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

SECOND TERM E-LEARNING NOTES

SUBJECT: CHEMISTRY

CLASS: SSS 3

 

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

 

WEEK 9

TOPIC: ETHNICAL, LEGAL AND SOCIAL ISSUES

CONTENT:

1. CHEMICAL WASTE

2. INDUSTRIAL POLLUTANTS

3. ROLES OF GOVERNMENS IN PREVENTING CHEMICAL DEGRATION REGULATION ETC

ETHNICAL, LEGAL AND SOCIAL ISSUES

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.

PERIOD 1: CHEMICAL WASTE

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.

EVALUATION

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

PERIOD 2: INDUSTRIAL POLLUTANTS

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.

EVALUATION

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

Period 3: ROLES OF LOCAL GOVERNMENT IN PREVENTING CHEMICAL DEGRADATION

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.

GENERAL EVALUATION

OBJECTIVE TEST:

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

ESSAY QUESTIONS

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.

WEEK 1

TOPIC: QUANTITATIVE AND QUALITATIVE ANALYSIS

CONTENT:

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.

1. STANDARDIZATION OF A SOLUTION

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.

EXAMPLES

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}

Solution

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⁺ SO₄^2-(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.]

2. DETERMINATION OF RELATIVE MOLAR MASS

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

EXAMPLES

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:

)

Solution

C_a =? Mole per dm³ V_a = 24.90cm³; n_a =1

C_b =? Mole per dm³; V_b = 25cm³; n_b = 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

But

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,

Substituting

Making C_b 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)

3. DETERMINATION OF MOLE RATIO

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

SOLUTION

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

SOLVED PROBLEM 2

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

Solution

1. i. amount n_y = 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 , n_z =

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

Mole ratio of the acid to Na2CO3 = n_y : n_z = 0.00124: 0.00138

n_y: n_z = 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⁺.

4. DETERMINATION OF DEGREE OF PURITY

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.

Mathematically

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.

Example;

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)

Solution

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.)

5. DETERMINATION OF WATER OF CRYSTALLISATION

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

Solution

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 ; V_a =13.10 cm³; n_a =2

Cb =? moldm^-3; V_b 25.0cm³ ; n_b =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.

6. DETERMINATION OF SOLUBILITY OF A SUBSTANCE BY TITRATION

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

Solution

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

C_a = 0.0905 mol dm^-3, V_a =24.90 cm³ ; n_a =2

C_b = ? moldm^-3 ; V_b = 25.0 cm³; n_b = 1

Making C_b 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

SUB-TOPIC 2: REDOX TITRATIONS

Introduction: Redox titrations involve reactions between oxidizing and reducing agents. The solutions used are: acidified KMnO_4, 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, no_A and that of the reducing agent, nr_A 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^– + 5C₂O₄^2- + 16H⁺ 10CO₂ + H₂O

Solution

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 C₂O₄^2- ≡ 2 moles of MnO₄^–

0.000755 mol C₂O₄^2- ≡ 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.