethe acid produced fromwater degradative reaction of
one gram of fat or oil
Acid no. = (VxM)KOHX MW KOH
gram fat or oil
Saponification no
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Saponification no.
Saponification no./Koettstorfer no. = mass of KOH reqdto saponify 1 gram fat or oil
Sap. no. = (Vblank, mLVsample, mL)(MHCl)(56.10)
gram fat or oil
Molar mass of Fat or Oil. = 168,300
Sap. Value
Example 32
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Example 32
The saponification no. of triglyceridesis 200. The average MW of the
triglycerides is:
A. 200 B. 280 C. 600 D. 840
Precipitation Titrations
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Precipitation Titrations
Common Technique:
Argentometric Titrations:
1. Mohr method
2. Volhard method
3. Fajans method
Method Titrant Indicator
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Mohr
AgNO3 K2CrO4
Volhard
KSCN ferric alum, NH4Fe(SO4)212H2O
Fajans
AgNO3
Fluorescein
Dicholro-fluorescein
End point: greenish-yellow to
pink
)()(: saq AgClClAgRxn )(42)(24)(2 saqaq CrOAgCrOAg
)()()(
)()(
:
saqaq
saq
AgSCNSCNAg
AgClClAg
ionBacktitrat
)(2
)(1
)(3 )( aqaqaq SCNFeSCNFe
)()( saq AgClClAg
Example 33
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Example 33
A 1.500-gram sample of impure aluminumchloride was dissolved in water andtreated with 45.32 mL of 0.1000 M
AgNO3 using K2CrO4 as indicator.Express the analysis in % AlCl3
Ans. 13.43 %
Titration with EDTA
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Titration with EDTA
EDTAEthylenediaminetetraacetic acid
combines w/ any metal ion in ratio of 1:1
Indicators:
1. Eriochrome Black T or Solochrome2. Calmagite
Example 34
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Example 34
An EDTA solution was prepared by dissolvingthe disodium salt in 1 L of water. It wasstandardized using 0.5063 gram of primarystandard CaCO3 and consumed 28.50 mLof the solution. The standard solution wasused to determine the hardness of a 2 Lsample of mineral water, which required
35.57 mL of the EDTA solution. Theconcentration (ppm) in terms of CaCO3is
Ans. 315.95 ppm
Example 35
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Example 35
The 300 mg sample of impure Na2SO4(142.04)was dissolved in sufficient water and the
sulfate was precipitated by the addition of
35.00 mL of 0.1022 M BaCl2. The precipitatewas removed by filtration and the remaining
BaCl2consumed 6.79 mL of 0.2467 M EDTA
for titration to the Calmagite endpoint.
Calculate the purity of the sample.
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Redox Reactions
Example 36
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Example 36
Balance the reaction:
2342 MnONOMnONO
RULES:
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U S An atom in its free or elemental form has oxidation equal to zero
For monoatomic ions, the oxidation number is equal to its charge
Metals have positive oxidation number such as alkali metals (+1),
alkaline
earth metals (+2), aluminum (+3), zinc (+2) and silver (+1)
Nonmetals usually have negative oxidation numbers:
Oxygen is usually2, except in peroxides (2) and superoxides (1) Hydrogen is usually +1, except in hydrides (1)
Fluorine has1 oxidation state; other halogens are usually in the
1 oxidation state, except when combined with oxygen, they are
positive; when different halogens are bound to each other, 1 is
assigned to the more electronegative halogen The sum of oxidation number of elements in a compound is zero.
The sum of oxidation number of elements in a polyatomic ion is the
charge of the ion
Example 37
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Example 37
What is the molarity of a KMnO4 solutionstandardized against 1.356 gram
Na2C2O4 (134 g/mol) requiring 25.1 mL
of the solution in acidic medium?
a. 0.161 M b. 0.403 M c. 1.008 M d. 0.856 M
Example 38
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Example 38
A sample of pyrolusite weighing 0.2400gram was treated with excess KI. The
iodine liberated required 46.24 mL of
0.1105 M Na2S2O3solution. Calculate %MnO2(86.94) in the sample.
a. 46.27% b. 30.85% c. 92.54% d. 76.12%
Example 39
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Example 39
A sample of iron ore weighing 385.6 mg wasdissolved in acid and passed through a
Jones reductor. The resulting solution 52.36
mL of 0.01436 M K2Cr2O7 for titration to thediphenylamine sulfonic acid endpoint.
Calculate % Fe3O4(231.55 g/mol) in the ore
sample.
a. 15.05% b. 45.15% c. 90.30% d. 67.98%
Example 40
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Example 40
A 10.00 gram sample of cooked-ham waspured with 200 mL of water, filtered and
the resulting solution containing dissolved
potassium nitrite was acidified. This solutionwas treated with 25.00 mL of 0.00514 M
KMnO4was back titrated with 14.97 mL of
0.01678 M FeSO4. Calculate the amount of
nitrite (46.01) in ppm.
Ans. 90 ppm
Masking
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Masking
Example 41:A 0.8521 gram sample of an alloy was found to contain Cu (63.55)and Zn (65.41) with small amounts of Pb (207.2) and Hg (200.59).The sample was dissolved in nitric acid and diluted to 500 mL. A 10mL aliquot was treated with KI to mask the Hg and the resultingsolution required 7.06 mL of 0.0348 M EDTA solution. A second 25mL aliquot was treated with ascorbic acid and the pH was adjustedto 2.00 to reduce Hg+2and the metallic Hg was removed from thesolution. To this solution, thiourea was then added to mask the Cuand the resulting solution required 8.58 mL for titration. The leadion was titrated in a 250 mL in the presence of NaCN to mask Cu,
Zn and Hg and required 3.11 mL for titration. Calculate thepercentage of Cu and Hg in the sample of alloy.
Ans. 47.08% Cu, 3.48% Hg