Ethical, Legal and Social Issues: Chemical waste, industrial pollutants, roles of government in preventing chemical degradation





Ethical, Legal and Social Issues: Chemical waste, industrial pollutants, roles of government in preventing chemical degradation









Human activities and industrialization have led to the occurrence of wastes in our environment today. These wastes pollute the air and negatively affect respiration and photosynthesis process. Water and land are also affected by the generation of chemical wastes resulting in various health hazards and at times death. Mostly affected are cities and highly industrialised environment. It is therefore important for government to put to check environmental pollution by laws regulations that will guide industries and individuals in waste generation disposal and management. These laws and regulations should be enforced to the letter so as to have environment and healthy citizens.


Unwanted materials that are generated from homes, offices industries, factories and are chemical in nature are called Chemical Waste. They are categorized into two namely: waste from combustion (ii) wastes from other sources.

1. Waste from combustion. This includes:

  1. Tobacco smoke produced by smoking tobacco products.
  2. Combustion of coal and solid fuel which generates, SO2 and polycyclic aromatic hydrocarbon as smoke.
  3. Combustion of liquid petroleum generating CO, oxides of nitrogen etc
  4. Industry and incineration can generate oxides of sulphur and nitrogen.
  5. Home generated combustion products such as NO2 (from gas cookers, formaldehyde (from building materials)

2. Wastes from other sources. They include:

  1. Fluoride addition to water. This has an unwanted effect like mottling of the teeth.
  2. Leaching of lead from pipes by soft water.
  3. Fertilizer leaching by water which increases the risk of methemoglobinemia (blue babies) due to the presence of nitrates in water.
  4. Deposition of solid hazardous waste.


  1. (a) what is a chemical waste? (b) list the two categories of chemical waste you know.
  2. Enumerate the sources of chemical waste.


Industrial pollutants are agents of pollution generated from different industries. These pollutants include: CO, NO2, CO2, ground level, Ozone (O3), particulate matter, SO2, hydrocarbon and lead.

1. CO: Carbon monoxide is a colourless and poisonous. About 80% of the gas is generated from road transport. It affects oxygen transportation round the body by the blood thus, resulting in heart diseases

2. NO2: Oxides of nitrogen omitted from vehicle exhaust gases to form NO2 which is harmful to the health. NO2 at high levels, causes irritation and inflammation on the lungs airways. NO2 dissolves rain water to form acid rain.

3. LEAD: Petrol combustion is the main sources of lead in the air is however been reduced by the phasing out of leaded petrol lead dust causes lead poisoning

4. GROUND LEVEL OZONE: ground level ozone is harmful to health unlike the upper level ozone which protects the earth. High level of the ground level irritate and inflame the lungs; it causes migraine and coughing and attack rubber, pigments and vegetation.

  1. PARTICULATE MATTER: sources of particulate matter in the air include sand, sea spray, construction dust or soot, coal burning etc. The presence of finer particles in the atmosphere poses more danger than larger particles because it is easily and deeply breathed into the lungs thus, having more toxic effect.

6. SO2: this gas results from burning materials or fuel containing sulphur. It dissolves in rain water in the atmosphere form acid rain causes skin irritation and attack buildings. A short term exposure to high level SO2 may cause coughing, tightening of the chest and lung irritation.

7. HYDROCARBONS: there are compounds of hydrogen and carbon such compounds include: 1,3 butadiene(primarily from vehicle exhaust) and beneze (from the combustion of petrol. Longer term exposure to these compounds has been linked to leukemia and other form of cancer.


Mention at least five (5) industrial pollutants and their respective harmful effect on the environment


The following measures are taking by government to prevent chemical degradation.

  1. Legislation by law makers by examining various degradation forms and suggesting possible way.
  2. Setting of minimum standards, minimum standards of waste generation and management are set for citizens and industries by government bodies.
  3. Government should also ensure the passage into law of the minimum standards set for citizens and, the legislation and, their subsequent enforcement. Industries and citizens should be made to abide by these laws.



  1. The following are major industrial pollution except. (a) CO2 (b) CO (c) SO2 (d) NO2 (e) lead
  2. The following diseases are caused by industrial pollutants except. (a) cancer (b) lead poisoning (c) leukemia(d) skin irritation (e) polio
  3. Which of the following gases can cause blood poisoning? (a)NO2 (b) CO2 (c) SO2 (d) CFC (e) CO
  4. The un covered raw food that is sold along major road is likely to contain some amounts of. (a) lead (b) copper (c) Argon (d) Sodium (e) Iron
  5. The gas that is most useful in protecting humans against solar radiation is (a) Chlorine (b) Ozone (c) CO2 (d) H2S (e) NO2


  1. Name three (3) industrial pollutants and suggest one way of reducing these pollutants in our environment.
  2. Give 2 examples of industrial pollutants which cause acid rain.
  3. Mention two gaseous and one solid impurity in the atmosphere.
  4. Explain briefly, the term chemical wastes.
  5. Outline any three (3) sources of wastes with accompany pollutants.




  1. Acid/base titration
  2. Redox titrations
  3. Test for oxidants and reductants

Sub-topic 1: Acid/base titration

Importance of acid –base titrations: Acid-base titrations are used for the following purposes:

  1. To standardize a solution of an acid or a base (standardization)
  2. To determines the molar mass of an acid or base.
  3. To establish the molar ratio of acid to base in a neutralization reaction.
  4. To determine the percentage purity or impurity of acid or base.
  5. To estimate the percentage of water of crystallization in an acid or base.
  6. To determine the solubility of a base.


A solution of an acid of unknown concentration can be standardized, i.e. its concentration determined, by the use of a standard solution of a base, vice versa. A balance equation of reaction is required, in order to obtain the mole ratio.


  1. A is a solution of tetraoxosulphate {vi} acid. B is a solution containing 0.0500 mole of anhydrous Na2CO3 per dm3. (a)Put A in the burette and titrate 20.00 or 25.00 cm3 portions of B using methyl orange as the indicator. Record the size of your pipette. Tabulate the burette readings, and calculate the average volume of the acid used. (b) From your result and data provided, calculate:
  2. the amount of Na2CO3 in 25.00 cm3 of B used
  3. concentration of A in moldm-3
  4. concentration of A in gdm-3
  5. number of hydrogen ions in 1.00dm3 of A. {Avogadro number = 6.02×1023 mol1}

The equation of reaction is: H2SO4 {aq} + Na2CO3{aq] Na2SO4 {aq} + H2O +CO2

{H =1,O=16, S=32}


  1. Volume of pipette: 25cm-3

Titration results {Hypothetical data}

Burette reading 1cm3 II cm3 IIIcm3
Final 24.75 49.15 25.70
Initial 0.00 24.75 1.35
Volume of acid used 24.75 24.40 24.35

Average volume of acid used from titrations II and III:

= = 24.38cm3

Note: Only the titre values from titrations I and II can be used in averaging, since they are within 0.20cm3 of each other. …Rough of first titre can also be used in averaging, if it is within 0.20cm3 of any other titre value, and is not crossed.—–Do not round up 24.38cm3 to 24.40cm3

(b)(i). To calculate the amount of Na2CO3 in 25.00CM3

Given: conc. of B = 0.050mol dm-3; Volume = 25/1000dm3

Amount = Conc. {mol dm-3 x Volume {dm3} = 0.050 x 25/1000 = 0.00125mol.

(ii).To calculate the concentration of A in mol dm-3: The various titration variables are:

CA = x mol dm-3; VA =24.38cm3; nA = 1, CB= 0.050mol dm-3, VB = 25cm3: nB =1

Method 1: proportion method {from the first principle}

From the balanced equation of reaction:

1mol of Na2CO3 = 1 mol of H2SO4

. 0.00125mol Na2CO = 0.00125mol H2SO4

i.e. 24.38cm3 of A contained 0.00125 mol H2SO4

A contained {0.00125 x 1000} /24.38 mol = 0.0513mol.Hence, concentration of A is 0.0513mol dm-3.

Method 2:

Mathematic formula method: From the data above, it is safe to use the mole ratio expression, in order to calculate the concentration CA of A, which the required variable.

Using CAVA/CBVB = nA/nB

Substituting; CA X 24.38/0.050 X25 =1/1

Making CA the subject of the formula

CA =1 X0.050X25 /24.38 X1 = 0.0513 mol dm-3

(iii) To calculate the concentration of A in gdm-3

Using conc. {gdm-3} x Molar mass {gmol-1}

Concentration of H2SO4, mol dm-3 =0.0513mol dm-3

Molar mass H2SO4, = 2 {1.0} + 32.0 + 4{16.0

= 2.0 +32.0 +64.0 = 98.0gmol-1

Substituting; Mass conc. = 0.0513×98 = 5.0274gdm-3.

=5.03gdm-3 {3 sig fig.}

(Iv) number of hydrogen ions in 1.00dm3 of A

1 dm3 of A contained 0.0513mol of H2SO4.

H2SO4 ionizes in water completely thus:

H2SO4 (aq) 2H+ SO42-(aq)

1mol 2mol

From the equation;

1 mole of H2SO4 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 x 6.02 x 1023 ions =6.02 x10 23 ions.{3 sig fig.]


In the determination of relative molar mass of an acid or bases by titration, the concentration of both acid and base, at least in gram per dm3, will be provided together with the balanced equation of reaction so as to establish the mole ratio


  1. A contains 1.6g of HNO3 in 250cm3of solution. B contains 10gdm-3 of XHCO3

25cm3 portion of B requires an average of 24.90cm3 of A for complete neutralization. Calculate the

  1. Concentration of acid in mol dm-3
  2. Concentration of XHCO3 in B in mol dm-3
  3. Mass of XHCO3
  4. Value of X

Equation of the reaction:



Ca =? Mole per dm3 Va = 24.90cm3; na =1

Cb =? Mole per dm3; Vb = 25cm3; nb = 1

  1. To calculate the molar concentration of A.

The given mass concentration of A is

1.6g of HNO3 in 250cm3

i.e. 250cm3 of the solution contained 1.6g of acid

Therefore, 1000cm3 of A contained

i.e. the mass concentration of A = 6.40gdm-3


Molar mass of HNO3 = (1×1) + (14×1) + (16×3) = 1+14+48 =63gmol-1

Molar conc.

  1. To calculate the molar concentration of B

Mass conc. Of XHCO3 in B = 10.0gdm-3 (given)

Since the value of X in the base XHCO3 is not known, the mole ratio expression must be used in order to find its molar concentration,


Making Cb the subject of the formula

  1. To find the mass of X in XHCO3
  2. To find the value of X in XHCO3

Molar mass of XHCO3 =98.0gmol-1

i.e X +1 + 12 + (16 x 3 ) = 98

X + 1 + 12 + 48 = 98

X + 61 = 98

X = 98-61

Relative atomic mass of X is X = 37 (no unit)


In order to determine the mole ratio of acid to base (or base to acid) by titration, the solution of the acid and the base provided for the titration must be known concentrations (standard solution). However, the equation of reaction will not be provided.

Example 1

A is a solution of 0.0500mol dm-3 hydrochloric acid

B is 0.0250mol dm-3 of a trioxocarbonate (iv) solution

25.00cm 3 portions of B required an average of 24.60cm3 of A for complete neutralization, using methyl orange as the indicator

  1. calculate:
  2. amount of acid in the average volume of A used
  3. amount of trioxocarbonate (IV) in the volume of B used
  4. mole ratio of the acid to trioxocarbonate (IV) solution in the reaction, express your answer as a whole number ratio of one
  5. state whether the PH of the following would be equal to 7, greater than 7 or less than 7
  6. solution A
  7. solution B
  8. titration mixture of A and B before end point


  1. (i) To calculate the amount of acid in A used

Amount of the acid = molar conc. of A x volume in dm3.

= 0.050 x 24.60/1000 mol

= 0.00123mol

(ii) To calculate the amount of trioxocarbonate (iV) in B used

Amount of Base = molar conc. Of B x volume in dm3

= 0.025 x 25/1000 mol.

= 0.000625 mol

(iii) Mole ratio of acid to trioxocarbonate (iV)

Mole ratio of acid to base, A: B = 0.00123: 0.000625

A : B =

A : B = 2 : 1

  1. (i) pH of A is less than 7

(ii) pH of B is greater than 7

(iii) pH is titration mixture before end point is greater than 7


In an acid – base titration, 24.80cm3 of 0.05000moldm-3 of a mineral acid Y neutralized 25.00cm3 of a solution containing 5.83g of Na2 CO3per dm3.

  1. From the information given above calculate:
  2. Amount (in moles) of acid Y used
  3. Amount in moles of Na2CO3 used
  4. Mole ratio of the acid to Na2CO3 I the titration
  5. i. What is the basicity of Y?

ii. Suggest what Y could be. Give reason for your answer.

  1. From your answer, write a balance equation of the reaction


  1. i. amount ny = conc (moldm-3) x volume (dm3)

= 0.0500 x

ii. Amount in mole of Na2CO3 used

first, calculate the concentration of Na2CO3 in moldm-3

Molar mass of Na2CO3 = 2 (23.0) + 12 + 3 (16)

= 46 + 12 + 48 = 106gmol-1

Conc of Na2CO3 =

Amount of Na2CO3 , nz =

  1. mole ratio of the acid to Na2CO3 in the titration

Mole ratio of the acid to Na2CO3 = ny : nz = 0.00124: 0.00138

ny: nz = 1: 1

  1. The basicity of Y is 2; Y since one mole of Y requires one mole of Na2CO3 .

ii. Y is H2SO4. Reason: one mole of the acid Y produces two mole of H+.


The degree of purity of an acid or a base can be determined by titration. For instance:

Suppose a sample of anhydrous Na2CO3 is contaminated by a salt, sodium chloride. A know mass of the impure sample is dissolved in a known volume of solution. The resulting solution of the impure Na2CO3 is then titrated against a standard solution of a pure acid.

It should be bore in mind that during the course of the titration, only the pure Na2CO3 in the mixture will be neutralized by the standard solution of the acid. The impurity remain in the solution, since it will not react will the acid hence the molar concentration obtained corresponding to that of the pure Na2CO3

Generally, the quantity of the pure substance is always less than that of the impure substance in a given solution, due to the presence of impurity.


Mass of pure substance = (mass of impure substance) – (mass of impurity)

  • In essence, you need more of an impure substance to complete a chemical reaction, than when the substance is 100% pure.


A is 0.100mol dm-3 hydrochloric acid

B contains 6.00g of a mixture of anhydrous sodium trioxocarbonate (iv) and sodium chlorides in 1.00dm3 of solution.25.00cm3 portions of B required an average of 22.65cm3 of A for complete neutralization

From the above data, calculate the

a amount in moles of Na2CO3 in one 1dm3 of B;

b mass of Na2CO3 in 250cm3 of B ;

c percentage by mass of NaCl in the mixture.

Equation of reaction;

2HCl (aq) + Na2CO3 (aq) 2NaCl (aq) + CO2 (g) + H2O (L)


A .To calculates the molar concentration of Na2CO3 in B.

The molar concentration corresponds to the pure anhydrous Na2CO3 that has been neutralized by the acid; hence, the mole ratio expression is to be used:

i.e. (CAVA)/NA = (CBVB)/Nb

The relevant variables are:

CA = 0.100mol dm-3; VA = 22.65 cm3; Na = 2 nB= 1; CB =? mol dm-3; VB = 25.00cm3;

Substitution: 0.100 x22.65 / 2 = CB X 25.0/1; making CB the subject of the formula;

CB = 0.100 X 22.65/2 X 25 = 0.0453mol dm-3

b. To calculate the mass of Na2CO3 in 250 cm3 of B.

First, calculate the concentration of B in g/dm3:

Mass conc. = molar conc. X molar mass

Molar mass of Na2CO3 = (23X2) + (12 X 1) +(16 X 3) = 46+12+48 = 106gmol-1

Molar conc. of B = 0.046 X 106 = 4.83gdm-3

i.e. 1000 cm3 of B contained 4.83g of Na2CO3

Therefore, 250cm3 of B contain 4.83 x 250/1000g = 1.21g of Na2CO3

i.e. B contains 1.21g of pure Na2CO3 in 250 cm3 .

c i.e. To calculate the % NaCl in the mixtures

% NaCl in the mixture corresponds to the % impurity.

Mass conc. of impure substance = 6.00gdm-3

Mass conc. of pure Na2CO3 = 4.83gdm-3.

Therefore, Mass conc. of impurity, NaCl = (6.00 – 4.83) gdm-3 = 1.17gdm-3

% impurity NaCl = 1.17/6.00 x 100/1 = 19.50% (3 sig.fig.)

Note; % purity = (100-19.50) % 80.50%

or % purity =4.83/6.00 x 100/1 =80.50% (3 sig.fig.)


When a hydrated substance is used in a volumetric analysis, it is only the anhydrous portion of the hydrate that will take part in the reaction. The water of crystallisation remains in the solution like an impurity.

Worked example

A is a solution containing 6.00g per dm3 of hydrochloric acid

B contains 12.0g per dm3 of Na2CO3.xH2O.

A student titrated 25cm3 of a portion of B with A using a methyl orange indicator, and found the average volume of A required to be 13.10cm3. Calculate the;

  1. concentration of A in moldm-3
  2. molar mass of Na2CO3.x H2O
  3. value of x
  4. Percentage of water of crystallisation in B. The equation of reaction is


  1. to calculate the concentration of a in moldm-3

mass concentration of HCl in A = 6gdm-3.

Molar mass of HCl = 1+35.5 =36.5g per mol

  1. to calculate the concentration of andydrous Na2CO3 in moldm-3

amount of HCL used = molar conc (moldm-3 ) x volume (dm3)

= 0.164 x 13.1/1000 = 0.002215 mole

From the equation of reaction:

2 mole of HCl requires 1 mole of Na2CO3

1 mole of HCl requires ½ (0.00215 mole) = 0.00108 mol of Na2CO3

That is : 25cm3 of B contains 0.00108 mol of Na2CO3

1000 cm3 of B contain 0.00108 x 1000/25 = 0.0432 mol

Hence the concentration of Na2CO3 in B is 0.0432 moldm-3

Alternative formula method

The relevant variables are

Ca =0.164 moldm-3 ; Va =13.10 cm3; na =2

Cb =? moldm-3; Vb 25.0cm3 ; nb =1

Using the relation

Making Cb the subject

  1. to calculate the molar mass of hydrated Na2CO3

mass concentration of B = 12.0 gdm-3 (given )

molar concentration of B = 0.04320 moldm-3

molar mass of hydrated Na2CO3= (mass conc)/(molar conc.).

=12.0/0.0430=279 g per mole

  1. To find the value of x

Molar mass of Na2CO3.x H2O = 279 gmol-1 (from c above)

106 + 18x =279

18x =279 – 106 = 173

X = 173/18 =9.61

X = 10 ( to the nearest whole number)

( note that the experimental value of x may differ significantly from the theoretical value,as a result of experimental errors, or due to the fact that hydrated substances are efflorescent

  1. To calculate the percentage of water of crystallization in B.

Mass concentration of Hydrated Na2CO3

= ( molar concentration of B) x (molar mass of anhydrous Na2CO3)

= 0.0430 x 106 gdm-3 = 4.56g dm-3.


The solubility of a solid in a liquid is the concentration of the saturated solution. It is defined as the maximum amount of the solid that dissolves in 1 dm3 of the solution at a given temperature. It is expressed in moles/dm3

The solubility of anhydrous substance, such as Na2CO3, K2CO3, NaHCO3. KHCO3 and Ca(OH)2 can be determine by titration. This is achieved by preparing a saturated solution of the substance at room temperature.

Preparation of saturated solution of Na2CO3 at room temperature

Place about 50cm3 of distilled water in a beaker, add powdered Na2CO3 little by little into it, and warm on a Bunsen burner, with stirring. Continue the addition of the salt with stirring and keeping the beaker warm at about 400C, until the salt can no longer dissolve. Allow the saturated solution to cool to the room temperature, and then filter the mixture. The filtrate is the saturated solution of salt at room temperature.

To determine the solubility of Na2CO3 at 250C;

  1. Pipette a known volume, say 25 cm3 of the saturated Na2CO3 solution at 250C, into a volumetric flask, and dilute it to a volume, say 1000cm3 with distilled water. Dilution will be necessary, if the saturated solution is of a high concentration
  2. Titrate 25 cm3 portions of the diluted solution with a standard solution of an acid using methyl orange indicator, and find the average volume of the acid used.

Worked examples

A is a solution containing 0.0905 mol dm-3 of trioxonitrate (V) acid

B was prepared by diluting 50.0cm3 of a saturated solution of Na2CO3 at room temperature to 1000 cm3

25 cm 3 portion of B were found to require an average of 24.9.0cm3 of A for complete neutralisation, using methyl orange as the indicator. From the data above, calculate the;

  1. Amount of HNO3 in the average volume of A used
  2. Concentration of B in mole per dm3
  3. Solubility of Na2CO3 in moles per dm3
  4. Mass of Na2CO3 that will be obtained by evaporating 1 dm3 of the saturated solution to dryness.

The equation of reaction is


  1. Amount of HNO3 in the average volume of A used.

1000 cm 3 of A contained 0.0905 mole of HNO3

Therefore, 24.90 cm3 of the solution will contain 0.0905 x 24.90/1000 mol HNO3

= 0.00225 mol

Or amount (mol) = conc. (mol dm-3) x volume (dm3)

= 0.0905 x 24.90/1000 = 0.00225 mol

  1. To calculate the concentration of B in moles per dm3

from the equation of reaction;

2 moles of HNO3 require 1 mole of Na2CO3

0.00225 mole will require ½ (0.00225 mol) of Na2 CO3 = 0.00113 mol of Na2CO3

That is; 25 cm 3 of B contain 0.00113 mol of Na2CO3

Therefore, 1000 cm3 (1 dm3) will contain 0.00113 x 1000/25 mol = 0.452 mol

Hence, the concentration of B is 0.452 mole per dm3

Alternative formula methods

The relevant variables are

Ca = 0.0905 mol dm-3, Va =24.90 cm3 ; na =2

Cb = ? moldm-3 ; Vb = 25.0 cm3; nb = 1

Making Cb the subject of the formula

  1. To calculate the solubility of Na2CO3 in mole per dm3.

Since B is obtained by diluting 50.0cm3 of saturated Na2CO3 solution t0 1000 cm3 then,

50 cm3 of the saturated solution contained 0.0451 mole.

Therefore 1000cm3 of saturated solution contained 0.0451 x 1000/50 mol = 0.902 mole

Hence, the solubility of Na 2CO3 is 0.902 mol dm-3

Alternative method: using diluting principle

50.0cm3 (V1) of saturated Na2CO3 solution of unknown concentration C1, diluted to 1000cm3 (V2) to obtain solution B, of concentrationC2, which has been found to be 0.0451 mole per dm3

  1. To calculate the mass of Na2CO3 in 1dm3 of the saturated solution

Molar conc = 0.902 moldm-3 ; molar mass of Na2CO3 = 106 gmol-1.

Mass conc = molar conc x molar mass of Na2CO3

Molar concentration = 0.902 x 106 = 95.6 g per dm3

Therefore, the mass of Na2CO3 obtained on evaporating 1 dm3 of the saturated solution is 95.6 g


Introduction: Redox titrations involve reactions between oxidizing and reducing agents. The solutions used are: acidified KMnO4, acidified K2Cr2O7, KIO3 and I2. The reducing agents are solutions of: iron (II) salts; ethanoic acid dehydrate, H2C2O4 . 2H2O; sodium ethanate, Na2C2O4; and sodium trioxosulphulsulphate (iv), Na2S2O3.

All the basic principles involved in acid – base titrations are also applicable to redox reactions. When the mole concept is applied to redox reagents, the amount of substance present in a given solution can be determined.

Worked Example 1

How many moles of KMnO4 are in 250cm3 of 0.012 moles per dm3 solution?

The calculate amount in moles.

Given: concentration = 0.012moldm3

Formula Method

Volume in dm3 = = 0.250dm3

Using: Amount = conc. (moldm-3) x vol (dm3)

= 0.102 x 0.250 mol

From the first principle

1000cm3 ( 1dm3 ) solution contain 0.102mole KMnO4

Therefore, 250cm3 solutions contain

= 0.0255mol KMnO4

End-point of a Redox titration

Indicators such as methyl orange and phenolphthalein are not used in redox titrations.

  • In a redox titration involving KMnO4 solution and a colourless reducing agent, an external indicator is required; KMnO4 acts its own indicator. The end-point is the first permanent pink colour.
  • In a redox titration involving iodine solution and a colourless reducing agent, starch solution is used. The end-point is signalled by a change from blue-black to colourless.

At the end-point, the amounts of the oxidizing agent, noA and that of the reducing agent, nrA are exactly in the same ratio as that required in the balanced equation of reaction. That is:

C = or =

Worked Example 2

25.0cm3 of a hot solution of 0.0302moldm-3 hydrated ethanoic acid, acidified with dilute H2SO4 required an average of 28.75cm3 of KMnO4 solution to complete the reaction.

  1. How many moles of the acid are therein 25.0cm3 of the acid?
  2. How many moles of KMnO4 react with 25.0cm3 of acid?
  3. How many moles of KMnO4 are in 1dm3 of solution?

The equation of the reactions is;

2MnO4 + 5C2O42- + 16H+ 10CO2 + H2O


a. To calculate the amount of acid in 25cm3 of solution:

Given: concentration = 0.0302moldm-3

Volume in dm3 = = 0.0250dm3

Using: Amount = conc (moldm-3) x vool (dm3)

= 0.0302 x 0.0250 = 0.000755mol.

b. To calculate the amount of KMnO4

From the balanced equation of reaction:

5 moles of C2O42- ≡ 2 moles of MnO4

0.000755 mol C2O42- ≡ MnO4

= 0.000302 mol KMnO4

  1. Amount of KMnO4 in 1 dm3 of solution:

The volume of KMnO4 solution used is 28.75cm3.

i.e. 28.75cm3 of solution contain 0.000302 mol KMNO4

Therefore, 1000cm3 (1dm3 ) contain

= 0.0105 mol KMnO4

Sub-topic 3: Test for Redox Reagents

Test for a Reducing agent

An oxidizing agent is used to test for a reducing agent. Any substance that reacts with one or more of the following reactions is a reducing agent:

  • To the solid substance in a test tube, add a few drops of bench dilute hydrochloric acid. There is effervescence; the gas given off is colourless, odourless, has no action on moist red or blue litmus paper, but gives a pop with lighted splint. The gas is hydrogen, while the substance is a metal above hydrogen in the activity series.
  • To the solution of the substance, add a few drops of purple acidified KMnO4. The KMnO4 solution is decolourised.
  • To the solution of the substance, add a few drops of yellow acidified K2Cr2O7 solution. The yellow colour of K2Cr2O7 solution turns green.
  • To the solution of the substance, add a few drops of yellow or brown solution of FeCl3. The solution turns green.

Test for an Oxidizing agent

A reducing agent is used to test for an oxidizing agent. Any substance that reacts in one or more of the following reactions is a oxidizing agent.

  • Allow H2S gas to come in contact with a solution of the substance on a strip of filter paper. There is a yellow deposit of sulphur on the filter paper.
  • To the solution of the substance, add a few drops of colourless acidified potassium iodide solution. The solution turns brown, due the evolution of iodide, which turns starch solution blue-black.
  • Heat a little of the substance with conc.HCl. A greenish-yellow gas with pungent smell is evolved. The gas, which bleaches litmus, is chlorine.
  • Heat the substance strongly in a test tube. The gas evolved is colourless, odourless, has no action on litmus paper, but relights a glowing splint. The gas is oxygen.
(Visited 3 times, 1 visits today)

error: Alert: Content is protected !!