Laboratory in Chemistrynd
2 semester / Spring 2021

ì

Gabor SAMJESKÉ/Quan PHUNG/Yoshiaki SHUKU

Lecture 2v Synthesis of Organic and Inorganic Compounds• Acetylsalicylic acid
• Melting Point (M.P.)• Potassium Trioxalate Ferrate (III) Trihydratev Volumetric Analysis and Titration• Titration of a strong monoprotic acid (HCl)Synthesis of Acetylsalicylic Acid: Introduction
3.1 Synthesis of Acetylsalicylic Acid (Experiment 4)
Main theme: “In this experiment, acetylsalicylic acid will be synthesized and its physical properties (inparticular, the melting point) will be determined.”.
Acetylsalicylic acid is the active component of aspirin, a widely used pain reliever and antipyretic.
Learning Objective: At the end of this experiment, you should be able to1. precipitate an organic compound from a reaction mixture by addition of water
2. isolate a solid by filtration
3. recrystallize a compound from water
4. determine melting points
5. calculate the yield of a productYou will learn basic procedures in organic and inorganic synthesis and determining physical properties

Be careful in handling acetic anhydride and concentrated (98%) sulfuric acid!!!
They are located under the hood. When sampling acetic anhydride and concentrated sulfuric acid,Synthesis of Acetylsalicylic Acid: Steps of the Synthesis

Synthesis Steps:
Reaction of ACTIVE CARBOXYLIC ACID DERIVATIVES with
ALCOHOLS, PHENOLS, PRIMARY AMINES or SECONDARY AMINESINTRODUCES THE ACYL GROUP (-COR) into the HYDROXYL OR AMINO GROUP,

thus forming the corresponding

CARBOXYLATE ESTERS or CARBOXYLIC AMIDES, respectively.
This reaction is known as the ACYLATION REACTION. The ACYL GROUP
derived from ACETIC ACID is called the ACETYL GROUP (-COCH3).

Synthesis of Acetylsalicylic Acid: Steps of the Synthesis
acetic anhydride

CARBOXYLIC ACID

O

O

R

Carbonyl group
HydroxylRgroup

C

C

Carboxylic group

R

C
O

OH

CARBOXYLATE ESTERS (ALSO A DERIVATIVE)
O

Carbonyl groupR

1

Hydrogenreplaced byradical R2 =(alkyl or
phenyl group

Carboxylic group
CH

R

2

PhenolsHO

C
O

Carboxylic acidderivative

O

O

C

OH

C

HC

CH

C

CH

Hydroxyl group

Synthesis of Acetylsalicylic Acid: Steps of the SynthesisTheinto yieldisas shown below
O

O

O

H2SO4

+

+

O
H 3C

OH

CH 3

OH

H 3C

OH

with

O

C

O
C

C

CH3

OH

O

O

Why is acetic anhydride used and not acetic acid?
Ä In case of acetic acid:
O
O

OH
OH

CH 3

+

O

+ H2O

O

C

C

OH

OH

O

CH3

Phenols cannot be acetylated
satisfactorily in aqueous
solution:
Acetylation of phenol proceeds
readily with acetic anhydride in
the presence of a little
concentrated sulfuric acid asSynthesis of Acetylsalicylic Acid: What is a Catalyst?
• A substance thatwithout itself undergoing any permanent chemical change.
• A catalyst is a substance that speeds up a given chemical reactionby• Catalysts may affect the reaction environment favorably, or, e.g., acid catalysts for reactionsof carbonyl compounds, or form specific intermediatesthat are not produced naturally.
Generic potential energy diagram showing theeffectof a catalyst in a hypothetical exothermic
chemical
reaction X + Y to give Z. The presence of thecatalyst opens
a different reaction pathway (shown in red)with a lower activation energy.
The product and the overall thermodynamics
are the same§
In case of isomers (more than one product), the catalyst
need to be selected so that the favored product will beformed with the highest yield.
§

https://en.wikipedia.org/wiki/Potassium_ferrioxalate

Synthesis of Acetylsalicylic Acid: Reaction Mechanism
One possible Reaction mechanism for the

O
H
H

H
O

O
H
H

O
H
H

O
H

H

HÊ

H 3C

H
O

O

H H 3C

CH 3

O
OH

O

HÊ

CH3

C

Ê
CH 3

O

HÊ

O
O

C

OH

Synthesis of Acetylsalicylic Acid: Measuring of the Melting Point
3.1.2 Measurement of the Melting Point, which is characteristic for the substance.
results in aof the mixture, even if both samples have the identical melting points.
Ä Any impurity has a similar effect!
• Accordingly, when an unknown sample is mixed with the known substance, the melting
point of the mixture can be used to identify the unknown sample (MIXED MELTING-POINT
DETERMINATION).
Some tips about measuring the melting point••••

Experimental melting points should be always reported as a range, for example, 42-44°
A pure organic compound usually melts over a range of two degrees or less
A higher melting point than stated in the literature cannot be measured in the lab!
A given sample is only used once to determine the melting point!

Synthesis of Acetylsalicylic Acid: Measuring of the Melting Point
Left: Probably rightMiddle: Good (tightly packed)
Right: Too much sample

Melting sequence (Ibuprofen sample)

Formation of tiny
drops (72°)

Liquid becomesvisible (73°)

The melting point(MP) of this sample
is reported as:
72 - 75°
Clear liquid (75°)© https://quondam.csi.edu/ip/physci/faculty/rex/MPTips.htm

Mixture of liquidand crystals (74°)

Synthesis of Potassium Trioxalate Ferrate (III) Trihydrate
3.2 Synthesis of Potassium Trioxalate Ferrate (III) Trihydrate (Experiment 5)
Main theme:“Potassium trioxalate ferrate (III) will be synthesized herein as an example of an inorganic compound
synthesis.
The complex decomposes upon irradiation with light, and you will study this process ofphotochemical decomposition.”.

Chemicals once transferred to other vessel should never be returned to the original bottle.

Synthesis of Potassium Trioxalate Ferrate (III) Trihydrate• Synthesis of Potassium Trioxalate Ferrate (III) Trihydrate /*/The Potassium trioxalate ferrate (III) trihydrate complex can be synthesized by heating
ferric hydroxide (Fe(OH)3) with an excess aqueous solution of potassium hydrogen oxalate.
Potassium trioxalate ferrate (III) trihydrate is then purified by recrystallization.

Fe3+ + 3 OH- ⇌ Fe(OH)33 KHC2O4 + Fe(OH)3 ⇌ K3[Fe(C2O4)3]  3 H2O• Compounds with two potential
coordination sites in a mostly organic
molecule are called “bidentate” ligands(e.g., H2NCH2CH2NH2, ethylenediamine).
• Oxalate is also a “bidentate” ligand!

Oxalic acid/*/ Trihydrate: The Trihydrate or “ 3 H2O” symbolizes “water of crystallization”, which is water incorporated

into the crystal during the recrystallization procedure.

Structure of Potassium Trioxalate Ferrate (III) Trihydrate• Structure of Potassium Trioxalate Ferrate (III) Trihydrate

Oxalic acid
The ferrioxalate metal ion complex exhibits optical activity since there are two non-superimposable
stereoisomers of the complex. In accordance with the IUPAC convention, the isomer with the left-handed
screw axis is assigned the Greek symbol Λ (lambda). Its mirror image with the right-handed screw axis isgiven the Greek symbol Δ (delta).

L-[Fe(C2O4)3]3https://en.wikipedia.org/wiki/Potassium_ferrioxalate

D-[Fe(C2O4)3]3-

Photochemical Reaction of Potassium Trioxalate Ferrate (III) Trihydrate
The Photochemical Reaction of Potassium Trioxalate Ferrate (III) Trihydrate• The trioxalate ferrate (III) complex ion undergoes the photochemical redox reaction
below. This reaction is used as the chemical actinometer, since the amount of theabsorbed light can be determined by quantifying the ferrous (II) ion generated.
• In this experiment, you will determine the change in color. (Not a quantitative analysis
just a qualitative analysis – description – of the observed changes)• Since this photochemical reaction occurs under visible light, the solution used as thechemical actinometer has to be prepared in the dark.

From Wikipedia: Actinometers are instruments used to measure the heating power of radiation. They are used inmeteorology to measure solar radiation as pyrheliometers.
An actinometer is a chemical system or physical device which determines the number of photons in a beam integrally
or per unit time. This name is commonly applied to devices used in the ultraviolet and visible wavelength ranges. For
example, solutions of iron(III) oxalate can be used as a chemical actinometer, while bolometers, thermopiles, andphotodiodes are physical devices giving a reading that can be correlated to the number of photons detected.
Chemical actinometry involves measuring radiant flux via the yield from a chemical reaction. It requires a chemicalwith a known quantum yield and easily analyzed reaction products.

4. Volumetric Analysis and Titration
3.1 Acid-base titration and the titration curve (Experiment 6)
Main theme: “In these experiments, you will use strong and weak acids as sample solutions, to whicha strong base is added as the titrant.
In order to obtain a pH titration curve, you will plot the measured pH values against the quantity of thetitrating reagent added.”Learning Objective:
By analyzing the pH titration curves, you will learn how concentration is related to dissociation in thesample and at what range of pH the given indicators change color

As there are large quantities of strong alkali solution on your shelf, be careful while conducting yourexperiment.
Use only your own reagents for your experiment because the reagents on other benches may haveslightly different concentrations. If, for some reason, you cannot avoid changing benches, confirm thatthe concentration of KOH standard solution at the new bench is the same as the one on the old benchVolumetric Analysis and Titration: Acid/Base Definitions
Definitions of acids and bases• Arrhenius
acid: generates [H+] in solution
base: generates [OH-] in solution
normal Arrhenius equation: acid + basesalt + water
example: HCl(aq) + NaOH(aq)NaCl(aq) + H2O(l)• Brønsted-Lowery
acid: anything that donates a [H+] (proton donor)
base: anything that accepts a [H+] (proton acceptor)
normal Brønsted-Lowery equation: acid + baseacid + baseexample: HNO2(aq) + H2O(aq)NO2-(aq)+ H3O+(aq)Each acid has a conjugate base, and each base has a conjugate acid. These conjugate
pairs only differ by a proton. In this example: NO2- is the conjugate base of the acid HNO2
and H3O+ is the conjugate acid of the base H2O.
• Lewisacid: accepts an electron pairbase: donates an electron pair
The advantage of this theory is that many more reactions can be considered acid-basereactions because they do not have to occur in solution.

Volumetric Analysis and Titration: Acid/Base Definitions
Examples of monobasic and dibasic acids
Monobasic acids

H1NO3(aq) + H2O(l) ⇌ [H3O]+(aq) + [NO3]--(aq)Nitric acid

Nitrate ion

H1NO2(aq) + H2O(l) ⇌ [H3O]+(aq) + [NO2]--(aq)Nitrous acidpKa = 3.37 (285 K)

Nitrite ion

H1OCl(aq) + H2O(l) ⇌ [H3O]+(aq) + [OCl]--(aq)Hypochlorus acidpKa = -1.64

pKa = 4.53

Hypochlorite ion

Dibasic acid

H2SO4(aq) + H2O(l) ⇌ [H3O]+(aq) + [H1SO4]--(aq)Sulfuric acid

Hydrogensulfate ion[H1SO4]--(aq) + H2O(l) ⇌ [H3O]+(aq) + [SO4]2--(aq)Hydrogensulfate ion

pKa ≈ -2.0

Sulfate ion

pKa ≈ 1.92

Volumetric Analysis and Titration: Principle• An

is the determination of theby exactly neutralizing the acid/base with an acidor base of known concentration.
• This allows forsolution.
• It makes use of the neutralization reaction that occurs between acids andbases and the knowledge of how acids and bases will react if theirformulas are known.

Before starting the titration, a suitable pH indicator must be chosen.
The point at which the indicator changes color is called the.
A suitable indicator should be chosen, preferably one that will experiencea change in color (an end point) close to theof thereaction.

Volumetric Analysis and Titration: pH Indicator

Indicator example:
Bromothymol blue

- H+
Ê H+

https://en.wikipedia.org/wiki/Bromothymol_blue

Highlyconjugatedstructured
Ä deepblue color

Volumetric Analysis and Titration: Equivalence Point• The
of the reaction, the point at which equivalent amounts of thereactants have reacted, will have a pH dependent on the relative strengths of the acidand base used.
• The pH of the equivalence point can be estimated using the following rules:
Ø A strong acid will react with a strong base to form a neutral (pH=7) solution.
Ø A strong acid will react with a weak base to form an acidic (pH<7) solution.
Ø A weak acid will react with a strong base to form a basic (pH>7) solution.
When a weak acid reacts with a weak base, the equivalence point solution will be basic if
the base is stronger and acidic if the acid is stronger. If both are of equal strength, thenthe equivalence pH will be neutral.
However, weak acids are not often titrated against weak bases because the color change
shown with the indicator is often quick, and therefore very difficult for the observer to see
the change of color.

Volumetric Analysis and Titration

HCl + NaOH ® Na+ + Cl- + H2O

A titration curve is a plot of pH vs. the amount of titrant added. Typically, the titrant is a strong(completely) dissociated acid or base. Such curves are useful for determining endpoints anddissociation constants of weak acids or bases.

Semi logarithmic plot (pH)pH

Amount of
titrant added

Volumetric Analysis and Titration: pH Change by Adding the Titrant

1. The pH starts out low, reflecting the high [H3O+] of thestrong acid and increases gradually as acid is neutralized
by the added base.
2. Suddenly the pH rises steeply. This occurs in theimmediate vicinity of the equivalence point. For this typeof titration, the pH is 7.0 at the equivalence point.
3. Beyond this steep portion, the pH increases slowly as morebase is added.
•

40 ml HCl, [HCl] = 0.1 M Ä [H+] = 0.1 M, pH = - log(10-1) = 1.00•

40 ml HCl + 10 ml NaOH [NaOH] = 0.1 M Ä [H+] = (30ml x 0.1 M HCl)/50 ml = 0.0600 M
Ä pH = - log(6.00 x 10-2) = 1.22 Ä ¼ of acidity neutralized but only small change in pH!

•

40 ml HCl + 30 ml NaOH Ä [H+] = (10ml x 0.1 M HCl)/70 ml = 0.0143 M
Ä pH = - log(6.00 x 10-2) = 1.85 Ä ¾ of acidity neutralized but only small change in pH!

Volumetric Analysis and Titration: Required Titrant Volume for a Given
pH
First step is as usual calculation of concentrations of all substances entered into thesolution. What we do next depends on the titrated substance.
If the titrated substance is strong (fully dissociated) we assume the neutralization reaction
to proceed quantitatively - and pH depends only on the amount of excess of titrated
substance (before the equivalence point) or excess of titrant (after the equivalence point).
Equivalence point pH is 7.00 (unless ionic strength and activities are considered). Veryclose to the equivalence point
must be accounted for.
If - still in the case of strong acid HA (HCl) and strong base BOH (KOH) - we want tocalculate the volume of titrant for a given pH, we will make the same assumption thatneutralization went to completion. We will also assume both acid and base to becompletely dissociated. That means that the solution is described by just two equations.
One is charge balance:
1: [H+] + [K+] = [Cl-] + [OH-]or[H+] + [B+] = [A-] + [OH-]{2 (+) = 2 (-)}

the other one the water ion product: (self dissociation constant of water, Kw)
2: [H+] x [OH-] = Kw = 1.00 x 10-14

Volumetric Analysis and Titration: Definition of the pH
Reminder: pH[H+] x [OH-] = Kw = 1.00 x 10-14
What is of interest in this reading, however, is the acid-base nature of a substance likewater. Water behaves both like an acid and a base. (Brønsted-Lowery) The acidity orbasicity of a substance is defined most typically by the pH value, defined as below:
pH = -log[H+]At equilibrium, the concentration of H+ is 1.00 × 10-7, so we can calculate the pH of waterat equilibrium as:
pH = -log[H+]= -log[1.00 × 10-7] = 7.00
Solutions with a pH of seven (7) are said to be neutral, while those with pH values below
seven (7) are defined as acidic and those above pH of seven (7) as being basic.

Volumetric Analysis and Titration: The Logarithmic pH Scale[H3O+]/mol L-1[OH-]/mol L-1

1

1 x 10-14

1 x 10-7

1 x 10-7

1 x 10-14

1

Not possible to print into one graph!
Ä Logarithmic mathematics to condense therange of [H3O+] and [OH] (1909, S. P. L. Sørensen)

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/ph.php

Søren Peter Lauritz Sørensen(født 9. januar 1868 i
Havrebjerg ved Slagelse, død
12. februar 1939 i
Charlottenlund) var en dansk
kemiker, kendt for
fastlæggelsen af pH-skalaen forsyrer og baser.

Volumetric Analysis and Titration: Volume of the Added Titrant
Relations between [H+], [A], [B], VA and VB&% ×!%!(#)!"!#$ =&!"!#$• Let VA be the initial volume of the acid, CA its initial concentration, VB the volume of
the titrant added and CB its concentration. If so, the concentrations of A- and B+ are[H+] + [B+] = [A-] + [OH-][H+] x [OH-] = Kw

• All above equations can be combined:• If we want to find the volume of the added titrant and its the only unknown
variable, the equation can be easily solved for:

Volumetric Analysis and Titration: Calculation of the Titrant pH
Practical Example: What is the pH of 0.1023 M HCl solution titrated 75% with a 0.08976 M
solution of NaOH?

After the solution has been titrated 75% there is 25% of original strong acid left. However, this acid isdiluted by the titrant addition. Let's assume we have started with VHCl mL of acid, and its concentration
was CHCl. Let's name the volume of the added base VNaOH and the base concentration CNaOH.
••

Initially there were
After 75% has been titrated•

The required volume is•

Total volume of the solution isand !!"#$ = #!"#$ $!"#$ moles of acid and baseand actual concentration of acid left is

25% of initial

Volumetric Analysis and Titration: Calculation of the Titrant pH#$%& ×0.25×$$%&=$#$%& × 1 + 0.75× $%&$!"#$• Note, that VHCl cancels out and we are left with

• Put in the known numbers…

• And the resulting pH isVolumetric Analysis and Titration: Weak Acid Strong Base Titration[NaOH] = 0

But pH ¹ -log [Pr-]= -log(KA [HA]/[A-])= -pKA + log([A-]/[HA])

Volumetric Analysis and Titration: pH Difference at the Equivalence Point

The four Major Differences Between a Strong AcidStrong Base Titration Curve and a Weak Acid-Strong
Base Titration Curve

1. The initial pH is higher
2. A gradually rising portion of the curve, called
the buffer region, appears before the steep riseto the equivalence point
3. The pH at the equivalence point is greater than
7.00
4. The steep part of the curve spans a narrower
pH rangeVolumetric Analysis and Titration: Titration of Weak Acid with Strong
Base
The following sample data models
the titration of a weak acid with astrong base.
The titration curve on the right,
pH vs Titrant Volume, represents
the sample titration data.

This method estimates the equivalence point from the inflection point of
the titration curve. To determine the equivalence point, draw two best-fit
lines along the portions of the curve where the pH values are fairly
constant. A third best-fit vertical line is drawn for the portion of the curve
where the pH changes rapidly. The midpoint of this vertical line represents
the equivalence point.

Inflection point: Point of a curve at which the sign of thecurvature changes

Volumetric Analysis and Titration: Titration of Weak Acid with Strong
Base
Once the volume at the equivalence
point is determined, then one-halfthat volume represents the volumeof strong base needed to neutralize
one-half the weak acid. The
equilibrium constant expression for
the weak acid is given by:
Ka = [H+][A-]/[HA]
When the acid is half neutralized,[HA] = [A-], and the concentration
terms cancel leaving
http://www.geocities.ws/junebug_sophia/AcBaTi.htm 2022.4.6

Ka = [H+]Taking the -log of both sides andrearranging results in-log[H+] = -log Ka or pH = pKa.

Volumetric Analysis and Titration: The Buffer EffectIf the titrated substance is weak, calculation of the pH before the end point is done using
the Henderson-Hasselbalch equation describing pH of buffer solutions.
To do so we assume that neutralization went to completion, and we calculate
concentration of the weak acid (base) and its conjugate base (acid) treating the titrant as alimiting reagent.
pH at equivalence point is calculated as effect of hydrolysis of the salt, and pH after theequivalence point depends solely on the titrant excess.
Practical example: Calculate pH of a 0.1012 M acetic acid solution after 5.14 mL of 0.1231
M NaOH have been added to 25 mL sample. pKa = 4.75.
First of all we will convert the given information to titration fraction.
We started with 25 mL of 0.1012 M acetic acid solution, that is 0.025×0.1012=0.00253 moles of acid(2.53 mmoles). To that we have added 5.14 mL of 0.1231 M NaOH solution, that is0.00514×0.1231=6.33×10-4 moles of base. As acetic acid reacts with sodium hydroxide 1:1

CH3COOH + NaOH → CH3COONa + H2O
we can easily calculate that 6.33×10-4/0.00253 = 0.25 (25%) of acid was neutralized.
As we have explained earlier, for a weak acid titration it is best to use the Henderson-Hasselbalch
equation:

Volumetric Analysis and Titration: The Henderson-Hasselbalch Equation
Henderson-Hasselbalch equation
We know 25% of the acid was neutralized, that means 75% is still intact. Note, that we
don't have to calculate exact concentrations of acid and conjugate base, as we areinterested only in their ratio. Even if we will express exact concentrations by initialconcentration, titration percentage and solution volume, everything but titration
percentage will cancel out (try to check it by yourself). Thus, we can just plug known
numbers into the equation:
And pH of 4.27 is the final answer

In chemistry, the Henderson–Hasselbalch equation describes the derivation of pH as ameasure of acidity (using pKa, the negative log of the acid dissociation constant) inbiological and chemical systems. The equation is also useful for estimating the pH of abuffer solution and finding the equilibrium pH in acid-base reactions (it is widely used to
calculate the isoelectric point of proteins).

Volumetric Analysis and Titration: What is a Buffer?
A buffer is anconsisting of aor a weak base and its conjugate acid.
• Its pH changes very little when a small amount of strong acid or base is added to it and thus it is used
to prevent any change in the pH of a solution• Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety ofchemical applications• Many life forms thrive only in a relatively small pH range, so they utilize a buffer solution to maintaina constant ph. One example of a buffered solution found in nature is blood (Do you know thebuffer?)§Buffer solutions achieve their resistance to pH change because of the presence of an
equilibrium between the acid HA and its conjugate base AHA

H+ + A-

When some strong acid is added to an equilibrium mixture of the weak acid and its
conjugate base, the equilibrium is shifted to the left, in accordance with
. Because of this, the hydrogen ion concentration increases by less than theamount expected for the quantity of strong acid added (Buffer effect!!)§ physoliogic buffer system: bicarbonate (pH 7.35 ~ 7.45)

H2CO3 ⇌ H+ + HCO3CO2 (dissolved) + H2O ⇌ H2CO3

Lecture 2: Summary & Remarks
Week 7 to 9 experiments will introduce you to two very important topics oflaboratory in chemistry:••

Synthesis
Quantitative Analysis

Learning those fundamental skills is required for almost all kind of laboratory work.
The correct handling of glass, chemicals and laboratory instruments is essential for
successful results in either synthesis or analysis.
••••

Always take detailed notes of every step you are doing in the lab. Note date,time, any observation you have made.
Having a detailed lab report does not only help you if you need to repeat
something but also others to understand your work.
Every scientific journal requires a detailed description of your experimental
work so that your results and conclusions become clear and understandable.
If something does not work out, think about where a mistake might have
happened and then try to avoid repeating the same mistake.

Lecture 2: Summary & Remarks“Pure logical thinking cannot yield us any
knowledge of the empirical world; allknowledge of reality starts fromexperience and ends in it.”Einstein, Albert. Ideas And Opinions (p. 271). ThreeRivers Press, NY (Kindle Edition)