Secondary LevelsCHEMISTRY STUDYfor

Practical Book

Complete

ai160266013218_Complete Chem Study For Sec Levels Practical Bk_Nigeria_Title Page.pdf 1 14/10/2020 3:22:14 PM

Complete Chemistry Study for Secondary Levels PRACTICAL BOOK has been written to conform to the new NERDC Senior Secondary School curriculum. In writing this book, the International Union of Pure and Applied Chemistry Naming Systems and International System of Unit (SI Unit) have been put into consideration.The book is well illustrated to make the identification and usage of chemistry laboratory apparatus easy for both the teachers and the students. The book has been divided into various topics that adequately cover the current practical chemistry curriculum. Each topic ends with adequate theoretical-based and practical-based standard questions. An appendix, which encompasses how to prepare chemistry bench reagents for qualitative analysis, is given at the end of the book to help teachers.In conclusion, this book is highly recommended for both the teachers and students of senior secondary schools and colleges to facilitate the effective teaching and learning of practical chemistry.

Preface

Topic 1: Identification of Basic Chemistry, Laboratory Apparatus, Uses and General Safety Precautions .. 1

Topic 2: Volumetric (Quantitative) Analysis ..................................................................................................... 20

Topic 3: Volumetric Analysis Calculations ......................................................................................................... 33

Topic 4: Volumetric Analysis—Redox Titrations ............................................................................................ 45

Topic 5: Qualitative Analysis (Inorganic Compounds) ................................................................................. 69

Topic 6: Qualitative Analysis of Organic Compounds .................................................................................... 88

Topic 7: Energy Changes in Chemical Reaction ............................................................................................... 97

Topic 8: Alternative to Practical—Practice Questions for External Examination .......................... 105

Answers of MCQs, Theoretical Questions Based on Practical and Practice Questions .... 114

How to Prepare Common Bench Reagents for Qualitative Analysis ........................................... 136

Index ................................................................................................................................................................................ 139

Contents

Topic Volumetric (Quantitative) Analysis2

2.1 INTRODUCTIONGenerally, a substance is analysed in the laboratory to establish its qualitative and quantitative chemical composition. Therefore, chemical analysis is categorised as qualitative analysis and quantitative analysis. In this topic, we will learn about quantitative analysis. In quantitative analysis, the concentration or amount of a particular species in a sample is determined accurately and precisely.For determining the amount of chemical substances in solution, there are two methods of analysis namely, volumetric (titrimetric) analysis and gravimetric analysis. In volumetric analysis, measurement of only volumes is involved while in gravimetric analysis, measurement of volumes as well as mass is involved.Volumetric analysis is the most widely used quantitative analytical method, which involves determination of the volume of a solution of known concentration that is required to react quantitatively with the measured volume of the solution of a substance whose concentration is to be determined.

2.2 ESSENTIALS OF VOLUMETRIC ANALYSISThe essentials of volumetric analysis are: 1. The chemical equation for the reaction should be known and must go to completion. 2. The reaction should be fast and proceed to completion over wide range of concentrations. 3. The substance to be determined should not react with the solvent dissolved in it. 4. The solution should be stable and should not undergo oxidation under the atmosphere.

2.3 SOME IMPORTANT TERMS PERTAINING TO VOLUMETRIC ANALYSIS 1. Titration: The process of addition of a known solution from the burette to the measured

volume of solution of another substance to be estimated until the reaction between the two is completed is called titration. The completion of reaction is generally indicated by a colour change.

2. Titrant: The reagent of accurately known concentration is called titrant. 3. Titrate: The substance which concentration is to be determined by titration is called titrate

or titrand. 4. Concentration mol/dm3 (Molar concentration): The molar concentration of a solution is

the number of moles of the solute present per dm3 of the solution. Its unit is mol/dm3 or M (molar).

Molar concentration = Number of moles of soluteVolume of solution (dm3)

OR Molar concentration = Number of moles of soluteVolume of solution (cm3)

× 1000

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i.e. M = nsV × 1000

⇒ M = wB × 1000

MB × V ∵ No. of moles of solute, nS = Mass of solute (wB)

Molar mass of solute (MB) 5. Concentration in g/dm3 (Mass concentration): The mass concentration of a solution is

the mass of the solute (g) present per dm3 of the solution. Its unit is g/dm3.

Mass concentration = Mass of solute (g)Volume of solution (dm3)

or Molar concentration = Mass of solute (g)Volume of solution (cm3)

× 1000

i.e. M = wBV × 1000

Relation between Molar concentration and Mass concentration Mass concentration (g/dm3) =

Molar concentration (mol/dm3) × Molar mass of solute (g/mol) 6. Unknown solution: The solution which concentration is not known and is to be estimated

via quantitative analysis is called unknown solution. 7. Standard solution: A solution which concentration is known is called standard solution. 8. Primary standard: Primary standard are substances that are of high purity therefore

suitable for preparing a standard solution. For example, anhydrous Na2CO3, ethanedioic acid, potassium heptaoxodichromate(VII) (K2Cr2O7) and sodium trioxodisulphate(II) (Na2S2O3).

Primary standard are used to standardise another solution by titration. A primary standard should possess the following properties:

It must not be hygroscopic, deliquescent and efflorescent; it must not be volatile; and should not react with air.

9. Secondary standard: The substance whose solutions need standardisation before use are known as secondary standards. Their standard solutions cannot be prepared directly by weighing. For example, sodium hydroxide, hydrochloric acid, tetraoxosulphate(VI) acid, potassium hydroxide, etc.

10. Equivalence point: A point at which the reaction between titrant and titrate is complete is called equivalence point. It is the theoretical end-point of titration.

11. Indicators: Indicators are weak organic acids or bases which colour change with the pH (acidity or basicity) of the solution. For example, methyl red, methyl orange, litmus solution and phenolphthalein.

2.4 PREPARATION OF STANDARD SOLUTION 1. Weigh the mass of the solute.

20.00 g

Fig. 2.1

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2. The solute is dissolved in distilled water in a beaker.

3. The solute is transferred into a volumetric flask.

4. More distilled water is added to obtain the required volume. The flask is stoppered and shaken.

2.4 TYPES OF INDICATORSOn the basis of their method of use, indicators are classified into three types: 1. Internal indicator: The indicators that are added in the reaction mixture directly are called

internal indicators. For example, methyl orange and phenolphthalein in acid-base titration. 2. External indicator: The indicators that are not added in the reaction mixture and are used

outside the system are called external indicators. For example, potassium ferricyanide [K3Fe(CN)6] in titration of double salt [(NH4)2Fe(SO4)2·6H2O] against potassium heptaoxo-dichromate(VII) [K2Cr2O7].

3. Self indicator: When the reagent itself shows colour change at the end of titration, it is called self indicator. For example, potassium tetraoxomanganate(VII) [KMnO4] (purple) in titration against a double salt (colourless).

Fig. 2.2

Fig. 2.3

Fig. 2.4

Solute

Filter funnel

Volumetric flask

Distilled water

Calibration mark

Stir

Solute

Distilled water

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2.5 TYPES OF ACID-BASE REACTION WITH SUITABLE INDICATOR, pH AND EXAMPLESAn acid-base titration is used to determine the unknown concentration of an acid or base by neutralising it with an acid or base of known concentration.The chemical reaction involved in acid-base titration is neutralisation reaction. The completion of the reaction is detected by using indicator, which undergoes a colour change at the end-point. Selection of the indicator depends on the strength of solutions involved in the reaction. 1. Titration of a strong acid and a strong base: A strong acid reacts with a strong base to give

a neutral salt (pH = 7). For example, hydrochloric acid reacts with sodium hydroxide to give sodium chloride and water. A strong acid-strong base titration is performed using any suitable indicator, e.g. phenolphthalein indicator. It will appear pink in basic solutions and colourless in acidic solutions. Other indicators like methyl orange and methyl red could be used.

2. Titration of a weak acid and a strong base: A weak acid reacts with a strong base to form a basic (pH > 7) solution. For example, ethanoic acid reacts with sodium hydroxide to form ethanoate ion and water. For weak acid and strong base titrations phenolphthalein indicator is used.

3. Titration of a strong acid and a weak base: A strong acid will react with a weak base to form an acidic (pH < 7) solution. For example, the reaction between ammonia (a weak base) and hydrochloric acid (a strong acid) in the aqueous phase. For titration of a strong acid and a weak base, methyl orange indicator is used which is yellow in basic medium and pink in acidic medium (methyl red can also be used).

There is no suitable indicator for the titration of weak acid and weak base.The point at which the colour change of the indicator becomes visible to the eye is called end-point. The pH of the solution at the end-point decides the choice of indicator. The different indicators with colour in acidic and basic media along with their pH range are tabulated below:

Indicator pH range Colour in acid solution

Colour in basic solution

Colour at end-point

Phenolphthalein 8.3–10.0 Colourless Pink Colourless

Methyl orange 3.1–4.4 Pink Yellow Orange

Methyl red 4.2–6.3 Red Yellow Orange

Litmus 5.0–8.0 Red Blue Purple

Bromothymol blue 6.0–7.0 Yellow Blue Green

Screen melthy orange 3.0–4.0 Purple Green Grey

2.6 PRINCIPLE OF DILUTIONDilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent like adding more water to an aqueous solution.

Diluted ConcentratedFig. 2.5

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Dilution calculations can be performed by using the relation:C1V1 = C2V2

Where, C1 = concentration of original solution.V1 = Volume of original solution.C2 = concentration of diluted solution.V2 = Volume of diluted solution.It should be noted that while making calculations, the units of both volume and concentration should be same on both sides.

2.7 PROCEDURE OF ACID-BASE TITRATIONThe following steps must be followed while performing titration: 1. Rinse the apparatus with distilled water, e.g. conical flask. 2. Rinse the burette with the solution to be taken in it and fix it in a clamp stand. 3. Fill the burette with the solution using a funnel. 4. Open the stopcock briefly to remove any air bubble in the nozzle of the burette. 5. Note the initial reading of the burette (read lower meniscus) only after removing the

funnel. 6. Rinse the pipette with the solution to be pipetted out. 7. Then, pipette out 20 or 25 cm3 of the solution to be titrated in a clean conical flask and add

2 to 3 drops of the indicator in it. 8. Place the plain white tile below the conical flask (to observe sharp colour change) and

adjust the burette in such a way that the nozzle enters the conical flask. 9. Operate the stopcock with left hand from back side of the burette to pour solution in the

titration flask drop by drop. Continuous swirling of flask should be done with right hand. 10. Close the stopcock when end-point is reached and carefully note the final reading of the

burette. 11. The difference between the final and initial reading will give the volume of solution used

for completion of the reaction. 12. Repeat the procedure till two or three concordant readings (differ by ±0.20 cm3) are

obtained and do the needful calculations.After attaining the end-point, the solution may be back titrated with titrand to avoid any error in result.

Burette filled with

acid

Alkali + 2 or 3 drops

of indicator

Conical flask

White tile

Fig. 2.6

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Burette reading Rough reading (cm3)

1st titre (cm3)

2nd titre (cm3)

3rd titre (cm3)

Final reading

Initial reading

Volume of solution usedOR

Titre value

Concordant reading of the burette = _______ cm3

The average of two or more readings that do not differ by more than 0.20 cm3 can be taken as concordant reading.

2.8 PRECAUTIONS DURING TITRATIONThe following precautions must be taken while doing titration: 1. Rinse the burette and pipette with the solution to be used in them. This is done to avoid

dilution of solutions with water sticking to their surfaces. 2. Air bubbles must not be there in the burette and pipette. This is to get correct volume of

the solutions. 3. Remove the funnel from the burette before taking the initial reading of the burette. 4. Note the reading of the burette up to two decimal places. 5. Never hold the pipette from the bulb otherwise heat of our body may cause expansion thus

introducing the error in results. 6. Do not blow out the last drop from the jet end of the pipette. 7. Never draw out corrosive solutions from a pipette. 8. The stopcock of the burette must be tightly closed to avoid any leakage of the solution. 9. Burette must be clamped exactly vertical so as to avoid parallax error while taking the reading. 10. Avoid saliva entry into the pipette to avoid dilution. 11. Use 2 or 3 drops of the indicators, in order to get a sharp end-point.Precautions to be taken when recording the result of volumetric analysis titration in the table 1. All units should be stated appropriately. 2. All readings must be recorded in two decimal places. 3. Arithmetic errors must be avoided during calculation. 4. Final volume (reading) must not be more than the capacity of the burette. 5. Concordant readings with values that differ not more than 0.20 cm3 should be used in

calculating average volume of acid used (average titre value). 7. Do not write with a pencil. 8. Cancellation of the table to make another one must be avoided. 9. Alteration in the table to agree with the teacher’s titre value must be avoided.Note: Always be sure of the correctness of your titre values before recording them in your table.

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2.9 CALCULATIONS INVOLVING VOLUMETRIC ANALYSISThere are two major methods of solving calculations involving titration: 1. First principle method 2. Formula (Mole ratio) methodExample 1: A is a solution of 0.25 mol/dm3 of H2SO4. B is a solution of NaOH of unknown concentration. (a) Put A into the burette and titrate it against 20 or 25 cm3 portion of B using methyl orange

as indicator. Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction involved in the titration is

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) (b) From your results and the information provided above, calculate the

(i) concentration of B in mol/dm3. (ii) concentration of B in g/dm3. [Na = 23, O = 16, H = 1, S = 32]SolutionAssuming that after titration the following results were obtained.Indicator used: Methyl orangeVolume of pipette used: 25 cm3

(a) Burette reading Rough reading (cm3)

1st titre (cm3)

2nd titre (cm3)

3rd titre (cm3)

Final reading 21.30 20.10 19.50 20.60

Initial reading 0.00 0.00 0.00 1.00

Volume of solution usedOR

Titre value

21.30 20.10 19.50 19.60

Note: The difference between 2nd and 3rd is not more than 0.20 cm3, hence the two readings are concordant and are used for averaging.

Average titre value (or volume of acid used) = 19.50 cm3 + 19.60 cm3

2 = 19.55 cm3

(b) First principle method Equation of reaction: H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) 1 mole 2 moles 1 mole 2 moles

If 1000 cm3 of H2SO4 contains 0.25 moles

19.55 cm3 of H2SO4 will contain = 19.55 cm3 × 0.25 moles1000 cm3

= 0.0048875 moles of H2SO4

From the above equation 1 mole of H2SO4 reacted with 2 moles of NaOH

∴ 0.0048875 moles of H2SO4 will react with = 0.0048875 × 2 moles1 mole

= 0.009775 moles of NaOH i.e. 25 cm3 of NaOH contains 0.009775 moles

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∴ 1000 cm3 of NaOH will contain = 1000 cm3 × 0.009775 moles25 cm3

Concentration of B in mol/dm3 = 0.391 moles/dm3 of NaOH Mass concentration (g/dm3) = Molar concentration (mol/dm3) × Molar mass Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol ∴ Mass conc. (g/dm3) of NaOH = 0.391 mol/dm3 × 40 g/mol = 15.64 ≅ 15.6 g/dm3

Formula (mole ratio) method Equation of reaction: H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) 1 mole 2 moles 1 mole 2 moles

Applying the formula

CAVACBVB

= nAnB

where CA = Concentration of acid in mol/dm3

CB = Concentration of base in mol/dm3

VA = Volume of acid VB = Volume of base nA = Number of mole of the acid nB = Number of mole of the base CA = 0.25 mol/dm3 nA = 1 VA = 19.55 cm3

CB = ? nB = 2 VB = 25 cm3

Substituting all the values into the formula, we get

0.25 mol/dm3 × 19.55 cm3

CB × 25 cm3 = 12

CB × 25 cm3 × 1 = 0.25 mol/dm3 × 19.55 cm3 × 2

CB = 0.25 mol/dm3 × 19.55 cm3 × 2

25 cm3 × 1 = 0.391 mol/dm3 of NaOH

(ii) Mass conc. (g/dm3) = Molar conc. (mol/dm3) × Molar mass Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol Mass concentration of NaOH (g/dm3) = 0.391 mol/dm3 × 40 g/mol = 15.64 ≅ 15.6 g/dm3

Example 2: The titration of an impure sample of KHP (potassium hydrogen phthalate) found that 36.0 dm3 and 0.100 mol dm–3 NaOH was required to react completely with 0.765 g of sample. What is the percentage of KHP in this sample?Note: The chemical formula of KHP is C8H5KO4. [C = 12, K = 39, H = 1, O = 16]SolutionThe equation for the reaction with NaOH is

KHC8H4O4 + NaOH KNaC8H4O4 + H2OKHP + NaOH KNaP + H2O

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Molecular mass of KHP (C8H5KO4) = 8 × 12 + 5 × 1 + 39 + 16 × 4 = 204 g/molMoles of NaOH = 0.0360 × 0.100 mol = 0.00360 mol When the moles of KHP neutralised by the NaOH.Moles of KHP = 0.00360 × 1 mol = 0.0036 molMass of KHP = 0.0036 × 204 g = 0.7344 g

Purity = Mass of pure KHPMass of impure KHP

× 100%

= 0.73440.765

× 100% = 96.1%

The KHP is 96.1% pure.Example 3: A is a solution of 0.050 mol dm–3 H2C2O4 (ethanedioic acid). B is a solution of KMnO4 [(potassium tetraoxomanganate(VII)], of unknown concentration. (a) Put B into the burette. Pipette 20.0 cm3 or 25.0 cm3 of A into a conical flask and add about

10.0 cm3 of dilute H2SO4. Heat the mixture to about 40°C–50°C and titrate it while still hot with B. Repeat the titration to obtain consistent titre values. Tabulate your results and calculate the average volume of B used. The equation of reaction is:

2MnO(aq) + 5C2O4(aq) + 16H+(aq) 2Mn2+(aq) + 8H2O(l) + 10CO2(g) (b) From your results and the information provided, calculate the:

(i) concentration of MnO in mol dm–3

(ii) concentration of KMnO4 in B g dm–3

(iii) volume of CO2 evolved at STP when 25.0 cm3 of H2C2O4 reacted completely. [O = 16.0, K = 39.0, Mn = 55.0, Molar volume of gas at STP = 22.4 dm3 mol–1]

[WASSCE 2016]Solution (a)

Burette reading Rough reading (cm3)

1st titre (cm3)

2nd titre (cm3)

3rd titre (cm3)

Final burette readings 36.50 30.40 41.70 26.80

Initial burette readings 12.30 5.90 17.10 1.90

Vol. of acid used 24.20 24.50 24.60 24.90

Average titre = 24.50 + 24.602

= 24.55 cm3

(b) (i) Molar concentration of KMNO4

C(MnO4

–) × V(MnO4–)

C(C2O42–) × V(C2O4

2–) = n(MnO4

–)n(C2O4

2–)

∴ C(MnO4

–) × 24.550.05 × 25.0 = 2

5

C(MnO4–) = 0.05 × 25.0 × 2

24.55 × 5 = 0.0204 mol dm–3

(ii) Mass concentration of KMnO4

Molar mass of KMnO4 = 39 + 55 + (16 × 4) = 158 g/mol

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Mass conc. KMnO4 = Molar conc. × Molar mass = 0.0204 × 158 = 3.22 g dm–3

(iii) Volume of CO2 (gas evolved)

n(CO2)

n(H2C2O4) = 105

n(CO2) = 105

× 0.05 = 2 × 0.05 = 0.100 mol

V(CO2) at STP = n × Vm

25 cm3 of H2C2O4

= 0.100 × 22.4 dm3 mol–1 × 251000

= 0.056 dm3 or 56.0 cm3

Example 4: Determine the volume of 0.125 mol dm–3 KMnO4 solution required to react completely with 25.0 cm3 of 0.25 mol dm–3 FeSO4 solution in acidic medium.SolutionThe balanced ionic equation for the reaction is

MnO + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2OFrom the balanced equation, it is evident that 1 mole of KMnO4 = 5 moles of FeSO4

Using the formula

CAVACBVB

= nAnB

CA = 0.125 mol/dm3 CB = 0.25 mol/dm3 VA = Unknown VB = 25 cm3

nA = 1 nB = 5Substituting these values in formula, we have

0.125 × VA0.25 × 25 = 1

5

⇒ VA = 0.25 × 25 × 10.125 × 5

= 10 cm3

Example 5: A is a solution of potassium tetraoxomanganate(VII). B is a solution of iron(II) chloride containing 4.80 g of the salt in 250 cm3 of solution. (a) Put A into the burette. Pipette 20.0 cm3 or 25.0 cm3 of B into a conical flask. Add 20.0 cm3

of H2SO4(aq) and titrate with A. Repeat the titration to obtain concordant titre values. Tabulate your results and calculate the average volume of A used. The equation of the reaction is:

MnO4(aq) + 5Fe2+(aq) + 8H+(aq) M2+(aq) + 5Fe3+(aq) + 4H2O(l) (b) From your results and the information provided, calculate the:

(i) concentration of B in mol dm–3; (ii) concentration of A in mol dm–3; (iii) number of moles of Fe2+ in the volume of B pipetted. [FeCl2 = 127 g mol–1]

[WASSCE 2018]

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SolutionEquation of the reaction:MnO4

– (aq) + 5Fe2+(aq) + 8H+(aq) Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

Burrette reading of the volume of the acid: (a)

Burette reading Rough reading (cm3)

1st titre (cm3)

2nd titre (cm3)

3rd titre (cm3)

Final burette readings 16.60 31.10 35.50 15.60

Initial burette readings 0.00 15.60 20.10 0.00

Vol. of acid used 16.60 15.50 15.40 15.60

Average titre = 15.50 + 15.40 + 15.603

= 46.503

= 15.50 cm3

(b) (i) Concentration of B in mol dm–3

250 cm3 of B contains 480 g of FeCl2

∴ 1000 cm3 of B will contain

4.80250

× 1000 g = 19.2 g dm–3

Concentration of B in mol dm–3

= Conc. in g dm–3

Molar mass = 19.2

127 = 0.151 mol dm–3

Concentration of A in mol dm–3 = ?

CAVACBVB

= nAnB

CA = Molar concentration in mol dm–3 of KMnO4 = ? VA = Volume of KMnO4 used = 15.50 cm3

CB = Molar concentration of FeCl2(aq) = 0.151 mol dm–3

VB = Volume of FeCl2(aq) = 25.00 cm3

nA = 1 nB = 5 Substituting these values in formula, we have

CA × 15.50 cm3

0.151 mol dm–3 × 25.00 cm3 = 15

CA = 0.151 × 1 × 25.005 × 15.50

= 3.77577.50

= 0.048709 ≅ 0.0487 mol dm–3

(ii) Number of moles of Fe2+ in the volume of B pipetted = ? 1000 cm3 of B contains 0.151 moles of Fe2+ 25 cm of B will contain

0.1511000

× 25 moles = 0.003775 mole of Fe2+

M ultiple Choice Questions 1. The colour of phenolphthalein indicator in dilute HNO3(aq) is [WASSCE] (a) Colourless (b) Orange (c) Pink (d) Purple

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T heoretical Questions Based on Practical Q1. Write the relation between molar concentration and mass concentration. Q2. Which indicator is used for titration of a strong acid and a weak base? Q3. What is primary standard? Give an example. Q4. Calculate the amount (in moles) present in 5.0 g of NaOH, given that the molar mass of

NaOH is 40.0 g mol−1. Q5. Calculate the mass of 0.01 moles of sodium hydrogen tetraoxosulphate(VI), NaHSO4.

(Molar mass of NaHSO4 = 120 g mol−1) Q6. A solution contains 1.05 g of anhydrous sodium trioxocarbonate (IV), Na2CO3 in

200 cm3 of solution. Calculate the concentration of the solution in grams per dm3. (Molar mass of Na2CO3 = 106 g mol−1)

2. In a titration, 25.00 cm3 of 0.100 mol dm–3 sodium hydroxide solution is exactly neutralised by 20.00 cm3 of a dilute solution of hydrochloric acid. Calculate the concentration of the hydrochloric acid solution. [WASSCE]

(a) 0.012 mol dm–3 (b) 1.25 mol dm–3 (c) 0.125 mol dm–3 (d) 12.5 mol dm–3

3. The most suitable indicator for the titration of hydrochloric acid with sodium trioxocarbonate(IV) is

(a) methyl red. (b) methyl orange. (c) litmus solution. (d) phenolphthalein. 4. In titration, end-point can be determined by change in colour by (a) measiring cylinder. (b) burette. (c) pipette. (d) indicator. 5. Consider the reaction represented by the following equation: Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g) What volume of 0.02 mol dm–3 Na2CO3(aq) would be required to completely neutralise

40 cm3 of 0.10 mol dm–3 HCl(aq)? (a) 200 cm3 (b) 100 cm3 (c) 40 cm3 (d) 20 cm3

R evision Exercises Q1. Define the following terms of volumetric analysis. (a) Titrant (b) Titrate Q2. Mention the precautions that must be taken care while doing titration. Q3. Burette readings (initial and final) must be given to two decimal places. Volume

of pipette used must also be recorded but no account of experimental procedure is required.

A is a solution containing 6.3 g dm–3 of HNO3. B is a soluion of Na2CO3.

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(a) Put A into the burette and titrate it against 20.0 cm3 or 25.0 cm3 portions of B using methyl orange as indicator. Record the volume of your pipette. Repeat the titration to obtain consistent titres. Tabulate your burette readings and calculate the average volume of A used.

The equation for the reaction involved in the titration is: 2HNO3(aq) + Na2CO3(aq) 2NaNO3(aq) + CO2(g) + H2O(l) (b) From your results and information provided above, calculate the: (i) concentration of B in mol dm–3; (ii) concentration of B in g dm–3; (iii) mass of sodium ions in 1.0 dm–3 of B. [H = 1; C = 12; O = 16; N = 14; Na = 23] [WASSCE] Q4. A is 0.100 mol dm–3 HNO3. B is a solution containing 2.50 g of a mixture of Na2CO3 and Na2SO4 in 25.0 cm3 of solution. (a) Put A into the burette and titrate it against 20.0 cm3 or 25.0 cm3 portions of B using

methyl orange as indicator. Repeat the exercise to obtain consistent titres values. Tabulate your burette readings and calculate the average volume of A used.

The equation for the reaction involved in the titration is: Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g) (b) From your results and information provided above, calculate the: (i) concentration of B in mol dm–3; (ii) concentration of Na2CO3 B in g dm–3; (iii) percentage of Na2CO3 in the mixture. [Na2CO3 = 106] [WASSCE] Q5. A is 0.100 mol dm–3 solution of an acid. B is a solution KOH containing 2.8 g per 500 cm3. (a) Put A into the burette and titrate it against 25.0 cm3 portions of B using methyl

orange as indicator. Repeat the titration to obtain consistent titres. Tabulate your burette readings and calculate the average volume of A used.

(b) From your results and information provided above, calculate the: (i) number of moles of acid in the average titre; (ii) number of moles of KOH in the volume of B pipetted; and (iii) mole ratio of acid to base in the reaction [H = 1.00; O = 16.0; K = 39.0] [WASSCE]