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Chemistry - FALL 2018
LAB EXPERIMENT 1: REACTIONS OF GROUP I AND GROUP II CATIONS
OBJECTIVES:
To differentiate between electrolytes and nonelectrolytes.
To get the idea of cation classification.
To get introduced to the chemistry laboratory.
To become familiar with test tube reactions.
To learn how to recognize Group I and Group II cations.
INTRODUCTION
Electrolytes and Nonelectrolytes
Electrolyte is a substance capable of conducting electricity when dissolved in water, while
nonelectrolyte dissolved in water does not conduct electricity. For example, a solution of kitchen salt in
water conducts, while the solution of sugar in water does not conduct electricity. It means that salt is
electrolyte, while sugar is not.
In order to conduct electricity, a substance has to be able to separate or dissociate into positively
(cations) and negatively (anions) charged ions in water. Electrolytes can generally be divided into two
categories, namely strong and weak electrolytes. While strong electrolytes give high ion concentration
when dissolved in water, weak electrolytes give low ion concentration. Figure 4.1 is giving an overview of
substance classification into electrolytes and nonelectrolytes.
Figure 1: Electrolytes and nonelectrolytes.
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Precipitation Reactions
If a reaction in which two liquid solutions are mixed is producing a precipitate, it is called a precipitation
reaction. The reaction product is slightly soluble in water. An example is a reaction between lead(II)
nitrate and sodium iodide, in which lead(II) iodide precipitate forms:
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)
Precipitate formation is marked by letter s next to the insoluble compound. (In a reaction, s stands for
solid, l for liquid, g for gas and aq for aqueous solutions.)
It is possible to predict the outcome of a precipitation reaction before it is experimentally performed. The
chemists are classifying substance in the following three groups: soluble, slightly soluble and
insoluble. A substance is considered to be soluble if a large amount of it can be dissolved in water. If it is
not the case, we are talking about slightly soluble or insoluble substances. There is a set of simple rules
which are used to predict precipitation reaction outcome when such a reaction is performed at 25⁰C
(Figure 4.2).
Figure 2: Solubility of the most important ions.
Molecular, Ionic and Net Ionic Equations
The following equation:
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)
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Represents an example of a molecular equation. It is called so because formulas of compounds are
written as neutral molecules. This type of equations is useful since it shows all compounds undergoing a
chemical reaction.
However, molecular equation is not showing what is happening at the level of particles, atoms, molecules
or ions. Therefore, all compounds should be written as they really appear in aqueous solutions, that is,
dissociated into ions, if they are electrolytes and non-dissociated, if they are nonelectrolytes. Such an
equation is referred to as ionic equation:
Pb2+
(aq) + 2NO3-(aq) + 2Na
+(aq) + 2I
-(aq) → PbI2(s) + 2Na
+(aq) + 2NO3
-(aq)
If we take a look at the left-hand side and the right-hand side of this equation, we can see that Na+
and
NO3- did not change their state during reaction. Therefore, they are called spectator ions. If we wish to
write an equation without these ions, we are emphasizing the chemical change that happened during the
course of the reaction. Such an equation is called net ionic equation:
Pb2+
(aq) + 2I-(aq) → PbI2(s)
Another example is a reaction between silver nitrate and hydrochloric acid.
Molecular equation: AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)
Ionic equation: Ag+
(aq) + NO3-(aq) + H
+(aq) + Cl
-(aq) → AgCl(s) + H
+(aq) + NO3
-(aq)
Spectator ions: H+ and NO3
-
Net ionic equation: Ag+
(aq) + Cl-(aq) → AgCl(s)
Group I Cations
Group I contains the following cations: Ag+, Pb
2+ and Hg2
2+. As shown in Figure 4.2, all chloride ions are
soluble, with an exception of silver, lead and monovalent mercury chlorides. This is used to distinguish
them from the other groups of cations, which means that a group reagent here is hydrochloric acid (HCl),
which is giving enough free chloride (Cl-) ions in a solution. In order to test Group I cations, we will use
AgNO3 as a source of Ag+ ions. Lead and mercury(I) ions are not used due to their toxicity.
Group II Cations
Group II contains cations whose chlorides are soluble, but whose sulfides are insoluble, namely Hg2+
,
Cu2+
, Bi3+
, Pb2+
, Cd2+
, As3+
, As5+
, Sb3+
, Sb5+
, Sn2+
and Sn4+
. Group reagent is H2S (hydrosulfuric acid).
This group is further divided in subgroups IIa and IIb on the basis of sulfide solubility in (NH4)2S. The first
five ions belong to IIa, while the remaining six belong to IIb group. We will use Cu2+
(from Cu(NO3)2) as a
subgroup IIa representative and Sn2+
(from SnCl2) and As3+
(from As2O3) as a subgroup IIb
representative.
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SAFETY MEASURES
HCl and H2S are acids and have to be handled in chemical hood. NaOH is a strong base and is a
corrosive substance. NH4OH is very hazardous in the case of skin contact (corrosive and irritant), eye
contact (irritant) or ingestion. It may cause damage to mucous membranes and respiratory tract. Handle
carefully in the chemical hood. All chemicals are to be handled with care and according to assistant’s
instructions. Wearing protective clothes (lab coat and gloves) is mandatory throughout whole exercise. In
the case of spills, inform your instructor and clean carefully. In the case of skin or eye contact,
immediately inform your instructor and wash with a plenty of water.
MATERIALS AND METHODS
Chemicals:
Silver nitrate (AgNO3)
Hydrochloric acid (HCl)
Sodium carbonate (Na2CO3)
Hydrosulfuric acid (H2S)
Potassium chromate (K2CrO4)
Sodium hydroxide (NaOH)
Copper(II) nitrate (Cu(NO3)2)
Ammonium hydroxide (NH4OH)
Tin(II) chloride (SnCl2)
Arsenic(III) trioxide (As2O3)
Potassium iodide (KI)
Lab equipment:
Test tubes
Test tube rack
Droppers
Chemical hood
Procedure:
1. Obtain 0.1 M solutions of compounds to be tested from your assistant. During this exercise, you
will be using 1 ml of reagent per reaction. Use separate dropper for every reagent. Note that HCl,
H2S and NH4OH are available in chemical hood.
2. Take AgNO3 solution and put 1 ml in each of six test tubes. Ag+ ion will be tested by the addition
of the following reagents: HCl, Na2CO3, H2S, KI, K2CrO4 and NaOH.
3. Observe reaction outcome and record observations on report sheet attached to this handout.
4. Take Cu(NO3)2 solution and put 1 ml into each of three test tubes. Cu2+
ion will be tested on the
following substances: H2S, NaOH and NH4OH. Record your observations.
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5. Take SnCl2 and place 1 ml of this reagent in two test tubes. Sn2+
ion will be tested using H2S and
NaOH. Record your observations.
6. Place 1 ml of As2O3 into a test tube. As3+
will be tested on H2S. Record your observations.
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REPORT SHEET
PRE-LAB QUESTIONS
1. What is a precipitation reaction?
2. How can the results of a precipitation reaction be predicted?
3. Which cations belong to the Group I? How can they be distinguished from other groups?
4. Which cations belong to the Group II? What is the group reagent and why?
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EXPERIMENT RESULTS
Group I Cations: Reactions of Ag+
Reaction with: Precipitating compound Precipitate appearance
HCl
Na2CO3
H2S
KI
K2CrO4
NaOH
Group IIa Cations: Reactions of Cu2+
Reaction with: Precipitating compound Precipitate appearance
H2S
NaOH
NH4OH
Group IIb Cations: Reactions of Sn2+
Reaction with: Precipitating compound Precipitate appearance
H2S
NaOH
Group IIb Cations: Reactions of As3+
Reaction with: Resulting precipitate Precipitate appearance
H2S
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POST-LAB QUESTIONS
1. Write molecular, ionic and net ionic equation for the following reactions of Ag+ ion.
Reaction with HCl
Molecular:
Ionic:
Net ionic:
Reaction with Na2CO3
Molecular:
Ionic:
Net ionic:
Reaction with K2CrO4
Molecular:
Ionic:
Net ionic:
2. Write net ionic equation for all reactions of Cu2+
.
Reaction with H2S:
Reaction with NaOH:
Reaction with NH4OH:
3. Write net ionic equation for all reactions of Sn2+
.
Reaction with H2S:
Reaction with NaOH:
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Chemistry - FALL 2018
LAB EXPERIMENT 2: REACTIONS OF GROUPS III, IV AND V OF CATIONS
OBJECTIVES:
To further understand cation classification.
To get introduced to the chemistry laboratory.
To learn how to recognize cations of Groups III, IV and V.
INTRODUCTION
Qualitative and Quantitative Chemical Analysis
The fact that certain ions will form hardly soluble precipitates of specific color when reacting under
controlled conditions is used to identify those ions. For example, Ag+ cation will form white AgCl
precipitate if HCl, NaCl or any other Cl--containing electrolyte is added to the solution. In other words, we
have used qualitative chemical analyses to identify certain substance (qualitative relates to the property
and type of substance). Qualitative chemical analysis usually refers to the systematic procedure of
proving the presence (or absence) of a substance, usually an ion.
Another important thing in chemistry is to determine the amount of substance present in a sample. In
order to do this, quantitative chemical analysis is used (quantity is the amount). It is important to note
that qualitative analysis has to precede quantitative, as is it first necessary to determine the type of
substance and then its amount.
Flame Tests
Flame test is another type of qualitative analysis. It was noted that the salts of some elements belonging
to the Group I and Group II of the periodic table (not Group I and Group II cations) color the flame in a
specific way. If an inert platinum wire is immersed in a concentrated solution of these salts and then
burned in flame, that flame will get characteristic color. For example, Ba gives yellowish-green color, Ca
gives orange-red and Sr carmine red color. This is another type of test which will be used in today’s lab.
This phenomenon happens as a consequence of the fact that when a particle absorbs light or any other
form of energy, it gets excited. In other words, if a substance is heated, enlightened or gets exposed to
mechanical energy, its own energy level gets higher. An example is (the only) hydrogen electron. In its
unexcited state, it is described by four quantum numbers: n=1, l=0, m=0 and s=1/2. If an H atom absorbs
enough energy to enable passage of its electron to a higher energy level, it will take up one of four
possible orbitals available at the higher level (2s, 2px, 2py or 2pz). If energy input was even higher, it
would pass to energy level 3 (n=3) and take up one of nine possible orbitals. If energy input was high
enough, electron will get free from nuclear attraction force and will leave an atom ionizing it (turning it into
cation).
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If energy absorption does not cause ionization, excitation state is unstable and will not last for long time
period, since an atom always tends to get stabilized by releasing excess energy. This energy is released
through light emission and released light wavelength corresponds to atomic energy levels. If the
wavelength belongs to the visible spectrum, it will be visible to the naked eye. Since excitation energy of
Group I and some Group II elements from the periodic table is relatively low, as these atoms contain one
or two electrons, light emission of these elements belongs to the visible spectrum. Different flame
coloration primarily depends on two factors; the number of valence electrons and total number of
electrons.
Group III Cations
The members of Group III are the following cations: Fe2+
, Fe3+
, Al3+
and Cr3+
. These cations can be
precipitated as hydroxides in the presence of ammonia. Therefore, their group reagent is NH4OH. This
group will be tested in today’s lab using FeCl3 (iron(III) chloride) as a source of Fe3+
cations.
Group IV Cations
The following cations belong to this group: Mn2+
, Ni2+
, Co2+
and Zn2+
. They are usually precipitated as
sulfides using (NH4)2S as a group reagent. ZnCl2 (zinc chloride) will be used as a source of Zn2+
ions to
test Group IV cations reactions.
Group V Cations
Group V cations are Ba2+
, Sr2+
and Ca2+
. These ions form soluble sulfides and insoluble carbonates in
neutral and alkaline solutions. Group reagent is (NH4)2CO3. We will use BaCl2 (barium chloride) as a
source of Ba2+
ions to test the reactions of Group V cations. Additionally, SrCO3 (strontium carbonate)
and CaCO3 (calcium carbonate) will be used for flame test.
Group VI Cations
Group VI cations are: Mg2+
, K+, Na
+ and NH4
+. These cations cannot be precipitated with any group
reagent mentioned for the previous five groups.
SAFETY MEASURES
H2SO4 is a strong acid and has to be handled in chemical hood. NaOH is a strong base and is a corrosive
substance. NH4OH and (NH4)2S are very hazardous in the case of skin contact (corrosive and irritant),
eye contact (irritant) or ingestion. They may cause damage to mucous membranes and respiratory tract.
Handle carefully in the chemical hood. All chemicals are to be handled with care and according to
assistant’s instructions. Wearing protective clothes (lab coat and gloves) is mandatory throughout whole
exercise. In the case of spills, inform your instructor and clean carefully. In the case of skin or eye contact,
immediately inform your instructor and wash with a plenty of water.
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MATERIALS AND METHODS
Chemicals:
Iron(III) chloride – FeCl3
Ammonium hydroxide – HN4OH
Potassium ferrocyanide – K4[Fe(CN)6]
Sodium hydroxide - NaOH
Ammonium sulfide - (NH4)2S
Sodium hydrogen phosphate – Na2HPO4
Zinc chloride – ZnCl2
Barium chloride – BaCl2
Ammonium carbonate – (NH4)2CO3
Sulfuric acid – H2SO4
Potassium chromate – K2CrO4
Strontium carbonate – SrCO3
Calcium carbonate – CaCO3
Lab equipment:
Test tubes
Test tube rack
Droppers
Chemical hood
Watch glass
Inoculating loop
Lab burner
Procedure:
Obtain 0.1 M solutions of compounds to be tested from your assistant. During this exercise, you
will be using 1 ml of reagent per reaction. Use separate dropper for every reagent. Note that
NH4OH, (NH4)2S and H2SO4 are available in chemical hood.
Take FeCl3 solution and put 1 ml in each of five test tubes. Fe3+
ion will be tested by the addition
of the following reagents: NH4OH, K4[Fe(CN)6], NaOH, (NH4)2S and Na2HPO4.
Observe the reaction outcome and record observations on the report sheet attached to this
handout.
Take ZnCl2 solution and put 1 ml into each of three test tubes. Zn2+
ion will be tested on the
following substances: NaOH, (NH4)2S and K4[Fe(CN)6]. Record your observations.
Take BaCl2 and place 1 ml of this reagent in four test tubes. Ba2+
ion will be tested using
(NH4)2CO3, H2SO4, Na2HPO4 and K2CrO4. Record your observations.
Perform flame tests for Ba2+
, Sr2+
and Ca2+
cations according to assistant’s instructions. Note the
color of each flame on the report sheet.
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REPORT SHEET
PRE-LAB QUESTIONS
1. Differentiate between qualitative and quantitative chemical analysis.
2. What is the goal of flame test?
3. What is the common property of all Group V cations?
4. What is the only non-metal cation mentioned in labs so far?
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EXPERIMENT RESULTS
Group III Cations: Reactions of Fe3+
Reaction with: Precipitating compound Precipitate appearance
NH4OH
K4[Fe(CN)6]
NaOH
(NH4)2S
Na2HPO4
Group IV Cations: Reactions of Zn2+
Reaction with: Precipitating compound Precipitate appearance
NaOH
(NH4)2S
K4[Fe(CN)6]
Group V Cations: Reactions of Ba2+
Reaction with: Precipitating compound Precipitate appearance
(NH4)2CO3
H2SO4
Na2HPO4
K2CrO4
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Group V cations: Flame tests
Cation Flame color
Ba2+
Sr2+
Ca2+
POST-LAB QUESTIONS
1. Write molecular, ionic and net ionic equation for the following reactions of Fe3+
ion.
Reaction with NH4OH
Molecular:
Ionic:
Net ionic:
Reaction with K4[Fe(CN)6]
Molecular:
Ionic:
Net ionic:
Reaction with Na2HPO4
Molecular:
Ionic:
Net ionic:
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Chemistry - FALL 2018
LAB EXPERIMENT 3: pH VALUE (CALCULATIONS AND EXPERIMENTAL DETERMINATION)
OBJECTIVES:
To get familiar with the concept of pH in chemistry.
To learn how to calculate pH of a solution.
To determine pH experimentally using indicator strips and pH meter.
INTRODUCTION
Acids and Bases
When talking about the concept of pH value, we usually think of strong acids and bases with toxic and
corrosive nature. However, the majority of acids and bases are weak. For example, Vitamin B and aspirin
are weak acids, while caffeine, nicotine and indigo are weak bases.
To decide if an acid or a base is strong or weak, one should consider how well it dissociates (ionizes) in
water. For example, any acid HA will react with water in the following way:
HA(aq) + H2O(l) → H3O+
(aq) + A-(aq)
If an acid is a stronger protein donor than hydronium ion (H3O+), it will give its proton to water and
dissociate completely. Such as acid is considered to be strong. On the other hand, weak acids are
weaker proton donors than hydronium ion, which means that a reaction between a weak acid and a water
molecule is reversible; acid HA will donate its proton to water to form H3O+, but H3O
+ will donate its proton
to A- to form HA at the same time.
If a substance HA is a weaker proton donor than H2O, it will receive a proton from water, as in the
following reaction:
HA(aq) + H2O(l) → H2A+
(aq) + OH-(aq)
Such compounds are termed strong bases. These bases are removing a proton from water and forming
OH- (hydroxide) ion, therefore completely ionizing in water.
The following are the general rules for classification of acids and bases:
1. Any protein donor stronger than H3O+ is a strong acid in water.
2. If a substance is a protein donor of strength between H3O+ and H2O, it is a weak acid.
3. Any protein acceptor stronger than H2O (weaker protein donor) is a weak base.
4. Any protein acceptor stronger than OH- is a strong base.
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Therefore, the concentration of H3O+ and OH
- ions is a measure of how strong acids and bases are.
pH Value
Water is an amphoteric substance, that is, it can act both as an acid and a base and is capable of self-
ionization, which is shown in the following reaction:
H2O(l) + H2O(l) → H3O+
(aq) + OH-(aq)
At 25⁰C, only two out of 109 water molecules self-ionize. The product of H3O
+ and OH
- is the ionic
product of water, Kw, which is 1.0 x 10-14
at 25⁰C:
[H3O+][OH
-] = 1.0 x 10
-14
[H3O+] = [OH
-] = 1.0 x 10
-7
If an acid is added to this solution, hydronium ion concentration goes above 1.0 x 10-7
, while hydroxide
ion concentration decreases, but never reaches zero. If base is added, OH- concentration goes above 1.0
x 10-7
, while H3O+ concentration decreases, but never reaches zero.
A good way of presenting such small concentrations of these two ions is pH value scale. pH is a
negative logarithm of hydronium ion concentration and is a dimensionless unit:
pH = -log[H3O+]
On the other hand, H3O+ concentration can be calculated if pH of a solution is known:
[H3O+] = 10
-pH mol/l
pOH is a negative logarithm of hydroxide ion concentration:
pOH = -log[OH-]
[OH-] = 10
-pOH mol/l
pKw is a negative logarithm of ionic product of water:
pKw = -logKw = -log(1.0 x 10-14
) = 14
pH + pOH = pKw
pH + pOH = 14
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Figure 3: pH value chart.
pH Calculations (Examples)
1. pH of a solution is 6.6. Calculate the concentration of H+ and OH
- ions. What is pOH of that
solution?
[H+] = 10
-pH
[H+] = 10
-6.6 = 2.5 x 10
-7 M
[OH-] = Kw / [H
+]
[OH-] = 1.0 x 10
-14 / 2.5 x 10
-7 = 0.4 x 10
-7 = 4 x 10
-8 M
pOH = 14 – pH = 14 – 6.6 = 7.4
2. What is the pH and OH- concentration in a solution of pOH = 1?
pH = 14 – pOH = 14 – 1 = 13
[OH-] = 10
-pOH = 10
-1 = 0.1 M
Measuring pH Value
pH value can be measured in different ways. Today, we will use two methods:
1. Universal pH indicator strips are using different color combinations to show pH of a solution.
2. pH meters are giving the most accurate measurements. They are using electric potential in a
glass electrode whose potential depends on hydronium ion concentration in a solution. The result
is shown on a display. When used, pH meter should first be calibrated with one or two buffers of
known pH value.
SAFETY MEASURES
All acids have characteristic unpleasant odors which is why they should be prepared in chemical hood.
Strong acids and bases are corrosive substances. Although they are available as diluted solutions, it is
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obligatory to wear lab coat and protective gloves. Any spills should be immediately reported and carefully
cleaned. In case of skin or eye contact, inform you assistant immediately.
MATERIALS AND METHODS
Chemicals:
100 mM HCl
1 mM HCl
100 mM CH3COOH
1 mM CH3COOH
100 mM NaOH
1 mM NaOH
100 mM Na2HPO4
1 mM Na2HPO4
100 mM KCl
1 mM KCl
Lab equipment:
Erlenmeyer flasks
pH indicator strips
pH meter.
Procedure:
Necessary solutions of HCl (hydrochloric acid), CH3COOH (ethanoic or acetic acid), NaOH
(sodium hydroxide), Na2HPO4 (sodium hydrogen phosphate) and KCl (potassium chloride) are
available in the lab.
Measure pH of every solution using pH indicator strips. Record values on the report sheet.
Measure pH of every solution using pH meter. Record values on the report sheet.
Obtain household products available in the laboratory. Measure their pH using indicator strips.
Record pH values and acid/base nature of those products on the report sheet.
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REPORT SHEET
PRE-LAB QUESTIONS
1. Differentiate between an acid and a base.
2. Calculate the pH and pOH of 1.2 x 10-3
M HCl solution.
3. Calculate pH, pOH and [OH-] of 0.1 M HNO3 solution.
4. If a solution X has pH = 5, which of the following is true:
a. Solution X is neutral.
b. H3O+ ion concentration is higher than OH
- concentration.
c. OH- ion concentration is higher than H3O
+ concentration.
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EXPERIMENT RESULTS
Table 6.1: pH values of acid, base and salt solutions.
Solution pH by indicator strips pH by pH meter Type of solution
100 mM HCl
1 mM HCl
100 mM CH3COOH
1 mM CH3COOH
100 mM NaOH
1 mM NaOH
100 mM Na2HPO4
1 mM Na2HPO4
100 mM KCl
1 mM KCl
Table 6.2: pH values of common household products.
Product pH by indicator strips Acid/base/neutral
Apple juice
Liquid soap
Clothes detergent
Table salt
Sugar
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Chemistry - FALL 2018
LAB EXPERIMENT 4: DETERMINING HCl CONCENTRATION (ACID-BASE TITRATION)
OBJECTIVES:
To get introduced to acid and base definitions by Arrhenius, Bronsted-Lowry and Lewis.
To understand what is acid-base titration.
To understand the role of indicators.
To grasp the concept of equivalence point and its position on the graph depending on the type of
acid and base used in titration.
To learn how to calculate the unknown concentration.
To perform titration.
INTRODUCTION
Definitions of Acids and Bases: Neutralization Reaction
Acids and bases are compounds which are usually studied and defined together. The first definition of
acids and bases was given by Swedish scientist Arrhenius, who explained that acids are compounds
that have extra H+ ions when dissolved in water, while bases donate extra OH
- ions in water, for
example:
HCl(aq) → H+
(aq) + Cl-(aq)
NaOH(aq) →Na+
(aq) + OH-(aq)
However, this definition gave an explanation of aqueous solutions only and did not explain the fact that H+
ion never exists alone, but is always paired with water molecule to give H3O+. Therefore, a new definition
of acids and bases was necessary.
Bronsted-Lowry definition states that both acids and bases are capable of reaction with water, whereby
acids donate a proton, while bases accept a proton:
HCl(aq) + H2O(l) → H3O+
(aq) – Cl-(aq)
NH3(aq) + H2O(l) → NH4+
(aq) + OH-(aq)
Due to a completely new definition of a base, many species that were not defined as acids/bases by
Arrhenius, were now included in these two classes. Acids and bases are connected to each other and
always act as a pair:
HCl(aq) H+
(aq) + Cl-(aq)
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acid proton + base
In this example, HCl/Cl- is an acid/base pair which is exchanging a proton and we call them conjugated
acid-base pair, where conjugated means common.
NaOH(aq) + H+
(aq) Na(H2O)+
(aq)
base + proton acid
Conjugated acid-base pair in this reaction is Na(H2O)+/NaOH, or simply Na
+/NaOH.
Neutralization is a reaction between two conjugated acid-base pairs. In a reaction between HCl and
NaOH, conjugated pairs are HCl/Cl- as a pair 1 and Na
+/NaOH as a pair 2. By consensus, acid is always
written first in an acid-base conjugated pair.
HCl(aq) + NaOH(aq) → [Na(H2O)]+ + Cl
-(aq)
acid 1 + base 1 → acid 2 + base 2
Acids and bases that can exchange more than one proton are said to be polyprotic, like H3PO4 and
H2CO3 (acids), or CO32-
and SO42-
(bases). Each ionization degree is making a new acid-base pair. For
example, carbonic acid is making two pairs: H2CO3/HCO3- and HCO3
-/CO3
2-, while phosphoric acid is
making three: H3PO4/H2PO4-, H2PO4
-/HPO4
2-, HPO4
2-/PO4
3-. Substances or ions that can act as acids and
bases are called amphoteric. From the previous examples, amphoteric ions are HCO3-, H2PO4
- and
HPO42-
.
Lewis gave an even newer definition of acids and bases in 1923 by expanding it to non-aqueous
solutions. He stated that a substance which is donating an electron pair is a base, while the
substance that is accepting an electron pair is an acid. Therefore, all compounds which have a free
electron pair and can give to an empty orbital of H+ are said to be bases (:N, :S, :Cl). In this way, many
metal ions and some oxides were recognized either as acids or bases (SOCl2, AlCl3, BF3).
:NH3 + H+
NH4+
Volumetric Analysis: Vocabulary
Volumetric analysis is a quantitative chemical method which is measuring volume in order to determine
the amount of substance. Volumetry is an analysis of solutions, mainly aqueous, in which the volume of a
solution of a known concentration necessary for a complete reaction of an analyzed substance is
measured. The basic vocabulary of volumetric analysis includes the following:
1. Standard solution – a solution of known concentration used for titration
2. Titration – careful addition of a standard solution from a burette to the analyzed substance until,
according to analyst’s opinion, the reaction is complete
3. Titrant – solution used to perform titration
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4. Equivalence point – a point in titration in which the amount of added titrant is chemically
equivalent to the amount of analyzed substance in a solution. If NaCl is reacting with AgNO3,
then:
NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)
Equivalence point is reached when 1 mol of NaCl reacts with 1 mol of AgNO3 (that is, 1 mol of Cl- reacts
with 1 mol of Ag+). In a reaction of H2SO4 and NaOH, equivalence point is reached when 2 mol of a base
react with 1 mol of an acid:
2NaOH(aq) + H2SO4(aq) → Na2SO4 + 2H2O
5. Titration endpoint – equivalence point is a theoretical point that cannot be experimentally
determined, but can be detected by a physical change, e.g. change of color or electric signal.
When detected, this point is called titration endpoint. The difference between equivalence point
and endpoint depends on analyst’s capability to detect the change and it should be as small as
possible to decrease the possibility of titration error.
6. Indicator – a substance which is added to the analyzed solution to detect endpoint; chosen
indicator should be able to make the difference between equivalence point and endpoint as small
as possible.
7. Primary standard – a compound or a solution of high purity and stability which is used as a
reference material in all volumetric analyses. The accuracy of the method depends on these
solutions (compounds) and their properties.
Volumetric Calculations
Titration can be used to determine concentration of unknown solution due to the basic premise that the
number of moles of a base added equals the number of moles of an acid at the equivalence point (where
titration ends). In other words, the volume of a base used to neutralize an acid can be used to precisely
determine concentration of an acid and vice versa. This is valid only if an acid and a base taking part in
the reaction have the ability to exchange the same number of protons.
n1 = n2; c1V1 = c2V2
1. How many milliliters of 0.1 M NaOH solution is necessary to neutralize 300 ml of 0.1 M HCl
solution?
Firstly, write the reaction: HCl + NaOH → NaCl + H2O
c1V1 = c2V2
0.1 M * V2 = 0.1 M * 300 ml
V2 = (0.1 M * 300 ml) / 0.1 M
V2 = 300 ml
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2. 50 ml of HCl has been neutralized with 29.71 ml of 0.01963 M Ba(OH)2 solution. What is HCl
concentration?
Reaction: 2HCl + Ba(OH)2 → BaCl2 + 2H2O
From the equation, it is obvious that 2 mol of HCl are reacting with 1 mol of Ba(OH)2. Therefore:
=
; 2Ba(OH)2 = 1HCl; 2nBa(OH)2 = 1nHCl
2 * cBa(OH)2 * VBa(OH)2 = cHCl * VHCl
cHCl = (2 * 0.01963 M * 29.71 ml) / 50 ml
cHCl = 0.02333 M
Volumetric Analysis: Titration Curves
When performing an acid-base titration, there are four possible combinations: strong acid-strong base,
strong acid-weak base, weak acid-strong base and weak acid-weak base (Figure 8.1). Each of these
combinations will give a specific titration curve which will differ in the position of an equivalence point
(equivalence point is found at the halfway of a vertical line which shows a dramatic increase in pH value).
From the shape of the titration curve, the type of acid and base used can be easily determined. Note that
each of four curves below has only one equivalence point.
Figure 8.1. Four possible titration curves depending on the nature of acid and base reacting.
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However, polyprotic acids are characterized by titration curves with more than one equivalence point. The
number of equivalence points in a polyprotic acid is equal to the number of protons it can release.
Therefore, H2SO4 with have two, while H3PO4 will have three equivalence points (Figure 8.2).
Figure 8.2. Top: Titration curve of a diprotic acid. Note that 200% (2 mol) of a base are necessary to neutralize 1 mol
of an acid. Bottom: Titration curve of a triprotic acid. 3 mol of a base are necessary to neutralize 1 mol of an acid.
Experimental Determination of HCl Concentration
HCl concentration is determined by titrating a known volume of an acid with a standard NaOH solution,
which is serving as a primary standard. Given below are chemical reactions that occur during the titration.
NaOH + HCl → NaCl + H2O
Na+
(aq) + OH-(aq) + H
+(aq) + Cl
-(aq) → Na
+(aq) + Cl
-(aq) + H2O(l)
OH-(aq) + H
+(aq) → H2O(l)
In other words, strong acid-strong base titration is a reaction between protons and hydroxide ions. In the
reaction, 1 mol of HCl reacts with 1 mol of NaOH, which is an important information to calculate HCl
concentration.
Since both solutions are colorless, it is necessary to add an indicator which will show equivalence point,
that is, the end of neutralization reaction. The indicators of choice in our lab are phenolphthalein and
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methyl orange. Phenolphthalein is colorless in acidic and pink in alkaline solution. When the
concentration of H+ and OH
- ions is equal, the change in color will happen. One more drop of NaOH will
cause indicator color to turn into light purple. If this happens, it means that titration is already behind its
equivalence point and that the result is not accurate. Another indicator, methyl orange, works on the
exactly same principle, but has different color; it is red in acidic, orange in neutral and yellow in basic
solutions (Figure 8.3).
Figure 8.3. Left: Change of color for litmus, methyl orange and phenolphthalein. Right: Both indicators are suitable
for strong acid-strong base titration.
SAFETY MEASURES
HCl and NaOH are present as diluted solutions. However, you are obliged to wear a lab coat and
protective gloves. Report any spills to your assistant immediately.
MATERIALS AND METHODS
Chemicals:
Standard NaOH solution of known concentration
HCl solutions 1 and 2 of unknown concentration
Phenolphthalein
Methyl orange
Distilled water.
Lab equipment:
Erlenmeyer flasks
20 ml glass pipette
Burette
Dropper
Wash bottle
Ring stand.
29
Procedure:
1. Pipette exactly 20 ml of HCl solution 1 of unknown concentration. Transfer it to 250 ml
Erlenmeyer flask (wide neck).
2. Carefully pour NaOH solution in the burette. Use funnel if necessary. Put a beaker under the
burette. Slowly open the pipe and release NaOH to bring its level in burette to 0 ml. Meniscus has
to be exactly on 0 mark.
3. In HCl-containing Erlenmeyer flask, add distilled water to make liquid level 1 to 1.5 cm high.
Added water will not affect titration results, since the amount (number of moles) of acid remains
the same.
4. Add 3 drops of phenolphthalein and swirl the flask to mix.
5. Place the flask under the burette and start titration. Put a piece of white paper under the flask to
notice the change in color easier.
6. Perform titration by continuous flask mixing. When you get close to titration endpoint, add titrant
drop by drop. After each addition, observe the color of the solution.
7. Titration endpoint is reached when the solution gets stable pink color that lasts for longer than 20
seconds. Record titrant volume used on the report sheet.
8. Repeat the same titration two more times.
9. Volume of titrant is the average value of three titrations. Calculate the concentration of solution 1.
10. Obtain HCl solution 2 from your assistant. Perform three titrations using methyl orange indicator.
Calculate the concentration of solution 2.
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REPORT SHEET
PRE-LAB QUESTIONS
1. Can you determine titration endpoint using pH meter instead of color indicator? How?
2. Why is equivalence point in strong acid-weak base titration below pH 7?
3. How many equivalence points do you expect H2CO3 titration curve to have? Why?
4. Show all conjugated acid-base pairs of H2SO4. Which ion(s) is/are amphoteric?
5. How many milliliters of 0.116 M H2SO4 will be needed to titrate 25.0 ml of 0.0084 M Ba(OH)2 to
the equivalence point? Write chemical equation and perform calculations.
6. 27.0 ml of 0.310 M NaOH is titrated with 0.740 M H2SO4. How many ml of H2SO4 are needed to
reach the end point? Show the chemical reaction.
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EXPERIMENT RESULTS
Table 8.1: Titration of HCl solution 1 with NaOH.
Titration Volume of NaOH used (ml)
1
2
3
Average value
VNaOH: _____________
cNaOH: _____________
VHCl: _______________
cHCl: _______________
Show how you calculated HCl concentration below.
Table 8.2: Titration of HCl solution 2 with NaOH.
Titration Volume of NaOH used (ml)
1
2
3
Average value
VNaOH: _____________
cNaOH: _____________
VHCl: _______________
cHCl: _______________
Show how you calculated HCl concentration below.
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