Autumn Lecture 5 (Volumetric Analysis)
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Transcript of Autumn Lecture 5 (Volumetric Analysis)
8/13/2019 Autumn Lecture 5 (Volumetric Analysis)
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Volumetric
Analysis
Term 1:
Week 3
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Outline
• The Mole
• Avogadro’s constant
• Reacting masses
• Concentration• Titration
• Standard solutions
• Rough and accurate titres• Significant figures
• Back titration
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How do we measure quantity of substance?
Element Symbol Relative atomic mass
Carbon-12 12C 12.000
Carbon C 12.011
Copper Cu 63.54
Hydrogen H 1.008
Magnesium Mg 24.312
Hydrogen = 1 g mol-1 1H 2H 3H - isotopes
Carbon-12 (12C) = 12 g mol-1
Before:
Now:
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Avogadro constant
The number of atoms in exactly 12grams of isotope 12C (1 mole) was
defined as Avogadro constant.
N A = 6.02 × 10 23
Macroscopic scale Atomic scale
Amadeo Avogadro
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One mole (n) is the amount of substance
which contains the same number of particles
(atoms, molecules or ions) as there are atomsin 12.00g of isotope 12C.
• The relative atomic mass in grams of all elements
contains the same number of atoms.
The Mole
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Molar mass
• Molar mass numerically equals to the atomic
mass but in units of gram per mole (g mol-1).
Molar mass of a molecular substance is equals to
its molecular mass.
The mass of one mole of a substance is called
molar mass (A or Mr ).
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Molar mass (or atomic weight for elements)
Atomic weight
in g/mol
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• The relationship below allow us to connectthe laboratory scale with the atomic scale
using our standard dimensional analysis.
General sequence of calculation
Grams of
substance A
Molar
mass of AMoles of
substance A
Avogadro’
s Number
Elementary
units of A
(atoms,
molecules,
ions)
During conversion take care of units (grams, not kilograms)
for mass and magnitude (×1023) for elementary units!
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Example of calculation:
The amount of substance n (in mole) and the number ofatoms (N) contained in it, are related to the Avogadro’s
number:
N/n = 6.02×1023 mol−1
Example: 2 g of magnesium ribbon
Number of atoms N) =
amount of substance (n) × Avogadro’s number (NA)
= 0.08226 × 6.02×1023 = 4.9523×1022 atoms
Amount of substance (n) =
mass (m) / molar mass (A)
= 2 / 24.312 = 0.08226 mol
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• Stoichiometry translates as “measure of the elements”.
• Practically, stoichiometry is used for converting chemical formulas
and equations that represent individual atoms, molecules, andformula units into to the laboratory scale units which are
milligrams, grams of these substances.
Stoichiometry
aA + bB cC
Grams of
substance
A
Molar
mass of A
Moles of
substance
A
Chemical
equation
Moles of
substance
B
Molar
mass of B
Grams of
substance
B
Stoichiometry
1 g ? g ? g
Stoichiometry is a quantitative relationship among the species
involved in a chemical process in a molecular or molar amounts.
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Stoichiometry
Mg(s) + O2(g) MgO(s)2 2
Magnesium ribbon with a mass of
2 g was burned in air. What is the
mass of obtained MgO?
nMg = mMg / AMg = 2 / 24.312 = 0.08226 mol of Mg
2 mol of Mg produces 2 mol of MgO. So, 0.08226 mol of
Mg will produce 0.08226 mol of MgO.
mMgO = nMgO x AMgO = 0.08226 x 40.312
= 3.31607g of MgO
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Limiting and excess reagents
Excess
Limiting
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Example:
H2 + O2 H2O2 2
m H2) = 20g
m O2) = 99.8gm H2O) = ?
= 20.0g ×1
2.016
= 9.92
Solution:
= 99.8g ×1
32.0
= 3.122
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2H2 + O2 2 H2OInitial (mol) 9.92 3.12
Stoichiometric
coefficient
Compare 3.124.96
Limiting reagent
(completely consumed)Excess reagent
(remained unreacted)
1
Finding limiting reagent
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2H2 + O2 2 H2O
Initial (mol) 9.92 3.12
Consumed (mol)
Result (mol)
-3.12-6.24
03.68 6.24
+6.24
Limiting reagent
(completely consumed)Excess reagent
(remained unreacted)
m = 6.24 mol ×18.04
1
= 112.57
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A solution is a homogeneous mixture inwhich the molecules or ions of thecomponents freely intermingle.
• The solvent is the medium intowhich the solutes are mixed ordissolved.
• A solute is any substancedissolved in solvent.
Concentration is the ratio of the amount ofsolute to the amount of solution.
Solution and concentration
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Molarity
• Chemists usually measure the concentration of solution
as the number of moles of solute per cubic decimetre ofsolution. Concentration in mol dm-3 is also called
molarity.1 cubic decimetre (dm3 ) = 1000 cm3 = 1 litre
Example: 11.7 g of NaCl was
dissolved in 500 ml of water. What is
the concentration of solution?
n = m/A = 11.7/58.5 = 0.2 mole of NaCl
M = n/V = 0.2/0.5 = 0.4 mol dm-3
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• Laboratory chemicals are usually purchased in
concentrated form and must be diluted withwater (made less concentrated ) before beingused.
• During dilution, the number of moles of solute
remains constant.
Dilution
× =
×
Moles of solute in
dilute solution
Moles of solute in
concentrated solution
× = ×
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Reacting masses
What if we don’t know the
concentration of HCl?
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Titration
• Titration (volumetric analysis) is a common laboratory
method of quantitative chemical analysis that is used todetermine the unknown concentration of a known
reactant.
Meniscus
39.3 mL
NaOH, 1 M (titrant)
HCl, 25 mL, 1 M (analyte)
+ pH indicator
TITRANT = STANDARD
SOLUTION (concentration& volume are known)
ANALYTE = TITRAND
(unknown concentration)
pH indicator
(colour of the solution changes
depending on the pH)
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Neutralization
reactants products
product of reaction
pH indicator
substance which changes the colour of the solution
depending on the concentration of H+ ions ( pH)
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pH indicators
Indicator Color on Acidic
Side
Color on Basic
Side
Methyl Violet Yellow Violet Bromophenol
BlueYellow Blue
Methyl Orange Red Yellow
Methyl Red Red Yellow
Litmus Red Blue
Bromothymol
BlueYellow Blue
Phenolphthalein Colorless Pink
Alizarin Yellow Yellow Red
A common way to detect the equivalence point in an acid-base
titration is to add an pH indicator, a compound that changes color asan acidic solution becomes basic (equivalence point), or vice versa.
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• Read the bottom of the meniscus, while your
eye is level with it. The meniscus can be seen
more easily against a white background.
Titration
Number of titration Rough Accurate 1 Accurate 2 Accurate 3
Initial burette reading (cm3) 10.0 10.0 10.0 20.0
Final burette reading (cm3) 35.3 35.0 35.1 45.0
Titre (cm3), Vin - Vfin 25.3 25.0 25.1 24.9
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Number of titration Rough Accurate 1 Accurate 2 Accurate 3
Initial burette reading (cm3) 10.0 10.0 10.0 20.0
Final burette reading (cm3) 35.3 35.0 35.1 45.0
Titre (cm3), Vin - Vfin 25.3 25.0 25.1 24.9
Rough and accurate titres
• Rough titration is the trial process (first titration)in order to determine the approximate end point
of the reaction.
• Accurate titration is the subsequent titration
performed with the high accuracy when the endpoint of the reaction has been already
established.
+ +
3 =25.0 ml
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ObservationVolume of 1.0
M NaOH (aq)
added/ mL
Colour of
indicator
0 Colorless
5 Colorless
10 Colorless
15 Colorless20 Colorless
21 Colorless
22 Colorless
23 Colorless
24 Colorless
25 Pink
26 Pink
27 Pink
28 Pink
When the two solutions just
react and neither is in excess,we have found the
equivalence point of the
titration.
=
× = .
+ → +
()
=
=
.
. = −
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Acid-base titration
http://www.youtube.com/watch?v=8UiuE7Xx5l8&feature=related
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Standard solution
• Standard solution is a solution containing a
precisely known concentration of an element
or a substance.• Standard solutions are usually prepared
from solid reagents.
• Standard solutions are used to determinethe concentrations of other substances, such
as analyte solutions in titrations.
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Back titration
• Back titration is useful if the endpoint of
the reverse titration is easier to identify
than the endpoint of the normal titration.
• Instead of titrating the analyte, a known
excess of standard reagent
(intermediate) is added to the solution,
and the excess is titrated.
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Back titrations are used when:
• one of the reactants is volatile, forexample ammonia.
• an acid or a base is an insoluble salt, forexample calcium carbonate
• a particular reaction is too slow
• direct titration would involve a weak acid- weak base titration (the end-point of
this type of direct titration is very difficultto observe)
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Example: Back Titration to Determine
the Amount of an Insoluble Salt
0.125g sample of chalk was placed in a 250mL beaker
and 50.00mL 0.200M HCl was added. The excess HCl
was then titrated with 0.250M NaOH. The average
NaOH titre was 32.12mL. Determine the mass of
calcium carbonate present in a chalk.
Step 1: Determine the amount of HCl in excess from the
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Step 1: Determine the amount of HCl in excess from the
titration results
a) Write the equation for the titration:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
acid + base → salt + water
b) Calculate the moles, n, of NaOH(aq) that reacted in the
titration:n = M x V
M = 0.250 mol L-1
V = 32.12 mL = 32.12 x 10-3 L
n(NaOH(aq)) = 0.250 x 32.12 x 10-3 = 8.03 x 10-3 mol
c) From the balanced chemical equation, 1 mole NaOH reacts
with 1 mole of HCl. So, 8.03 x 10-3 mol NaOH reacted with 8.03 x
10-3 moles HCl. The amount of HCl that was added to the chalk
in excess was 8.03 x 10-3 mol.
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Step 2: Determine the amount of calcium carbonate in chalk
a) Calculate the total moles of HCl originally added to the chalk:
n(HCl total added) = M x VM = 0.200 mol L-1
V = 50.00 mL = 50.00 x 10-3 L
n(HCl total added) = 0.200 x 50.00 x 10-3 = 0.010 mol
b) Calculate the moles of HCl that reacted with the calcium
carbonate in the chalk.
n(HCl titrated) + n(HCl reacted with calcium carbonate) =
n(HCl total added)
n(HCl total added) = 0.010 mol
n(HCl titrated) = 8.03 x 10-3 mol
8.03 x 10-3 + n(HCl reacted with calcium carbonate) = 0.010
n(HCl reacted with calcium carbonate) = 0.010 - 8.03 x 10-3 =
1.97 x 10-3 mol
) f
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c) Write the balanced chemical equation for the reaction
between calcium carbonate in the chalk and the HCl(aq).
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
d) From the balanced chemical equation, calculate the moles of
CaCO3 that reacted with HCl.
From the equation, 1 mol CaCO3 reacts with 2 mol HCl so, 1
mol HCl reacts with ½ mol CaCO3. So, 1.97 x 10-3 mol HCl had
reacted with ½ x 1.97 x 10-3 = 9.85 x 10-4 mol CaCO3 in the
chalk.
e) Calculate the mass of calcium carbonate in the chalk.
n = mass / An = 9.85 x 10-4 mol (moles of CaCO3 that reacted with HCl)
MM(CaCO3) = 40.08 + 12.01 + (3 x 16.00) = 100.09 g mol-1
mass = n x A = 9.85 x 10-4 x 100.09 = 0.099g
The mass of calcium carbonate in the chalk was 0.099g.
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Reliability
Chemistry involves the interpretation ofquantitative measurements, usually made as apart of experiment.
Each measurement has 4 aspects:
1) Object of measurement
2) Value
3) Units
4) Reliability
Example:
The mass of iron was 4.05 grams.
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• The accuracy of a measurement is indicated bythe number of digits used to represent it.
• Digits that result from a measurement such thatonly the digit farthest to the right is not knownwith certainty are called significant figures.
• The number of significant figures in ameasurement is equal to the number of digitsknown for sure plus one that is estimated.
Significant figures
This digit has some uncertainty This digit has some uncertainty
These two digits are known for sure These three digits are known for sure
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When are zeros significant?
Rules for determining number ofsignificant figures:
• All nonzero digits are significant.
• Zeros between nonzero digits are
significant.
• In a number with no decimal point, zeros
at the end of the number (tracing zeros)
are not necessarily significant.• If a number contains a decimal point,
zeros at the beginning (leading zeros)
are not significant, but zeros at the end
of the number are significant.
125.3
Example:
402
33,000
0.00340
Wh d i ifi fi ?
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Why we need significant figures?
33.100 mL
20.256 mL
0.0005 mL
0.3 mL
+
+
+
53.6565 mL
Analytical pipette
Measuring cylinder
Significant figures express the accuracy of experiment.
53.7 mLrounding
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What to write in log book
Lab report template
• Changes from
procedure (if any)• Results
• Observations
• Calculations
Complete lab report (long)
• Risk assessment
• Short intro• Procedure
• Results
• Observations
• Calculations
• Short conclusion