UNIT 14 ELECTROLYTIC CONDUCTION
Structure
14.1 Introduction
Objectives
14.2 Electrolysis
14.3 Comparison of Galvanic and Electrolytic Cells
14.4 Faraday's Laws of Electrolysis and Electrolytic Conduction
14.5 Measurement of Conductance of Aqueous Solutions of Electrolytes
14.6 Results of Conductance Measurements and their Interpretation
14.7 Kohlrausch's Law and its Usefulness
14.8 Conductometric Titrations
14.9 Acids and Bases
14.10 Acid-Base Equilibria
14.11 Common Ion Effect
14.12 Acidity and Alkalinity of Aqueous Solutions
14.13 Buffer Solutions and Hydrolysis
14.14 Hydrolytic Equilibria
14.15 Precipitation Equilibria
14.16 Applications of the Solubility Product
14.16.1 Solubility and IGp
14.16.2 The Effect of Added Electrolytes
14.16.3 Applications in Qualitative Analysis
14.17 Summary
14.18 Glossary
14.19 Answers to SAQs
14.1 INTRODUCTION
In the previous unit, sources of electrical energy like the primary and secondary cells
were discussed. Electrical energy can be transported through matter by the
conduction qf electric charge from one point to another in the form of an electric
current. This can take place if there are charge carriers in the matter. These charge
carriers can be electrons, positive ions and /or negative ions. Substances like
diamond, silica etc., ujhich offer high resistance to the flow of electric charge are
called insulators. Metals like copper, silver etc. which offer very little resistance to
the passage of electricity are called conductors. The conduction in metals is referred
to as metallic or eltctronic condition since the electrons of the metal are able to
migrate freely and carry the negative charges through the metal. A inetal can be
. considered to be a rigid lattice of positive ions around which are smeared an equal
number of mobile electrons. On applying an electrical potential, the electrons are
forced to move in one direction while the positive ions remain stationary. The
electronic conduction. however, decreases with an increase of temperature as a
consequence of an increase in the thermal vibrations of the lattice. In the case of
semiconductors like silicon, germanium etc., the electronic conduction increases with
an increase of temperature. The electrons, rather tightly bound to local centres at
room temperature, become free to move as the temperature is increased. Both the
conductors have one common characteristic, i.e., there is no chemical change
accompanying the charge transport.
,
Equiiibria
-demW
Objectives
After studying this unit, you should be able to :
* use Faraday's laws of electrolysis to calculate the amounts of material
deposited or dissolved,
* predict the electrode reactions in electrolytic cells,
* differentiate between galvanic and electrolytic cells.
* explain the dependence of conductance of strong and weak electrolytes on
concentration,
* appreciate the usefulness and applications of conductometric titrations,
* define acids and bases,.
* calculate the pH of solutions of acids or bases,
* apply the Henderson equation to calculate the pH of buffer solutions,
* explain how salts can be acidic or basic using hydrolytic equilibria, and
* appreciate the usefulness of the solubility product principle in qualitative
analysis.
14.2 ELECTROLYSIS
Electrolytes like sodium chloride, either in the molten state or in an aqueous solution
conduct electricity but are decomposed by the passage of an electric current. A
current of electricity can be passed into an aqueous solution or a fused salt via two
electrodes dipping into the solution. The electrode connected to the negative pole of
the battery is a source of electrons and is the cathode of the electrolytic cell, since
reduction can occur at this eleceode. The other electrode connected to the positive
pole of the battery is deficient in electrons and so can accept electrons. It is the anode
of the electrolytic cell. Under the influence of an applied EMF the positive ions
(cations) of the electrolyte move towards the cathode and get reduced. The negative
ions(anions) move towards the anode and get oxidised. This movement of ions of an
electrolyte under the influence of an applied EMF is called electrolytic conduction
and its chemical decomposition caused by the passage of electricity is called
electrolysis. As a specific example. we consider the electrolysis of molten sodium
chloride using inert electrodes like Pt or graphite that do not react with either the
electrolyte or the products of electrolysis. Of the two ions present, Na+ and C1-, only
the Na+can be reduced. At the cathode of the electrolysis cell, the reduction reaction,
Na+ ( 1 ) + e- = Na ( 1 ), occurs and the sodium metal collects around the electrode.
The C1- anion on reaching the anode, loses the electron (oxidation) to the anode and
a neutral chlorine atom is formed. Two such atoms combine to give a chlorine
molecule, which bubbles off as a gas. The net chemical change that takes place in the
electrolytic cell is called the cell reaction. It is obtained by adding together the anode
and cathode reactions in such a way that the same number of electrons are gained and
lost. Thus, in the electrolysis of molten NaCl,
cathode reaction :
2Na'+2e-=2Na
anode reaction :
2Clm- 2 e=C12
Cell reaction :
2 Na++2 Cl-->
2 Na + CI,
(electrolysis)
In the simple case of molten sodium chloride, the current-carrying ions are found to
get discharged at the electrodes. When many electro-active species (capable of
getting reduced or oxidised at the electrode) are present and also the electrode
material is susceptible to be attacked, complications arise. Reactions of several types
are possible at each electrode. The reactions that may occur at the anode include
i) dissolution of the anode by oxidation
e.g., Ag--> Ag' + e- ;Cu ->
cu2++ 2 e-, etc.
.
EleclrdytlcConduction
ii) discharge of anions, and
e.g., 2 Cl- --+ C12+ 2 e'}
iii) oxidation of water in aqueous solutions
!
I
I
Among the conceivable oxidation reactions at the anode, the reaction with the more
negative (less positive) reduction potential is likely to take place. If the applied EMF
is gradually increased, the oxidation process occur in the order of increasing
reduction potentials (most negative first and least negative last). After all, this is a
prediction based on thermodynamics and so one should actually know how rapid
these electrode processes are in order to predict the electrode reaction.
At the cathode, reduction of a cation to give metal takes place in a few cases.
Ag++ e ' j Ag
In the case of cations of highly reactive metals like Na, K.etc., it is water that is
preferentially reduced to give hydrogen
Examples :
(1)
Electrolysis of an aqueous solution of Na Cl- - using Pt electrodes
+
Cathode Reactions
Na++e =Na
Remarks
Not possible because of negative EO value
possible. Products at the cathode are H2
and OH-. Na+ ions move towards the.
cathode to preserve elecmcal neutrality.
Concentration of Wis too small for this
reaction to occur
Anode Reaction
Remarks
2C1--2e=C12(g)
Actually observed
240->02+4W+4e-
Oxygen evolution is not observed
H2 ( g ) + C12( g ) + 2 O K
Therefore, the net Cell Reaction : 2 C1- + 2 H, 0 ->
' electrolysis
(2)
Electrolysis of an aqueous solution of sodium sulphate using Pt
electrodes to give hydrogen at the cathode and oxygen at the anode :
Cathode:
2 H 2 0 + 2 e m = H 2 ( g )+ 2 0 H -
Anode :
2H2O=O2(g) + 4 H + + 4 e -
Therefore, the cell reaction is
If the H+and O H ions are allowed to mix, the cell reaction will be
What is the function of sodium sulphate in this case? The anode reaction results in the
formation of H+in the anode. The sulphate anions migrate to the anode to maintain
electrical neutrality. Simil'arly positive ions ( Na') move into the cathode region to
compensate for the presence of O H ions. It will be noticed that the current-carrying
ions need not necessarily be discharged at the electrode. The ability to carry current
depends on the concentration and speeds of the ionic species. But the ability to get
reduced or oxidised depends on the electrode potential.
14.3 COMPARISON OF GALVANIC AND
ELECTROLYTIC CELLS
In a galvanic cell, electrons released during oxidation at the anode push the electrons
of Lhe anode material into the external circuit. The anode, being a source of negative
charge, is the negative electrode. At the cathode, electrons are used up for the
reduction reaction and as a result of the electron deficiency it is called the positive
electrode.
In electrochemistry. electrodes are designated as anode or cathode depending on
whether oxidation or reduction takes place at an electrode. In an electrolytic cell, the
anode is the positive electrode since it is connected to the positive pole of the
galvanic cell. Anions attracted by the anode, get oxidised and the electrons released
to the anode are drawn by the cathode of the galvanic e l l to be used in the reduction
reaction in the galvanic cell. The cathode of an electrolytic cell is a negative electrode
since it is comected to the negative pole (anode) of the galvanic e l l . The electrons
released during the oxidation at the anode of the galvanic cell are available for the
reduction of catiws or other species at the cathode of an electrolytic cell. Thus the
galvanic cell may b& linked to an electron-circulating pump, taking electrons from the
anode to the cathode of an electrolysis cell (Figure 14.1).
I
Galvanic
I
I
Electrolysir
cell
Figure 14.1 :Galvanic and E l e d y s i s cells
14.4 FARADAY'S LAWS OF ELECTROLYSIS AND
ELECTROLYTIC CONDUCTION
-
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-
-
-
p-
-
--
Michael Faraday was the first to describe quantitatively the ralationship between the
quantity of electricity passed (current in ampere x time in seconds or amp. sec or
coulombs and the extent of chemical decomposition brought about by electrolysis.
His investigations are sumrnarised in the form of two laws of electrolysis.
First law : The amount of any substance deposited at or dissolved from an
electrode as a result of the passage of an electric current is proportional
to the quantity of electricity passed.
If W grams of a substance are deposited or dissolved by passing a current of i
-amperes for t seconds, according to the first law
In Eqn. 14.1, e is called the electrochemical equivalent of the substance. From Eqn.
14.1, it is seen that when i t or the quantity of electricity passed is one coulomb,
W = e. Thus, the weight of the substance deposited or dissolved by the passage of one
coulomb of electricity is the electrochemical equivalent, It is no longer used.
Second Law :The masses of different species deposited at or dissolved from
electrodes by the same quantity of electricity are proportional to their
chemical equivalent weights.
If W, and W2 are the amounts of two products of electrolysis obtained by passing q
Coulombs of electricity, according to the second law
i
In Eq. 14.2, E, and E2 are the chemical equivalent weights of substances 1 and 2
respectively. Equations 14.1 and 14.2 can be combined to give Eqn. 14.3
In Eqn. 14.3, F is a constant of proportionality. It will be seen that W = E when i
t = F. In other words F, called the Faraday, is the quantity of electricity required to
deposit or dissolve 1 gram equivalant of a substance. The value of F has been found
to be 9.6487 x 184 C/mol of electrons or 96500 C/mol of electrons. The equivalent
weight is equal to the molecular or atomic mass divided by the number or electrons
transferred per molecule, atom or ion.
Significance of Faraday :
It is found experimentally that 1 coulomb of electricity deposits 0.001 119 gm of
silver. The amount of electricity needed to deposit 108 gm or 1 gm equivalent (or 1
log
= 96500 coutombs. This is the value of 1
gm atomic wei&%%ofsilver =
0.001 119
Farad ay .
Thus Faraday is the quantity of electricity needed to deposit or liberate one
gram-equivalent weight or one gram atomic wt. or 6.023 x lo2, number of atoms
(Avogado number). To deposit one gram-equivalent weight, 1 Faraday is needed.
Example 14.1
A current of 0.5 amp., when passed through a solution of a chloride of formula
ACl,, for 3520 seconds, gave at the cathode 1.28 g of a metal. (a) If the atomic
mass of A is 197.0 g mol-', find n (b) If the cell is c o ~ e c t e din series with
another electrolysis cell in which water is electrolysed. what will he volume at
S.T.P. of oxygen collected at the anode?
Solution:
I
i
(a) No. of faradays passed = 4 = OS5 3520 = 0.01824
F
96500
A grams of A ,96500 C required ,
For depositing n
197 gram
or one Faraday = n
Therefore , the formula of the chloride is ACl,
E q u l l l b ~ ~ & ~ ~ t r ~ h e m l ~b)
r g The evolution of oxygen at the anode of another cell can be represented as
1 mol of 02 = 22.4 lit (at S.T.P.) = 4F
22
- 4 lit x 0.01824 F = 0.102 litres.
vol. of 0, = 4F
Faraday's laws are exact and are applicable to aqueous solutions as well as fused
salts. If there are several processes taking place at an electrode, Faraday's laws are
applicable to all the electrode processes taken together. With reference to each
electrode reaction, it is usual to associate the term current efffcfency, to indicate how
much of the total current is made use of in a process. It is defined as the ratio of the
actual amount of the material deposited at or dissolved from an electrode and that
expected theoretically on the basis of Faraday's laws.
Example 14.2
By the passage of 0.06 faradays of electricity 1.057 g of nickel (atomic mass.
58.71) was deposited at the cathode. If hydrogen is also evolved
simultaneously what is the volume at S.T.P. of hydrogen evolved at the
cathode?
What are the current efficiencies for Ni deposition and hydrogen evolution?
I
Equivalentweight of Ni = 2
5871
Solution :
58.7 grams of Ni. The current
0.06F electricity should deposit 0.06 x 2
x 100= 60.0 8.The rest of the
efficiency for Ni deposition = 1-057
0.06 x 58.71
current utilised for hydrogen evolution is equal to 0.06 - 0.036 = 0.024F. The
volume of hydrogen evolved at S.T.P. can be calculated from the fact that 2F
of electricity is required to evolve 22.4 of H2 at S.T.P.
22.4 x 0.024 F = 0.2688 1or 268.8 ml at
Therefore ,volume of hydrogen liberated = 2F
S.T.P.
SAQ 1
What volume of oxygen would be liberated from an aqueous solution of NaOH
by a curreht of 2 amperes flowing for 1; hrs at 2 7 C and 1 am pressure.
SAQ 2
1.
An aqueous solution of sodium sulphate containing sulphuric acid is
elecuolysed between Pt electrodes. What will be the products obtained
at each electrode ?
2.
calculate the amount of products liberated or dissolved at each electrode
if a current of 2.68 amp is passed for one hour through an aqueous
solution of CuSO, (atomic weight of Cu = 63.54)
1
Aqueous solutims of electrolytes, like metallic conductors, obey Ohm's law (Eqn.
14.4) which enables us to calculate the resistance (R) from the EMF (E) and current
(i)
E=iR
(14.4)
The passage of direct current (d.c.) through the solution causes electrolysis and
consequently there will be a decrease in the concentration. The resistances of aqueous
solutions of electrolytes are always measured using an alternating current (a.c). The
resistance of any elecuical conductor is proportional to its length (I), and inversely
proportional to its area of cross section (A),
I
R a - I or
A
In Eqn. 14.5, the constant of proportionality, p, is called the specific resistance or
resistivity of the conductor. It is the resistance offered to the passage of electric
current by a conductor of length 1.0 cm having a cross sectional area of 1 cm2. From
~ q n14.5,
.
it is seen that the units of p are ohm cm (C G S) or ohm m (S I). The
resistivity, being the resistance between opposite faces of a one cm3 of material, is
very useful in comparing the resistances of different conductors.
In electrolytic conduction, it is the conductance (G) which is reciprocally related to
the resistance, that is more useful. For the comparision of conductances, the spec#ic
conductance or conductivity (K) which is the reciprocal of resistivity is useful. The
conductivity is given by Eqn. 14.6
' '
-
' '.
.
The units of K are ohm- c m or o h m m- The reciprocal ohm is some times
called mho and sometimes the siemens (S). This quantity G is thus the conductance
between the opposite faces of a one cm3of the material. Conductivities of a few
substances are given in Table 14.1
Table 14.1 :Conductivities of a few substances at 298 K
Substance
G/ohm cm-
Silver
6.38 x 1 6
copper
5.80 x 1 6
Aluminium
3.80 x 1 6
1.0 M KC1 (aq)
1 . 1 0 ~lo-'
0.01 M KC1 (aq)
1.41 x l u 3
0.1 M acetic acid (aq)
5 . 2 0 ~1 r 4
water
4.0 x 10-*
'
In table 14.1, the conductivities are listed in the decreasing order. It will be seen that
in the case of aqueous solutions of electrolytes, conductivity depends on
concentration. For the sake of comparison of conductivities we need to use the
Equilibria & Electrochemistry
solutions of electrolytes of the same concentration. One molar solutions of sodium
~ each
+ ion transports
chloride and magnesium sulphate may be used. In ~ g sG',
two units of charge whereas in Na+ Cl- each ion transports only one unit of charge.
However, if the concentration is expressed in equivalents per litre (normality), one
equivalent of any electrolyte gives rise to positive ions with a total charge of IF and
negative ions with a total charge of 1F. The conductance of a volume of a solution
containing 1 equivalent of an electrolyte confined between two large parallel
electrodes 1 cm apart is called the equivalent conductance A
!
Let 1 gram equivalent of an electrolyte be dissolved in V cm3 of the solution. Since
the electrodes are 1 cm apart, the solution will cover the electrode surfaces upto an
area of V cm2. From Eqn. 14.6.
Under these conditions the conductance G is the equivalent conductance, A.
Substituting for A = V c d ,I = 1 cm, we have Eqn. 14.8. Here KV is numerically
equal to the equivalent conductance.
A=KV
(14.8)
In Eqn. 14.8, V is the volume of the solution in cm3 containing 1 gram equivalent of
the electrolyte. If the concentration of the aqueous solution is c equiv. lit-': 1.gram
equivalent of the electrolyte will be present in l/c lit or 1000/c cm3. Thus V is 1000
cm3/c. Substituting in Eqn 14.8, we get a more useful form of this equation.
'
'.
The units of (A) will thus be 1000cm3lit-' ohm" cm-' or ohm- cm2equiv- The
equiv. litlast term is usually dropped since it is implied in expressing A as the equivalent
conductance. The molar conductivity A,, is defined by Eqn. 14.10
'
Where M is the concentration in mol dm3and cmis the concentration in mol m-3. It
will be seen that for electrolytes like magnesium sulphate.
It is convenient to use equivalent conductance, since it is a measure of the
current-carrying ability of all the ions from 1 gram equivalent of any electrolyte.
14.5 MEASUREMENT OF CONDUCTANCE OF
AOUEOUS SOLUTIONS OF ELECTROLYTES
The specific and equivalent conductances of aqueous solutions of electrolytes are
calculated by determining the resistance of the electrolytic conductor contained in a
conductance cell of known dimensions I and A. As indicated earlier, an alternating
current (a.c) is to be used to prevent elec~olysis.Usually an a.c of 1000 to 2000
cycles per second is used so that the small amount of electrolysis taking place in one
half of the cycle is reversed in the opposite half of the cycle. Platinum electrodes, on
which platinum black is deposited. are used so that the electrode reactions occur
rapidly and there is no accumulation of the products of electrolysis.
A Wheatstone bridge arrangerncnt is normally used to measure the resistance of the
solution (Figurel4.2). The point of contact P is moved along the wire AB till no
current is detected in D. The resistance is so adjusted that the balance point is almost
midway between A and B. Since a.c. of this frequency is audible, an earphone
detector is used. A minimum nohe in the earphone indicates the balance point under
these conditions. In place of an earphone, analog or digital detectors can also be used.
i
Resistance
Figure 142 :Muwement d &tma of m aqueous dutim of an
ele~lolyleusing 8 Whewume bridge unmganau.
Resistanceof the cell - -AP
(14.1 1)
Known resistance ( R ) PB
The resistance of the cell can be calculated using Eqn. 14.1 1. Once the resistance of
the cell is known, the value of K and A are calculated by using Equations 14.6 and
14.9 respectively. Calculation of K requires a knowledge of the dimensions 1and A of
the cell. The direct measurement of 1 and A is rather difficult,once the cell is
assembled. For a given cell, both 1and A are constants and the ratio 1/A, called the
cell constant (k), can be calculated by measuring the resistance of a solution of 0.1N
KCI, whose specific conductance is accuratelly known. The conductance cell is now
standardised and can be used to calculate conductivity of any solution containing an
electrolyte. 14.6 can be' written in terms of the cell constant as Eqn. 14.12
Since ordinary distilled water has an appreciable conductance, specially prepared
water called conductivity water, with a resistance of nearly lo6ohms, is used to
prepare all the solutions. Since the resistance varies with temperature, the cell must
be kept in a constant temperature bath. Conductance or specific conductance is an
additive property. In an aqueous solution containing several electrolytes, the total
conductance is given by Eqn. 14.13.
In Eqn. 14.13,,,G
is the conductance of water used in preparing the solutions and
the summation applies to all the electrolytes present. In the case of strongly
is so small that it can be neglected in compais~nwith
conducting solutions,,,G
the conductance of the electrolyte solution. For weakly conducting solutions,
For maximum sensitivity in measuring high conductances (low resistances) a cell
with a high value of k is used. A cell with small electrodes separated by a large
distance is used (Figure 14.3a). Conversely, for the measurement of low
conductances (high resistances), 1/A should be as small as possible by having large
electrodes separated by a short distance (Figure 14.3b).
Example 14.3
A conductance cell, when filled with 0.1 molar solution of KCI, has a
resistance of 33.2 ohms. The same cell after washing, rinsing etc., has a
resistance of 300 ohms when filled with 0.1 molar solution of acetic acid
(HOAc), and 55.2 ohms when filled with 0.03 molar solution of sodium
sulphate. The specific conductance of 0.1M KC1 at this temperature is 0.01 164
ohm- cm' The conductivity of water used to prepare these blutions is
' '.
Equilibria & Eleetrochemlstry
1
Fig. 14.3 :&e
of conductance cell for the mcasumnent of
(a) low resistana (high conductance)
(b) high resistance (lowconductance)
7.6~
1r40hm-' cm-'at this temperature. Calculate (a) the cell constant, and
(b) equivalent and molar conductances of the solutions of (i) acetic acid and
(ii) sodium sulphate.
Solution:
Therefore,
(b)
In order to calculate A, the specific conductances are required.
(i)
A,
,,=1000-CK
This includes h e contribution from h e water used for preparing
the solution.
Therefore, K due to HOAc alone
= 1.287 x 10-~-7.6 x 1 r 4= 5 . 2 7 ~lo4= 5.27 x 10-4ohm" cm-'
Therefore,
In case of acetic acid, equivalent weight = molecular weight, and
hiOAc
= AHOAc
= 5.27 o h m cm2
'
Therefore,
K ~ sSO,
, --6 . 9 9 2 ~1r3-7.6 x 1r4=6.233x 1 r 3o h m 1cm-I
1 moleofNa$304= 2molofNa+= 1 molof s @ - = ~ F
Equivalent weight of Na2S04= molecular weight / 2 or 0.03 M
solution of Na2S04= 0.06 N Na2S04
Therefore ,
= 2 X 103.8 or 207.6 ohm- l c d
14.6
RESULTS OF CONDUCTANCE MEASUREMENTS
AND THEIR INTERPRETATION
The specific and equivalent conductances of all electrolytes are found to vary with
concentration. As the solution is diluted. the specific conductance increases and
reaches a limiting value in the case of many electrolytes. This limiting value is called
the equivalent conductance at zero concentration (A0) or at infinite dilution (A?,
since the value of A0 does not increase further with dilution. The results are
represented as plots of A against 6 ,
where c is the concentration in equivalents per
litre (Figure 14.4). From Figure 14.4, it is seen that in the case of solutions of strong
acids, bases and their salts, the plots are'reasonably linear and the values of A do not
vary appreciably with dilution. Tfiese are called strong electrolytes and in these
cases it is possible to obtain thevalues of A0 by extrapolation. the case of weak
electrolytes like acetic acid, ammonium hydroxide etc., A increases rapidly as the
solution is diluted, especially at low concentrations. In these cases, value of A0 cannot
In
Figure 14.4 :The variation of equivalent conductance of a few elecirolyles with ancmration (Figure not to rcale)
Arrhenius attempted to explain the variation in K and A with concentration using his
electrolytic dissociation theory. According to this theory, an electrolyte dissociates
.
into ions when dissolved in water and the degree of dissociation (a ) increases with
dilution. The conductance of a solution of a given electrolyte depends on the number
ions, and also on their velocities. Arrhenius assumed that dilution does not affect the
velocities, but increases the number of ions available from an electrolyte. At infinite
dilution, a tends to be unity and at all other concentrations, a < 1. When the solution
containing the electrolyte is diluted, though the extent or degree of dissociation (a)
increases, the number of ions per cm3 of the solution decreases. The specific
conductance is the conductance between the opposite faces of a one cm3 of material
and since 1 cm3 of material contains a lesser number of ions than before, K decreases
with dilution. The increase in (A ) with dilution was attiibuted to an increase in the
number of ions available from one equivalent of an electrolyte at high dilutions. At
infinite dilution, since a = 1, the value of (A ) does not vary and becomes equal to A',
which is the limiting value of A. Since a = 1 corresponds to A', Arrhenius proposed
A
that the ratio -should be a measure of a, at any concentration. Thus,
A0
The values of a experimentally determined from colligative properties of aqueous
solutions of electrolytes agreed quite satisfactorily with those obtained from the
conductance measurements. Ostwald verified Arrhenius theory by applying the law
of mass action to the equilibrium existing between an electrolyte and its constituent
ions. Consider a solution of normal acetic acid (a weak electrolyte) in equilibrium
with H+and OAc' ions. If a is the degree of dissociation, the equilibrium
Electrolytic Condudlon
&-kanis(~
.
concentrations of H+ OAc- ,and HOAc are ac,c a and c ( 1 - a ) respectively. If V
is the volume of the solution containing one equivalent of an electrolyte. V = l/c.
The expression for the equilibrium constant (K), given by Eqn. 14.15
is known as the Ostwald's dilution law. The value of K was found to be
reasonably constant only for weak electrolytes and not for strong electrolytes.
Strong electrolytes have been shown to consist of ions even in the solid state and
in aqueous solutions the electrolytes are completely ionised (a= 1) at all
concentrations.
According to the Debye-Htikel theory, which is applicable to both strong and weak
electrolytes, as a result of the interionic attraction the velocitiek of the ions are
influenced. Each ion is surrounded by an ion cloud of oppositely charged ions (ion
atmosphere). This ion cloud retards the movement of the central ion. In the case of
strong. completely dissociated electrolytes, the variation in conductivity with
concentration can be traced to this retardation which increases with concentration. At
very high concentrations, ions associate into ion pairs, and so, the effective number of
conducting species will be less than in the absence of ion association. As the solution
is diluted, the ions are father apart, and the density of the ion cloud decreases. The
interionic forces of attraction will be less, the speed of the ion increases, and so the
equivalent conductance increases. At infinite dilution, the ions are so far apart that
there are no interionic forces of attraction. Thus, the equivalent conductance reaches a
maximum value. For weak, incompletely dissociated electrolytes, changes in
conductivity occur as a result of an increase in the degree of dissociation with
increasing dilution.
14.7 KOHLRAUSH'S LAW AND ITS USEFULNESS
At infinite dilution, in the absence of any interionic forces of attraction, each ion of
an electrolyte can be expected to behave independently. Kohlrausch found that the
difference. A, ( KX ) - A, (NaX) was the same irrespective of the nature of the anion,
X. Similarly, a constant value for the difference A0 (KX)- A0 (KY), where X and Y
are anions, was observed. These results could be explained if it is assumed that A,, for
each salt is considered to be the sum of the equivalent conductances of the constituent
ions (ion conductances) at infinite dilution. He put forward his law of independent
conductance of ions as, "at infinite dilution, each ion makes a definite contribution to
the equivalent conductance of an electrolyte, whatever be the nature of the other ion
of the electrolyte". This can be expressed as Eqn. 14.16
A0= A!+ k:
(14.16)
In this equation, k: and 5: are the ion conductances at infinite dilution of the cation
and the anion respectively. TI4e ion conductances of a few ions are given in Table 14.2
From this table, it will be seen that for polyvalent ions, the values are only for one
equivalent and not for a mole. The conductances reflect the amount of current that
these ions can carry. Thus, one should compare the conductances
. ( f ~2 ) . ( f Fe3 ) etc.. since in each case. we are referring to
)10 ( Na+) )10
+
)10
+
the amount of species associated with 1 mole of electrons.
'
1
Also. A' ( MgSO, ) = 53.1 + 80 = 133.1 ohm- cm2 equiv7
5
'
but A: ( MgSO, ) = 106.2 + 160= 266.2 ohm- cm2mol-
'
'
Table 14.2 :Ionic equivalent conductances a t infinite dilution of a few ions a t 298 K*
Cations
1:
Anions
71:
H+ .
349.8
OW
197.6
Rb+
77.8
( 1/4 ) Fe( CN -:)
111.0
Cs+
77.3
( 1/3) ~ eC( N ) ~ -
99.1
NH:
73.4
( 1/3 ) PO:-
92.8
K+
73.5
( 1/2) 0:-
83.0
Electrdytk Condudlon
If
NO;
Formate
Acetate
It was seen earlier (Figure 14.3) that values of A, for strong electrolytes can be
obtained by extrapolation. In order to obtain A0 values in the case of weak
electrolytes like acetic acid. we can use A0 values of suitable strong electrolytes as, _
shown below:
=kO(H')+ ~'(oAc-)
In the case of sparingly soluble salts. the saturated solution is so dilute that further
dilution does not change the value of A. Hence it is assumed that A = A'.
Thus,
'1
1
AO(f B ~ s ~ ~ ) = A ~ ( ~ B ~ c ~ ~ ) + A ~ ( ~ N ~ s ~ ~ ) - A ~ ( N ~ c ~ )
Example 14.4
'
1
1
At 298K. the A0 values of NH4CI ;- BaC12 and - Ba ( OH ), in o h m cm2 are
2
2
149.9, 139.9 and 262.2 respectively.
Calculate A0 of NH40H.
I
Equlllbrla & Electrochemistry
Solution:
= 149.9 + 262.2 - 139.9
= 272.2 ohm1cm2
Example 14.5
4
1
The h0values of - BaCl - N$S04 and NaCl in o h m c d are 139.9, 130.1
2'2
2
1
and 126.5 respectively. Calculate the he0 of y BaSO,.
Solution:
= 143.5 ohm1 c d
Example 14.6
When water with a specific conductance of 1.1 x
o h m cm- is saturated
with BaSO,( s ), the saturated solution is found to have a specific conductance
o h m cm- I. Calculate the solubility in n~ole/litof BaSO, in
of 4.6 x
water at 298 K.
Solution:
concn. of BaSO, in equiv/lit =
ld x 3.5 x 1 r 6 = 2.44
J
143.5
1 mol of BaSO, = 2 equiv.
Therefore, concn. of BaSO, in mol/lit = 1.22 x lo-'
SAQ 3
1.
A decinormal(0.1 normal) solution of sodium acetate, when placed
between two electrodes each of area 1.5cm2, and placed at a distance of
0.72 cm, has a resistance of 524 ohm. Calculate the specific and
equivalent conductances of this solution.
2.
The equivalent conductances (A0) on ohm' cm2 at infinite dilution of
NaCl, HCOONa and HCl are 126.4, 109.6 and 426.1 respectively.
Calculate (Ao) (HCOOH). The equivalent conductance of 0.01 N
HCOOH is 50.7. Calculate the degree of dissociation and [H'].
14.8 CONDUCTOMETRIC TITRATIONS
The conductance of a solution is the summation of the contributions from all the ions
present. It depends on the number of ions per unit volume (concentration), and also
on the ion conductances. During the course of a titration, the concenuations of the
ions change, and so conductance also changes. In this method, the variation of the
conductance of the solution during the course of a titration is followed. It is not
necessary to determine the actual specific conductance of the solution, and any
quantity proportional to it can be used. For a given cell with a known cell constant, K
is inversely proportional to resistance R, and so one can use the reciprocal of
resistance or conductance to follow the titration. The conductance changes with
dilution, and so it is usual to add a concentrated solution of the titrant from a burette.
Since this procedure involves the measurement of resistance, one can use this method
in the case of coloured solutions, colloidal solutions and even dilute solutions. The
use of conductance to locate h e end points in titrations is based on the changes in the
slope of the plot of conductance versus volume of titrant (titration curves). This
change of slope may be due to the presence of ions of high conductance in the
solution, or weakly ionised or insoluble compounds after the equivalence point. Thus,
it is possible to follow the progress of neutralization and precipitation titrations.
Acid-Base Titrations
The conductometric titration curve for the neutralisation of a strong acid (H' C1-) by
a strong base (Na' OH-) is given in Figure 14.5(a). The reaction can be represented as:
H+ + C1-
+
Na+ + OH-
+--+
+--+
Species present
in the conductance
cell initially
Titrant
--+ Na++ Cl- + H, 0
The highly conducting hydrogen ions initially present in the solution are replaced by
the sodium ions having a much smaller ionic conductance. Hence, the conductance of
the solution decreases as the base is added (the descending position of the curve in
the figure). Beyond the equivalance point, further addition of a strong base inuoduces
highly conducting OH- ions in solution. Since these are not used up in the reaction,
the conductance increases in direct proportion to the excess of base added ( the
ascending portion of the curve). Normally, the conductance of the solution is
determined after successive additions of the alkali, such that points covering the
range before and after the end point are obtained. Two straight lines are drawn
through these points. The point of intersection of the two straight lines gives the
volume of the alkali required to neutralise the given volume of the acid. If a weak
base like an aqueous solution of ammonia (NH40H) is used as a titrant in determining
the concentration of the strong acid, the descending portion of the curve is similar,
since this represents the decrease in conductance as a result of replacing the hydrogen
ions by the ammonium ions. After the equivalence point, the conductance will remain
almost constant because of the low conductance of NH40H compared to NH,CI
[Figure 14.5(b)].
The neutralisation titrations of weak acids or bases are rather difficult. A weak acid
like acetic acid (HOAc) is present mostly as unionised molecules and the solution has
a low conductance due to the small amounts of H+ and OAc- ions. The neutralisation
reaction can be represented as
HOAc
+-+
Species present in the
conductance cell initially
Na'0I-I-
+ t-+ -+Na++OAC+K,O
Titrant
As neutralisation proceeds, the common ion formed, i.e.. the acetate ion, supresses
thc ionisation of the acetic acid (Sec 14.11). Hence, initial addition of small amounts
of an alkali result in a decrease in conductance. With further addition of sodium
Electrolytic Conduction
Volume o f T i t r a n t
F ~ U14.5
R : Carductmetric T~uaticmcurves
(a) HCI against NaOH
(b) HCI against N H 3
hydroxide. the cohductance of Na+ and OAc- ions exceeds that of acetic acid and
conductance of the solution increases. After the equivalence point, thcre is a sharp
increases in conductance due to the Na+ and OH- which are no more required for
neutralising HOAc [Figure 14.6(a)]. If NH,OH is used as a titrant, the initial portion
of the curve is similar to that in Fig 14.6(a). After the equivalence point, there is
practically no change in conductance because of the very small conductance of
NH,OH [Fig.14.6(b)l.
Volune o f T i t r a n t
Figure 14.6 :Conductanetrictitration curves
(a) Acetic acid against NaOII
(b) Acetic add against NH3
Precipitation tiuations are based on the fact that the contribution of the ions of the
sparingly soluble salt to the conductance of the solution is rather small. The reaction
between K+ Cl- and Ag+NO; can bc represented as
te-9
Species present in the
conductancecell initially
+
-++
->
AgCl(s)+K++NOj
Tiuant
The net result being the replacement of C1- by NO; of almost equal conductance,
there is practically no change in conductance as the titration is performed. However,
beyond the equivdlence point the addition of Ag' NO; increases the number of ions in
solution and so the conductance increases [ Figure 14.7 ]
Volume o f Titrant
Figun 14.7 :Condudmetric urntion KCl against A D @ (tilrant)
14.9 ACIDS AND BASES
-
-
-
-
- -
-
-
Acids and bases were originally defined by Arrhenius as substances capable of giving
rise to hydrogen ions and hydroxide ions, respectively, in solution. This definition
applies only to aqueous solutions. In order to account for the acid-base behaviour in
solvents like liquid ammonia, liquid sulphur dioxide, glacial acetic acid etc., Lowry
and Bronsted, independently put forward the proton transfer theory. According to this
theory, an acid is defined as a substance with a tendency to loose a proton and a base
is a substance with a tendency to gain a proton. This definition applies to all solvent
systems. For an acid to donate its proton, some proton acceptor (a base) must be
present. Similary a potential base will accept a proton only in the presence of a proton
donor. When an acid loses a proton, the remaining portion of the molecule will have a
tendency to accept a proton and hence it is a base. In general one can write
The acid and the base which differ by a proton are said to be conjugate to one
another. When a potential acid like hydrogen chloride gas is dissolved in water, the
solvent functions as a base to accept the proton to form the hydronium ion, H30+.
This means that the hydrogen ion in aqueous solution is not the bare proton, H+,but a
solvated proton of the general formula H+(H20 ,) The simplest formula, H30+,is
obcained when n is unity. This reaction can be represented as
HCl
+ H,O
t
-$
H30t
+C1(1)
Acid- 1
Base-2
Acid-2
Base-1
It will be seen that C1- is the conjugate base of HCl and H30+is the conjugate acid of
the base, H20. The dissolution of a weak acid like acetic acid (HOAc) in water to
give a i acid solution can be represented as
HOAc + $0 3 H30++ OAc(2)
The acidity of an aqueons solution of an acid is attributed to the presence of H,O+.
Greater the concentration of this hydronium ion, stronger is the acid. Since HCl is a
stronger acid compared to acetic acid, the position of equilibrium for reaction (1) is
almost to the right, and in the reaction (2), it is almost to the left of the equation.
Alternatively, one can consider that the tendency of C1- to accept a proton is far less
compared t that of OAc'. Thus, acetate ion is a stronger conjugate base than CT.In
general, the conjugate base of a strong acid is weak and vice versa. If a solute like
ammonia gas is dissolved in water, the solvent water acts as an acid and the reaction
1 %
can be represented as
ElectrolyticCoadudlon
I
1
Equilibria & Electrochemistry
Solvents Eke water which can either take up protons or donate protons are said to be
amphiprotic. Solvents like liquid HF, liquid HOAc which can give protons are called
protogenic solvents. Solvents like liquid NH,,pyridine etc., which act as bases are
said to be protophilic. Solvents like benzene, toluene etc. which cannot donate or
accept protons, are called aprotic solvents.
Strong electrolytes are completely ionised in aqueons solutions and the
concentrations of the ionic species in solution can be calculated from the
concentration of electrolyte. For example, in a 0.05 M solution of barium chloride,
[Ba2'] = 0.05 M and [Cl- ] = 0.1 M. In the case of weak electrolytes, whatever is
dissolved may not be in the form of ions. Only a small fraction of the solute gives rise
to ionic species in solution. In such cases, the concentrations of the ionic species can
be calculated from the application of law of mass action to such equilibria. Equilibria
involving electrolytes like acids, bases and sparingly soluble salts in aqueous
solution, are included in ionic equilibria.
14.10 ACID-BASE EOUILIBRIA
The ionisation equilibria in the case of weak acids (HA) and bases (B) can be
represented as
HA + H20 2 H30' + A-
(3)
and
B+H20 2 BH'+OH
(4)
respectively. The equilibrium constants for these equilibria are given by equations
14.17 and 14.18 respectively.
The weight of one litre of water at 298 K is 997 g and this corresponds to 55.4 mol of
water in a litre. In its reaction with solutes, the change in the concentration of water is
rather small and hence one can consider [ H20 ] to be a constant in these equations.
Consequently, these can be written as Equations 14.19 and 14.20 respectively.
In Eqn 14.19, H30' is represented as I? for the sake of simplicity. The equilibrium
constant K, is called the dissociation or the ionisation constant of the acid and Kb is
that for the base. The molar concentrations of the various species can be calculated
from the degree of dissociation and concentration (Sec. 14.6). If a is the degree of
dissociation of the acid HA of concentration c,, the concentrations of the various
species in equilibrium are
[H']=[A-]=c,a
and [ H A ] = c , ( l - a ) .
Similarly, if P is the degree of dissociation of the base B of concentration cb, we can
write
It is assumed that [ H+] or [ BH' ] are available only from the dissolved solutes. If
there are other sources for these ions, the concentrations from these sources must also
be included (Sec.14.11).
It will also be seen that K, and K,, in terms of the degree of dissociation, are given by
Equations 14.20 and 14.21 respectively.
Since a and p can be measured conductometrically~Sec.14.6), it is possible to
evaluate K, and K,. If the ionisation constant is known, it is possible to calculate he
degree of dissociation and also the various concentrations (Example 14.8).
Example 14.7
'
A 0.1 M solution of NH3 has an equivalent conductance of 3.60 o h m cm2 at
298 K. At the same temperature. the ion conductances in ohm'. c d of N
e
and O H ions are 73.4 and 197.6 respectively. Calculate (1) the degree of
dissociation, (2) the concentration of O H ions, and (3) the dissociation
constant, Kb.
'
Solution:
= 1.788 x lo-'
Since,
( l - $ ) ~ l , ~ , ~( iP. e .~) 1c. 7 6 8 ~ 1 0 - '
The latter value is within 1.2%error of that in (3).
Example 14.8
The ionisation constant of formic acid is 2.0 x
at 298 K. Calculate
the degree of dissociation and [ H+]of a 0.01M solution of fonnic acid at
298 K.
Solution:
From Eqn. 14.20,
on simplification we get a quadratic equation in a,i.e., a2+ 0.02 a - 0.02 = 0
Solving this equation we get,
Therefore,
The ionisation constant is a measure of the strength of an acid or a base. Thus,
nionochloroacetic acid (K, = 1.4 x 10- 3, is stronger than formic acid
(K,= 2.0 x 1 r 4 ) , which is itself stronger than benzoic acid (K,= 6.3 x lo-').
Similarly, among the amines, the base-strength decreases as
diethyamine > ethylarnine > ammonia > aniline
(Kb= 1 . 0 0 ~
(Kb=4.66x
(Kb= 1.77 x lo-')
(Kb=3.8x 1F14 '
ElechPlgtk Condudlam
Equllibtia & k % c h d m i s t r y
SAQ 4
1.
Fill in the blanks :
Conjugate Acid
Conjugate Base
~ ~ o ; y ~
2.
HCOOH
A decimolar solution of acetic acid is 1.33% ionised. Calculate the
degree of dissociation (a).[H'] and the ionisation constant (K,). At the
same temperature, enough solid sodium acetate is dissolved in one litre
of 0.1 M acetic acids so as to make the concentration of the salt equal to
I .O M. What will be the [H+]of the resulting solution ? (Hint a << 1)
14.11 COMMON ION EFFECT
Equilibria involving weak electrolytes are affected by the presence of salts having the
same cation or anion as those derived from the electrolyte. The LeChatelier's
principle (sec. 12.6) predicts that the degree of dissociation of the weak electrolyte
should decrease in the presence of a common ion. This is referred to as the common
ion effect and is very useful in controlling the concentration of ions derived from the
weak electrolyte. For example, one can make use of this effect to decrease the [ O m
available from an aqueous solution of NH, by adding enough ammonium chloride, so
that only the hydroxides of iron aluminium and chromium are precipitated
(Sec.14.16). It is possible to calculate the concentrations of the various ionic species,
if the equilibrium constant for the dissociation of the electrolyte is known.
For example,
With the addition of sodium acetate, the value of [CH3C007 is increased and in
order to maintain the constant value of K, at a given temperature, [H30+]must
decrease and [CH3COOH]must increase. Hence the ionization of CH3COOH(a weak
electrolyte) is suppressed.
Example 14.9
At 298K, the [H+]of a 0.01M solution of an organic monobasic acid (HA) is
1.732 x
M (a) calculate the ionisation constant of the acid at this
temperature. (b) To a litre of this solution of the acid enough of solid NaA is
added, so that the concentration of NaA is 0.02M.What will be the value of
[Ht] after the addition of NaA? Assume that there is no volume change as a
result of the dissolution.
'
~olution:
(a) H A 2
~(1-a)
Since,
H+ + Aac
ac
a c = [ H+ 1 = 1.732 x 1 r 3 , c being 0.01 M ,a = 0.1732
Therefore.
(b)
L
The addition of the common ion A- is expected to decrease the degree of
dissociation. If a' is the degree of dissociation, the dissociation
equilibrium can be represented as
, H A 2 H++Ac ( 1 -a')
,.I
(a'c) ( a ' c + x )
Where x is the concentration of the added common ion
Since, ( a' )' c << a' x
.
the equation can be written as
solving, a' = 0.0178
' c = 0.0178 x 0.01
and so, [ H+] = a
= 1.78 x
The common ion effect is also relevant in the dissociation of weak polybasic or
plyprotic acids. These are acids capable of donating more than one hydronium ion.
For example, %C204, H2C03are weak dibasic acids, whereas H,PQ ,H3As04are
weak tribasic acids. In aqueous solutions, these acids lose protons' in a stepwise
fashion and each step is characterised by a dissociation constant. Thus, in the case of
H3P04, the dissociation equilibria can be represented as
H W - .H~+ + P O ~ -; K,= 1.2%1 r 1 2
The values of the dissociation constants for successive stages of a polyprotic acid
always decrease steadily, as shown in the case of phosphoric acid. This is because of
the difficulty with which a proton must be lost from ionic species like H2POi and
H*- in the second and third stages, respectively. Compared to the loss of a proton
from a neutral molecule, these processes are much more difficult. The [ H+ ] is given
by [ H+1= [ H+ 1from step 1 + [ H+] from step 2 + [ H+ ] from step 3. However, the
[ H+] from steps 2 and 3 are very small as a result of the common ion effect due to
the [ W ] from the first step. It therefore the [ H+] is equal to that from the first step.
It will be noticed that species like H2PO; and H @-can function both as an acid
and a base. Such species are said to amphiprotic.
14.12 ACIDITY AND ALKALINITY OF AQUOIUS
SOLUTIONS
Since the solvent water is capable of losing as well as gaining a porton, there must be
a proton transfer equilibrium as shown below:
The very small electrical conductivity of pure water indicates the presence of ions in
low concentrations. This indicates that the dissociation of water to give H+and O H
ions must be taking place to a small extent. Applying the law of mass action to the
self-ionisation of water, we have
Since most of the water remains undissociated, [ H20 1 may be considered to be
constant, and so, Eqn.(14.22) simplifies to Eqn.14.23
The equilibrium constant K, being the product of ionic concetrations, is called the
ionic product of water and has a value of 1.0 x 1W l4 at 298 K. In any aqueous
solution, equation 14.23 should hold good. Since pure water is neither acidic nor
basic but neutral, [ H+ ] = [ O H ] = 1.0 x 1 r 7 . This condition of,equality is also a
condition for neutrality in aqueous solutions. If [ H+] is greater than 1.0 x
M,
the solution is acidic and if it is less than 1 r 7 , the solution is alkaline.
A convenient scale of acidity and alkalinity. which avoids the use of negative
exponents in expressing [ H+ 1, was introduced by Sorenson. He defined the pH of a
solution as the negative logrithrn of the hydrogen ion concentration (Eqn. 14.24)
In this scale, a pH equal to 7 represents the condition of neutrality. For an acidic
solution of [ H+ ] = 1.0 x 1 r 4,pH = 4 and for an alkaline solution in which
[ O H ] = l . O x 10-~or[H+]=-- Kw - 1.0 x l r 9 , the pH will be 9. This method
IO HI
has the advantage that all states of acidity and alkalinity upto one molar solution can
be expressed by a series of numbers from 0 to 14. Thus it will be seen that, for acidic
solutions, pH < 7, and for alkaline solutions. pH will be > 7 .This kind of p(X)
notation can be applied to express the concentrations of other ions, ionisation
constants and other equilibrium constants. Thus from Eqn 14.23
-log[H+]-log[OH]=14or
pH+pOH=14
(14.25)
Equation 14.25 enables one calculate pOH from pH and vice versa. The same
reciprocal relationship that one finds between [ H+ ] and pH holds in all these cases.
; pK, = 3.70) is a stronger acid than benzoic
Thus, formic acid ( K, = 2.0 x
; pK, = 4.21 ) . Similarly, among bases, ammonia
acid ( K, = 6.3 x
( K, = 1.77 x 1 r 5 ;pK, = 4.75 ) is stronger than anline ( K, = 3.8 x 10;
pK, = 9.42). The pH of solution can be measured using a pH meter (Sec.13.10).
Though the pH scale covers a range of 6 to 14, in very strongly acidic solutions the
pH may be negative and in very strongly alkaline solutions it may be greater than 14.
Example 14.10
Calculate the pH of the following solutions (1) 1.0 M HClO, (2) a solution in
which [ H+] = 3.564 x 1 r 4and (3) a solution in which [ O K ] = 2.0 x 1W3M.
i
Electr~blle
Condudion
Solution:
i
i
pOH = - log[ O H ] = ( -5.301 ) = 2.691
pH= 14- 2.69= 11.31
Example 14.11
(3)
Calculate the [ H+] in the following
(1) a solution of pH = 4.75, and (2) pH = 12.31
Solution:
(1) log[H+]=-pH=-4.75= 5.25
-.
Therefore,
(2)
log [ H+ I = - 12.31= i3.69
Therefore,
[ H+]= 4 . 8 9 ~
M ; [OH-] = 1 . 0 lo-"
~
=O.O~M
4.89 x 10- l3
SAQ 5
1. a) An analysis of water in a lake indicated that [W] = 3.2 x 1 r 5mol/lit.
What is the pH of the water in the lake ? Is it acidic or basic ?
b) Calculate the [
H'] that corresponds to each of the following values of
pH.
i) lemon juice pH = 2.25
ii) pH of milk of magnesia = 10.50
The ionic product of water is 1.0 x 1W14
2. a) Given that K, of acetic acid is 1.8 x 1w5,calculate the pH of 0.15 M
solution of sodium acetate.
b) If the dissociation constant of ammonium hydroxide is 1.8 x 1v5. What
will be the pH of 0.15 M solution of NH4N03?
Equllibrln & ElortroehmWr~
1
14.13 BUFFER SOLUTIONS
1
A careful control of pH is important in life processes, in industry and in certain
analysis. For this purpose, suitable buffer solutions are used. When reasonable
amounts of an acid or an alkali are added to an aqueous solution of ammonium
acetate or a mixture of ammonium chloride and ammonium hydroxide, there is
practically no change in the pH of the solution. The resistance that a solution exhibits
to changes in pH upon the addition of acid or alkali is called buffer action and
solutions exhibiting this property are called buffer solutions. Buffer soultions usually
consist of (1) a weak acid (acetic acid) and its conjugate base (acetate ion), or (ii) a
weak base (ammonia) and its conjugate acid (ammonium ion), or (iii) a weak base
(ammonia) and a weak acid (acetic acid). Buffer solutions are used mainly for
resisting changes in pH or for providing solutions of known pH.
The buffer action will be explained using two examples. Consider the buffer solution
consisting of HOAc and NaOAc. If H+ions are added, these are removed by OAc- to
form the unionised acid.
If O H ions are added, these are removed by the reaction with the unionised HOAc to
give OAc- ions.
HOAC+ OH- -+
H20 + OAC-
The pH of the solution thus remains practically constant. In the case of a buffer
solution consisting of NH3 ( NH40H ) and NH4C1the following reactions
account for the buffer action. In a similar fashion, if a buffer solution of ammonium
acetate is used, the buffer action can be explained as follows
H++ OAC- (from buffer) -+
O H + NH: (from buffer) -+
HOAc
NH3 + H20
The pH of a buffer solution can be calculated by considering the effect of adding a
common ion (Sec.14.11) to the weak acid or the weak base equilibria. Consider a
buffer solution consisting of a weak acid (HA) and its salt like NaA. This is a special
case of the ionisation of the acid in the presence of a common anion, A-. The
dissociation constant K, of the acid is given by Eqn 14.19
If the initial concentration of the acid is c, and that of the salt NaA is c, and a is the
degree of dissociation in the presence of NaA, the equilibrium concentrations of the
various species can be expressed as follows.
[H+]=c,a, [A-]=(c,a+c,)and [HA]=c,(l - a )
Substituting these values in Eqn 14.19, we have
As explained in Example 14.9, a < < 1 and so c, ( 1 - a ) G c,. Considering the [ A- 1,
it will be realised that it is only the added salt NaA, that is the major source of A- in
solution, since [ A- ] available from HA will be very small as a result of the common
ion effect. So c, a <<c, or ( c, a + c, ) c,. Thus Eq. 14.26 can be written as
=
j
i
4
!
I
1
i
K. = [*I
[ Salt ]
[Acid]
The above equation can be rearranged to give Eq. 14.27
=K,W
[~ t . 1
[ Salt ]
Taking logarithms and changing the sign throughout, we have
- l o g [ ~ ] = - l o g ~ . - lAcid
og~or
[ Salt ]
pH = pK, + log [ Salt ]
[ Acid ]
Equation 14.28, known as the Henderson equation, is useful in calculating the pH of
buffer mixtures and also the amounts of the buffer components required to prepare
the buffer solution. A similar equation for the system B - BIF) ( NH,, - N@ ) is
obtained from the expression for the ionisation constant of the base .(Eq.14.20)
I
b
As in the case of A
:i - A- system, invoking the approximations, [ BH' ] 5 [ Salt 1, and
[ B I e [ Base 1. we have
'1
K ~ = [ O H I( S l l r l o r
[ Base ]
=K,W
[ Salt 1
Taking logarithms and changing the sign throughout equation 14.29, we get
I
Buffer solutions show buffer action only if limited quantities of H+or O H ions are
added. The buffer capacity of a mixture may be regarded as the amount of acid or
alkali that a buffer solution can use up without suffering a change in pH. It is found
[Salt] or [Salt] is
that buffer solutions have their greatest capacity when the ratio
[acid] [base]
equal to unity and that the concentrations used are about 1.OM. For satisfactory buffer
action or if the pH is to remain constant within f 1 ,these ratios must lie between 10
and 0.1: Thus, the useful ranges of buffer solutions lie between pH = pK, f 1 or
pOH = p Kb I 1. For example, p K, of acetic acid is 4.75 and the useful range of the
acetic acid-sodium acetate buffer is 3.75 to 5.75. Similarly, for NH, - NH4 C1 buffer
mixture, the useful range is 8.25 to 10.25.
Example 14.12
A solution is prepared by dissolving 0.20 mole of sodium formate and 0.25
mole of formic acid in water and making the total volume upto 250 ml exactly.
The ionisation constant of formic acid is 2.0 x
Calculate (i) the pH of this
solution, (ii) the change in pH by adding 1.0 ml of concentrated HCl
(normality = 10.0) to 250 ml of this solution, and (iii) the change in the pH by
adding 1.0 ml of 10 N NaOH solution to another 250 ml portion of the buffer
solution.
Solution:
i,
pH = pK, + log [ Salt ]
[ Acid 1
The pK, of formic acid = -log ( 2 x 1V ) = 3.70
[Salt1 =
Om2 'O0O = 0.8 moll lit
250
ElectrolyUcConduction
Equilibria & J3kIroehemLFtry
Similarly, [Acid] = Oe2'
lMX)
= 1.0 molPit
250
Therefore, pH = 3.70+ log @! = 3.60
1.o
(ii)
no. of moles of HCl added = 1.0 ml x
lit x 10 mol lit1000ml
'
= 0.01 mol
[HCl] added = 0.04 momit ( Since, .O1 mol is contained in 250 ml)
The added HCI reacts with the conjugate base HCOO- as
HCOO- + H+ ---4 HCOOH
Hence the concetration of HCOOH resulting from this reaction is equal
to the concentration of HCl added.
Thus [HCOOH] = 1.0 + 0.04 + 1.04 mol/lit.
This reaction between HCOO- and H+ also decreases the concentration
of the salt. sodium formate by 0.04 momit.
[Salt] = 0.8-0.04 = 0.76 molllit
Therefore, pH after the addition of HCl = 3.70 + log-076,- 3.56
1.04
(iii)
Therefore, Change in pH is only by 3.56-3.60 or -0.04 units. The pH
decreases by 0.04 units.
no.of moles of NaOH = 0.01
-
[ O H ] added = 0.04 momit
The added O H reacts with HCOOH as shown below.
O K + HCOOH
0.04 momit
%O + HCOO-. Thus the [HCOW increases by
Therefore, [ HCOO- ] = [Salt] = 0.8 + 0.04 = 0.84 molPit
Simultaneously [HCOOH] decreases by 0.04 molpit
[HCOOH] =1.0 - 0.04 = 0.96 momit
0.84 pH after adding the alkali = 3.70 + log - 3.64
0.96
Therefore. change in pH = 3.64 - 3.60 = 0.04
The pH of the solution increases by 0.04 units
Now, let us define hydrolysis and consider different types of salts formed by
combinations of acids and bases. The details of hydrolytic equilibria will be
considered in the next section.
Hydrolysis of Salts
Definition :
The hydrolysis of a salt is defined as the interaction of the ions of a salt with either
the H+or the O W ions of water to form a basic or an acidic solution.
If the ions of the salt interact with O W of water, the salt is said to be hydrolysed. The
resulting solution is acidic. If the ions of the salt, dissolved in water react with H+
ion of water, the salt is said to be hydrolysed. The resulting solution is basic. Here,
we sunmarise the formulae used and derive them in the next section.
Types of salts
Based on this definition of hydrolysis, thc salts arc classified as follows into four
types :
(i)
Salts of strong acids and strong bases
eg. NaCl
NaCl is the salt of a strong acid HCI and the strong base NaOH.
NaCl
2 Na++Cl-
(ii) Salts of weak acids and strong bases :
eg. CH3COONa
sodium acetate is formed by the combinalion of CH3COOH, a weak acid and NaOH,
a strong base.
H20 -+
H++ OH-
The acetate ions interact with the H+ ions to Torn1 uniot~iscdor wcakly ionised acetic
acid. These salts are hydrolysed and ihc rcsul ling sol 111ion is basic.
(iii) Salts of strong acids and weak bases :
eg. NH4Cl
Ammonium chloride is formed by thc inrcrnc~ionof thc strong acid HCI and weak
base, NH,OH.
H20 --+
H++ OH-
NH; + OH- ---+ NH,OH
The ammonium ion interacts with the hydroxyl ion to form ammonium hydroxide, the
salt is hydrolysed andathe resulting solution is acidic.
(iv) Salts of weak acids and weak basis
eg. CH,COONH,
Ammonium acetate is formed by rhc in~craclionol' ammonium hydroxide, a weak
base and acetic acid.
NH: + OH- ---+ NH40H
Acetate ions interact with. H+ ions lo for111 aci~icacid and hydroxyl ions and
Ilydroxidc. TIlc salt is hydrolysed and
ammonium ions interact to form ammonii~n~
!he resulting solution is neutral.
14.14 HYDROLYTIC EOUILIBRIA
Salts are formed by the neutralization of an acid by an cquivalcnt amount of a base or
vice versa. However, aqueous solutions of salts likc sodium acetate, sodium cyanide,
ammonium chloride, etc., are not neutral and ~ h pll
c is nor cqual to 7.0 in these cases.
Electrochemistry
Only in the case of salts formcd from strong acids and bases, the aqueous solutions
are neutral. An aqueous solution of sodium acetate has a pH greater than 7.0
(alkaline). whereas that of ammonium chloride has a pH less than 7.0 (acidic). In the
case of sodium acetate, thc acctate ion present in solution is a strong base and
abstracts protons from water as shown below
OAc-+ H20 2 HOAc + O H
The hydroxide ions formed in the reaction are responsible for the solution having a
pH > 7.0. In the case of ammonium chloride, the N q being the conjugate acid of a
weak base, is capable of donating a proton to water to give rise to H30+ions.
The formation of H,O+ explains to acidic nature of the solution. This interaction of
ions of the electrolyte leading to Lhe abstraction of a proton from the solvent, or the
donation of a proton to Lhc solvcnt is called hydrolysis. It will be noticed that only
strong conjugate acids and bases get hydrolysed in aqueous solution. Weak conjugate
bases like Cl- ,NO; do not undcrgo hydrolysis. This interaction leads to a decrease in
the concentration of a conjugatc bases like cG- ,s2-,C2 @- etc., which are
normally uscd to prccipitatc cations. Hcnce ways and means of preventing hydrolysis
are important.
The extcnt to which thc hydrolysis rcaclion proceeds is determined by the equilibrium
constant, called the hydrolysis constant (K, ) of the salt undergoing hydrolysis. From
Kh it is possible to calculate the pH of such aqueous solutions. Consider the salt of a
wcak acid (HA) and a strong basc (MOH), for example NaOAc. The anion, Aundergocs hydrolysis as
.
The reaction can be considcred as hydrolysis of the salt MA or dissociation of the
conjugate base, A-. For this equilibrium, the hydrolysis constant (Kh ) can be written
as Eqn. 14.29 in which [ H20 ] is considered as a constant.
K, = [ H A ] [ O H ]
[A-I
The above'cquation can bc written as
[ H A ][ H I ] [ o K ~ --Kw
(14.30)
[H+l[A-l
Ka
From the hydrolysis equilibrium, it is sccn that [ HA ] = [ OH-] and if Kh is rather
small, the extcnt of hydrolysis will also be small. Thus the equilibrium concentration
of A- can bc taken to bc equal to 111coriginal concentration of A' i.e., Equation 14.29
can bc writtcn as Eqn. 14.31
Kh-
Since, [ H+] [ OH- ] = K,, Eqn. 14.31 becomes
From equations 14.30 and 14.32, we get
Taking logarilhms and multiplying by -1 throughout, we get
1
1
1
pH=-pK,+-pK,+-1og.c
2
2
2
Similarly, for the salt (BHX) of a weak base (B) and a strong acid (HX) like NH,Cl.
the hydrolysis reaction can be written as
The hydrolysis constant K,,can be writlen as follows.
Since [ B ] = [ H30+] and [ BH+ ] s c ,one can wrilc
From Equations 14.35 and 14.36 we get
For the salt (BHA) of a weak acid (HA) and a weak basc(B), thc hydrolysis reaction is
The hydrolysis constant is
Considering the dissociation equilibirum of HA
since [A-] = c, Eqn (14.40) can be written as
Substituting for [HA] in Eq. 14.39, we get Eq. 14.40
If a is the degree of hydrolysis, Eq. 14.29 and 14.35 can be written interms of a as
eqn. 14.42.
Here K represents K, or K,, as the case may be. If a << I,Eqn. 14.42 simplifies to
Eqn. 14.43.
Electrolytic Condudlon
From Eq. 14.43 ,it will bc seen that extent of hydrolysis increases as (i) K, increases,
i.e., as the temprature increases, (ii) as K decreases, i-e.. the weaker the acid or the
base, and (iii) as c decreases i.e., as the solution gets diluted. Thus it is seen that at a
given temperature, the degree of hydrolysis (a)can be decreased if a high
concentration of the salt is used. For a salt of the type BHA, if is the degree of
hydrolysis, we can write the equilibrium as
~(1-PI
From Eqn. 14.38
~(1-PI
cP
cP
.
it will be seen that K, will be large since K, . Kb a product of two small quantities, is
rather small. This means that hydrolysis will take place extensively and is
independent of concentration. At higher tempratures, the extent of hydrosis will be
quite considerable.
Practically, all the cations except those of alkali and alkaline earth metals undergo
hydrolysis in aqueous solution to give acidic solutions as shown in the case of salts of
~ 1 ,
+
and so on. As in the case of other equilibria, it will be seen that the hydrolysis
reaction can be supprcsscd by adding one of the products of the equilibrium. In this
case, the hydrolysis reaction can be suppressed by adding acids. Thus, while
preparing aqueous solutions of salts of polyvalent cations, it is advisable to add acids
to prevent hydrolysis.
Example 14.13
Calculate (i) the hydrolysis constant (K,) (ii) the degree of hydrolysis (a)and
(iii) the pH of a 0.1M solution NH,CI. The ionisation constant of NH3 ,K,,, at
this temperature is 1.80 x 10-5. What will be the degree of hydrolysis if 1.OM
solution of NH,CI is used? The ionic product of water is 1.0 x 10- l4 at this
temperature.
Solution:
The hydrolysis reaction is
NH: + H,O 2 NH, + H30+
From Eqn. 14.35,
,
K, = a2
a2c ( Because, a h rather small)
(1-a)
I
Elecbolytic Conduction
iii) From Eqn. 14.37,
14.15 PRECIPITATION EOUILIBRIA
In chemical analysis, precipitation is an important process. In qualitative analysis, the
separation and identification of the cations and anions is bascd on precipitation
reaction. In quantitative analysis, one should make sure whcthcr the desired
constituent is quantitively precipitated or not. An understanding of thc solubility
product, which is based on the application of thc law of mass action, is t h e f o r e
quite essential.
When a saturated solution of an electrolytc is in contact with thc solid solute, there
exist an equilibrium between the ions in the solution and the solid. Consider such an
equilibrium between MA and its ions M+ and A-.
Applying the law of mass action to this hctcrogcncous systcm in cquilibriurn. we can
write the expression for the equilibrium constant, K.
In a saturated solution, [MA(s)] can be considerd to be a constant, since it is not
likely to change as a result of the dissolution proccss. Hcncc Eqn.14-44 can be writte'n
as Eqn 14.45
Ksp=[M+1LA-]
(14.45)
Whatever has dissolved is assumed to bc cxisting as ions. The ionic product
represented by Eqn. 14.45 is called the solubility product and applies to any
saturated solution. However, it is of practical inlcrcst only in lhc case of sparingly
soluble salts.
The solubility product expression (Eqn. 14.45) being an equilibrium constant,
depends on the stoichiometric coefficients of thc ionic spccics, is., the formula of the
spearingly soluble electrolyte. For a 1.1 (AgC1) or 2.2 (BaSO, ) or $3 ( Fe PO, )
electrolytes of the formula MA, the expression for K,, can bc writtcn as follows. Let
S be the solubility in water in mol/dm3 or molllit of thc salt, at a given temperature.
Since whatever is dissolved
exists as ions, the ionic concentration arc
[Mn+]=[An-]=S
Therefore,
K,=[M"+] [A"-]=s'
It is assumed that MA(s) is the only source lor thc ions Mn+and An-. .
(14.46)
For a 1.2 (Ag2Cr04) or 2.1 (CaF, ) electrolyte, the equilibria can be expressed as
M2A
2 2 W + A%
MA,
3 ~ ' + + A-2
and
respectively. The solubility product constants for h$ A and M 4 are given by
equations 14.47 and 14.48 respectively
K , = [ M + ] ~[A2-]
(14.47)
[M"] [ A-1'
(14.48)
K,=
If S is the solubility in moVdm of the salt M2A, then [ M+] = 2s and [ A'- ] = S.
Substituting these values in equation 14.47 we get Eqn. 14.49 as the expression for
the solubility product of M, A
K~~=[M']'[A~-]=(~S)~(S)=~S~
(14.49)
Similarly for M A,, if X is the solubility in movdm3 ,
[ M2 + ] = X and [ A- ] = 2 X.The expression for the solubility product of MA, is given
by Eqn.14.50
[ M 2 + ][ A - ] ~ = ( x ) ( ~ x ) ~ = ~ x ~
(14.50)
Example 14.14
Write down the solubility product in terms of the solubility (S movdm3 ) of
the elecirolyle Sb, S3
Solution:
Sb,S3(s)
2 2sb3++3s22s
3s
K,,=[Sb 3 + I 2 [ s2-13o r ( 2 ~ ) ' ( 3 S ) ~
= 108 s5
14.16 APPLICATION OF THE SOLUBILITY PRODUCT
14.16.1 Solubility and Ksp
The solubility product is a measure of the solubility of the salt in water at a given
temperature. From equations 14.46, 14.49, 14.50 and Examplel4.14, it will be
realised that for the sake of comparision one should consider b e IZ,values of salts of
similar formulae. Thus Agl (K,, = 1.0 x 10- 16)is less soluble than AgCl (
KT= 1.1 x 10-lo). Also, Ca F2( K,, = 3.2 x 10- l1 ) is less soluble than
Mg F2 ( K, = 7.0 x 10- ). Howcvcr, instead of K,,, one can also compare the molar
solubilities (S) of salts. Thus Ag2CrQ ( K, = 1.7 x 10' l2 and S = 7.5 x 10'' M ) is
more soluble than Ag CI ( KT= 1.1 x 10-''and S = 1.05 x 10"M ).
14.16.2 The Effect of added electrolytes on the equilibrium
between the sparingly soluble electrolyte and its constituent ions.
The added electrolyte, irrespective of its nature, influences this equilibrium. When
the added eleclrolyte has an ion in common with the salt MA, we have b e common
ion effect (sec 14.11). which is relevant to the precipitation equilibria. The constant
K,, defines the maximum concentration of the ions in solution bat can remain in
equilibrium wiih ihe sparingly soluble salt. Consider the effect of adding
H+C1-to Ag C1 ( s ) .
2
AgCl ( s )
Ag++ ClAs indicated earlier, the [ Cl- ] in equilibrium with AgCl(s), in the absence of any
other source.of Cl- ,is approximately 10-5 M. If [ Cl- ] exceeds this value, [ Ag' 1
also should decrease. (Le Chatalier principle). In other words, the solubility of the
sparingly soluble salt will decrease in the presence of a common ion, or the added Ct
combines with Ag+ to form AgCl(s). Thus, we can say that precipitation will occur if
the product of the concentrations of the ions (ionic product) exceeds the solubility
product. This is the condition for precipitation to occur. It will also be seen'that if the
concentrations of one or both the constituents are decreased either by formation of a
complex or a weakly ionised substance, th solubility product equilibrium is
disturbed. Hence more of the salt dissolveshQis is the basis for the dissolution of
AgCl in NH,. Here Ag' forms a complex [ Ag ( NH, )2 ]+and if the [ NH, 1 is quite
high, practionally all the Ag' is converted into to the complex. Hence, more of AgCl
dissolves in order to restore the equilibrium condition. Ultimately, all the solid can be
dissolved in NH,. This can be represented as
AgCl (s )
2 Ag++ Cl-
14.16.3 Applications in Qualitative Analysis
Among the sulphides. those of Hg, Cu, Pb, Cd, Bi. As, Sb and Sn are less soluble
than those of Co, Ni, Mn, and Zn. Hydrogen sulphide is used as a source of sulphide
ions to precepitate these sulphides. Being a weak electrolyte, its overall dissociation
can be represented as
,
k
)
H2S
2 2H++s2-
constant
For a given concentration of dissolved H2 S. we can write [ s2-1 = [ H+ I .Thus,
.
we decrease [s2-] by increasing [ H+] so that only the less soluble sulphides of
group II get precipitated. Once these are removed, the other cations can be
precipitated by increasing [ s 2 - ] available from H,S by making the medium very
weakly acidic or alkaline. Thus h e group IV sulpliides are precipitated by passing
H2S into an arnmoniacal medium containing the cations of his group.
Similarly. by decreasing [ O K ] available from a weak electrolyte like aqueous
NH3 ( NH40H ) by adding NH4CI, it is possible to precipitate the less soluble
hydroxides of Fe. Al and Cr in group 111. Under these conditions. the hydroxides of
Mg. Ni, Mn, Co and Zn are not precipitated.
Example 14.15
The specific conductance of conductivity water, at 298 K, is
1.61 x 10d ohm1cm-'. When saturated with pure AgCl(s). its conductivity
' '.The ionic equivalent conductances at
increases to 3.41 x l ~ ~ o h mc-minfinite dilution of Ag+and Cl- in ohm- cm2 are 61.9 and 76.3 respectively.
Calculate (i) the solubility of AgCl(s) in moll lit. (ii) the solubility product of
AgCl(s) and (iii) the solubility of AgCl in 0.001 N NaCl solution.
Solution:
'
As indicated earlier (Sec 14.7), A G A' in this case.
h 0 ( ~ g ~ 1 ) = 6 1 . 9 + 7 6 . 3138.2ohm-'
=
cm2
Where S is the solubility of AgCl(s) in cquiv / lit or moll lit.
lr6=
1 . 3 0 2 ~l(r%ovlit
Thercforc, S = lWOx
138.2
ii)
Ksp= [ Ag' ] [ Cl- ] = s2(Eqn. 14.46)
=( 1 . 3 0 2 ~lo-')'
iii)
In the prcsencc of 0.001 NaCl, the [ Cl- ] available from NaCl is 0.001
and this is >> [ Cl- ] available from AgCl, in the absence of a common
ion. One would expect the [ Cl-] available from AgCl(s) to be
decreased by h e c o m n ion effect-(Sec 14.4.31). Lets' be the mhbi
of AgCl(s) in the prescnce of added C t . One would expect S' << S.
Ag Cl ( s ) $ Ag'
+ Cl-
s
I
(S+O.001)
s'(s'+0.001)= 1 . 7 0 ~10-lo
S' is rather small and so S' x S' will be <<0.001 S'
Therefore, 0.001 S' = 1.70 x 1
moll lit
s'=1 . 7 0 ~l w 7 mol/lit
The value of S in thc absence of C1- is 1.302 x 10" momit
Example 14.16
To a solution containing 0.01 mol of Fe3' and 0.01 mol of ~ g " and
' 2 mol of
NH, C1 in a litrc of the solution, enough ammonia is added with stirring till the
concentration of NH, is 2.OM. The values of KPfor Fe ( OH ) and Mg ( OH )2
are 3.8 x
and 3.4 x 10- respectively. Which cation gets precipitated?
The dissociation constant for thc rcaction,
,
NH, + H,O
2 N@ + O H
is 1.8 x lo-5 at this Lcmpcraturc. Assume that there is no volume change.
Solution:
In order to find out which cation gets precipitated we have to determine the
ionic products for cacll hydroxide and find out which product is greater than
the solubility product.
Fe(OH),
F ~ ~ ' + ~ o; H
K,= [Fe3'] [ 0 H 1 3
2
The conccnlrations of' [ FC, ] and [ ~8'] are 0.01M each. In order to find [ O H ]
we have lo considcr thc dissociation equilibrium of NH,, that is the source of O H
+
[
1 in solution = [ NH: ] from added NH, C1+ [ NK ] derived from.added NH3.
= [ NH: ] from NH, Cl (since the other quantity is very small)
= 2.0 M
NH:l
1 . 8 lvs=
~
[ O H ] --2 . 0 ~[ O H ]
2.0
[ NH, I
:. Thcrcforc, [ O H ] = 1.X x lo-'
The ionic products arc
(i)
[ ~ e ~ [' o] H ] ~ = o . o (~ 1 . 8 1
~0 . ' ) ~ = 5 . 8 3 2 ~ 1 0 - "
(ii)
[~~~+][0~]Z=0.01(1.8~10~~)'=3.24~1~'~
The ionic product in the case of Fe ( OH ) viz.,
,
5.832 x 10- I' is >> K, ( 3.8 x 1r3*
) and so it is Fe ( OH )3 that gets precipitated.
In the case of Mg ( OH ),. the ionic product ( 3.24 x 10- l2 ) is less than its
K, ( 3.4 x 10-I'). and so. it is not precipitated.
SAQ 6
1.
The solubility of Mg F2 at room tempcraturc is 0.0012 moll lit
Calculate the solubility product of Mg F,. What will be the solubility of
Mg F2 in 0.1 M MgSO, ?
2.
The solubility product of lcad chloridc ( PbC12) is 1.6 x 1r5.
If 500
ml of 0.03 M NaCl is mixed with 500 ml of 0.3 M lcad nitrate. will there
be a precipitation of PWI, ?
14.17 SUMMARY
Eleclrolytes, either in the molten state or in aqueous solution. conduct electricity and
undergo chemical reactions at the elcctrodcs. This process is callcd electrolysis.
Faraday's laws of electrolysis summarise thc quantitative rclationship between the
quantity of electricity (ampere-seconds or coulombs) passcd and the extent of
chemical reaction that takes place. One Faraday, equal to 96500 C, is Lhe quantity of
electricity required to deposit or dissolve onc cquivalcnt of a substance.
If instead of a direct current. an altcmating currcnl is uscd. it is possible to measure
the resistance (conductance) of electroryrrz conductors. In the casc of aqueous
solutions of electrolytes, the conduclivily dcpcnds on conccnlration. For a
comparison of conductivities. equivalcnt conductanccs arc q u itc useful. The variation
of specific and equivalent conductances with diluiio~lof aqucous solution of
electrolytes could be accounted for by lllc Arrhenius thcory of electrolytic
dissociation. The Debye-Huckel theory of intcriot~icartraclion provides a more
reasonable explanation of this bchaviour. Kohlrausch's law of independent
ion-conductances at infinite dilution cnablcs oric lo calculate thc cquivalent
conductance at iflinite dilution of wcak clcctrolytcs and sparingly soluble
electrolytes. The change in the conduclancc wilh concentralion forms the basis for
carrying out acid-base and precipitation titrations conductometrically.
An important application of conductomclry is thc calculalion of the degree of
dissociation of weak acids and bascs. By applying Ihc law of mass action to equilibria
involving weak acids and weak bascs, onc can calculalc rhcir ionisation constants.
Similarly, one can calculate the solubilities and solubiliry products of sparingly
Equlllbrlr & Ekdroehaaldq
soluble salts. The term "common-ion effect" refers to the influence of electrolytes,
having the same cation or anion as those derived from the weak acids, weak bases or
sparingly soluble electrolytes, on equilibria involving such species.'
Systems involving weak acids (bases) in the presence of common ions show buffer
action. They are useful in resisting the changes idpH of such a buffer system. The pH
can be calculated using the Henderson equation. The conjugate b$es of weak acids
are strong enough to abstract a proton from water which results intan alkaline
solution. Similarly, the conjugate acid of a weak base is a strong acid, capable of
donating protons to water. The interaction of the ions of a salt with either the Ht or
the OW ions of water is called hydrolysis. The solubility product principle, common
ion effect and hydrolysis find extensive applicatio& in chemical analysis.
14.18 GLOSSARY
: Chemical decomposition caused by the passage of
elcctricity.
Electrolysis
Faraday
: The charge contained in one mole of electrons.
Electrolytic conduction : The movement of ions of an electrolyte under the
influence of an applied EMF.
Conductivity water
: Watcr obtained by repeated distillations, where
conductivity is only due to the conductivities of H+and
OH- ions.
a2c
: K =where K = equilibrium constant, a = degree
Oswalds dilution law
(1 -a)'
of dissociation and c, the concentration of ttle
electrolyte.
: The theory wherein the molecules of acids, bases and
salts arc dissociated into constituent ions.
Arrhenius theory
A, = A", + Loa. where. the terms refer to the equivalent
conductances of the solution. cations and anions
respectively at infinite dilution.
~itratibnswherein the conductance of the electrolyte
: solution is measured as a function of the volume of the
titrant added.
Kohlrausch's law
:
Conductometric
titrations
Acids & bases
: An acid is a substance capable of accepting electrons,
while a base is a molecule capable of donating electrons.
:
- log [H']. The Henderson equation pH = pKa +'
is useful for calculating the pH.
[Acid]
Solutions of a weak acid and its salt or a weak base and
its salt whish resist changes in pH when acids or bases
are added to them.
Thc interaction of the ions formed from a salt with
or
OH-ions of water to yield basic or acidic solutions.
log
Buffer solutions
:
Hydrolysis
:
Solubility product
:
In MA $ Mt + A-, KSp =
14.19 ANSWERS TO SAQs
SAQ 1
40H- ---+
02
+ 2H20
1-21 hrs. = 5400 scc.
[m]
[A7
4 x 96500 Coulombs liberate = 32 g of 0,
2 x 5400 Coulombs liberate = 32 x 5400x 2
96500, g
nRT
Volume of O, V = -
P
SAQ 2
1)
2)
H, at cathode and 0,at anode
3.177 Cu is deposited at cathode and 3.177 g of Cu is dissolved from
anode
SAQ 3
1)
2)
SAQ 4
1)
Conjugate base :s@- and HCOOConjugate base :NH; and Hz PO,
2)
a=0.0133; [ ~ + ] = 1 . 3 3 x 1 0 - ~ ; ~ = 1 . 7 7 x 1 O[-~~+; ] = 1 . ? 7 x l O - ~
SAQ 5
1.
a) pH = 4.49. Water in the lakc is acidic
b)i)5.6x1w3 ii)3.2~10"'
2.
a) 8.97 b) 5.03
SAQ 6
1)
-2)
'
Yes ,Since ion product [ pb2 1 [ C1- ] = ( 0.15 ) ( 0.015 ) is
' greater than Kip.
+
FURTHER READINGS
1.
James E Brady and John R Holum, Fundamentals of Chemistry, John
Wiley and Sons. (1981 and later editions)
2.
Leonard W Fine and Herbert Beall, Chemistry for Engineers and
Scientists, Saunders College Publishing (1990)
3.
J.C Kuriacose and J. Rajaram. Chemistry in Engineering and
Technology Volumes 1 and 2, Tata McGraw-Hill Publishing Company
Ltd., (New Dclhi (1988)
4.
Atkins, P.W., Physical Chemistry, ELBS and the Oxford University
Press. 5th Edition (1993)
Caslcllan Gilbert, W.; Physical Chemistry, Addison-Wesley, Reading
(Mass). 5th Edition.
5.
6.
Cottrell A. H., Introduction to Metallurgy, Edward Arnold, London.
7.
Daincls Farrington and Robert A. Alberty. Physical Chemistry, John
Wilcy, Ncw York.
8.
Ferguson F.D. and Jones T.K. Phase Rule. Butterworths, London.
9.
Findlay Alcxandcr; Phase Rule and its Applications, Dover
Publications, New York.
Glastonc Samual and David Lewis; Elements of Physical Chemistry;
Macmillan, London.
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11.
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