Content

Module Name

Volumetric Analysis

Class

SS2

Objectives

After studying this unit, students should be able to:

a.      Explain the roles of different tools used for volumetric analysis

b.     Take proper precautions while carrying out Acid-base titration.

Introduction

Volumetric Analysis is an aspect of quantitative analysis, which involves measuring the volumes of solutions of reactants in a chemical reaction, so as to determine the amounts of reactants in such solutions. Volumetric methods of analysis comprises of techniques in which the volume of a solution of known concentration is measured.

The most popular aspect of volumetric analysis is acid-base reactions. The technique of determination is by TITRATION. Titration is the procedure by which a solution of known concentration is added to another solution until the chemical reaction between the solutes is complete. The solution whose concentration is known is called the STANDARD SOLUTION OR THE TITRANT.

In titration, the standard solution is slowly added from a burette to a solution, which contains a known mass of solute. The latter solution is commonly referred to as the UNKNOWN. The point at which equivalent or stoichiometric amount of acid is added to the alkali is known as the EQUIVALENCE POINT of titration. This point signifies when the reaction must have come to completion.

Thus, in the process of titrating an unknown solution with a standard solution, there must be some way to determine when the equivalence point of the titration has been reacted. This is because, in most of the cases, the two solutions being titrated against each other are colorless and determining the equivalence point will be a guessing game and thereby introduce error. To avert such inappropriateness, the end point of the equivalence point is usually done by using what is called an INDICATOR.

The indicator in most cases is an organic substance which changes color in the titrated solution when an amount of the standard solution (titrant) added is Equivalent to the amount in the sample. The moment at which the indicator changes the color of the titrated solution is called the END POINT of the titration.

A successful titration, a titration with good quantitative results, is possible only if the end point as established by the indicator occurs at the equivalence point of the titration.

Volumetric Tools and Techniques

These are tools and techniques used during volumetric analyses. The most important pieces of apparatus in acid-base titration are the burette and pipette. Others are conical flask, beaker, volumetric flask, white tile and wash bottle.

1.     Burette: Is a long glass tube fitted with a tap through which the liquid contained in it is delivered via a small opening at one end. It is designated to measure accurately the volume of liquid delivered during a titration. The commonest one has capacity of 50cm³

2.     Pipette: Is used for transferring liquids or solutions from one container to another. It is a long tube with a short cylindrical bulb at the middle. They have varying capacities between 1cm³ to 50cm³. The most commonly used for quantitative analysis are the 10cm³, 20cm³, 25cm³ and 50cm³ capacity pipettes.

3.     Conical Flask: Is used in volumetric analysis when good mixing of liquids is required for example, during titration. It has a narrow neck to prevent solution for spurting when being shaken. The commonly one used in Volumetric work is the 250cm³ conical flask.

4.     Beaker: Is used for collecting/holding the solution to be added to the burette and the solution to be pipetted.

5.     Volumetric Flask: This is a flat-bottom flask with a long narrow neck. It is mostly used for preparing standard solutions. The commonly used in volumetric analysis is the 250cm³ capacity flask.

6.     Wash Bottle: This is used to add small quantities of distilled water to a volumetric flask, and also to rinse down any drops of solution on the sides of a conical flask.

7.     White Tile:  It is placed underneath the conical flask to spot the end point color change. It makes it easier to observe a color change.

General Procedure for Titration

1.     Rinse the apparatus with distilled water (e.g. conical flask).

2.     Rinse the burette with the solution to be taken in it.

3.     Fill the burette to near the top with the titrate solution of known quantity and concentration.

4.     Use a pipette to measure out the other solution (unknown concentration) into the conical flask.

5.     Add 2-3 drops of Indicator.

6.     Note the burette reading and write it down.

7.     Add the titrant from the burette slowly to the flask swirling the flask to ensure mixing. Once the indicator color change remains for longer periods of time, reduce the flow.

8.     Stop the addition of the titrant as soon as the color change remains permanent. Record the burette reading.

9.     The difference between the initial and final burette reading is the amount of titrant added.

10.  Repeat the procedure till three concordant readings are obtained and do the needful calculations.

Precautions During Acid-Base Titration

i.         To get accurate values, one must take special care to clean all the apparatus with distilled water. Slight presence of any other chemical will lead to mistakes in the result.

ii.       While adding acid/base, it should be done drop-wise.

iii.     Contamination should be avoided as much as possible.

iv.      While taking the readings, make sure that the markings are at eye level to avoid parallax error.

v.        In the case of colorless solutions, the lower meniscus is read whereas the upper meniscus is read for colored solutions.

vi.      For pipetting any solution, always use the index finger.

vii.    Make sure that the burette is not leaking and that there is no air bubbles trapped inside it.

viii.  The indicator should never be used in excess, that is, use two or three drops.

ix.      Take extra care to fill the pipette to get accurate readings, do not allow saliva to enter into the pipette to avoid dilution

x.        Remove the funnel before titration to avoid increase in the volume of the solution in the burette.

xi.      Do not blow the last drop at the top of the pipette to avoid increase in the volume of the base.

Conclusion

In this unit, the learners have been exposed to Volumetric analysis, it’s tools and techniques and also the general procedure for titration and Precautions during acid-base titration.

Exercise

1.     Define the following terms used in volumetric analysis

a.      Standard Solution

b.     Equivalence point

c.      End point

2.     List five (5) volumetric analysis tools

3.     List five (5) precautions necessary to ensure accurate results during titration.

Content

Module Name

Units of Measurement

Class

SS2

Objectives

After studying this unit, students should be able to:

A.    Get familiar with S.I units of mass, volume, unit of amount and unit of Molar Mass.

B.    calculate the molar mass of acids and bases.

C.    prepare a standard solution

Introduction

Weighing an object is a practice common in a Chemistry laboratory. The object in this case is usually a salt, a compound, a reagent, water or glass container. The result of weighing is the mass in weight of the object. Since Chemistry is a scientific discipline, weighing must be definitive and comparable with other weighing’s done in other laboratories.

Thus, measured or weighed quantities must be expressed in appropriate and comparable unit for proper understanding of the object in relation with other quantity. If we forget to give the units when reporting the result, the measurement will be meaningless as it will be difficult to interpret. It is therefore a must that all measurements must be accompanied with the appropriate unit.

Units of Measurement

Measured quantities such as the mass, volume and length must be expressed in suitable units, if not, the results of such measurements will be of no use. For instance, it is useless and meaningless to say that the quantity of water in a container, is simply 10.1, without attaching the unit of volume.

Unit of mass (m)

The base unit of a measured solid material say a bottle or a salt is the kilogram, abbreviated as kg. However, it’s lower fractional unit, the gram, g, is commonly used.

For conversion purposes:

                                         1.00kg = 1000g (or 1.00×10³g)

                                         1.00g = 0.001kg (or 1.00×10^-3kg)

Unit of Volume (v)

The base unit of measured liquid, which is called volume, is the cubic meter, abbreviated as m³. However, the lower units, such as cubic decimeter, dm³ and cubic centimeter, cm³, are more popular.

Some of the pieces of apparatus that are designed for accurate measurement of volumes are the burette, pipette and measuring cylinder. They are available in various sizes.

For conversion purposes:

                                       1.00dm³ =1000cm³ (or 1.00×10³cm³)

                                       1.00cm³ = 0.001dm³ (1.00×10^-3dm³)

Unit of Amount (n)

The base unit of amount of a chemical substance is the mole; abbreviated as mol. A mole is the amount of Substance which contains as many elementary chemical units as there are atoms in 12.0g of carbon -12. These elementary chemical units maybe atoms, molecules, ions and electrons.

In chemistry, one mole of an element or a compound is equal to its relative atomic mass, formula mass, or molecular mass expressed in grams, or any unit of mass, i.e.

  1 mole of carbon weighs 12.0g and contains 6.02×10²³

  1 mole of H₂O weighs 18.0g and contains 6.02×10²³ molecules

  1 mole of NaOH = 40.0g = 6.02×10²³ molecules.

Unit of Molar Mass (M)

The base unit of Molar mass, M of a chemical substance is grams per mole (g/mol or gmol^-1). When the relative atomic, molecular or formula mass of a chemical substance is expressed in grams, it is called MOLAR MASS (M); that is, mass of one mole. Molar mass = mass of one mole.

Hence, the molar mass of NaOH = 40.0g/mol; that of H₂O = 18.0g/mol.

Amount of Substance

Concentration of Solutions

This is a term used to refer to amount of solute in the given volume of solution. It is defined as the amount, n, of the solute divided by the volume, v, of the solution. That is

    Concentration, C, = n(mol)/V(dm³) or number of moles, n = CV

The concentration term is, therefore, expressed in

      mol per dm³ (mol.dm^-3).

For example,

1 moldm^-3 sodium hydroxide solution contains 1 mole (or 40g) sodium hydroxide in 1dm³ of solution.

Consider the reaction represented by the equation

                 aA + bB …….. Products

‘a’ mole of the substance A reacts with ‘b’ moles of another substance B to give the products.

Amount of A consumed by the reaction = C_AV_A = a ……(1)

Amount of B consumed by the reaction = C_BV_B = b ……(2)

Divide (1) by (2)

C_AV_A/C_BV_B = a/b

The ratio a/b is called the mole ratio of the reactants A to B

Concentration of a solution, at times may be expressed in terms of the mass in (grams) of the solute dissolved in a given volume (in dm³) of the solution. For example, 2.0grams per dm³ (or 2.0gdm^-3) sodium hydroxide solution is prepared by dissolving 2.0g of sodium hydroxide in 1dm³ of solution. When concentration term is expressed in grams per dm³, it is called MASS CONCENTRATION.

Hence,

Mass concentration of a solution = mass in grams of solute/volume of solution in dm³.

Standard Solution

This is a solution whose concentration is known. That is, it is a solution which contains an accurately known mass of solute dissolved in a known volume of solution.

It is prepared, if the chemical composition of the substance is known, and can be obtained in a pure state. For instance: a liquid solution known to contain 1.00g of pure sodium hydroxide, NaOH, in say, 250cm³ of solution, is a standard solution.

A standard solution of a known pure substance is prepared by weighing accurately a known mass of the substance, dissolving it in water, and making up the solution to a definite Volume in a volumetric flask.

 Volumetric flask is used in preparing a standard solution to a specific volume. A volumetric flask should not be heated, in order to maintain its capacity.

Dilution of Solution

This is based on the argument that the amount of solute in a given volume of solution does not change by increasing the volume of the solution. This statement will become clearer if you perform the experiment below.

Fill a tea cup with water, add two cubes of sugar into it and stir. Taste the solution. Transfer the sugar solution into a bowl and add water to fill. Stir the solution and taste it.

You will notice that the solution in the tea cup is sweeter than that in the bowl. One can say that the solution in the tea cup has a higher concentration (sweeter) than that in the bowl. By adding more water to the solution in the tea cup both the volume and sweetness (concentration) of the solution will change but the (2cubes) of sugar remains the same. We therefore say that the original solution is diluted.

                         Tea cup                                 Bowl

                C₁V₁ = n (2 cubes).                C₁V₁ = (2 cubes)

Concentration of solution = C₁

Volume of solution = V₁

Amount of solute (sugar) = n (2cubes)

Since C₁V₁ = n ………. (1)

and C₂V₂ = n ………….. (2)

C₁V₁ = C₂V₂ ………….. (3)

Equation (3) is called the dilution formula. The number of cubes of sugar is equivalent to the number of moles of solute.

Summary

In this unit, the module discussed the various units of measurement students are likely going to use in the chemistry practical class of Volumetric analysis.

Exercise

a.      State the S.I unit of (a) amount. (b) molar mass

b.     How many molecules are in one mole of HNO₃?

c.      Calculate the molar mass of:

(a) H₂SO₄ (b) NaHCO₃

[ H = 1; C = 12; O = 16; Na =23; S = 32]

Content

Module Name

Acid-Base Titration

Class

SS2

Objectives

After studying this unit, students should be able to:

1.     List the terms used in acid-base titration.

2.     Explain the pH scale

3.     Relate the choice of indicator to type of reaction

4.     Carry out acid-base titrations using appropriate indicators

5.     Enumerate the different types of acid-base reactions

6.     State the variables in acid-base titrations

Introduction

Acid-base reaction is the most popular aspect of volumetric analysis. It is determined by titration with an appropriate base solution. Conversely bases are determined by titration with an appropriate acid solution.

Acid-base titration is the technique employed to determine the amount of acid required to neutralize completely a given amount of base and vice versa. In titration process, one solution, usually the standard, is slowly added from the burette to another solution placed usually in a conical flask that contains a known volume or a known mass of solute, until the chemical reaction between the two reactants is complete.

Acid-Base Titration

Terms Used in Acid-Base Titration

1.       Acid: An acid is a substance which when dissolved in water produces hydrogen ion as the only positive ions. The process is known as IONIZATION

a.      Strong and Weak Acids

i.                 STRONG ACID: Ionizes completely in solution to produce very high concentration of hydrogen ions. Examples are

_–             Hydrogen Chloride Acid – HCl_(aq)……..H⁺_(aq) + Cl^–_(aq)

_–             Trioxonitrate (v) acid – HN0_3(aq)….H⁺_(aq) + N0_3-(aq)

ii.               WEAK ACIDS: Ionizes partially or completely in solution producing very low concentration of hydrogen ions. Examples are

–        Trioxocarbonate(iv)acid – H₂CO_3(aq)……2H⁺_(aq) + CO₃^2-_(aq)

–        Tetraoxophosphate(v)acid – H₃PO_4(aq)……3H⁺_(aq) + PO₄^3-_(aq)

–        Ethanoic Acid – CH₃COOH_(aq)……H⁺_(aq) + CH₃COO^–_(aq)

2.     BASE: A base is a metallic oxide or hydroxide. Soluble bases are called alkalis. Example

Na₂O_(s) + H₂O_(l) …. 2NaOH_(aq)

K₂O_(l) + H₂O_(l) ……. 2KOH_(aq)

Strong And Weak Alkalis

i.                 Strong Alkalis: Ionize completely in solution to give very high concentration of negatively charged hydroxide ions. Example

Sodium Hydroxide: NaOH_(aq) …. Na⁺_(aq) + OH^–_(aq)

Potassium Hydroxide: KOH_(aq) ….K⁺_(aq) + OH^–_(aq)

ii.               Weak Alkalis: Ionize partially or completely in solution producing very low concentration of hydroxide ions. Example

Calcium Hydroxide. – Ca(OH)_2(aq) ……. Ca^2-_(aq) + 2OH^–_(aq)

3.     Neutralization: Is a process in which an acid reacts completely with an equivalent amount of alkalis or base to form a salt and water only. Example

            H⁺Cl^–_(aq) + Na⁺OH^–_(aq)….Na⁺Cl^–_(aq) + H₂O_(l)

             Acid            Base            Salt           Water

pH Scale

The pH of a liquid is measure of its hydrogen ion, [H⁺] concentration. It is measured by the use of a device called pH meter, which has an inbuilt scale that ranges from 0 to 14. A liquid with a pH value of 7 is neutral. If the pH  value of a liquid is less than 7, then, it is acidic while if it is greater than 7, it is alkaline pure (distilled). Water has a pH value of 7, that is, it is neutral.

Common Acid-Base Indicators, Their Colour Change and pH-Range

Indicator

An indicator is an organic compound which shows an abrupt change in colour with acidic or basic solutions. It dissociates slightly in solution to produce ions and undissociated molecules.

There are two types of indicators:

1.          Common acid-base indicators: These indicate whether a solution is acidic or basic. Litmus, methyl orange and phenolphthalein are common acid-base indicators. 

2.          Universal indicators: It is a mixture of organic indicators, which indicate pH values by the changes of definite colours depending on H⁺ ion concentration. The universal indicator is better than an acid-base indicator as it not only shows whether the solution is acidic or basic but also gives the relative strength of acidic or alkaline solution.

The point at which the stoichiometrically equivalent quantities of substances have been brought together is known as equivalence point of titration. To determine when the equivalence point of titration is reached, an indicator is used. In acid-base titrations, the indicators such as methyl orange, phenolphthalein and bromothymol blue which are usually organic dyes that change color according to the hydrogen ion concentration of the solution or liquid to which they added or found themselves, are used.

Types of Acid-Base Titration and Choice of Indicator

The choice of indicator for a particular set of an acid-base titration depends on the strength (strong or weak) of the acid and base.

Wrong Use of Indicator

The accuracy of a titration depends on the use of the correct indicator.

Wrong choice of indicator will lead to wrong result. For instance, in the titration of a solution of a strong acid (HCl) with that of a weak base (Na₂CO₃), methyl orange should be used. If Phenolphthalein is used instead, the end point will appear when only half of the weak base Na₂CO₃ has been used up, as shown below

1. HCl + Na₂CO₃ ….. NaHCO₃ + NaCl
2. H₂SO₄ + Na₂CO₃ …. NaHCO₃ + NaHSO₄

The reason is that Phenolphthalein is sensitive to weak acid such as NaHCO₃. For instance, if you add say methyl orange indicator, the colour you will notice is yellow. As you run in the acid from the burette, the neutralization reaction occurs. The colour in the conical flask will still be orange until complete neutralization when there is stoichiometric equivalence of the acid and base.

The next drop of acid into the conical flask wil not have a base to react with. Therefore there will be an extra drop of acid. At this point, the colour in the flask will now have to change because the indicator now finds itself in a new environment that is, acidic environment. In the particular case of methyl orange, the colour will now change to pink thus marking the end point.

Variables in an Acid-Base Titration

Conclusion

Exercise

The relevant variables in acid-base titration are:

C_a = Concentration of acid , in moldm^-3

C_b = Concentration of base, in moldm^-3

V_a = Volume of acid used in cm³ (or dm³)

V_b = Volume of base used in cm³ (or dm³)

n_A = Amount of the acid

n_B = Amount of the base

The units of C_a and C_b on one hand, and V_a and V_b on the other hand must be correct  and the same.

The ratio of these amounts must be thesame as the ratio of the reacting amount n_A and n_B.  

In titrations designed to analyze solutions, the equation for the reaction is given so that the ratio n_A/n_B is known. The concentration of one of the solutions is also known. The volumes V_a and V_b are measured during the titration. Substituting all the known quantities in the titration formula allows the concentration of the unknown solution to be calculated.

In this unit, the learners have been exposed to acid-base titration, pH scale, choice of indicator and variables in acid-base titration.

1.     Explain Acid-base titration

2.     List the terms used in acid-base titration

3.     Explain the pH scale

4.     List the types of indicators and relate the choice of indicator to type of reaction.

Content

Module Name

Acid-Base Determination of the  Concentration (Standardization) of a Solution

Class

SS2

Objectives

After studying this unit, students should be able to:

1.       Use a pipette

2.       Use a burette

3.       Carryout a titration correctly

4.       Calculate the concentration of a solution from a titration data

Introduction

In this unit, the module is set to determine the concentration of a solution. The requirements, procedure, results and the treatment of the results shall be discussed.

Experiment

Acid-base determination of the concentration (standardization) of a solution

Requirements

Procedure

1.     Rinse the burette with a few drops of solution A and fill it with same solution above the zero mark. Drain to the marking, making sure the burette is full. Note the initial readings.

2.     Using a pipette, transfer 25cm³ of solution B into a conical flask. Add two to three drops of methyl orange. Place the conical flask on a white tile.

3.     Run solution A from the burette into the conical flask whilst shaking vigorously until a permanent faint pink coloration is observed. Note the final reading.

4.     Repeat the titration steps 3 and 4 to get a set of three concordant readings.

5.     Tabulate your burette readings

6.     Determine the average titre value.

7.     Calculate the mass of the concentration of a solution by using

Mass concentration = Molar concentration x Molar mass

The equation of reaction in the titration is

H₂SO₄ + 2NaOH …. Na₂SO₄ + 2H₂O

Problem

A is a solution containing 0.125 mol dm^-3 of H₂SO₄. B is a solution containing 2.8g per 250cm³ of NaOH. 25cm³ portion of B was titrated against solution A.

From your average titre value,

(a)   Determine the average titre value.

(b)  Suggest with reasons a suitable indicator for the reaction.

(c)   Calculate

i.                 Concentration of B in mol dm^-3

ii.               Concentration of A in g dm^-3

[H = 1, S = 32, O = 16, Na = 23]

Results

Assuming the following Results were obtained after titration

Treatment of Result

(a)    Average titre –

(b)  Any indicator because it is a reaction between a strong acid and a strong alkali, i.e. Methyl orange or methyl red or phenolphthalein.

(c)    

i.                 250cm³ of NaOH contain 2.8g

Therefore 1000cm³ of NaOH will contain

Mass concentration = Molar concentration x Molar mass

Molar mass of NaOH = 23 + 16 + 1

= 40g mol^-1

11.2 gdm^-3 = Molar concentration of B x 40g mol^-1

               Molar concentration (in mol dm^-3) =

0.280mol dm^-3 of NaOH

ii.               To determine concentration of A in gdm^-3

Mass concentration (gdm^-3) of A =

Molar concentration (moldm^-3) x Molar mass

Molar mass of H₂SO₄ = [1×2 + 32 + 16×4] = 98g mol^-1

Therefore concentration of A in (gdm^-3)

= 0.125mol dm^-3 x 98g mol^-1

= 12.25 = 12.3g dm^-3

Conclusion

The calculations done in this unit has exposed the students to the determination of the concentration of a solution. It also showed the students that any indicator can be use in this experiment because it is a reaction between a strong acid and a strong alkali i.e. methyl orange or methyl red or phenolphthalein.

Exercise

A is a solution containing 10.0g of sodium hydrogen tetraoxosulphate (iv) (NaHSO₄) per dm³ (weak acid). B is a solution of sodium hydroxide (strong base) containing Xg of NaOH per dm³

(a)   Put A into the burette and titrate 25cm³ or 20cm³ of B with A, using methyl orange as indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of A required to neutralize the stated volume of B.

(b)  From your results, calculate:

i.                 The concentration of B in mol dm^-3

ii.               The value of X

The equation for the reaction is

NaHSO_4(aq) + NaOH_(aq) …….. Na₂SO_4(aq) + H₂O_(l)

[S = 32.1, Na = 23.0, O = 16, H = 1.0]

Content

Module Name

Acid-Base Determination of Percentage Purity of a Substance

Class

SS2

Objectives

After studying this unit, students should be able to:

1.     Use a pipette

2.     Use a burette

3.     Carryout a titration experiment correctly

4.     Calculate the purity of a substance

Introduction

Percentage purity is the percentage of the material which is the actual desired chemical in a sample. It is expressed as

Percentage purity =   x 100

The module uses volumetric analysis to determine the percentage purity of a substance. Sometimes one of the substances used in the titrations maybe impure. This impurity maybe due to improper preparations of the substance.

Experiment

Determination of the percentage purity of a substance

Requirements

Chemicals

·       Solution of HCl acid containing 7.0g dm^-3

·       Solution of impure base, B, KOH containing 8.55 g dm^-3

·       Methyl orange

Apparatus

^·            Pipette 25cm³

^·            Burette 50cm³

^·            Volumetric flask 250cm³

^·            Conical flask 250cm³

^·            White tile

Procedure

1.         Rinse the burette with a few drops of solution A and fill it with same solution above the zero mark. Drain to the mark, making sure the burette is full

2.         Using a piptte, tranfer 25cm³ of solution B into a conical flask. Add two to three drops of methyl orange. Place the conical flask on a white tile.

3.         Run solution A into the conical flask whilst shaking vigorously until a permanent faint pink colouration is observed.

4.         Repeat te titration steps using fresh portions of solution B to get a set of three concordant readings.

5.         Tabulate your burette readings

6.         Find the average titre value.

The equation of reaction in the titration is

HCl + KOH ….KCl + H₂O

Problem

A is a solution of HCl containing 7.0gdm^-3. B is a solution of impure KOH containing 8.55gdm^-3. 25cm³ portion of B was titrated against solution A using methyl orange as indicator.

From your average titre value, determine

i.                 Concentration of A in mol dm^-3

ii.               Concentration of B in mol dm^-3

iii.             Percentage purity of KOH in B

           [ H = 1, Cl = 35.5, KOH = 56.0g mol^-1]

Results

Assuming the following results were obtained after titration.

Average titre = = 16.60 cm³

Treatment of Result

Note that: The titre value a student used may be different from the titre value another student will obtain.

i.                 To calculate the concentration of solution A

     Molar mass of HCl = 1 + 35.5

                    = 36.5 g mol^-1

              Concentration of A [C_A] =   = = 0.192mol dm^-3

ii.               To calculate  the concentation of solution B

                        Apply  

                         C_A =0.192mol dm^-3                          C_B = ?

                            V_A = 16.60cm³                           V_B = 25cm³

                                                 n_A = 1                                           n_B =  1

Substituting the values in the formula, We have

C_B x 25 x 1 = 0.192 x 16.60 x 1

C_B =  = 0.127488 = 0.128mol dm^-3

iii.             To calculate the percentage purity of KOH in B

        Molar mass of KOH = 56 g mol^-1

Concentration [g mol^-1] = Concentration [ mol dm^-3 ] x molar mass

                      = 0.128mol dm^-3 x 56g mol^-1

                          = 7.168 = 7.17 g mol^-1

Percentage purity =  x  

                                        x  = 83.8596

83.86%.

Conclusion

This experiment has demonstrated one of the most important application of volumetric analysis in the industry. The determination of percentage purity of a substance is a quantitative method of ascertaining how pure the substance in question is as a preliminary investigation of the quality of the substance.

Exercise

1.       Solution A contains 3.650g of HCl per dm³. B is a solution of impure sodium trioxocarbonate (iv) containing 1.500g of the impure salt per 250cm³

a.      Put A into the burette and titrate with 20 or 25cm³ of B using phenolphthalein as indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of A used.

b.     From your results and the information provided, calculate the:

i.                 Concentration of A in mo ldm^-3

ii.               Concentration of B in mol dm^-3

iii.             The percentage purity of sodium trioxocarbonate (iv) in the impure solid. The equation for the reaction of hydrogen chloride acid with sodium trioxocarbonate (iv) is

2HCl_(aq) + Na₂CO_3(aq)  ……. 2NaCl_(aq) + CO_2(g) + H₂O_(l)

H = 1, Cl =35.5, Na = 23, C = 12, O = 16

Content

Module Name

Acid-Base Titration of Solution of a Weak Base and a Strong Acid

Class

SS2

Objectives

After studying this unit, students should be able to:

–        Use a burette

–        Use a pipette

–        Carryout a titration correctly

–        Calculate the concentration of acid in mol dm³

Introduction

In this unit, the module is design to conduct the titration of a solution of a weak base, Na₂CO₃ and a strong acid, H₂SO₄. In this type of titration, the acid and base would react to form an acidic arrangement.

Experiment

Acid-base titration of solution of a weak base and a strong acid

Requirements

Chemicals

–        25cm³ of 0.0500 mole of anhydrous Na₂CO₃ per dm³

–        H₂SO₄ acid

–        Methyl orange

Apparatus

·       Burette 50cm³

·       Pipette 25cm³

·       Volumetric flask 250cm³

·       Conical flask 250cm³

·       White tile

Procedure

1.     Rinse the burette with a few drops of solution A and fill it with same solution above the zero mark. Drain to the mark, making sure the burette is full. Note the initial readings.

2.     Using a piptte, tranfer 25.00cm³ of solution B into a conical flask. Add two to three drops of methyl orange. Place the conical flask on a white tile.

3.     Add the solution (acid) from the burette to the solution in the titration flask. Note the final reading.

4.     Repeat the titration steps 3 and 4 to get a set of three concordant readings.

5.     Tabulate your burette readings.

6.     Calculate the average volume of acid used.

Problem

A is a solution of tetraoxosulphate (vi) acid. B is a solution containing 0.0500 mole of anhydrous Na₂CO₃ per dm³. 25.00cm³ portions of B was titrated against solution A using methyl orange as indicator.

From your results and the data provided, calculate the

i.                 amount of Na₂CO₃ in 25.00cm³ of B used;

ii.               concentration of A in mol dm^-3

iii.             concentration of A in gdm^-3

iv.             number of hydrogen ions in 1.00 dm³ of A.

[Avogadro number = 6.02 x 10²³ mol^-1]

[H = 1, O = 16, S = 32]

Results

Assuming the following results were obtained after titration.

Average volume of Acid used = =  = 24.38cm³

^i.                           To calculate the amount of Na₂CO₃ in 25.00cm³

Given: Concentration of B = 0.050 mol dm^-3; Volume =

Amount = Concentration (mol dm^-3) x Volume (dm³)

= 0.050 x  = 0.00125mol

ii.               To calculate the concentraton of A in moldm^-3

The various titration variables are:

C_A = x mol dm^-3; V_A = 24.38cm³; n_A = 1

C_B = 0.050 moldm^-3; V_B = 25cm³; n_B = 1

Apply  

Substituting:  

Making C_A the subject of the formula

C_A =  = 0.0513moldm^-3

iii.             To calculate the concentration of A in moldm^-3

Using the  the expression:

      Mass concentration (gdm^-3) = (moldm^-3) x Molar mass (gmol^-1)

Concentration of H₂SO₄, in moldm^-3 = 0.0513 moldm^-3

Molar mass H₂SO₄ = 2(1.0) + 32.0 + 4(16.0)

       2.0 + 32.0 + 64.0 = 98.0gmol^-1

Substituting: Mass concentration = 0.0513 x 98 = 5.0274 gdm^-3

5.03gdm^-3 (3d.p)

iv.             Number of hydrogen ions in 1.00dm³ of A

1 dm³ of A contained 0.0513 mol of H₂SO₄

   H₂SO₄ ionizes in water completely thus

H₂SO₄ + aq ….. 2H⁺_(aq) + SO₄^2-_(aq)

   1 mol                  2 mol

       From the equation:

1 mole of H₂SO₄ produces 2 moles of H⁺

Therefore 0.0513 mole produces 2 x 0.0513 moles of H⁺

= 0.103 mol of H⁺

But 1 mole of H⁺ contains 6.02 x 10²³  ions;

Therefore 0.103 mol of H⁺ contain 0.103 x 6.02 x 10²³ ions

 6.20 x 10²² ions. (3 sig. Fig.)

Conclusion

The experiment and calculation done in this unit exposed us to acid-base titration of solution of a weak base and a strong acid.

Exercise

25cm³ of 0.10 moldm^-3 solution of sodium trioxocarbonate (iv) were titrated against dilute hydrogen chloride acid.

1.     Tabulate your burette readings

2.     Calculate the average volume of acid used and hence determine the concentration of the acid in mol per dm³.

3.     Give a suitable indicator for the titration and give its colour change at end point.

4.     State three precautions necessary to ensure that accurate results are obtained for the titration.

The equation for the reaction is

Na₂CO_3(aq) + 2HCl_(aq) ….. 2NaCl_(aq) + H₂O_(l) + CO_2(g)