Chemistry Chemistry A Level(Form Six) Chemistry Notes Form Six (A Level)


Written by

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.

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)
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

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:
Cathodic reaction:
Cell reaction:
Electrochemical cell can also be represented in an abbreviation way known as cell notation.
Anode cathode
I.e. for Daniel cell
1. Given the overall reactions, write their corresponding cell notations

2. From the cell notation give the cell reactions


Cu(aq) + 2Ag+(aq)→Cu2+(aq) +2Ag(s)

The difference between electrode potential of the two electrodes constituting an electrochemical cell is known as electromotive (emf) or cell potential.
This acts as a driving force for a cell reaction and it is expressed in volts.
The emf of a cell is calculated by subtracting the standard electrode potential of the left electrode from that of the right electrode.
For Daniel cell
= 0.34 – (-0.76)
= 1.1v

Since standard electrode potentials are in reduced form, for oxidation half reaction the sign of electrode potential should be reversed.
= 0.76 + 0.34
= 1.1v

1.Prediction of the occurrence of chemical reaction.
If the emf of the cell calculated is negative the reaction is non spontaneous i.e. the reaction does not occur in the way it is written unless external forces apply but the reaction is spontaneous in the opposite direction.
If the emf of the cell is positive the reaction is spontaneous
Example: given the following reaction
Predict the direction of the reaction given that reduction potential of nickel (Ni2+/Ni) Eo = -0.25v of mercury Hg+/Hg = 0.14v
Cell notation
The reaction is non spontaneous since emf is negative hence backward reaction is favored.

For each of the following reactions at standard condition, decide if it occurs spontaneously in the direction written.
Given for .

=-0.14 – (-0.76)
= 0.62V

II. Cell notation:
=1.09 – (1.36)
= -0.27V

III. Cell notation:
=-0.4 – 0.34
Therefore reaction (1) is spontaneous while reaction (2) and (3) are non-spontaneous

Briefly explain what happen when
I.Fe is dipped in CuSO4 solution
II.Cu is dipped in FeSO4 solution
Given the following values
a)State which species are the strongest oxidant and which oxidant and weakest reductant.
b)State which species is the strongest oxidant and weakest reductant.
c)The lead rods are placed in a solution of each CuSO4, FeSO4, AgNO3 and ZnSO4. In which solution do you except coating of another metal on lead rod. Explain.

2. (i) since Fe has a negative reduction potential then it has a positive oxidation potential and hence will displace copper metal from CuSO4 solution
(ii)Since Cu has positive reduction potential, then it has a negative oxidation potential and hence it will NOTE displace Fe from FeSO4solution therefore no reaction will be possible.
3. (a) strongest oxidant
Weakest reductant
Maximum and minimum emf of a cell
It is possible to decide which cell has to be constructed either of great or smallest emf for a given electrodes. A cell with greatest emf is obtained by
choosing two electrodes of greatest cell reactivity difference. A cell with minimum emf, choose two electrodes with closest reactivity.
Study the following electrodes.
4. Explain how you can construct a cell that will yield
i. Maximum emf
ii. Minimum emf
For maximum emf, we choose two electrodes of greatest cell reactivity difference.
= 2.87 – (-3.04)
= 5.91v
For minimum emf, we choose two electrodes with closest reactivity
= -0.4 – (-0.44)
= 0.04v

We have been considering electrode potential under standard conditions of molar solution, pressure of 1 atom and 298K. When the conditions are altered he values of electrode potential changes thus we have to define the potential of the cell under non-standard conditions.
The Nernst equation shows the relationship between emf of the cell at standard conditions and emf under non-standard conditions.

Where; – emf of a cell at any conditions
– emf of a cell under standard conditions
R – Universal gas constant (8.314 Jmol-1K-1)
n – Number of moles electrons being transferred
F – Faraday’s constant (96500c)
At standard temperature (298K)
The above equation can be applied to half reactions and overall reactions

Always write a balanced cell reaction to obtain ‘n’ and position of ions, either reactants or products together with their stoichiometry
Calculate the emf of the given cell at standard temperature.
Alternative Solution 1
= -0.13 – (-2.56)
= 2.43v
Alternative Solution 2
n = 6
Example 2
Calculate the emf of Daniel cell at using 2M ZnSO4 solutions and 0.5M CuSO4 solution
Alternative Solution 1

= 0.34 – (-0.76)
= 1.1v
Alternative Solution 2
Half reactions
Here n = 2
Now using
Ecell =1.08v

Example 3
Calculate the emf for the following voltaic cells
Alternative Solution 1
n = 2
= 0.8 – (-0.76)
= 1.56v

Alternative Solution 2
= 1.47v

d) Solution
Alternative Solution 1
= 0.8 – (0.77)
= 0.03v

Alternative Solution 2
Half reactions
n = 1
Ecell =0.0122v
c) Solution
= -0.14 – (-2.56)
= 2.42v

Half reactions
2Al + 3Sn2+ ® 2Al3+ + 3Sn
n = 6
Ecell =2.418v

Consider to what happens to a Daniel’s cell if we use it to do some electrical work i.e. it can be connected to a small electric motor. After sometimes the motor will stop. The cell will run down. When this happens and there is no overall transfer of electricity from one half cells to the other. When there is no overall change taking place in a chemical reaction the equilibrium has been established. At equilibrium, electron density of both electrodes is equal and there is no transfer of electricity between the two half cells.
Applying the Nernst equation to a Daniel’s cell
Since the reaction is at equilibrium, the ratio
KC = 1.67 x 1037

The large value tells us that, the equilibrium lies almost entirely in favour of copper metal and zinc ions.
Generally, the cell at equilibrium at standard temperature is given by the following expression;

Calculate KC for the following voltaic cell
Cell reaction
= -0.12 – (-0.46)
= 0.34v

KC = 3.20 x 1013

What concentration of will emf of the cell be zero at if concentration of is
Cu + 2Ag+ ® Cu2+ + 2Ag
n = 2
= 0.8 – (0.338)
= 0.462v
3. Given and . Calculate the equilibrium constant for the reaction State the significance of equilibrium constant.
= 0
Cu + 2Cu2+ ® 2Cu+
n = 2
= 0.53 – 0.34
= -0.187v
= -6.3282
KC = 4.69 x 10-7
(a) Given the which is thermodynamically feasible, the reduction of Cu2+ by Ag or reduction of Ag+ by Copper.
(b) Calculate the standard emf of the cell and the equilibrium constant.
A half concentration cell which consists of two half cells with identified electrode that differs in ion concentration because the electrodes are identified. for oxidation is numerically equal and opposite in sign to for reduction. As a result
The reaction in the cell takes place in order to reduce the difference in concentration until (when the two concentrations are equal. The higher concentration is reduced and the lower concentration is increased.)
Oxidation reduction
Cu ® Cu2+ + 2eCu2+ + 2e ® Cu
Electrochemical cell: Anode (-)
Cathode (+)
Electrolytic cell: Anode (+)
Cathode (-)
Calculate the emf of the following concentration cell

Using Nernst equation
= 0.0591v
pH is the degree of alkalinity, or acidity of a substance. It is obtained by using the hydrogen ion concentration pH = –
The pH of a solution is determined by using any electrode provided its standard electrode potential is known and the concentration of ions in that electrode should be 1M. This will be one of the half cells another half cell is made up hydrogen electrode dipped into the solution whose pH is to be determined
H2(g) + 2 H+(aq) +2e E 0 = 0.00v
2Ag+ (aq)+ 2e Ag(s) E = 0.8v
H2(g) + 2Ag+(aq) 2H+(aq0+ Ag(s) E0 = 0.8V
Applying Nernst equation at standard temperature
E cell = cell
= 0.8 –
= 0.8 + 2(-
E cell = 0.8 + 0.0591
pH= =
Question: Calculate the of the following cell and hydrogen ion concentration.
Zn/Zn2+ // H+ /H2, pt
Zn2+ /Zn = -0.76v
E cell = 0.115v

= 0 – (-0.760)
= 0.76v
Zn(s) → Z(aq) + 2
2H+(aq) + 2(g)
Zn(s) + 2H+(aq) Zn2+(aq)+ (g)
= + x 2
0.115 = 0.76 + 0.0591
-0.0591) = 0.645
= 10.913
= 10.91
Whenever a redox half equation involves H+ or OH ion, its redox potential depends on the of the solution.

(aq)+ (aq)+ (aq)+ 4O(l) E = 1.52V
The Nernst equation at standard temperature is

Under standard conditions, all effective concentrations are 1M
= 1.52 –
= 1.52 – 0
= 1.52v

Suppose the is changed from O to 5 i.e. the concentration changes from 1M to 1 x 10-5 M
= 1.52 – )-8
= 1.52 +
= 1.04 v
Due to change of from O to 5, the electrode potential decreases from 1.52 to 1.047v.
That means, permanganate (vii) becomes a less powerful oxidizing agent when PH increases

Calculate of the following cell;-
Zn/Zn2+ //H+ /H2, pt
Zn2+ /Zn = -0.76v
E cell = 0.115v

If E for Zn /Zn2+//Cn2+/Cn is 1.1v
i.Calculate the E cell when concentration of Zn2+ is 2M and Cu2+ is 0.5M
ii.What is E cell when concentration of Zn is 0.4M, = 0.1M

Write down the expression for the cell emf for the following reaction;
(aq)+ (aq)+ (aq)+ 4H2O(l)
Briefly explain why the oxidizing power of permanganate (Vii) ion is quite sensitive to the concentration of H+ in the solution.
Is the series of standard electrode potential with respect to their elements from more negative standard electrode potential to the more positive standard electrode potential.
Is the arrangement of electrodes of elements in order of reducing power

It is a good guide for predicting reaction that takes place in solution especially displacement reaction.

Displacement reaction
Is a types of reaction in which an atom or element displace another element or atom in a compound
What will happen when magnesium ribbon is added to a solution of AgNo3?
= -2.37v
= 0.8v

The more negative the electrode potential the greater is the reducing power of that element i.e. the more likely it is to give out the electrons and acts as a reducing agent therefore Mg will reduce Ag+ions to Ag (s).
2Ag+(aq)+ 2 2Ag(s)
Mg (s) + 2Ag+(aq) Mg2+(aq)+ 2Ag (s)
= 0.8 – (-2.37)
= 3.17v
Thus, the element higher in electrochemical series will displace the one lower in the series.
2. Displacement of hydrogen from mineral acids
Metals which are higher in the electrochemical series than hydrogen reacts with acids and replaces hydrogen but metals below hydrogen have no action with mineral acids.
Complete the following reaction:-
i. Zn(s) + 2HCl (aq) ZnCl2(s) + H2(g)
ii. Cu(s) + HCl(aq) No reaction
3. The knowledge of electrochemical series helps as on choosing method for extraction of metals.

Higher most metal can’t be extracted from the oxides by chemical reduction process. This is because they are strong reducing agent hence they can be reduced easily from their oxide by electrolysis.

For electrolysis, fused or molten metal should be used and not aqueous solution. Why?
In aqueous solution, there are H+ ions. Hence metals prefer to react with element in lower electrochemical series than the metal itself, hence for electrolysis we use fused or molten metal and not aqueous solution.
Corrosion is the deterioration of the metals due to the chemical reactions taking place on the surface. Usually, the process is due to the loss of metal to a solution in some forms by a redox reaction (unwanted redox reactions)
For corrosion to occur on the surface of a metal there must be anodic area where a metal can be oxidized to metal ions as electrons are produced.
Anode area:-
M(s) Mn+ + ne+ And cathodes area where electrons are consumed by any of all of several half reactions.
Cathodic reaction:-
2H+(aq) + 2e H2 (g)
2H2 O(l) + 2e 2OH(aq)+ H 2(g)
4e+ O2(g) +H2 O(l) 4OH(aq)
Anodic reactions occur at cracks or around the area with some impurities.

The most common corrosion process is rusting of iron
Rust is hydrated iron (III) Oxide (Fe2O3. XH2O) which appears as a reddish brown substance on the surface of the iron bar.
Both water and air (oxygen) are required for rusting to occur.
The presence of dissolved salts and acid in water increases its conductivity and speed up the process of rusting.
If the iron object has free access to oxygen and H2O as in flowing water, a reddish brown ion (III) oxide will be formed which is a rust.
For iron:
Anode: Fe(s) Fe2+ (aq)+ 2e
Cathode:O2(g) + 2H2O(l) + 4e 4OH(aq)
2Fe(s) + O2(g) + 2H2 O(l) 2Fe2+(aq)+ 40H(aq)
4Fe(OH)2(s) + O2(g) 2H2O(l) + 2Fe2 O3.XH2 O
If oxygen is not freely available: The farther oxidation of iron (II) hydroxide is limited to formation of magnetic iron oxide.
6Fe(OH)2(s) + O2(g) 2Fe3 O3.H2O + 4H2O(l)
Fe3O4 + H2O(l) Fe3O4 + 4H2O
Black magnetite

Prevention of rusting
Corrosion can’t be made non-spontaneous but it can be prevented by making the rate of the reaction negligible. This can be done by covering the metal surface with protective coating or by providing alternative redox pathways (oxidation-reduction pathways).

Protective coatings are usually of 3 types:
i.Painting: – is the simplest and most common method where a metal surface is properly cleaned and then applied with several layers of rust – proofing paint.
ii.Corrosion inhibitors: – these interfere with flow of charges needed for corrosion to take place i.e. phosphate (II) coating on a surface of iron of steel. Using phosphoric acid serves that purpose.
iii.Galvanization: (sacrificial protection) a metal can also be protected by coating with a thin film of second metal where the second metal is oxidised instead of the 1st metal.
Often iron is coated with another metal like zinc, tin or chromium for protection on the surface.
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).
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?
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

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 = ρx
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. ()
From the above expression high resistance means low conductance and vice versa.
Conductivity is the reciprocal of resistivity and is also called specific resistance (kappa)
= (
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 is referred as cell constant (K)
= K
R = ρ =
= , = =

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

1. What is the dilution of 0.2M, NaOH solution?
Given l = 0.2mol/dm3
V =
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.
= 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
= 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.
λ∞= 1.303

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
i) R = ρ
= =
K = 0.215cm-1
= 6.6 x 10-4 s
= K
= 6.6 x 10-4 s x 0.215cm-1
= 1.419 x 10-4 Ð…cm-1

ii) =
= 14.195

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 CaC, the resistance was 876Ὡ. Calculate.
i.Conductivity of the solution
ii.Molar conductivity of the solution if given of 0.01M KCl is 0.4114Ð…m-1

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.
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 .

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.

Question 1
At infinity dilution, will the molar conductivity of strong and weak electrolyte of same concentration be the same?
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?
Because at infinity dilution the molecule must have to dissociate into free ions.

The molar conductivity of weak electrolyte is proportional to the degree of dissociation.
= K ……………. (i)
At infinity dilution, the weak electrolyte is completely ionized.
= 1 or 100%
= K (Constant)……….. (ii)
Taking ratio of equation (i) and (ii)
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/).
AB A+ (aq) + B(aq)
1mole 0 0 start
At equilibrium
But V =
AB A+ (aq) + B(aq)
K =
For a weak electrolyte, degree of dissociation is very small thus the expression
1 – 1
Ka =
K a = C

1. At 250c the solution of O.1M of ethanoic acid has a conductivity of 5.0791×10-2m-1mol-1.
calculate the pH of the solution and dissociation constant Ka (Answer Ka=1.69 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 (C) is 390.7Ð…to calculate :-
i. Calculate degree of dissociation of acid
ii. Equilibrium constant
3.The resistivity of 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 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

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. = +
E.g. = + (C)
λ(Al2(S) = 2(AL3+) + 3(S)

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 (CCOOH) can be determined from of potassium ethanoate (CCOOK) hydrochloric acid (HCl) and potassium chloride (KCl)
λ(CH3COOH) = (CH3COOK) + (HCl) – KCl)
= (CH3COO) + (K+) + (H+) + (Cl) – (K+) – (Cl)
λ(CH3 COOH) = (CH3COOH) + (H+)

1. Calculate (NH4OH given that 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 (NaF), (NaoH) and (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.

Leave a Comment