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I. CONSTRUCTION OF ELECTROCHEMICAL CELLS
Electrochemical cells are devices that use chemical reactions to produce electrical power. These are sometimes known as Galvanic or Voltaic cells.
e.g. dry cells, car batteries.
It contains two half cells connected together by external circuits.
Half cell is an arrangement which consists of an electrode dipped into a solution containing its ions. When the two half cells are connected, the resulting component is called electrochemical power. A good example is Daniel cell. It is constructed from Zinc and copper electrode.

PROCEDURE FOR CONSTRUCTION OF ELECTROCHEMICAL CELL
1. Identify between the electrodes, which electrode is supplying or gaining electrons by studying their electrode potential.
II. An electrode with negative standard electrode potential is more reactive (supplies electrodes) than the positive one.
III. If both are negative the one which has more negative electrode potential is more reactive.
E.g. Sn = -0.16
Mg = -0.25 (more reactive)
If both electrodes are positive, the one which has less positive value is more reactive.
E.g. 0.07v (more reactive)
0.2v
The electrode which supplies electrons should be placed on the left and electrode which gains electrons should be on the right hand side.
E.g. construct a Daniel cell and shows the direction of flow of electrons and current given that EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
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The electrode which supplies electrons oxidation takes place and the electrode is oxidized. This electrode is called ANODE. The electrode which receives electrons reduction takes place and the electrode is REDUCED. It is called CATHODE.
Zn – anode
Cu – cathode
Overall reaction is obtained by adding the two half reactions and is called cell reaction.
Anodic reaction: EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
Cathodic reaction: EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
Cell reaction: EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
Electrochemical cell can also be represented in an abbreviation way known as cell notation.
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
Anode cathode
I.e. for Daniel cell EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
Questions
1. Given the overall reactions, write their corresponding cell notations
a) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
b) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
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Note
zinc is preferred to tin because zinc protects iron
against rusting when its coating has broken down. This is because it has a more negative reduction potential than iron: it acts as a cathode hence it is not changed. This is called cathodic protection.
Tin protects iron only as long as coating is intact. Once the coating is broken down, tin actually promotes corrosion of iron as iron has more negative reduction potential than tin. Thus iron acts as anode and dissolves while tin acts as cathode and does not change.
Factors which affect corrosion:
i. Position of metal in the electrochemical series(E.C.S). the reactivity of metal depends upon its position in the electrochemical series. More the reactivity, the more likely it is to be corroded.
ii. Presence of impurities in the metal. The impurities help to set up the voltaic cells which increase the speed of corrosion.
iii. Presence of electrolytes: Presences of electrolytes in water also increase the rate of corrosion. E.g. Corrosion of iron in sea water takes place to a large extent than in distilled water.
iv.Presence of CO2 in water: water containing CO2 acts as an electrolyte and increases the flow of electrons from one place to another. (CO2 + H2O) form carbonic acid which dissolves into ions and hence acts as an electrolyte).
Examples
1. Why do you think zinc on iron is sometimes called sacrificial anode?
2. Explain why blocks of Mg can be attached to wills of ship or irons pipes with the aim of preventing rusting.
3. Tin cans are made of Iron coated with thin film of tin. After a crack occurs in the film, a can corrodes much more rapidly than Zinc coated with Iron. Explain this behavior.
4. Why is it that with enough time, corrosion will always defect the protection applied to iron?
ANSWERS
Zinc or Iron is sometimes called sacrificial anode because it after it wears off, the metal can get exposed and hence starts to undergo rust.
Conductivity in solutions

Electrolytes
These are substances which allow electricity to pass through them in their molten state or in form of their aqueous solution, and undergo chemical decomposition e.g., acids, bases and salts.

Classification of electrolytes
All electrolytes do not ionize by the same extent in the solution. According to this we have strong and weak electrolytes
Strong electrolytes are those that ionize completely into ions in the solution
e.g. Salts, mineral acids, some bases.
Weak electrolytes are these that ionize partially into ions in the solution
e.g. in organic acids. HCN, Na4OH.

Electrolytic conduction
When a voltage is applied to the electrodes dipped into an electrolytic solution, ions of the electrolyte move towards their respective electrodes and therefore electric current flows through the electrolytic cell. The process of the electrolyte to conduct electric current is termed as conductance or conductivity.
Like metallic conductors, electrolytic solution also obey ohm’s law which states that “the strength of the current flowing through a conductor is directly proportional to the potential difference applied across the conductor and inversely proportional to the resistance of the conductor
i.e. V = RI.
Resistance of any conductor is directly proportional to the length L and inversely proportional to the area of cross – section.
R EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
R = ρx EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
ρ=Resistivity
Resistivity is the resistance of a conductor having unit length and unit area of cross-section. SI unit Ohm-meter (Ὡm).
Conductance is the measure of the ease with which the current flows through a conductor a (λ orΛ)
Conductance is the reciprocal of electric resistance. (EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2))
From the above expression high resistance means low conductance and vice versa.
Conductivity is the reciprocal of resistivity and is also called specific resistance EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(kappa)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) (EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
For an electrolytic cell, l is the distance apart between the two electrodes and A is the total area of cross-section of the two electrodes. Therefore, for a given cell, l and A are constant. If the dimensions of the cell are not altered, the ration EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) is referred as cell constant (K)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = K
R = ρEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2),EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)

Molar conductance (λm)
Is the conductivity of volume of a solution which contains 1 mole of the solute.
Or
It is the conducting power of the ions produced by dissolving 1 mole of an electrolyte in a solution.
Molar conductance is given by EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) where v = volume containing 1 mole of a solute and is called dilution.
Concentration of an electrolyte depends on the volume i.e.
V =EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
Where C is the concentration in mol/dm3 and EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) is EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)m-1
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = Ѕm2mol-1 or m2Ὡ -1 mol-1


Example
1. What is the dilution of 0.2M, NaOH solution?
Solution:
Given l = 0.2mol/dm3
V = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
V = 5dm3mol-1
2.0.0055M silver nitrate has a molar conductivity of 2.98 x 10-3m2 Ω-1 mol-1 . Calculate conductivity of that solution.
Solution
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) x c
= 2.98 x 10-3 m2Ω-1 mol-1 x 0.0055moldm-3
= 2.98 x 10-3 m2 Ω-1 x 5.5cm-3
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = 0.01529 Ὡ-1 m-1

3. Calculate the molar conductivity of 0.3M, KOH solution which has a conductivity of 391Ὡ1-1 m-1.
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
= EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
= EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
= EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
λ∞= 1.303EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)

5.The cell constant of the conductivity cell was stated as 0.215 cm-1.
The conductance of the 0.01 moldm-3solution of KNO3was found to be 6.6 x 10-4 Ð…
i. What is the conductivity of the solution?
ii.What results does this give for the molar conductivity of KNO3
Solution:
i) R = ρEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
K = 0.215cm-1
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = 6.6 x 10-4 s
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)K
= 6.6 x 10-4 s x 0.215cm-1
= 1.419 x 10-4 Ð…cm-1

ii) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
= EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)= 14.195EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)

Questions
1. How many grams of acetic acid must be dissolved in 1dm3 of water in order to prepare a solution with a
conductivity and molar conductivity of 575cm-1 and 9255cm2mol – 1 respectively?
2. 0.05M NaOH solution offered a resistance of 31.6Ὡ in a conductivity cell at 298K. If the cell constant of a cell is 0.367 cm-1. Calculate the molar conductivity of NaOH solution.
3. The conductivity cell filled with 0.01M KCl has a resistance of 747.5Ὡ at 250c,when the same cell was filled with aqueous solution of 0.05M CaCEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2), the resistance was 876Ὡ. Calculate.
i.Conductivity of the solution
ii.Molar conductivity of the solution if given EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)of 0.01M KCl is 0.4114Ð…m-1

VARIATION OF MOLAR CONDUCTIVITY WITH CONCENTRATION
The intensity of electricity that can pass through the solution depends on
i.The number of concentration of free ions present in the solution.
ii. Speed with which ions move to their respective electrodes.

An increase in concentration gives an increase in total number of solute particles in a given volume of solution and this might well be expected to give an increased conductivity.
Conduction in strong electrolytes.

The molar conductivity is high since strong electrolytes ionize completely into free ions and the molar conductivity increases slightly in dilution. WHY?
In strong electrolytes, there are vast numbers of ions which are close to each other. These ions tend to interfere with each other as they move towards their respective electrodes. The positive ion are held back by the negative ions and vice versa which in turn interrupts their movement to the electrodes (reduce the speed with which they move)
Dilution ions get separated from each other and at an average distance they can move freely or easily.
NOTE:-
As the dilution increases, there comes a time for amount of interference become small so that further dilution to has no effect.
At this point the molar conductivity remains constant and it is known as molar conductivity at zero concentration EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2).

Conduction in weak electrolyte
The molar conductivity is less because there is less number of particles as the minority of particles are dissociated into ions. On dilution the molar conductivity increases as the molecules dissociate more into ions which increases the number of free ions. Therefore for weak electrolytes the molar conductivity depends on the degree of dissociation of molecules into ions.
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)

Question 1
At infinity dilution, will the molar conductivity of strong and weak electrolyte of same concentration be the same?
Answer
NO, the molar conductivity will be different because it also depends on the size of the ions which would either increase or decrease the speed of ions.

Question 2
Why at infinity dilution, the molar conductivity of weak electrolyte remains constant?
Answer
Because at infinity dilution the molecule must have to dissociate into free ions.


MOLAR CONDUCTIVITY AND DEGREE OF DISSOCIATION
The molar conductivity of weak electrolyte is proportional to the degree of dissociation.
i.e.EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = KEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) ……………. (i)
At infinity dilution, the weak electrolyte is completely ionized.
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = 1 or 100%
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = K (Constant)……….. (ii)
Taking ratio of equation (i) and (ii)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
Ostwald’s dilution law
It states that
“For a weak electrolyte the degree of dissociation is proportional to the square root of reciprocal of concentration”
Consider a weak binary electrolyte AB (i.e. ethanoic acid) in solution with concentration C (mol/EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)).
AB EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) A+ (aq) + B(aq)
1mole 0 0 start
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) At equilibrium
But V = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
AB EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) A+ (aq) + B(aq)
K = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
But AB = CH3 COOH
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
For a weak electrolyte, degree of dissociation is very small thus the expression
1 – EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) 1
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) Ka = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
K a = C EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)

Example
1. At 250c the solution of O.1M of ethanoic acid has a conductivity of 5.0791×10-2EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)m-1mol-1.
calculate the pH of the solution and dissociation constant Ka (Answer Ka=1.69EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) 10-2M)
2. A 0.001 moldm-3 solution of ethanoic acid was found to have molar conductivity of 14.35Sm2mol-1. Use this value together with molar conductivity at infinity dilution of ethanoic acid EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) (CEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)) is 390.7Ð…EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)to calculate :-
i. Calculate degree of dissociation of acid
ii. Equilibrium constant EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
3.The resistivity of EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)M KCl solution is 361Ω and conductivity cell containing such a solution was found to

have a resistance of 550Ω
a.Calculate the cell constant
b.The same cell filled with 0.1M ZnSO4 solution had resistance of 72Ω. What is the conductivity of this solution?

KOHLRAUSCH’S LAW OF INDEPENDENT IONIC MOBILITY
Kohlrausch notes that the difference between molar conductivity at infinity dilution λ∞ values for the two salts which were strong electrolytes and of the same cation and anion was always constant.
Using the λ∞ values in D.. cm2 mol-1
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
This observation shows that each type of ion (caution or anion) contribute a definite amount of molar conductivity of an electrolyte of infinity dilution independently from other ions present in the solution i.e. a fraction of current that an ion carry is always constant and it doesn’t depend on the compound in which it is contained.
Hence Kohlrausch’s law
States that;”The molar conductivity of an electrolyte at infinity dilution is equal to the sum of the molar conductivities of the caution and anion”.
OR State that the molar conductivity at infinity dilution of the solution equal to the sum of molar conductivity at infinity dilution of its components ions.
i.e. EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) + EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)
E.g. EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)= EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) + EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(CEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2))
λ(Al2(SEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)) = 2EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(AL3+) + 3EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(SEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2))

Application of Kohlrausch’s law
In direct determination of molar conductivity at infinity dilution for weak electrolytes. Thus by using strong electrolytes we can easily. Calculate the molar conductivities at infinity. Dilution for weak electrolytes.
E.g.: The EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) (CEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)COOH) can be determined from EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) of potassium ethanoate (CEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)COOK) hydrochloric acid (HCl) and potassium chloride (KCl)
CH3COOH + KCl EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) CH3COOK + HCl
λ(CH3COOH) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(CH3COOK) + EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(HCl) – EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)KCl)
= EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(CH3COO) + EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(K+) + EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(H+) + EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(Cl) – EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(K+) – EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(Cl)
λ(CH3 COOH) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(CH3COOH) + EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(H+)

REVIEW QUESTIONS
1. Calculate EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(NH4OH given that EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2) of three strong electrolyte NaCl, NaOH and NH4Cl in Ð…cm2 mol-1 are 126.4, 248.4 , 149.8 respectively.

2.The molar conductivity of 0.093 CH3COOH solution at 298k is 536 x 10-4 Sm2mol-1 .The molar conductivity at infinity dilution of H+ and CH3COO are 3.5 x 10-2 and 0.41 x 10-2Sm2 mol-1.what is are the dissociation constant of CH3COOH

3. A 0.05M HF solution has a conductivity of 91.81mol-1 m-1 at 298k.At the same temperature EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(NaF), EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(NaoH) and EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - ORGANIC CHEMISTRY 1.1 -POLYMERS (2)(H2O) are 493360 and 162Sm2mol-1 respectively. Calculate dissociation constant.
4. The λ(NaI), λ(CH3COONa) and λ(CH3COO2Mg) are 12.69,9.10 and 18.78Sm2 mol-1 respectively at 250C. What is the molar conductivity of MgI2 at infinity dilution.




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