Experiment: 1

Preparation of solutions from solids

Objective: To prepare solutions from solid reagents

Theory: A solution is a homogeneous (composition and properties are uniform) mixture of

two or more substances. Solutions are commonly made up in the laboratory from solid
materials, from liquids or from other solution.

Solution is made from two components. These are:

Solute – is the substance present in a smaller amount

Solvent – is the substance present in a larger amount

If both components in a solution are 50%, the term solute can be assigned to either
component. When gas or solid material dissolve in a liquid, the gas or solid material is called
the solute.When two liquids dissolve in each other, the major component is called the solvent
and the minor component is called the solute.

Preparation of Stock Solutions

A stock solution is prepared by weighing out an appropriate portion of a pure solid or by

measuring out an appropriate volume of a pure liquid and diluting to a known volume.
Exactly how this is done depends on the required concentration units. For example, to prepare
a solution with a desired molarity you would weigh out an appropriate mass of the reagent,
dissolve it in a portion of solvent, and bring to the desired volume. To prepare a solution
where the solute’s concentration is given as a volume percent, you would measure out an
appropriate volume of solute and add sufficient solvent to obtain the desired total volume.

There are several methods of describing solution concentration such as mole fraction,
molality, molarity, normality, and percentage composition.

1.Mole fraction (X): is the ratio of the number of moles of one component to the total
number of moles in a solution. For instance, the solution made up of two components, let the
number of moles of one component be (nA) and the other component be (nB). If the mole
fraction of A and B in the solution are denoted by XA and XB respectively can be expressed
as:

1
 nA
XA = Where, nA=¿ Number of moles of solute A ,
nA+ nB
nB
XB = nB = Number of moles of solute B
nB+nA
Mole fraction is a dimensionless quantity; that is, it has no units.

2.Molality (m) : is the ratio of number of moles of solute to kilogram of solvent. Usually
denoted by the letter (m),

Number of moles of solute n( mol)

Molality (m) = =
Kilogram of solvent m( Kg)

3. Percentage composition (%): The percent of a solute is a frequently

designation of solutions.
A) Percent by weight (w/w): is the percent of total weight of solution contributed by the
weight of the solute. Or, the ratio of the weight of the solute to the total weight of the
solution multiplied by 100%.

weight of solute x
Percent by weight (%w/w) = ∗100 %
Total weight of solution

B) Percent by volume (%V/V) : is the percent of final volume represented by the

volume of the solute used to make the solution.

volume of solute
Percent by volume (%V/V) = ∗100 %
Total volume of solution

C) Percent by weight volume (%w/v) : the percent of solute and volume of solvent used
to make the solution.

weight of solute
Percent by weight-volume (%w/v) = ∗100 %
Volume of solv ent
4. Molarity (M): is the most commonly used concentration unit in chemistry and defined as
the number of moles solute per litre of solution. It is designated by the letter (M).

mole of solute n(mol)

Molarity = =
Litre of solution V (L)
5. Normality (N): is the number of equivalents (No of Equiv.) of a solute per volume of the
solution expressed in litres (V)

Number of equivalents
Normality (N) =
Litre of solution

mass of solute
Therefore, Normality (N) =
equivalent weight x Volume solution( L)

Determining the number of equivalent weight of a substance is not straight forward, since it
depends on the reaction unit, which is that part of a chemical species involved in a reaction.

a) In an acid–base reaction

The reaction unit is the number of H+ ions donated by an acid or accepted by a base.

The equivalent weight of an acid or base (Eqv. Wt acid or base) is the ratio of its molecular
weight (M.Wt) to the number of H+ ions donated or accepted per molecule of the acid or
base in the reaction.

𝐄𝐪𝐯. 𝐰𝐭 =
M .Wt
H+ accepted∨donated

b) For an oxidation–reduction reaction

For an oxidation–reduction reaction the reaction unit is the number of electrons released by
the reducing agent or accepted by the oxidizing agent

The equivalent weight of an oxidizing agent /reducing agent is its molecular weight (M. Wt)
divided by the number of electrons it accepted or donated (n) per molecule

𝐄𝐪𝐯. 𝐰𝐭 = , Where n is the number of electrons released by the reducing

M . Wt
n
agent or accepted by the oxidizing agent

c) Precipitation / Complexation reactions

The reaction unit is the charge of the cation or anion involved in the reaction; the equivalent
weight of the substance furnishing the cation or anion is its molecular weight divided by the
total positive charge or negative charge per molecule.

3
𝐄𝐪𝐯. 𝐰𝐭 =
M . Wt
, Where z is the number of total positive charge or negative charge
z

For solution preparation:

To prepare a solution with a desired normality from a pure solid, we weigh out the correct
mass of reagent and dissolve it in a volumetric flask.

Apparatus and Chemicals

Chemical: NaCl, distilled water

Apparatus: beaker, wash bottle, volumetric flask, Spatula, measuring balance

Procedure:

Preparation of 1 M NaCl in 50 ml of solution

1. Calculate the Mass of NaCl that required to prepare 1 M NaCl in 50 mL of Solution

2. Weight out the calculated amount of NaCl on the balance.
3. Transfer it to beaker and add 30 ml of distilled water to the beaker containing the salt.
4. Stir the mixture until all the salt dissolves.

5. Transfer the solution in to the volumetric flask and adjust the volume exactly 50 mL
mark by adding enough distilled water. Then stopper the flask and swirl the solution
vigorously to mix well.

6. Express the concentration of your solution interms of mole fraction, molality,

molarity, and normality, percent weight by weight and percent weight by volume.

(M.wtNaCl = 58.5 g/mol)

Question:

1) In a biochemical assay, a chemist needs to add 4.07 g of glucose (C 6H12O6) to a reaction

mixture. Calculate the volume in millilitres of a 3.16 M C 6H12O6 solution she should use for
the addition. Answer: 7.16 mL

2) explain how you would prepare aqueuous solutions of

a. 250 mL of a 3.00 N KOH solution?

b. 250 ml of CaCO3
3) How many gram-equivalents are contained in?
 a. 1.8909 gm of chemically pure oxalic acid H2C2O4.2H2O

b. 20 mL of 0,12 N NaOH solution.

4) Calculate the ppm (parts per million) concentration of 2.5*10-4 M solution of

a.CaCl2 b. HNO3

5) calculate the equivalent weight and normality for a solution of 6.0 M H3PO4

given the following reactions:

(a) H3PO4 (aq) + 3OH– (aq) ⇌ PO43– (aq) + 3H2O (l)

(b) H3PO4 (aq) + 2NH3 (aq) ⇌ HPO42– (aq) + 2NH4+ (aq)

(c) H3PO4 (aq) + F– (aq) ⇌ H2PO4– (aq) + HF (aq)

5
 Experiment: 2

Preparation of solution from liquids

Objectives: To prepare solutions from liquid reagents (from stock solutions)

Theory: Preparing Solutions by Dilution

Dilution is the procedure for preparing a less concentrated solution from a more concentrated
one (stock solution). Stock solution is a solution of known concentration from which other
solutions are prepared.

Solutions with small concentration are often prepared by diluting a more concentrated stock
solution.
A known volume of the stock solutions is transferred to a new container and brought to a
new volume. Since the total amount of solute is the same before and after dilution; That
means the number of moles of solute before and after dilution is constant, we know that

𝑫𝒊𝒍𝒖𝒕𝒊𝒐𝒏 𝒇𝒐𝒓𝒎𝒖𝒍𝒂:

M concentrated* V concentrated = M dilute*V dilute

moles taken from moles placed in dilute
concentrated solution solution

To find the concentration of stock solution we can use the following formula
density ( sp . gv )wt % 1000
Molarity =
molar mass

Chemicals and apparatus

Apparatus: Pipette, beaker, wash bottle, volumetric flask.

Chemicals: NaCl, Concentrated H2SO4, distilled water.

Procedure:
To prepare solution from a liquid or other solution

I. Preparation of 0.05 M NaCl in 250 ml of solution

a) Measure 12.5 mL of NaCl solution prepared in the above procedure (Experiment 1)

b) Introduces it into 250 mL volumetric flask

c) Dilute the solution by adding water to the mark.

Calculate the amount in molarity, in g/L, % (w/v) (M.wtNaCl = 58.5 g/mol).

Repeat with 5, 15, 20, 25, 30 mL NaCl solution.

Calculate the amount in molarity, in g/L, % (w/v).

II. Preparation of 250 ml of 0.2 M H2SO4 from a more concentrated stock solution
Concentrated H2SO4 purchased from a chemical supply company known to be 98.0% (mass
percent) and has a solution density of 1.84 g/mL.

Caution: Never add water to concentrated acid. Addition of concentrated acid to water
causes heating (very exothermic) and cause splattering. Always add the concentrated acid
to water slowly with stirring. Place the beaker or the flask in an ice bath to help cool the
resulting solution and prevent spattering

1) First calculate the molarity of the concentrated acid (stock solution or liquid)

(H2SO4) required for preparing the desired solution for the experiment.

(2) Prepare 0.2 M H2SO4 in 250 mL volumetric flask. (Show your calculation.)
Use a pipet to measure the amount of solution or liquid calculated in step (1).

(4) Add the solution or liquid to a volumetric flask of appropriate volume.

(5) Fill the volumetric flask approximately two thirds full and mix.

(6) Carefully fill the flask to the mark on the neck of the flask.

Use a wash bottle or medicine dropper if necessary.

(7) Mix the solution thoroughly by stoppering the flask securely and inverting it ten to
twelve times.

Repeat with 10, 15, 20 mL of 0.5 M H2SO4 solution prepared.

7
 Calculate the amount in molarity, in g/L, % (w/v).

Question:

1. Calculate how many mL of 15.8 M nitric acid should be

diluted to 0.250 L to make 3.00 M HNO3. Answer: 47.5 Ml
2. Preparing 0.100 M HCl the molarity of “concentrated” HCl purchased for
laboratory use is approximately 12.1 M. How many milliliters of this reagent should
be diluted to 1.000 L to make 0.100 M HCl? Answer: 8.26 mL
3. How can you prepare 50 ml of 0.1 M HCl solution using 6.0 M HCl solutions that is
on the laboratory bench?
4. Describe the preparation of 100 ml of 0.1 M H2SO4 from concentrated solution that
has density of 1.831 g/cm and is 98 % (W/W) H2SO4 ( M.wt = 98. g/mol).
5. Calculate stock concentration in molarity of HCl that contains 35.4 wt% HCl with
relative density of 1.18.

Experiment: 3

Estimating the solubility of an ionic compound

Objectives: To estimate the solubility of an ionic compound in water

Theory: The extent to which a substance may be dissolved in water, or any solvent, is
quantitatively expressed as its solubility, defined as the maximum concentration of a
substance that can be achieved under specified conditions. In another case solubility is the
ability of a substance, the solute, to form a solution with another substance, the solvent.
The extent of the solubility of a substance in a specific solvent is generally measured as the
concentration of the solute in a saturated solution, one in which no more solute can be
dissolved. At this point, the two substances are said to be at the solubility equilibrium. For
some solutes and solvents, there may be no such limit, in which case the two substances are
said to be "miscible in all proportions"
The solute can be a solid, a liquid, or a gas, while the solvent is usually solid or liquid. Both
may be pure substances, or may themselves be solutions. Gases are always miscible in all
proportions, except in very extreme situations and a solid or liquid can be "dissolved" in a gas
only by passing into the gaseous state first.

Substances with relatively large solubility’s are said to be soluble. Substances with relatively
low solubility’s are said to be insoluble, and these are the substances that readily precipitate
from solution (Insolubility is the opposite property, the inability of the solute to form such a
solution).

The solubility mainly depends on the composition of solute and solvent (including their pH
and the presence of other dissolved substances) as well as on temperature and pressure.

Ionic compounds dissolve in water if the energy given off when the ions interact with water
molecules compensates for the energy needed to break the ionic bonds in the solid and the
energy required to separate the water molecules so that the ions can be inserted into solution.

Solubility rule

1) All common salts of the group 1A elements (such as Na +, K+ and Li+) and ammonium
(NH4+) are soluble.
2) All common salts containing acetate (CH3COO-) or nitrate (NO3-) are soluble.
3) F- are soluble except those of silver (Ag), mercury (Hg), and lead (pb).
4) All compound containing sulfate (SO42-) are soluble except those of barium
( Ba),strontium (Sr), lead (pb), calcium (Ca), silver (Ag), and mercury (Hg).

9
 5) Compound following rule 1containing carbonate (CO 32-), Hydroxide (OH-), oxide
(O2-), and phosphates (PO43-) are soluble.

Chemicals and Apparatus:

Apparatus: Test tube, pencil, rough paper, stirrer

Chemicals: Distilled or de-ionized water, Copper (II) sulphate, Copper (II) carbonate,
Calcium carbonate, Calcium chloride, Calcium nitrate, Silver nitrate

Procedure

1) Take five test tubes with about 3 ml of de-ionized water to the reagent bench.
2) With a pencil label the test tubes with the formulas of the following compounds.
3) Add a small amount (about the size of a pea) of the compounds to the appropriate test
tube.
4) Stir Ionic compounds in water

Fill the following table from your experiment or demonstration

Substance Formula of the Soluble or insoluble?

Compound
Soluble insoluble

Copper (II) sulphate in water

Copper (II) carbonate in water

Calcium carbonate in water

Calcium chloride in water

Calcium nitrate in water

Silver nitrate

Questions:

1) Explain the difference between the hot solution and the cold solution.
2) How does temperature affect the solubility of a solid in water?
3) Identify whether the following substances are soluble or insoluble in water
a) group 1 metal cations (Li+, Na+, K+, Rb+, and Cs+) and ammonium ion (NH4+)
b) the halide ions (Cl−, Br−, and I−)
 c) sulfates of Ag+, Ba2+, Ca2+, Hg22+ , Pb2+, and Sr2+
d) carbonate (CO2−3) , chromate (CrO2−4) , phosphate (PO34-) , and sulphide (S2−) ions
hydroxide ion (OH−)

4) The solubility product of silver carbonate (Ag2CO3) is 8.46 × 10−12 at 25°C. Calculate the
following:

a) The molarity of a saturated solution

b) the mass of silver carbonate that will dissolve in 100 mL of water at this
temperature Answer: a) 1.28 × 10−4 M b) 3.54 mg

Fundamentals of Titrimetry
Titration- is the process in which a standard reagent( titrant) from the burette is slowly added
to the solution of an analyte (titrand) until the reaction is judged to be completed. A standard
solution (or a standard titrant) is a reagent of known concentration that is used to carry out
a titration.

Equivalence point in a titration is a theoretical point reached when the amount of added
titrant is chemically equivalent to the amount of analyte in the sample.
For example
The equivalence point in the titration of NaCl with AgNO3 occurs after exactly one mole of
Ag+ has been added for each mole of Cl- inthe sample.

Cl- + Ag+ ⇌ AgCl(s)

The equivalence point in the titration of H2SO4 with NaOH is reached after introducing

11
 2 moles of base for each mole of acid.

H2SO4 + 2NaOH → Na2SO4 + 2H2O

The end point is the point in a titration when a physical (colour) change occurs that is
associated with the condition of chemical equivalence.
In a titration, the primary systematic error is the endpoint determination. The difference
between the equivalence point and the measured end point is called the titration
errorIndicators are often added to the analyte solution to produce an observable physical
change (signaling the end point) at or near the equivalence point.

A primary standard is a highly purified compound that serves as a reference material in

titrations and in other analytical methods.
The accuracy of a method critically depends on the properties of the primary standard.
Important requirements for a primary standard are the following:
High purity
• Atmospheric stability
• Absence of hydrate water so that the composition of the solid does not

change with variations in humidity

• Modest cost
• Reasonable solubility in the titration medium
• Reasonably large molar mass so that the relative error associated with
weighing the standard is minimized
The number of substances proposed as primary standard is quite large.
The most commonly used ones are:
A. Acid-Base reactions
Sodium carbonate (Na2CO3), Borax (Na2B4O7.10H2O), Benzoic acid (C7H6O2), Potassium
hydrogenphthalate (KHC8H2O4), potassium hydrogen iodate KH(IO3)2, Sulphamic acid
(NH2SO3H), Oxalic acid (H2C2O4.2H2O)
B. precipitation formation reactions
Silver, Silver nitrate (AgNO3)
C. Oxidation-reduction reactions
Potassium dichromate (K2Cr2O7), Potassium bromate, KBrO3, Potassium iodate (KIO3),
potassium hydrogen iodate (KH(IO3)2), Iodine (I2), Sodium oxalate (Na2C2O4), Arsenic oxide
As2O3, pure iron.

A secondary standard is a compound whose purity has been determined by chemical

analysis. The secondary standard serves as the working standard material for titrations and for

many other analyses.

Ideal Requirements for Standard Solutions

The ideal standard solution for a titrimetric method will

 be sufficiently stable so that it is necessary to determine its concentration only once;

 react rapidly with the analyte so that the time required between additions of reagent is
minimized;
 react more or less completely with the analyte so that satisfactory end points are
realized;
 undergo a selective reaction with the analyte that can be described by a balanced
equation.
Few reagents completely meet these ideals.

Volumetric Calculations
Most volumetric calculations are based on two pairs of simple equations that are derived from
definitions of the mole, the millimole, and the molar concentration.For the chemical species
A, we can write
mass A(g)
ammount A (mol) =
molar mass A (g/mol )
The second pair of equations is derived from the definition of molar concentration, that is,
ammount A (mol) = 𝐕(𝐋) 𝐱 𝐌𝐀(𝐦𝐨𝐥/𝐋)
ammount A (mmol) = 𝐕 (𝐦𝐋) 𝐱 𝐌𝐀(𝐦𝐦𝐨𝐥/𝐦𝐋)

Any titration process involves at least four stages.

1) Initial stage: A pure solution of analyte (in titration flask) before any volume of reagent is
added
2) Before equivalence (or pre-equivalence) point: The volume of reagent added to analyte
does not make the reaction complete (when there is still excess of analyte).
3) At equivalence point: The amount of reagent added is chemically equivalent to the
13
amount of substance being determined (analyte).
4) After equivalence (or post-equivalence) point: The amount of reagent added is higher
than the amount of substance being determined.

Precautions

Following precautions should be taken while performing the experiment - Precautions:

1. Clean all the apparatus with distilled water before starting the experiment and then
rinse with the solution to be taken in them.

2. Rinse the pipette and burette before use and take out the bubbles at the nozzle of the
burette.

3. Use antiparallex card or autoparallex card while taking the burette readings.

4. Take accurate readings once it reaches the end point and don’t go with average
readings.

5. Never use a burette with a broken nozzle.

6. The pipette should be used carefully.

7. The strength of the unknown solution should be taken upto two decimal places only.

8. While noting reading at the endpoint, no drop should be hanging at the nozzle of the
burette.

9. Endpoint should be detected carefully and precisely.

 Experiment 4

Standardization of sodium hydroxide solution

Objective: To standardize NaOH using a secondary standard HCl.

Theory: A standard solution is one whose concentration is known. The concentration is

usually expressed in normality (equivalents) and molarity. The process by which the
concentration of a solution can be determined is said to be standardization. Because of the
availability some of substances known as primary standards, in many instances the
standardization of a solution are not necessary. Primary standard substances are analytically
pure, and by dissolving a known amount of a primary standard in a suitable medium and
diluting to a definite volume, a solution of known concentration is readily prepared. This
experiment consists of three parts. Most standard solutions, however, are prepared from
materials which are not analytically pure and they have to be standardized against a suitable
primary standard.

A substance should meet the following requirements to be taken as a primary standard;

1. It should be easily obtained in an analytically pure state (˃99.99%)

2. The composition of the primary standard should be an unalterable in the air at
ordinary or moderately high temperature
3. Its equivalent weight ought to be high in order to reduce the effect of small
weighing error
4. It should be readily soluble under the given conditions of analysis
5. On titration no interfering product should be present
6. The primary standard should be colorless, before and after titration, to avoid
interference with indicators
7. The prepared standard solution should be one that does not change
concentration with time

A solution that has been standardized employing a primary standard is considered as

secondary standard. Such a solution could be used to determine the concentration of solutions

15
that require standardization. The number of substances proposed as primary standards is quite
large. The most commonly used standards are:

Acidimetric standards: sodium carbonate (Na2CO3, eq.wt=53.00) and Borax

(Na2B4O7.10H2O, eq.wt=190.72)

Alkalimetric standards: sulphamic acid (NH2SO3H, eq.wt= 97.098); potassium

hydrogeniodate (KH (IO3)2, eq.wt =389.928); potassium hydrogenphthalate (KHC 8H4O4,
eq.wt= 204.22); oxalic acid (H2C2O4.2H2O, eq.wt=63.03).

Sodium hydroxide usually contains Na2CO3 as impurity, due to a reaction with CO 2 from the
air. Thus, it becomes expedient to standardize it either by using any of the alkalimetric
primary standards or a standardized acid solution to determine its exact concentration. In
order to remove dissolved carbon dioxide gas from water solvent, the distilled water used
should be previously boiled and cooled. Unless it is stored in plastic bottles, NaOH readily
attacks glass walls of a reagent bottles forming sodium silicate.

Chemicals and Apparatus:

Apparatus: Burette, pipette, conical flasks, beaker, wash bottle, volumetric flask

Chemicals: HCl, NaOH, sodium carbonate (Na2CO3, eq.wt=53.00), Methyl red indicators,
ethanol, distilled water

Preparation of solution

1) Prepare approximately 0.2 N solution of hydrochloric acid by diluting 1 ml of

concentrated hydrochloric acid (usually between 10 N and 11 N) to 100 ml.
Caution: if any acid is split on the skin or eyes, thoroughly rinse the affected area
with water, or dilute solution of sodium hydrogen carbonate, or milk. For severe
cases, go to the nearest clinic
2) Weight accurately 5.3 g of dried Na2CO3 and transfer to a 1000 mL volumetric flask,
add about 300 mL of distilled water, and shake until dissolved.
Adjust the volume to the mark and mix well. The concentration will be 0.05 M.
3) Prepare approximately 0.1 M NaOH solution by dissolving 4.0 g of NaOH in boiled
and cooled distilled water. Transfer it to 1000 mL volumetric flask and dilute it to the
mark.
 4) % methyl red indicator: Dissolve 0.1 g of the free acid in 1000 mL hot water or
dissolve in 60 mL of ethanol and dilute with 40 mL of water.

Procedure

I. Using primary standard Na2CO3 standardize HCl solution as follows:

Pipette out 10 mL of 0.05 M Na2CO3 solution in to 250 mL titration flask and add 2-3 drops
of methyl red indicator. Titrate the solution with HCl until it changed to colorless that
indicates the end point of titration. Repeat the titration three times.

Calculate the concentration of HCl using the following expression:

NHCl x VHCl = NNa2CO3 x V Na2CO3

N Na 2CO 3 x V Na2 CO 3
NHCl =
VHCl

II.Using the secondary standard HCl standardizes NaOH solution as follows:

Pipette out 10 mL of the NaOH solution in to 250 mL titration flask and add 2-3 drops of
phenolphthalein indicator. Titrate it with previously standardized HCl solution until the color
of the solution changes to yellow. Repeat the titration three times.

Calculate the concentration of NaOH using the following expression:

NNaOH x VNaOH = NHCl x VHCl

NHCl x VHCl
NNaOH =
VNaOH

Questions:

1. A 9.5 ml NaOH is titrated with 0.5 M HCl requiring 14.5 mL in the presence of
phenolphthalein indicator end point. How many milligram per liter of NaOH are in the
solution ? ( f.wt NaOH = 40g/mol)

Experiment 5

Title: Determination of NaOH and Na2CO3 in the same solution

17
Objective: To determine the exact amount of NaOH and Na2CO3 in the same solution

Theory: All alkalis absorb CO2 from the air and are converted into carbonates.

2NaOH + CO2 ------- Na2CO3 + H2O

Thus, a solution of NaOH always contains Na 2CO3. Due to the fact that one has the option of
titrating Na2CO3 as a mono functional base or as di functional base depending upon the
indicator chosen, the amount of the compound as NaOH and Na 2CO3 in solution can be
determined as follows.

If the mixture is titrated with HCl in presence of phenolphthalein or bromothymol blue

indicator, at the equivalence point of the titration all the NaOH is converted to NaCl and H 2O
and all the Na2CO3 into NaHCO3.

NaOH + HCl → NaCl + H2O ……………….. (1)

Na2CO3 + HCl→ NaCl + NaHCO3…………….. (2)

When the solution of NaOH and Na2CO3 is titrated in the presence of methyl orange (methyl
yellow) the indicator turns pink only after all the salt is converted into H2CO3 and NaCl.

NaHCO3 + HCl → H2CO3 + NaCl ……… (3)

As we can be seen from the equation (2) and (3), the amount of HCl consumed to reach the
end point of the titration when using phenolphthalein is half the amount required in the
presence of methyl orange. In other way, half of the Na 2CO3 is titrated in the presence of
phenolphthalein, and all of it in the presence of methyl orange.

Chemicals and Apparatus:

Chemicals: Mixture of Na2CO3 and NaOH, 0.1 N HCl, 0.5 % Methyl orange, 0.1 %
phenolphthalein.

Apparatus: Burette, pipette, conical flasks, beaker, wash bottle, volumetric flask.

Procedure:

Estimation of NaOH and Na2CO3 in the same solution

1) Pipette out 10 ml of the solution mixture containing unknown concentration NaOH and
Na2CO3 in to a clean conical flask.

2) Add 3 drops of 0.5 % phenolphthalein indicator solution and titrate the contents of the
conical flask against the previously standardized HCl solution

3) Stop the titration when the light pink color just disappears and note down the volume of
HCl (VX mL) for this end point

5) To the same titration mixture then add 3 drops of 0.05% solution of methyl orange
indicator and resume the titration.
6) Continue the titration until the pale yellow color just changes to pale red and note down
the volume of HCl (say VY mL) consumed for this color change.
7) Repeat the titration to obtain at least two equal titrant volumes

Calculate the amounts of NaOH and Na2CO3 as given below.

Determination of NaOH

NNaOH x VNaOH = NHCl x VHCl

VNaOH = 10 ml VHCl = (VX-VY) mL

NNaOH =? NHCl = 0.1N Known from Experiment (6)

N HCl x V HCl
NNaOH=
V NaOH

Amount of NaOH in g/L = NNaOHx40(Eq.wt of NaOH)

Determination of Na2CO3

N Na2CO3 x V Na2CO3 = NHCl x VHCl

VNa2CO3 = 10 ml VHCl = 2(VY) mL

NHCl = 0.1N Known from Experiment (6)

N Na2CO3=?

N HCl x V HCl
N Na2CO3 =
V Na2 CO3

19
Amount of Na2CO3 in g/L = N Na2CO3 x 53 (Eq.wt of Na2CO3)

Question

1. Why alkaline solution should be protected from the atmosphere.

2. Why sodium hydroxide contaminated with sodium carbonate?
3. Why is it important to rinse the burette with the titrant before beginning the titration?
4. What is acidimetry ?Alkalimetry?
5. A mixture of NaOH and Na2CO3 will titrated with 0.25 M HCl requiring 26.2 mL for
the phenolphthalein end point and an additional 15.2 mL to reach the modified
methyl orange end point . How many milligrams of NaOH and Na2CO3 are in the
mixture ? ( f.wt Na2CO3= 106g/mole ,F.wt NaHCO3=84g/mole)

Experiment 8

Estimation of Na2CO3and NaHCO3 in the same solution

Objective: To determine the exact amount of Na 2CO3and NaHCO3 in the same solution
Theory: This experiment consists of two parts.

i) Standardization of HCl solution

ii) Determination of NaHCO3 and Na2CO3

If a mixture of Na2CO3 and NaHCO3 is titrated with standard HCl solution in the presence
of phenolphthalein indicator, the light pink color disappears when Na2CO3 is changed in to
NaHCO3. The volume of HCl solution (say Vx mL) used up will account for half
neutralization of Na2CO3 .

Na2CO3 + HCl → NaCl + NaHCO3……………. (1)

To the same titration mixture, if methyl orange indicator is added and the titration is
continued, the color change from straw yellow to pale red indicates the total neutralization
reaction.
NaHCO3 + HCl → NaCl + CO2 + H2O ……… (2)
The volume of HCl required for this reaction (say Vy mL) will account for the second half
neutralization of Na2CO3 and complete neutralization of NaHCO3 originally present in the
given analyte solution. From the volume of HCl solution (Vx and Vy mL) and reactions (1)
and (2),it is possible to calculate the amounts of Na2CO3 and NaHCO3 in the given solution.

Chemicals and Apparatus:

Chemicals: Mixture of Na2CO3 and NaHCO3, 0.1 N HCl, 0.5 % Methyl orange, 0.1 %
phenolphthalein.

Apparatus: Burette, pipette, conical flasks, beaker, wash bottle, volumetric flask

Procedure:

1. Pipette out 10 ml of the analyte solution containing Na 2CO3 and NaHCO3 in to a clean
conical flask.
2. Add 3 drops of 0.5% phenolphthalein indicator solution and titrate against the
previously standardized HCl solution taken in burette until the pink color just
disappears.
Note down the volume of HCl required for this end point (say Vx mL)
3. Add 3 drops of 0.05% solution of methyl orange indicator to the same titration
mixture and resume the titration,the end point is indicated by color change from straw
yellow to pale red. Note down the volume of HCl required for this end point (say Vy
mL)
4. Repeat the titration to obtain at least two equal titre volumes.
Calculate the amount of Na2CO3 and NaHCO3 as given below:

Determination of Na2CO3

N Na2CO3 x V Na2CO3 = NHCl x VHCl

VNa2CO3 = 10 ml VHCl = 2(Vx) mL

N Na2CO3=? NHCl = 0.1N

N HCl x V HCl
N Na2CO3 = , therefore Amount of Na2CO3 in g/L = N Na2CO3 x 53(Eq.wt of
V Na2 CO 3
Na2CO3)

Determination of NaHCO3

N NaHCO3 x V NaHCO3 = NHCl x VHCl

21
VNaHCO3 = 10 ml VHCl = (Vy - Vx) mL

N NaHCO3=? NHCl = 0.1N

N HCl x V HCl
N NaHCO3= , therefore Amount of NaHCO3in g/L = N NaHCO3 x 53(Eq.wt
V NaHCO 3
of NaHCO3 in g/eq)

Question
1) Titrant of 25ml of the solution containing a mixture of Na 2CO3 and NaHCO3 took
9.46 ml of 0.12 NHl solution in the presence of phenolphthalein and 24.86 ml in the
presence of methyl orange (15.4 ml beyond the first end point ). Find the normality or
the solution with respect to Na2CO3 and NaHCO3 respectivelly .(given F wt
Na2CO3=106 g/mole, F.Wt NaHCO3 = 84g/mole).How many grams of Na2CO3 and
NaHCO3 are in 250 ml of this solution?