Chapter 2 – volumetric Analysis (test by measurement of volume)

· Introduction to volumetric analysis
· Mole concept
· Chemical equations
· Acids
· Bases
· pH Measurement
· Indicators
· Definition of terms in volumetric analysis
· Titration
· Calculations in volumetric analysis
2.1 INTRODUCTION
Volumetric analysis which is also called quantitative analysis refers to the determination of how much of a given component is present in a solution of a sample.
This is achieved or estimated by determining what volume of one substance that reacts with a known volume of another and it is done through titration. Titration is a method employed in volumetric analysis in which one solution is added to another solution such that it reacts under conditions in which added volume may be accurately measured. Titration is an easy and quick way to find the concentration of an element or compound in solution. Not much is required for a titration, needed are pipette, burette, retort stand and conical flask.
There are two common types of titration

• Acid-Base titration
• Oxidation-Reduction titration.

Acid-Base titration is described here in correspondence with the WASSCE syllabus.
But before discussing this in full, a good knowledge of acids and bases, mole concept and chemical equations will be useful and so it is illustrated below.

2.2MOLE CONCEPT
A mole (abbreviated mol) is the amount of substance which contains the same number of specified units as there are atoms in 12grams of carbon (12C). These specified or elementary units could be atoms, molecules, ions etc.
Results show that 12grams of carbon-12 contains 6.02 X 1023 atoms.
One mole each of different compounds or elements or molecules contains the same number of fundamental units which is the constant 6.02 X 1023 referred to as the Avogadro’s constant though they have different masses.

Molar Mass (M)
The molar mass, M of a substance is it’s relative atomic, molecular or formular mass expressed in grams. It is the mass of one mole of that substance. It is expressed in grams per mole. (e.g. the molar mass M of NaCl is 58.5g/mol)
The weighed mass, m of a chemical substance is directly proportional to its
Amount, n (in mole).
Mathematically; mass (m) ~ amount (n)
mass = Constant x amount
the constant of proportionality is the molar mass M
mass = Molar mass x amount.
Exercise: use the above derivation to find the mass of 0.015mol of Baking soda (NaHCO3) given that the molar mass of baking soda is 84g/mol [ 23+1 +12+(16x3)]

Percentage by mass of an Element in a Compound
Here, the knowledge of molar mass and atomic mass of various elements is employed to find the % by mass of any element in a compound.
Example: Find the % by mass of i. sodium
ii. Oxygen
in sodium hydrogen trioxocarbonate (iv) (baking soda).
Solution: mass of NaHCO3 is 23+1+12+(16x3) = 84
i. % by mass of sodium = 23 x 100%
84
ii. % by mass of Oxygen = 48 x 100%
84

Exercise 1 : calculate the percentage by mass of sodium in common salt (NaCl).

2.3CHEMICAL EQUATIONS
Chemical equation is a paper representation of a chemical reaction. Chemical reactions often occur when two or more chemicals are mixed together. Here, potassium iodide is mixed with lead nitrate to form a solid yellow chemical, lead iodide.



In writing a chemical reaction, the substances participating in the reaction [reactant(s)] are written on the left while the outcome of the reaction [product(s)]
are written on the right hand side of the equation.

Example: aA + bB = sS + wW
Acid + Base = Salt + Water
(reactants) (products)
The numerical coefficient in an equation represents the number of moles of reactants and products.
In the above representation of a chemical reaction a and b are the coefficient of the reactants the ratio a is the mole ratio.
b

2.4ACIDS
Acid is defined in different ways.
An acid could be defined as substance that reacts with water to produce excess hydroxonium ions (H3O+).
It is also refered to as a substance that ionize in water to produce H+ or H3O+ as the only positive ions in solution.

Strong and Weak Acids
A strong acid is one that ionizes almost completely in water. A strong acid releases a large number of H+ when in solution.
On the other hand, a weak acid is one that ionizes only slightly. It releases only very few hydrogen ions in solution.

EXAMPLES OF ACIDS

• Hydrochloric acid (HCl) in gastric juice
• Sulphuric acid (H2SO4)
• Nitric acid (HNO3)
• Carbonic acid in softdrink (H2CO3)
• Uric acid in urine
• Ascorbic acid (Vitamin C) in fruit
• Citric acid in oranges and lemons
• Acetic acid in vinegar
• Tannic acid (in tea and wine)
• Tartaric acid (in grapes)

Basicity : The basicity of an acid is the number of ionizable hydrogen in one mole of the acid.

Acid	Formula	Ionization	Basicity
Hydrochloric acid	HCl	HCl + H2O = H3O+(aq) + Cl-(aq)	1
Nitric acid	HNO3	HNO3 + H2O = H3O+(aq) + NO3-(aq)	1
Sulphuric acid	H2SO4	H2SO4 + 2H2O = 2H3O+(aq) + SO4-(aq)	2

Concentrated and Dilute Acids
If there is only a little water present with the acid, the solution is concentrated. And if a lot of water is present, the acid is said to be dilute.
Caution: it is dangerous to add water to concentrated acids. A dilute solution can be made by pouring the acid carefully into water.


PHYSICAL PROPERTIES OF ACIDS

• Corrosive ('burns' your skin)
• Sour taste (e.g. lemons, vinegar)
• Contains hydrogen ions (H+) when dissolved in water
• Has a pH less than 7
• Turns blue litmus paper to a red colour

Chemical Properties of Acids

• Reaction with Metals: all acids except Nitric acid (HNO3) react with Metals like Zinc, Magnesium and Iron to form salt and hydrogen gas.

Acid + Metal = Salt + Hydrogen.

• Reaction with Bases: acids react with bases to form salt and water only. This reaction is called neutralization reaction.

Acid + Base = Salt + Water

• Reaction with Trioxocarbonate(iv)

Acid + trioxocarbonate(iv) = Salt + Water + Carbon(iv)oxide

2.5BASES
Bases are substances that cancel the effect of an acid. More appropriately, a base is defined as a substance that neutralizes the effect of an acid to form a salt and water.
Bases are usually metal oxides or metal hydroxides.

EXAMPLES OF BASES AND ALKALIS

• Sodium hydroxide (NaOH) or caustic soda
• Calcium hydroxide ( Ca(OH)2 ) or limewater
• Ammonium hydroxide (NH4OH) or ammonia water
• Magnesium hydroxide ( Mg(OH)2 ) or milk of magnesia
• Many bleaches, soaps, toothpastes and cleaning agents

Alkalis: alkalis are bases that are soluble in water. Examples of alkalis are sodium hydroxide NaOH, Potassium hydroxide KOH, Calcium hydroxide Ca(OH)2 and Ammonia NH3
Alkalis release hydroxide ions(OH-) in solution. On this note then, when acid (which release H+ or H3O+) and alkalis are mixed, the following reaction takes place.
H+ + OH- = H2O
OR
H3O+ + OH- = 2H2O


Strong and Weak Alkalis
Strong alkalis are those that ionize completely in water to release many hydroxide ions into solution. E.g. NaOH and KOH.
Ammonia and calcium hydroxide are said to be weak acids because they ionize to form only few hydroxide ions in solution.

NH3(aq) + H2O = NH+4(aq) + OH- (aq)
PROPERTIES OF BASES AND ALKALIS

• Corrosive ('burns' your skin)
• Soapy feel
• Has a pH more than 7
• Turns red litmus paper to a blue colour
• Many alkalis (soluble bases) contain hydroxyl ions (OH-)

Reacts with acids to form salt and water

2.6 pH Measurement
pH is a measure of the acidity or alkalinity of a substance.
pH is measured using the pH meter.

The pH scale ranges from 0 to 14 and it is a convenient way to express hydrogen ion concentration in a given solution.

THE pH SCALE



SUMMARY TABLE OF COMMON ACIDS/BASES WITH pH

Common Acids	pH	Common Bases	pH
Hydrochloric acid Sulphuric acid Stomach juice Lemons Vinegar Apples Oranges Grapes Sour milk White bread Fresh milk	0.1 0.3 1-3 2.3 2.9 3.1 3.5 4 4.4 5.5 6.5	Human saliva Distilled water Blood plasma Eggs Seawater Borax Milk of magnesia Ammonia water Limewater Caustic soda	6-8 7 7.4 7.8 7.9 9.2 10.5 11.6 12.4 14

2.7INDICATORS
Indicators are substances which change colour when acids or alkalis are added to them. They are often made from vegetables or flowers.



The table below shows common indicators

Indicator	Colour in alkaline solution	Colour in neutral solution	Colour in acidic solution	Type of titration
Methyl orange	Yellow	Orange	Pink	Strong acid vs. strong base
Phenolphthalein	Pink	Colourless	Colourless	Weak acid vs. strong base
Litmus	Blue	Purple	Red

• Universal indicator

Universal indicator is a mixture of different indicators which gives a range of colours. These colours indicate a degree of acidity on a numerical scale called the pH scale (See pH scale above)

This means that when universal indicator is added to a solution, one of the colours above will be observed. For instance if the solution changes blue, it implies the pH of that particular solution is either 9, 10 or 11 and so on.


































2.8 DEFINATION OF TERMS IN VOLUMETRIC ANALYSIS
In volumetric analysis, terms like concentration, molarity, molar solution, standard solution, titration and endpoint are often used.

Concentration
The concentration of a solution is the amount of solute in a given volume of a solution. Concentration can be expressed in mol/dm3 or g/dm3

*Molar concentration (Molarity): this is the concentration of a solution in mol/dm3. Molar concentration of a solution is the amount of the substance in mole present in one dm3 or 1L of solution.
Mathematically,
Molar conc. = Amount (in mole)
Volume( in dm3)

*Mass concentration: this is concentration of solution in g/dm3. It is the amount of substance in grams present in one dm3solution.
Mathematically,
Mass conc. = mass (in g)
Volume( in dm3)

The mass conc. and molar conc. are related by the formular

Mass conc. (g/dm3) = Molar mass(g/mol) X Molar conc. (mol/dm3). This implies that Molar conc. = mass conc. Molar mass

Note: This is obtained from the formular mass =Molar mass(M) X amount (n) that was established in mole concept.

Working Example 1: a solution of sodium hydroxide contains 5g per dm3 of sodium hydroxide. Calculate the molar conc. [the mass of 1mole of sodium hydroxide in 40g]
Solution:
Molar conc. = mass conc.
Molar mass

= 5g/ dm3
40g/mol

= 0.125mol/dm3




Standard solution
A standard solution is a solution whose concentration is known. For example a solution containing 5g of NaOH in 1 Litre of water is a standard solution.

Molar solution
Molar solution is a solution that contains exactly one mole (the molar mass) of a substance in 1dm3 or 1L of solution. For example, a molar solution of sodium hydroxide is made by dissolving 40grams of sodium hydroxide pellets to make 1 dm3 of solution.


Working Example: calculate the number of moles present in 250cm3 of 0.1Molar (mol/ dm3) sodium hydroxide.
Solution:
First the volume unit is converted to dm3 250 = 0.25 dm3
1000
Using the relationship
Molar concentration = Amount in mole
Volume in dm3
Amount in mole = Molar concentration X Volume in dm3
= 0.1 mol/ dm3 X 0.25 dm3 = 0.025mol

Exercise 2

1. Give the mass of
1. one mole of Na2CO3
2. two mole of HNO3
2. Calculate the volume of 0.251mol/dm3 solution of tetraoxosulphate(vi) acid, H2SO4 that will contain a mass of 4.5g of the raw acid

[mass of 1mole of H2SO4 = 98g]















2.9TITRATION
A step by step description of the titration process is given below.
Required for Titration:
Burette Acid solution
Pipette Base(alkali) solution
Retort stand indicator
Conical flask

It is usual in volumetric analysis to put the acid in the burette and to measure out the alkali with a pipette into a conical flask. The acid is then run form the burette into the flask to neutralize its contents.
Procedure:
Step1 Wash and dry all apparatus
Step2 Collect some acid solution in a beaker and the alkali in another beaker or a flat
bottom flask. (remember to label them)
Step3 Pipette 25ml (20ml) of the base into a conical flask and add a few drops
(1-3drops) of the indicator required for the type of titration.






Step4 Clamp the burette to the retort stand (it should be just high enough for the conical flask to be placed underneath). Fill the burette with the acid by means of a funnel. Remove the funnel and adjust the acid level to the zero calibration mark.










Step5 Place the conical flask carefully under the burette. Do this with your left hand and
raise as well as swirl the conical flask with your right hand. Keep doing this until
there is a colour change in the solution in the conical flask.




This stage when the colour changes is called the endpoint of the titration. At this endpoint, the mole ratio of acid to base are exactly in the same ratio as required in the balanced equation of reaction. The end point is also called neutralization point. From the fore going,at the endpoint, the following relationship holds.
Mole of acid used from the burette(CaVa) = Mole of acid in equation of reaction(na)
Mole of base taken by pipette (CbVb) Mole of base in equation of reaction(nb)

Such that

Ca Va = na
CbVb nb

Ca = concentration of the acid in mol/dm3
Cb = concentration of base in mol/dm3
na = no of moles of the acid from the equation of reaction
nb = no of moles of the base from the equation of reaction.
Va = volume of acid read from the burette
Vb = volume of base (size of the pipette)






















CALCULATIONS IN VOLUMETRIC ANALYSIS
In calculating, the solution being sought is to solve a particular challenge. The various challenges that will require titration and the calculations to arrive at a convincing result are summarized in the table below and details are illustrated with experiment thereafter.

Type of titration calculation	Procedure	Application/Aim
Standardization	A standard solution is made and used to find concentration of the solution in question.	To determine the concentration of a solution whose concentration is unknown.
Formular mass of a compound Mr	A standard solution is made of the compound. It is titrated with another standard solution and its concentration in mol/dm3 is found. Using mole = g/Mr , Mr can be known.	Simply to associate your experimental result to what it will be in laid down theory.
Purity	A standard solution of the impure solid is made. Titration enables you to find the mass of the pure solid.	High level of purity is required for needed for pharmaceutical purpose.
Back titration	Usually, a solid is reacted with an excess of acid. The acid left remaining after the reaction is found by titrating with a base. It is therefore possible to find how much of the acid reacted with the solid and more about the solid.	· It is used to determine the equivalent weight of a metal · It is employd to test how safe yoghurt is in later part of this book.
Solubility	A saturated solution of the substance is made and then titrated with a standard solution of another. From this the solubility in mol/dm3 is determined.	Solubility is of fundamental importance in a large number of scientific practical application ranging from ore processing to the use of medicine
Water of crystallization	A standard solution of the substance with water of crystallization is made and then titrated against another standard solution.	The determination of water of crystallization is of great importance to pharmaceutical analysis because the water content of any medicine strongly influence their quality , how long it stays before it expires and its stability.
Mole ratio	Make a standard solution of both acid and base, then, titrate as usual. When the volume of acid that reacts with a known volume of the base is found by titration, the mole ratio na:nb can be calculated.	To determine through titration the mole ratio of acid to base in a chemical reaction.
Temporary hardness of water.	The hard water is titrated with a standard acid solution. The concentration of the causative agent in the hard water can be calculated after titration.	To determine the extent of hardness in water and to use this result obtained to specifically treat the water.

1 ] Standardization

Experiment 1:

Title: Determination of concentration of Hydrochloric acid using standard sodium
hydroxide. Aim: To determine the strength(concentration) of Hydrochloric acid. Provided: 0.1M(mol/dm3) NaOH
Procedure:

• Wash and Dry your apparatus
• Collect the acid in a reagent bottle and some sodium hydroxide in a beaker
• Pipette 20-25 ml NaOH into a conical flask and add a few drops of Methyl Orange indicator.
• Clamp the burette-(It should be just high enough for the conical flask to be placed underneath) and fill it with acid by means of funnel. Remove the funnel and adjust the level of the acid to zero mark.
• Place the conical flask carefully under the burette.

• Titrate the acid against thr alkali by opening the tap of the burette. Do this with

your left hand and raise and swirl the conical flask with your right hand. Keep titrating until the endpoint is reached. Then read perform two more titrations.

Recording
Do not write out any experimental procedure record your results as shown below.

For a pipette size of 25ml(cm3)

	Rough Titre(cm3)	1st Titre (cm3)	2nd Titre (cm3)
Final Reading Initial Reading	23.50 0.00	24.50 0.00	24.70 0.00
Volume of acid used	23.50	24.50	24.70

Average volume of acid used = 24.60 cm3 / ml

Equation for the Reaction
NaOH + Hcl == NaCl +H2O
Mole ratio = 1: 1
na = 1 nb = 1

Calculation
Calculate: I) the concentration of the acid in mol/dm3
II) the concentration of the acid in g/ dm3

Solution
I) Using
CA VA = na
CB VB nb

CA = CB VB na
VA nb
CB = 0.1 mol/dm3 VA= 24.60cm3 VB=25cm3
CA = 0.1 x 25 x 1
24.60 x 1
= 0.102mol/dm3

II) Concentration in g/dm3 = Concentration in mol/dm3 x molar mass
= 0.102 x 36.5
= 3.709g/dm3

Task 1
25cm3 of NaOH reqired 27.5 cm3 of 0.1M HCl for complete neutralization.
Calculate i) the molar concentration of NaOH
ii) the concentration of NaOH in g/dm3

Experiment exercise
Aim: To determine the concentration of Na2CO3 solution using standard HCl acid
Required: Na2CO3 solution, 0.1M HCl acid, titration apparatus and
methyl orange indicator
Procedure:
Collect Na2CO3 solution and HCl
Titrate 25cm3 of Na2CO3 solution against the acid using methyl orange as indicator
Record your results
Calculate average volume of acid used
Equation for the reaction:
Na2CO3 + 2HCl == 2NaCl + H20 + CO2

Calculation
Calculate i) the number of moles per dm3 of Na2CO3
ii) the number of g/ dm3 of Na2CO3
2] Mole Ratio
The mole ratio of acid to base in a chemical reaction na : nb can be estimated from titration results carried out using standard solutions of both acid and base. When this mole ratio is found, an equation for the reaction can be conveniently written.

Working Example
A is a solution of 0.095 mol/ dm3 HCl while B is a solution containing 0.0475 mol/dm3 of Na2CO3. 25cm3 of B needs an average volume of 25.50 cm3 of A for complete neutralization.
a) What indicator do you think was used for the above titration? Why?
b) Calculate the mole ratio of HCl to Na2CO3 to the nearest whole number ratio and hence write the equation for the reaction.

Solution
CA = 0.095 mol/dm3
CB = 0.048 mol/dm3
VA = 25.50 cm3
VB = 25.00 cm3
CA VA = na
CB VB nb
na ­­ = 0.095 x 25.50
nb 0.048 x 25.00
= 2.08
(Approximately = 2/1)

Equation of Reaction
Na2CO3 + 2HCl == 2NaCl + CO2 + H2O

















3]Relative Molecular Mass

The relative molecular mass of a base or an acid can be found by titration as follows:

A known mass of base is dissolved in a known volume of water (i.e. the concentration in g/dm3). If the base solution is titrated with a standard acid in a usual titration order, then, the concentration of the base in mol/dm3 can be known.


The relation:
Concentration in g/ dm3 = Concentration in mol/ dm3 x Molar mass in g/mol
can be used to find the molar mass and the relative molecular mass is the molar mass without unit.

Working Example (WASSCE 2009)
A is 0.095 mol/ dm3 HCl
B is a solution containing 13.50g/ dm3 of X2CO3.10H2O.
a) Put A into the burette and titrate it against 20.0cm3 or 25cm3 portions of B using methyl orange as indicator. Tabulate your readings and calculate the average volume of A used.
b) From your results and the information provided above, calculate the
I, Concentration of B in mol/ dm3
II, Molar mass of X2CO3.10H2O.
III, Percentage by mass of X in X2CO3.10H2O.
[Hint: H=1, C=12, O=16]
The equation for the reaction involved in the titration is
2HCl + X2CO3.10H2O == 2XCl + 11H2O + CO2

Solution
A,)

	Rough (cm3)	1st titre value (cm3)	2nd titre value (cm3)
Final Reading Initial Reading	30.10 5.00	24.90 0.00	25.00 0.00
Volume	25.10	24.90	25.00

Average volume of acid = (24.90 + 25.00) / 2 = 24.95cm3

I) Using
CA VA = na
CB VB nb

CB = CA VA nb
VB na
Ca = 0.0950 mol/ dm3
Va = 24.95 cm3
nb = 1
na = 2
Vb = 25 cm3

Therefore CB = 0.0950 x 25.2
25 x 2
= 0.049 mol/dm3

II) Mass concentration in g/dm3 = molar mass in g/mol x molar concentration in mol/dm3
Molar mass in g/mol = mass concentration in g/dm3
Molar concentration in mol/dm3

= 13.50g/dm3
0.047215

= 285.93 g/mol

III) Mass of X2CO3.10H2O = 285.93
(2 x X) + [12 + (3 x 16)] + [10 x ((2 x 1) + 16)] = 285.93
2X + 12 + 48 + 180 = 285.93
2X + 240 = 285.93
2X = 285.93 – 240
2X = 45.93
X = 22.965
Percentage mass of X in X2CO3.10H2O =

Mass of X in X2CO3.10H2O. x 100%
Molar mass of X2CO3.10H2O.

= 45.93 x 100% = 16.06%
285.93














4]Treatment of Temporary Hardness in Water
People living in areas where there is large deposit of limestone or gypsum(calcium tetraoxosulphate(VI)) find that it is very difficult to get a lather with soap while washing clothes. Their water supply is said to be hard.

Temporary hardness of water is caused by the presence of Ca(HCO3)2. To make hard water suitable for both domestic and industrial purposes, hardness has to be removed. These can be achieved by:

• Boiling: Boiling removes temporary hardness but it is expensive on a large industrial scale.
• Adding Lime(Calcium Hydroxide): This is an effective and cheap method but an exactly calculated amount of the lime(Ca(OH)2) is to be carefully added. This implies that the amount of Calcium Hydroxide (lime) to be added must be correct, because too much of the lime will itself cause further hardness.

Determination of Quantity of Lime
To be able to determine this exact amount, the extent of the hardness is to be found, this is basically the concentration of Ca(HCO3)2 in g/dm3. Since CaHCO3 is the causative agent. It is the concentration in g/dm3 of Ca(HCO3)2 that we will be finding by titration.

The Experiment example given below depicts this and the exact amount needed is calculated from the result of the titration.

Example: 25cm3 portions of sample of hard water containing Ca(HCO3)2 were titrated with 0.050 mol/dm3 HCl using methyl orange as an indicator and the following results were obtained.

	Rough (cm3)	1st Titration (cm3)	2nd Titration (cm3)
Final Initial	24.00 0.00	48.10 24.00	24.08 0.00
Volume Used	24.00	24.10	24.08

Average Volume of acid used VA = 24.10 + 24.08
2
=24.09cm3
Equation for the reaction:
Ca(HCO3)2 + 2HCl == CaCl2 + 2CO2 + 2H2O

From the results and other information’s provided, calculate the:
I Concentration of Ca(HCO3)2 in mol/dm3
II Concentration of Ca(HCO3)2 in g/dm3

Solution

I, mole ratio (na:nb) = 2:1
Given CA = 0.050 mol/dm3
VA = 24.09 cm3
VB = 25 cm3

Applying CA VA = na
CB VB nb

CB = CA VA nb
VB na

= 0.050 mol/dm3 x 24.09cm3 x 1
25cm3 x 2

= 0.024 mol/dm3

II, Concentration in g/dm3 = Concentration in mol/dm3 x molar mass in g/mol
= 0.024 x 162
= 3.903 g/dm3

From our experiment calculations, we have been able to tell the concentration of Ca(HCO3)2 in both mol/dm3 and g/dm3 which is the extent of hardness of the water sample.

The next step is to use these information to work out the exact amount of lime to be added to a given volume if the hard water to soften it.

When lime is added to hard water, it reacts with Ca(HCO3)2 through the following equation.

Ca(HCO3)2 +Ca(OH)2 == 2CaCO3 +2H2O

From the equation 1 mole of Ca(HCO3)2 reacts with1 mole of Ca(OH)2.
1 mole of Ca(HCO3)2 has mass 162g while
1 mole of Ca(OH)2 has mass 74g.
From results obtained 0.024 mol/dm3 of Ca(HCO3)2 was present in the water and the reacting mole ratio of Ca(HCO3)2 to Ca(OH)2 is 1 :1, it means that for the given sample of water used in the exercise, 0.024 moles of Ca(OH)2 must be added to 1dm3 of the water to treat hardness.
Or 1.776g of lime must be added to 1dm3 of the water to soften it.








5. Purity
A common problem in Chemistry is determining whether or not a substance is pure and to what extent is it pure/ impure. Chemicals used often need to be pure to ensure efficiency and safety during reactions but determining whether or not a given substance is pure can be difficult. Impurities may harm people using such things as drugs if it exceeds certain limits in the formulation. When chemist make something from a chemical reaction and separate it from the final mixture it will still have small amount of other substances mixed with it and this makes the desired product impure.


%Purity = Mass of pure product x 100%
Mass of impure product

Example
in an experiment to produce aspirin, 121.2g of solid was obtained as total yield but analysis showed that only 109.2g of this total yield was actually aspirin. Calculate the percent purity of the product.
Solution: %purity = 109.2g x 100% = 90%
121.2g


If the impure chemical is an acid or a base, titration with a standardized solution can be used to assess its purity.

This titration method is useful in determining purity only if a standardized solution is available for the titration. If a known volume of a standardized solution is used in a titration, then the moles of both acid and base can be determined, from the moles, the mass of the pure substance can be determined and compared to the mass of the impure substance to find the percentage purity.

In the experiment below, the purity of an acid H2SO4 is being determined with a standard NaOH solution.

Experiment:
Title: Percentage purity of substance
Aim: To determine the extent of purity of an impure H2SO4.

Procedure:
A is a solution of an impure tetraoxosulphate (VI) acid with concentration of 16 g/dm3. B is a standard solution of sodium hydroxide containing 15mol/dm3
Titrate 25ml of B with A in the usual procedure for acid – base titration using methyl orange as indicator.

Calculation:
The average volume of acid required to neutralize 25ml of standard sodium hydroxide solution is 15.52ml.
Equation for the reaction
H2SO4 + 2NaOH == Na2SO4 + 2H2O
na = 1, nb = 1
Calculate i) the concentration of the pure acid in mol/dm3
ii) the concentration of the pure acid in g/dm3
iii) the percentage purity of the acid

Solution

i) During titration, the base reacts with only the pure acid for neutralization. The impurity does not affect the neutralization reaction.
When: CA VA = na is employed it’s concentration of the pure acid that will be found.
CB VB nb
CA = CB VB na
VA nb
= 0.15 x 25
2 x 15.52

= 0.121 mol/dm3

ii) Mass concentration of pure acid in g/dm3 = molar mass of acid x molar concentration of acid in mol/dm3
= 98g/mol x 0.121 mol/dm3
= 11.86g/dm3
iii) Since the mass concentration of the impure acid is 16g/dm3 and that of the pure acid is 11.86gdm3, the concentration of impurity is 16 – 11.86 g/dm3 = 4.14g/dm3
Percentage of impurity = mass concentration of impurity x 100%
Mass of impure acid
= 4.14 x 100%
16
= 25.88%
Similarly,
Percentage of purity = mass of concentration of pure acid x 100%
Mass of impure acid
= 4.14 x 100%
16
= 74.12% OR
The percentage purity = 100% - percentage of impurity
= 100% - 25.88%
= 74.12%
6) Water of Crystallization
Crystallization is a separation technique employed to recover back salt from solution when the salt to be separated is destroyed if it is heated to dryness as in evaporation.
In crystallization, the solution containing the salt is first heated to remove most of the water by evaporation. The concentrated solution gotten from this initial evaporation is allowed to cool slowly, then crystals of the salt begin to form on cooling.
Crystal of Alum

As the crystal is being formed some water molecules remain and occur in the crystal structure giving if its shape and colour but it is not chemically bonded to the crystal. The water is refered to as water of crystallization. Copper II Sulphate Crystals
Such salts containing water of crystallization are called hydrous substances or specifically called hydrates.

The water of crystallization of hydrates can be found by titration. During the titration, only the dry (anhydrous) part of the hydrate is actually neutralized. Which means the water of crystallization remain in the titration mixture as impurity.
The following experiment exercise describes how water of crystaliztion is determined by titration.

Experiment title: to find the number of molecules of water of crystallization of sodium carbonate
Required: standard solution of sodium carbonate containing 60g of the hydrated salt
Per dm3, 0.5M hydrochloric acid, titration apparatus and methyl orange indicator
Procedure: pipette 25cm3 of the sodium carbonate solution and titrate with the standard
hydrochloric acid. Use methyl orange as indicator. Record two accurate
titrations.

Calculations
· Find the number of moles of HCl used.
· Write down the equation of reaction then find the number of moles of sodium carbonate in 25cm3 of solution.
· Find the molar concentration of the sodium carbonate solution,
· Calculate the concentration in g/ dm3 of the anhydrous sodium carbonate, Na2CO3. (Say E g/ dm3)
· Let the formular of the hydrated sodium carbonate be Na2CO3.xH2O (x is the number of molecules of water of crystailzation). The following holds mass of one mole of Na2CO3 = 106 = E
mass of one mole of Na2CO3.xH2O 106 + 18x 60

and since E is already found, calculate x to the nearest whole number



























7) Solubility
Solubility expresses how well a solute form solution in a solvent.
Solvent is a substance (mostly liquid) which dissolves other substances to form solution. The substance which dissolves is called solute and the resulting mixture is called solution.
By definition, solubility is the quantity in grams or number in moles of a given solute needed to saturate 1 dm3 of a solvent at a given temperature. Solubility is expressed in mol/ dm3. * A saturated solution is one which contains as much solute as it can dissolve at that temperature in the presence of excess undissolved solute in the mixture.
For example, at 250C, 400g of potassium trioxonitrate(V) will saturate I dm3of water. Whereas, Lead (II) Iodide is only slightly soluble at 250C; at this temperature the solution is saturated when only 0.08g have dissolved in 1 dm3


Home Activity
Take ¼ glass of water then continuously add salt to it spoon by spoon with constant stirring. What do you observe after sometime?
You will observe that while adding salt initially, the salt dissolved completely but upon continuous addition at some point the salt stopped dissolving. At this point, the salt solution is said to be saturated.

Laboratory work
Experiment Title: to determine the solubility of Calcium Hydroxide
Required: Saturated solution of calcium hydroxide, 0.05M hydrochloric acid, titration
apparatus, methyl orange indicator.
Procedure: pipette 25cm3 of the saturated solution of calcium hydroxide (freshly prepared) into a conical flask and titrate it with the standard hydrochloric acid using methyl orange as indicator. Record two accurate titrations.

Calculations
· Write down the equation for the reaction.
· Find the concentration of the calcium hydroxide in mol/dm3 which is the solubility.














BACK TITRATION (INTRODUCTION TO A’LEVELS)
There is no precise defination of back titration. It can simply be said to be the reverse of normal titration.


There are so many ways of looking at back titration and they are based on the purpose of the investigator.
Reasons for Back Titration
The test substance (analyte) may be in solid for: usually in this case the solid is reacted with an excess of another reagent (say acid) first. Solid + Reagent (in excess) = Product + remaining reagent. The excess reagent left after the reaction is complete is then determined by titrating the mixture. It is therefore possible to find how much of the excess reagent that reacted with the solid from the titration calculation. A simple analogy of this can be related to a child who goes to the shop to buy chocolate. The child is sent to the store with a known amount of money you gave (excess reagent) in order to purchase the chocolate. Now when the child comes back you will subsequently determine the amount she used to buy from the remainder money brought back. This approach is used to determine the equivalent weight of a metal as will be seen in the experiment below.
The analyte reacts slowly with the other reagent in a forward titration: In this case if the test is subjected to back titration, better result is obtained. This case is employed in the determination of Lactic acid in Yoghurt in Science appreciation project in the next chapter.
The anaylte may contain impurities which may interfere with direct titration: This approach is used to determine how much of acidic impurity is present in a system.

Equivalent weight of a Metal
The equivalent mass (or weight) of an element is the number of parts by mass of the element that will combine with or displace one part by mass of hydrogen or eight parts by mass of oxygen.
Mathematically, we can define the equivalent weight of an element to be equal to the molecular weight of the substance divided by its valence.





Experiment Title: to determine the equivalent weight (or mass) of Magnesium
From reason 1 above, if a known mass of magnesium reacts with a known volume of acid, Mg + 2HCl = MgCl2 + H2 (g) the remaining acid left after the reaction can be titrated with a standard base and the volume of the acid that reacted with the magnesium can be found.
Required:
1M Hydrochloric acid, 1M Sodium hydroxide solution, a piece of magnesium metal, titration apparatus, methyl orange indicator.
Caution: handle both solutions with care as they are highly concentrated.
Procedure:
Clean the piece of magnesium metal, weigh about 0.1g of it accurately into a conical flask.
Fill the burette with the sodium hydroxide solution.
Use a 25ml pipette to measure out 25ml of the 1M hydrochloric acid solution into the conical flask containing the 0.1g Magnesium.
When the magnesium is completely dissolved, titrate the remaining acid in the reaction mixture with the base solution in the burette using methyl orange as an indicator.
Record your results as usual.
Note: before washing the burette, carefully rinse it twice with some of the acid.

Example of Calculation
Equation of reaction: Mg + 2HCl = MgCl2 + H2 (g)
From the equation, 1mol of Magnesium reacts with 2mol of Hydrochloric acid.
Remember that from the definition of equivalent mass, the equivalent mass of magnesium should react with 1mole of HCl because 1mole of HCl contains one part by mass of hydrogen. And 1mol of the acid in 1dm3 (1000ml) total volume of solution is 1M solution. Therefore, the equivalent mass of magnesium should react with 1dm3 of 1M solution of the acid.
Suppose 16.8ml was the average volume of base used during the titration process. NaOH + HCl = NaCl + H2O
from the equation, 1mol of NaOH reacts with 1mol of HCl. Then,
25ml of 1M NaOH reacts with 25ml 0f 1M HCl. This implies that,
16.8ml of 1M NaOH reacts with 16.8ml of 1M HCl.
a. what was the volume of acid that reacted with 0.1g of Magnesium.
Answer: since 25ml was added initially and it was found by titration that 16.8ml reacted with the base then (25ml ₋ 16.8ml) 8.2ml of the acid reacted with 0.1g of Magnesium.
b. find the number of moles of the acid in 8.2ml HCl.
Answer: 1mol is contained in 1000ml of solution
Xmol is contained in 8.2ml of solution. Solving for xmol
Xmol = 8.2ml X 1mol = 0.0082mol
1000ml
c. 0.0082mol of the acid reacted with 0.1g of magnesium. What mass of magnesium should react with one mole of the acid?
Answer: 0.0082mol reacted with 0.1g,
1mol should react with x
X = 0.1g x 1mol = 12.2g (approx.)
0.0082mol
d. From the above calculation, what is the equivalent mass of Magnesium?
Answer: the equivalent mass is that which will react with 1mole of acid. And as calculated above it is 12.2g.
You can also confirm if this experiment is correct by solving it for we also said the equivalent weight is the molecular mass/valence which is 24g/2 = 12g
The 0.2g addition is due to unavoidable experimental error.
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