Chapter 7 Solutions of Electrolytes

1
Chapter 7 Solutions of Electrolytes
Electrical Units (????)
The electrostatic force (???) F between two
charges Q1 and Q2 separated by a distance r in a
vacuum is
The SI unit of charge Q is the coulomb (??),
C. The constant ?0 is the permittivity of a
vacuum (??????) and has the value 8.854?10-12 C2
J-1 m-1.
If the charges are in a medium having a relative
permittivity (??????), or dielectric constant
(????), of ?, the equation for the force is
For example, water at 25 ?C has a dielectric
constant of about 78, with the result that the
electrostatic force between ions are reduced by
this factor.
The electric field (??) E at any point is the
force exerted on a unit charge (1 C) at that
point. The field strength at a distance r from a
charge Q, in a medium of dielectric constant ?,
is thus
? is electric potential (??)
F newton, NJ m-1 E N C-1J C-1 m-1V m-1 ?
volt, Vkg m2 s-3 A-1
2
7-1 Faradays Laws of Electrolysis (???????)
The mass of an element produced at an electrode
is proportional to the quantity of
electricity (??) Q (SI unitcoulomb, C) passed
through the liquid. The quantity of
electricity is defined as equal to the current I
(SI unitampere, A) multiplied by the time t
(SI unitsecond, s) QIt The mass of an
element liberated at an electrode is proportional
to the equivalent weight (??) of the
element. Faradays law is independent of P, T,
electrode materials, and electrolyte solution.
CAs
Note
1mol??????????1mole?1/2H2, 1mole?1/2Cu, 1mol?Na?
3
The proportionality factor that relates the
amount of substance deposited to the quantity
of electricity passed through solution is know as
the Faraday constant and given the symbol F.
In modern terms, the charge carried by 1 mol of
ions bearing z unit charges is zF.
According the latest measurements, the
Faradays constant F is equal to 96485 C
mol-1. In other words, 96485 C will liberate 1
mol of Ag, 1 mol of 1/2H2 (i.e., 0.5 mol of
H2), etc. If a constant current I is passed
for a period of time t, the amount of substance
deposited is The charge on one electron is
4
Example 7.1
An aqueous solution of gold (III) nitrate,
Au(NO3)3, was electrolyzed with a current of
0.0250 A until 1.200 g of Au (atomic weight
197.0) had been deposited at the cathode.
Calculate (a) the quantity of electricity passed,
(b) the duration of the experiment, and (c) the
volume of O2 (at NSTP) liberated at the anode.
The reaction at the cathode is 3e-Au3?Au or
e-1/3Au3?1/3Au3 The equation for the
liberation of O2 is 2H2O?O24H4e- or
1/2H2O?1/4O2He-
Solution
a.
b.
c.
New standard temperature and pressure, viz. 25.0
?C and 1 bar at NSTP 1 mol of gas occupies 24.8
dm3.
5
Problems 7.3
Electrolysis of molten KBr generates bromine gas,
which can be used in industrial bromination
processes. How long will it take to convert a
500.00-kg batch of phenol (C6H5OH) to
monobromophenol using a current of 20000 A?
Solution
The chemical reaction involved are
Two moles of electrons are involved in the
generation of each mole of bromine gas, which
reacts with one mole of phenol. Each batch
consists of 500.00 kg or 5313 moles of phenol
(molecular weight 94.114 g mol-1), which
requires 5313 moles of bromine, or 10626 moles of
electrons for the reaction. Therefore, since
ItnF, where n is the numbers of electrons
exchanged in the reaction, we get
6
7-2 Molar Conductivity (?????)

• Nonelectrolyte - solution can not dissociate into
  ions, e.g. sucrose.
• Electrolyte - solution can be dissociated into
  ions, e.g. NaCl, CH3COOH
• Strong electrolyte - greater extent of
  dissociation , e.g. NaCl, CuSO4, HCl
• Weak electrolyte - weak extent of
  dissociation, e.g. CH3COOH, NH3

Figure 7.1a
According to Ohms law (????), the resistance R
of a lab of material is equal to the electric
potential difference V divided by the electric
current I
The SI unit of potential is the volt (V) and that
for current is the ampere (A). The unit of
electrical resistance is the ohm(kg m2 s-3
A-2), given the symbol ?.
The reciprocal of the resistance is the
electrical conductance (??) G, the SI unit of
which is the siemens (S?-1). The electrical
conductance of material of length l and
cross-sectional area A is proportional to A and
inversely proportional to l
The proportionality constant ?, which is the
conductance of a unit cube (see Figure 7.1a), is
known as the conductivity (???) for the
particular case of a solution of an electrolyte,
it is known as the electrolytic conductivity
(?????). Its SI unit is ?-1 m-1 (S m-1), but
the unit more commonly used is ?-1 cm-1 (S
cm-1).
7
The electrolytic conductivity is not a suitable
quantity for comparing the conductivities of
different solutions. If a solution of one
electrolyte is much more concentrated than
another, it may have a higher conductivity simply
because it contains more ions. The concept of
what is now called the molar conductivity and
given the symbol ? is employed. It is defined as
the electrolytic conductivity ? divided by
concentration c,
The unit of ? is ?-1 cm2 mol-1S cm2 mol-1
Suppose that we construct a cell having parallel
plates a unit distance apart, the plates being
of such an area that for a particular solution
of 1 mol of electrolyte (e.g., HCl, 1/2CuSO4)
is present in it (see Figure 7.1b). The molar
conductivity ? is the conductance (i.e., the
reciprocal of the resistance) across the plates.
We need not actually construct cells for each
for which we need to know the molar
conductivity, since we can calculate it from
the electrolytic conductivity ?. If the
concentration of the solution is c, the volume of
the hypothetical molar conductivity cell must
be (1 mol)/c in order for 1 mol to be between
the plates (see Figure 7.1b). The molar
conductivity ? is therefore the electrolytic
conductivity ? multiplied by 1/c.
Figure 7.1b
8
Problem 7.7
The quantity l/A of a conductance of cell is
called the cell constant. Find the cell constant
for a conductance cell in which the conductance,
G, of a 0.100 M KCl solution is 0.01178 S at 25
?C. The molar conductivity for 0.100 M KCl at 25
?C is 128.96 S cm2 mol-1. If a 0.0500 M solution
of an electrolyte has a measured conductance of
0.00824 S using this cell, what is the molar
conductivity of the electrolyte?
Solution
Example 7.2
The electrolytic conductivity of a 0.1 M solution
of acetic acid (corrected for the conductivity of
the water) was found to be 5.3?10-4 ?-1 cm-1.
Calculate the molar conductivity.
Solution
This result can also be written as 5.3 S cm2
mol-1, where S stands for siemens, the unit of
conductance, which is equivalent to the
reciprocal ohm.
9
Figure 7.2
?o, the molar conductivity at infinite dilution,
or zero concentration.
??????????????
The importance of the molar conductivity is that
it gives information about the conductivity of
the ions produced in solution by 1 mol of a
substance. In all cases the molar conductivity
diminished as the concentration is raised, and
two patterns of behavior can be
distinguished. strong electrolytes the molar
conductivity falls only slightly as the
concentration is raised. (????) weak
electrolytes produce fewer ions and exhibit a
much more pronounced fall of ? with increasing
concentration. (????)
10
7-3 Weak Electrolytes The Arrhenius Theory
(??????)
There exists equilibrium in solution between
undissociated molecules AB and the ions A and
B-
At very low concentrations this equilibrium lies
over to the right, and the molar conductivity
is close to ?o. As the concentration is
increased, this equilibrium shifts to the left
and the molar conductivity decrease from ?o to
a lower value ?. The degree of dissociation
(???), that is, the fraction of AB in the form
AB-, is ?/?o, which is denoted by the
symbol ?
At high concentrations there is less
dissociation, while at infinite dilution there is
complete dissociation and the molar conductivity
then is ?o.
This theory explains the constant heat of
neutralization of all strong acids and bases,
which are dissociated to a considerable extent,
the neutralization involving simply the
reaction HOH-?H2O. However, Arrheniuss
explanation of the decrease in ? with
increasing concentration is valid only for weak
electrolytes.
11
The osmotic pressures of solutions of
electrolytes were always considerably higher than
predicted by the osmotic pressure equation for
nonelectrolytes.
modified equation
where i is known as the vant Hoff factor.
For strong electrolytes, the vant Hoff factor
is approximately equal to the number of ions
formed from one molecule thus for NaCl, HCl,
etc., i2 for Na2SO4, BaCl2, etc., i3 and
so forth. However the value of i decrease with
the increasing ionic strength. For weak
electrolytes, the vant Hoff factor i involves
the degree of dissociation. Suppose that one
molecule of a weak electrolyte would produce, if
there were complete dissociation, ? ions. The
number of ions actually produced is thus ??, and
the number of undissociated molecules is 1-?,
so that the total number of particles produced
from 1 molecule is
(CH3COO)2Zn with ?0.5
12
Ostwalds Dilution Law (???????)
Consider an electrolyte AB that exists in
solution partly as the undissociated species AB
and partly as the ions A and B-
The equilibrium constant, on the assumption of
ideal behavior, is
Suppose that an amount n of the electrolyte is
present in a volume V and that the fraction
dissociated is ? the fraction not dissociated is
1-?. The amounts of the three species present at
equilibrium, and the corresponding
concentrations, are therefore
The equilibrium constant is
Dilution Law
??????????
13
The lower the concentration c, the larger the
degree of dissociation. As the concentration
approaches zero, the degree of dissociation ?
approaches unity that is, dissociation
approaches 100 as infinite dilution is
approached.
c0.1 ? ?0.916 c0.5 ? ?0.732 c1 ? ?0.618
K1
14
Problems 7.4
The following are the molar conductivities ? of
chloroacetic acid in aqueous solution at 25 ?C
and at various concentrations c
c, 10-4 M 625 312.5 156.3 78.1 39.1 19.6 9.8
?, ?-1 cm2 mol-1 53.1 72.4 96.8 127.7 164.0 205.8 249.2
Plot ? against c. If ?o362 ?-1 cm2 mol-1, are
these values in accord with the Ostwald dilution
law? What is the values of dissociation constant?
Solution
c 10-4 M ? ?-1 cm2 mol-1 ? 1-? Kc?2/(1-?) 10-3 M
625 53.1 0.147 0.853 1.583
312.5 72.4 0.200 0.800 1.563
156.3 96.8 0.267 0.733 1.520
78.1 127.7 0.353 0.647 1.504
39.1 164.0 0.453 0.547 1.467
19.6 205.8 0.569 0.431 1.472
9.8 249.2 0.688 0.312 1.487
average K1.51?10-3 mol dm-3
15
c 10-4 M ? ?-1 cm2 mol-1 c?2
625 53.1 176.2
312.5 72.4 163.8
156.3 96.8 146.5
78.1 127.7 127.4
39.1 164.0 105.2
19.6 205.8 83.0
9.8 249.2 60.9
K1.7?10-3 mol dm-3 ?o347 ?-1 cm2 mol-1
16
7-4 Strong Electrolytes (????)

• Arrhenius Theory
• Some of the results are seriously inconsistent
  with Ostwalds dilution law.
• Constants K calculated from ?/?o values vary,
  sometime by several powers of 10,
• as the concentration is varied.
• Values for the conductivity ratio (?/?o) are
  sometimes significantly different from
• those obtained from the vant Hoff factor (i).
• The fall in ? with increasing concentration must
  be attributed to some cause other
• than a decrease in the degree of dissociation.
• The solvent plays a more active role (solvation
  theory).
• Solvent molecules are sometimes bound quite
  strongly to ions.
• The importance of the electrostatic force
  between ions are also taken into account.

17
Dubye-Huckel Theory
The decrease in the molar conductivity of a
strong electrolyte is attributed to the mutual
interference of the ions, which becomes more
pronounced as the concentration increases. The
arrangement of ions in the solution is not
completely random because of the strong
attractive forces between ions of opposite
signs. In the immediate neighborhood of any
positive ion, there tend to be more negative than
positive ions. The small amount of
ordering that does exist is sufficient to exert
an important effect on the conductivity of
the solution.
Figure 7.3
When the NaCl is dissolved in water, this
ordering is still preserved to a very slight
extent. The ions are much farther apart than in
the solid the electrical attraction are
therefore much smaller and the thermal motions
cause irregularity.
In solid NaCl there is a regular array of Na and
Cl-.
18

• Two effects are pronounced in the strong
  electrolytes
• 1.relaxation or asymmetry effect (????)
• Ions motion under potential applied is retarded
  by the interaction of
• opposite ions due to the asymmetry of ionic
  atmosphere.
• If an electric potential is applied, a positive
  ion will move toward the
• negative electrode and must drag along with it
  an entourage of negative
• ions. The more concentrated the solution, the
  closer these negative ions
• are to the positive ion under consideration,
  and the greater is the drag.
• The ionic atmosphere around a moving ion is
  therefore not symmetrical
• the charge density behind is greater than that
  in front, and this will result
• in a retardation in the motion of the ion.
• 2.electrophoretic effect (????)
• Ions motion under potential applied is retarded
  by the attractive force
• between solvent molecules and ions due to the
  tendency of ionic
• atmosphere to drag the solvent molecules.

19
The Ionic Atmosphere (???)
Figure 7.4
Figure 7.4 shows a positive ion, of charge zce,
situated at a point A e is the unit positive
charge and zc is the valence of the ion. On the
average, the charge density will be negative
because of the electrostatic forces. In
other words, the probability that there is a
negative ion in the volume element is greater
than the probability that there is a positive
ion. The negative charge density will be
greater if r is small than if it is large.
Suppose that the average electric potential in
the volume element dV is ?, and let z and z-
be the positive numerical values of the ionic
valencies. The work required to bring a
positive ion of charge ze from infinity up to
this volume element is ze?, a positive
quantity. Because the central positive ion
attracts the negative ion, the work required to
bring a negative ion of charge z-e is z-e?,
the negative quantity.
Note for Na?z1 for Cl-?z-1
20
The time-average numbers of the positive and
negative ions present in the volume element are
given by the Boltzmann principle
N and N- are the total numbers of positive and
negative ions, respectively, per unit volume of
solution kB is the Boltzmann constant T is the
absolute temperature
The charge density in the volume element (i.e.,
the net charge per unit volume) is given by
a univalent electrolyte zz-1, NN-
Ni and zi represent the number per unit volume
and the positive value of the valence of the
ions of the ith type.
This equation relates the charge density ? to the
average potential ?. In order to obtain these
quantities separately it is necessary to have
another relationship between ? and ?.
It means that the net charge is negative.
21
Figure 7.5
The direction of the vector field due to a
positive charge at the origin is the same as the
direction of the vector from the origin to the
point P. To make the connection we will be
interested in determining the flux of the
electric field (i.e., the number of lines of
force) through an element of area on a surface
perpendicular to the electric field vector.
Consider a closed surface S surrounding a charge
Q. At any point on the surface the field strength
is E, and for an element of area dS the scalar
product E.dS is defined as
? is the angle between E and the normal to dS
flux of the electric field through dS
divergence theorem
Poisson Equation
22
For a spherically symmetrical field,
The general solution of Eq. () is
A and B are constants
In a infinitely dilute solution,
The potential at a distance r will simply be the
potential due to the ion itself, since there is
no interference by the ionic atmosphere.
Therefore, the potential at a distance r due to
an ion of charge zc is
23
The potential produced by the ionic
atmosphere There is no independence on r this
potential is uniform and exists at the central
ion.
If the ionic atmosphere were replaced by a charge
zce situated at a distance 1/? from the central
ion, the effect due to it at the central ion
would be exactly the same as that produced by the
ionic atmosphere. The distance 1/? is therefore
referred to as the thickness of the ionic
atmosphere.
L is the Avogadro constant
The thickness of the ionic atmosphere is
inversely proportional to the square root of the
concentration the ionic atmosphere moves further
from the central ion as the solution is diluted.
AB
AB2
AB
AB3
24
Example 7.3
Estimate the thickness of the ionic atmosphere
for a solution of (a) 0.01 M NaCl and (b) 0.001 M
ZnCl2, both in water at 25 ?C, with ?78.
Solution
(a)
(b)
25
Problem 7.14
The radius of the ionic atmosphere (1/?) for a
univalent electrolyte is 0.964 nm at a
concentration of 0.10 M in water at 25 ?C (?78).
Estimate the radius of the ionic atmosphere for
(a) in water at a concentration of 0.0001 M, (b)
in a solvent of ?38 at a concentration of 0.1 M
and (c) a uni-bivalent electrolyte in a water a
concentration of 0.10 M.
Solution
?constant
(a)
(b)
(c)
26
Mechanism of Conductivity
Relaxation or Asymmetry Effect (????)
Figure 7.6
The effect of the ionic atmosphere is to exert a
drag on the movement of a given ion. If the
ion is stationary, the atmosphere is arranged
symmetrically about it and does not tend to move
it in either direction. However, if a potential
that tends to move the ion to the right is
applied, the atmosphere will decay to some
extent on the left of the ion and build up
more on the right. Since it takes time for these
relaxation processes to occur, there will be
an excess of ionic atmosphere to the left of the
ion (i.e., behind it) and a deficit to the
right (in front of it). This asymmetry of the
atmosphere will have the effect of dragging the
central ion back.
Relaxation force (???) fr is
where V is the applied potential gradient and w
is a number whose magnitude depends on the type
of electrolyte for a uni-univalent electrolyte,
w is 2- 0.586.
27
Electrophoretic Effect (????)
Ions are attracted to solvent molecules mainly
by ion-dipole forces(??- ???) therefore, when
they move, they drag solvent along with them.
The ionic atmosphere, having a charge opposite
to that of the central ion, moves in the
opposite direction to it and therefore drags
solvent in the opposite direction. This means
that the central ion has to travel upstream,
and it therefore travels more slowly than if
there were no effect of this kind.
Electrophoretic force (???) fe is
where Kc is the coefficient of frictional
resistance of the central ion with reference to
the solvent and ? is the viscosity of the medium.
The viscosity enters into this expression because
we are concerned with the motion of the ion past
the solvent molecules, which depends on the
viscosity of the solvent.
28
Debye-Huckel-Onsager Equation
Based on the assumption of complete dissociation
where P and Q can be expressed in terms of
various constants and properties of the system.
For the particular case of a symmetrical
electrolyte i.e., one for which the two ions
have equal and opposite signs (zz-z), P and
Q are given by
Figure 7.7
For aqueous solution of uni-univalent
electrolytes, the theoretical equation is
found to be obeyed very satisfactorily up to a
concentration of about 2?10-3 mol dm-3 at
higher concentrations, deviation are found.
The corresponding equations for other types of
electrolytes in water are also obeyed
satisfactorily at very low concentrations, but
deviations are found at lower concentration
than with uni- univalent electrolytes.
29
Problem 7.10
The molar conductivity of KBr solutions as a
function of concentration at 25 ?C is given in
the following table. By a linear regression
analysis of suitable variables find the value of
?o for KBr.
c, 10-3 M 0.25 0.36 0.50 0.75 1.00 1.60 2.00 5.00 10.00
?, S cm2 mol-1 150.16 149.87 149.55 149.12 148.78 148.02 147.64 145.47 143.15
Solution
c0
?o151.14 S cm2 mol-1
30
Ion Association ( ????)
????????????????????,?????????????????????????????
????,?ZnSO4???????,Zn2?SO4-2????????,????????????
????????????????,??????????,??????????,???????????
????
Bjerrum presented a theory that ions of opposite
signs separated in solution by a distance r form
an associated ion pair held together by coulombic
forces. The Bjerrum distance r is
For the case of a uni-univalent electrolyte
(zczi1) the value of r at 25 ?C, with ?78.3,
is equal to 3.5?10-10 m0.358 nm. The expression
for the electrostatic potential energy of
interaction between two univalent ions separated
by a distance r is
The electrostatic potential energy at this
distance r is thus four times the mean kinetic
energy per degree of freedom. The energy at this
distance is therefore sufficient for there to be
significant ion association, which will be
dynamic in character in that there will be a
rapid exchange with the surrounding ions. At
distance smaller than 0.358 nm the probability of
ionic association increase rapidly. Therefore, to
a good approximation one can say that if the ions
are closer than 0.358 nm, they can be considered
to be undissociated and to make no
contribution to the conductivity.
31
However, in aqueous solution univalent ions can
rarely approach one another as closely as 0.358
nm, and ion association is therefore of little
importance for such ions. In solvents having
lower dielectric constants, however, there can be
substantial ion-pair formation even for
uni-univalent electrolytes.
As shown in Figure 7.8, the formation of ion
pairs was negligible in solution of high
dielectric constant, but very considerable in
solutions of low dielectric constant. Dioxane
has a dielectric constant of 2.2 at 25 ?C, and
the r value for a uni-univalent electrolyte is
12.7 nm ion association is therefore expected
to be important. Much more ion association is
found with ions of higher valence. With salts
having ions of different valences, such as
Na2SO4, ion association will lead to the
formation of species such as NaSO42-. Again
there will be reduction in conductivity, since
these species will carry less current than the
free ions. Bjerrums theory has been extended
to deal with such unsymmetrical electrolytes
and also with the formation of triple ions.
Figure 7.8
Ka, the association constant for ion-pair
formation with tetraisoamylammonium nitrate,
against the dielectric constant of the solvent
ranging from 2.2 to 78.6.
32
Conductivity at High Frequencies and Potentials
Suppose that the alternating potential is of
sufficiently high frequency that the time of
oscillation is small compared with the time it
takes for the ionic atmosphere to relax. There
will then not be time for the atmosphere to relax
behind the ion and to form in front of it the
ion will be virtually stationary and its ionic
atmosphere will remain symmetrical. Therefore, as
the frequency of the potential increases, the
relaxation and eletrophoretic effects will become
less and less important, and there will be an
increase in the molar conductivity.
If the applied potential is 20000 Vcm-1, an ion
will move at a speed of about 1 ms-1 and will
travel several times the thickness of the
effective ionic atmosphere in the time of
relaxation of the atmosphere. Consequently, the
moving ion is essentially free from the effect
of the ionic atmosphere which doe not have time
to build up around it to any extent.
Therefore, at sufficiently high voltages, the
relaxation and eletrophoretic effects will
diminish and eventually disappear and the molar
conductivity will increase. This effect is known
as the Wien effect. However, molar
conductivities of weak electrolytes at high
potentials are anomalously large, and it
appears that very high potentials bring about a
dissociation of the molecules into ions. This
phenomenon is known as the dissociation field
effect.
33
7-5 Independent Migration of Ions
The plots of ? against concentration can be
extrapolated back to zero concentration to give
the ?o value. This extrapolation can only
satisfactorily be made with strong
electrolytes. With weak electrolytes there is a
strong dependence of ? on c at low
concentrations and therefore the extrapolations
do not lead to reliable ?o values.
Kohlrauschs law of independent migration of
ions (????????????) Each ion is assumed to make
its own contribution to the molar conductivity,
irrespective of the nature of the other ion with
which it is associated.
This will be the same whatever the nature of the
anion.
34
Problem 7.5
The electrolytic conductivity of a saturated
solution of silver chloride, AgCl, in pure water
at 25 ?C is 1.26?10-6 ?-1 cm-1 higher than that
for the water used. Calculate the solubility of
AgCl in water if the molar ionic conductivities
are Ag, 61.9 ?-1 cm2 mol-1 Cl-, 76.4 ?-1 cm2
mol-1.
Solution
Problem 7.16
The molar conductivity at 25 ?C of a 0.01 M
aqueous solution of ammonia is 9.6 ?-1 cm2 mol-1.
For NH4Cl, ?o129.8 ?-1 cm2 mol-1 and the molar
ionic conductivities are OH-, 174.0 ?-1 cm2
mol-1 Cl-, 65.6 ?-1 cm2 mol-1, respectively.
Calculate ?o for NH3 and the degree of ionization
in 0.01 M solution.
Solution
35
Problem 7.6
The electrolytic conductivity of a 0.001 M
solution of Na2SO4 is 2.6?10-4 ?-1 cm-1. If the
solution is saturated with CaSO4, the
conductivity becomes 7.0?10-4 ?-1 cm-1. Calculate
the solubility product for CaSO4 using the
following molar conductivities at these
concentrations ?(Na)50.1 ?-1 cm2 mol-1
?(1/2Ca2)59.5 ?-1 cm2 mol-1 .
Solution
The increase in conductivity, 4.4?10-4 ?-1 cm-1,
is due to the CaSO4 present thus
where c is the concentration of CaSO4 2c is the
concentration of 1/2CaSO4.
The value of ?(1/2SO42-) is obtained from the
conductivity of the conductivity of the NaSO4
solution
Note that since the concentration of Na2SO4 is
0.001 M, that of 1/2Na2SO4 is 0.002 M
36
(No Transcript)
37
Ionic Mobilities (??????)
The mobility of an ion, u, is defined as the
speed with which the ion moves under a unit
potential gradient. Suppose that the potential
drop across the opposite faces of a unit cube
is V and that the concentration of univalent
positive ions in the cube is c. If the mobility
of the positive ions is u, the speed of the
ions is uV. All positive ions within a
distance of uV of the negative plate will
therefore reach that plate in unit time, and
the number of such ions is uVc. The charge
they carry is FuVc, and since this is carried
in unit time, the current is FuVc.
Figure 7.9
u is the mobility of the positive ions
The electrolytic conductivity due to the positive
ions, ?, is
The SI unit of mobility is (m s-1)/(V m-1)m2 V-1
s-1, but it is more common to use the unit cm2
V-1 s-1.
38
Example 7.4
The mobility of a sodium ion in water at 25 ?C is
5.19?10-4 cm2 V-1 s-1. Calculate the molar
conductivity of the sodium ion.
Solution
Problem 7.20
If a potential gradient of 100 V cm-1 is applied
to a 0.01 M solution of NaCl, what are the speeds
of the Na and Cl- ions? The molar ionic
conductivities are Na, 50.1 ?-1 cm2 mol-1
Cl-, 76.4 ?-1 cm2 mol-1.
Solution
The velocities in a gradient of 100 V cm-1 are
thus
Na 5.19?10-2 cm s-1 Cl- 7.92?10-2 cm s-1
39
Problem 7.21
A solution of LiCl at a concentration of 0.01 M
is contained in a tube having a cross-sectional
area of 5 cm2. Calculate the speeds of the Li
and Cl- ions if a current of 1 A is passed. The
molar ionic conductivities are Li, 38.6 ?-1 cm2
mol-1 Cl-, 76.4 ?-1 cm2 mol-1.
Solution
The resistance of a 1-cm length of tube is thus
The potential gradient is thus 173.9 V cm-1.
The velocities in a gradient of 173.9 V cm-1 are
thus
Li 0.070 cm s-1 Cl- 0.138 cm s-1
40
7.6 Transport Numbers
Consider an electrolyte of formula AaBb, which
ionizes as follows
The quantity of electricity carried by a mol of
the ions Az, whose charge number is z, is
aFz, and the quantity of electricity crossing a
given cross-sectional area in unit time is
aFzu, where u is the mobility. This quantity
is the current carried by the positive ions when
1 mol of the electrolyte is present in the
solution. Similarly, the current carried by b
mole of the negative ions is bFz-u-. The
fraction of the current carried by the positive
ions is therefore
transport number of the positive ions
transport number of the negative ions
the solution is electrically neutral
41
The individual molar ion conductivities are given
by
The molar conductivity of the electrolytes is
If t and t- can be measured over a range of
concentrations, the value at zero concentration,
to and t-o, can be obtained by extrapolation.
These values allow ?o(AaBb) to be split into the
individual ion conductivities ?o(Az), ?o(Bz--)
by use of the above equations.
42
Problem 7.19
The transport numbers for HCl at infinite
dilution are estimated to be t0.821 and
t-0.179 and the molar conductivity is 426.16 ?-1
cm2 mol-1. Calculate the mobilities of the
hydrogen and chloride ions.
Solution
43
Faradays laws, according to which equivalent
quantities of different ions are liberated at the
two electrodes, can be reconciled with the fact
that the ions are moving at different speeds
toward the electrodes. Figure 7.10 shows how
theses two facts can be reconciled. The diagram
shows in a very schematic way an electrolysis
cell in which there are equal numbers of
positive and negative ions of unit charge.
Figure 7.10a shows the situation before
electrolysis occurs. Suppose that the cations
only were able to move, the anions having zero
mobility after some motion has occurred, the
situation will be as represented in Figure
7.10b. At each electrode two ions remain
unpaired and are discharged, two electrons at the
same time traveling in the outer circuit from
the anode to the cathode. Thus, although only
the cations have move through the bulk of the
solution, equivalent amounts have been
discharged at the two electrodes. It is easy
to extend this argument to the case in which
both ions are moving but a different speeds.
Thus Figure 7.10c shows the situation when the
speed of the cation is three times that of the
anion, four ions being discharged at each
electrode.
Figure 7.10
44
Hittorf method
Figure 7.11
A very simple type of apparatus used in the
Hittorf method is shown in Figure 7.11. The
solution to be electrolyzed is placed in the
cell, and a small current is passed between the
electrodes for a period of time. Solution then is
run out through the stopcocks, and the samples
are analyzed for concentration changes.
Suppose that the solution contains the ions M
and A-, which are not necessarily univalent but
are denoted as such for simplicity. The faction
of the total current carried by the cations is
t, and that carried by the anions is t-. Thus,
when 96485 C of electricity passes through the
solution, 96485 tC are carried in one direction
by t mol of M ions, and 96485 t-C are carried
in the other direction by t- mol of A- ions. At
the same time 1 mol of each ion is discharged at
an electrode.
45
Figure 7.12
Figure 7.12 represents that the cell containing
the electrolyte is divided into three compartment
by hypothetical partitions. One is a compartment
near the cathode, one is near the anode, and the
middle compartment is one in which no
concentration change occurs.
At the cathode A loss of 1-tt- mol of M, and
a loss of t- mol of A-
(i.e., a loss of t- mol of the electrolyte MA) In
the middle compartment No concentration
change At the anode Loss of 1-t-t mol of A-
loss of t mol of M
(i.e., a loss of t mol of the electrolyte MA)
In the preceding discussion we have assumed that
the electrodes are inert (i.e., are not attacked
as electrolysis proceeds) and that ions M and
A- are deposited. If instead, for example , the
anode were to pass into solution during
electrolysis, the concentration changes would be
correspondingly different the treatment can
readily be modified for this and other
situations.
46
Problem 7.17
A solution of LiCl was electrolyzed in a Hittof
cell. After a current of 0.79 A had been passed
for 2 h, the mass of LiCl in the anode
compartment had decreased by 0.793 g. a.
Calculate the transport numbers of the Li and
Cl- ions. b. If ?o(LiCl) is 115.0 ?-1 cm2 mol-1,
what are the molar ionic conductivities and the
ionic mobilities?
Solution
(a)
Quantity of electricity 2 h ? 3600 s h-1 ? 0.79
A 5688 C Amount deposited 5688/96485
0.05895 mol Loss of LiCl in anode compartment
0.793/42.39 0.01871 mol
(b)
47
Problem 7.18
A solution of cadmium iodide, CdI2, having a
molality of 7.545?10-3 mol kg-1, was electrolyzed
in a Hottorf cell. The mass of cadmium deposited
at the cathode was 0.03462 g. Solution weighing
152.64 g was withdrawn from the anode
compartment and was found to contain 0.3178 g of
cadmium iodide. Calculate the transport numbers
of Cd2 and I-.
Solution
48
Moving Boundary Method
Figure 7.13a
Suppose that it is necessary to measure the
transport numbers of the ions in the
electrolyte MA. Two other electrolytes MA and
MA are selected as indicators each has
an ion in common with MA, and the electrolytes
are such that M move more slowly than M and
A- moves more slowly than A-. The solution of
MA is placed in the electrolysis tube with the
solution of MA on one side of it and that of
MA on the other the electrode in MA is the
anode and that in MA is the cathode. As the
current flows, the boundaries remain distinct,
since the M ions cannot overtake the M ions
at the boundary a, and the A- ions cannot
overtake the A- ions at boundary b. The slower
ions M and A- are known as the following
ions. The boundaries a and b move, as shown
Figure 7.13a, to positions a and b. The
distance aa and bb are proportional to the
ionic velocities, and therefore
49
The movement of only one boundary may be
followed, as shown in Figure 7.13b. If Q is the
quantity of electricity passed through the
system, tQ is the quantity of electricity
carried by the positive ions, and the amount
of positive ions to which this corresponds is
tQ/F. If the concentration of positive ions is
c, the amount tQ/F occupies a volume of
tQ/Fc. This volume is equal to the area of
cross section of the tube A multiplied by the
distance aa through which the boundary
moves. Thus it follows that
Figure 7.13b
Simple apparatus for measuring the transport
number of H, using Li as the following cation.
A clear boundary is formed between the receding
acid solution and the LiCl solution and is
easily observed if methyl orange is present.
50
7-7 Ions Conductivities
If MCl, NaA, and NaCl are all strong
electrolytes, their ?o values are readily
obtained by extrapolation and therefore permit
the calculation of ?o for MA, which may be a
weak electrolyte.
This procedure is also for a highly insoluble
salt, for which direct determinations of ?o
values might be impractical.
51
Problem 7.15
The molar conductivities of 0.001 M solutions of
potassium chloride, sodium chloride and potassium
sulfate (1/2K2SO4) are 149.9, 126.5 and 153.3 ?-1
cm2 mol-1, respectively. Calculate an approximate
value for the molar conductivity of a solution of
sodium sulfate of the same concentration.
Solution
Ionic Solvation
There is no simple dependence of the
conductivity on the size of the ion. The results
show that K moves faster than either Li or
Na. The explanation is the Li, because of its
small size, becomes strongly attached to about
four surrounding water molecules by ion-dipole
and other forces, so that when the current
passes, it is Li(H2O)4 and not Li that
moves. With Na the binding of water molecules
is less strong, and with K it is still weaker.
52
Mobilities of Hydrogen and Hydroxide ions
Hydrogen ions has an abnormally high conductivity
in a number of hydroxylic solvents such as
water, methanol, and ethanol, but is behaves more
normally in nonhydroxylic solvents such as
nitrobenzene and liquid ammonia.
At first sight the high values might appear to be
due to the small size of the proton. However,
there is a powerful electrostatic attraction
between a water molecule and the proton, which
because of its small size can come very close to
the water molecule. As a consequence the
equilibrium
Figure 7.21
lies very much to the right. In other words,
there are very few free protons in water the
ions exist as H3O ions, which are hydrated by
other molecules. Thus there remains a difficulty
in explaining the very high conductivity and
mobility of the hydrogen ion. By virtue of its
size it would be expected to move about as fast
as the Na ion, which in fact is the case in the
nonhydroxyl solvents.
53
In order to explain the high mobilities of
hydrogen ions in hydroxylic solvents such as
water, a special mechanism must be invoked. As
well as moving through the solution in the way
that other ions do, the H3O ion can also
transfer its proton to a neighboring water
molecule
The resulting H3O ion can now transfer a proton
to another H2O molecule. In other words,
proton, although not free in solution, can be
passed from one water molecule to
another. Calculations from the known structure
of water show that the proton must on the
average jump a distance of 86 pm(10-12m) from an
H3O ion to a water molecule but that as a
result the proton moves effectively through a
distance of 310 pm. The conductivity by this
mechanism therefore will be much greater than by
the normal mechanism. The proton transfer must
be accompanied by some rotation of H3O and
H2O molecules in order for them to be positioned
correctly for the next proton transfer.
CH3
CH3
CH3
CH3
In methyl alcohol,


O
O
O
O
H
H
H
H
H
H
54
Ionic Mobilities and Diffusion Coefficients
Ions also migrate in the absence of an electric
field, if there is a concentration gradient.
The migration of a substance when there is a
concentration gradient is known as
diffusion. The tendency of a substance to move
in a concentration gradient is measured in terms
of a diffusion coefficient D. Nernst showed
that the relationship between the diffusion
coefficient and the mobility u of an ion is
These relationships show that diffusion
experiment, as well as conductivity measurements,
can provide information about mobilities and
hence about molar conductivities. Diffusion
experiments with uni-univalent electrolyte AB-,
such as NaCl, will be concerned with the
diffusion of the two ions A and B-. The overall
diffusion coefficient D must then relate the
individual diffusion constants DA and DB-, and
Nernst showed that for uni-univalent
electrolytes the proper average is
55
Waldens Rule
Walden noticed that the product of the molar
conductivity, ?, and the viscosity ? of the
solvent was approximately constant
Stokes-Einstein equation presented that the
diffusion coefficient of a particle is inversely
proportional to the viscosity of the solvent.
Since ionic conductivities and diffusion
coefficients are proportional to one another,
Waldens rule is easy to understand.
Stokes-Einstein equation
56
7.8 Thermodynamics of Ions
Due to the absolute thermodynamic values for
individual ions cannot be measured directly, they
can be estimated on the basis of theory.
Unfortunately, owing to the large number of
interactions involved, the theoretical treatment
of an ion in aqueous solution is very difficult.
The conventional procedure is to set the value
for H as zero.
The following absolute values are generally
agreed to be approximately correct, for the
proton
Note Gibbs energy of hydration is the change in
Gibbs energy when an ion is transferred from the
gas phase into aqueous solution.
57
Example 7.5
Calculate the absolute enthalpies of hydration of
Li, I-, and Ca2 on the basis of a value of
-1090.8 kJ mol-1 for the proton, using the values
listed in Table 7.4.
Solution
The enthalpy of hydration of a uni-univalent
electrolyte MX- is independent of where
conventional or absolute ionic values are used.
Thus,
The absolute value for I- will be
The absolute value for Li will be
Consider CaI2, since we have raised the value for
each I- ion by 1090.8, we must lower the Ca2
value by 2?1090.8 thus
The absolute value for Ca2 will be
58
Problem 7.24
The following are some conventional standard
enthalpies of ions in aqueous solution at 25 ?C
Ion ?fHo/kJ mol-1
H 0
Na -239.7
Ca2 -543.1
Zn2 -152.3
Cl- -167.4
Br- -120.9
Calculate the enthalpy of formation in aqueous
solution of 1 mol of NaCl, CaCl2, and ZnBr2,
assuming complete dissociation.
Solution
NaCl -239.7-167.4 -407.1 kJ mol-1 CaCl2
-543.1-(2?167.4) -877.9 kJ mol-1 ZnBr2
-152.3-(2?120.9) -394.1 kJ mol-1
59
7-9 Theories of Ions in Solution
Many theoretical treatments of ions in solution,
especially in aqueous solution, have been put
forward. Some of the treatment are quantities,
attempts being made to obtain expressions for
the strength of the interactions between ions and
the solvent. Other treatments are qualitative,
but based on the mathematical treatments.
Another classification of theories of ions in
solution is based on whether the solvent is
treated as if it were a continuum, or whether the
molecular nature of the solvent is taken into
account. The latter theories are obviously more
realistic, but unfortunately they are difficult
to develop in any detail. The continuum
theories, such as the electrostriction treatment
of Nernst and Drude and the treatment of Max
Born, are much simpler, and in spite of their
lack of realism they give a surprisingly
useful interpretation of ionic behavior in
solution.
60
Drude and Nernsts Electrostriction Model
Drude and Nernst called the electrostriction of
solvent molecules by ions. The strong
electrostatic field due to an ion causes neighbor
polar solvent molecules, such as water
molecules, to become closely packed and to occupy
less volume. The solvent is regarded as a
continuous dielectric for the mathematical
convenience. Electrostriction plays an important
role in connection with the mobilities of ions
and greatly affects the entropies of ions in
solution. The recognition by Drude and Nernst
that ions bring about electrostriction of ions
played an important role in leading to a general
acceptance of the theory.
61
Borns Model
Born suggest a simple interpretation of the
thermodynamic quantities for ions in
solution. The solvent is assumed to be a
continuous dielectric and the ion a conducting
sphere. The work of charging such a sphere is
the Gibbs energy change during the charging
process.
Based on Borns model, the total reversible work
in transporting increments until the sphere has a
charge of ze is
This work, being non-PV work, is the
electrostatic contribution to the Gibbs energy of
the ion
The electrostatic Gibbs energy of hydration is
therefore
for water at 25 ?C, ?78
62
Figure 7.14
Figure 7.14 shows a plot of absolute ?hydGo
values against z2/r. In views of the simplistic
nature of the model, the agreement with the
prediction of the treatments is not
unsatisfactory the theory accounts quite well
for the main effects.
Note, however, that the agreement with the theory
is worse the higher the charge and the smaller
the radius. Figure 7.14 also shows the prediction
for ?2, which is approximately the dielectric
constant of water when it is subjected to an
intense field. The line for ?2 is closer to the
points for ions of high charge and small radius.
The electrostatic entropy of hydration is
The reason that the theory leads to the same
expression for the absolute entropy of the ion
and for its entropy of hydration is that it gives
zero entropy for the ion in the gas phase.
63
for water at 25 ?C, ?78
Figure 7.15
Figure 7.15 shows a plot the experimental ?hydSo
values, together with a line of slope -4.10.
The line has been drawn through a value of -42
J K-1 mol-1 at z2/r0, since it is estimated
that there will be a ?hydSo value of about -42
J K-1 mol-1 due to nonelectrostatic effects.
In particular, there is an entropy loss
resulting from the fact that the ion is
confined in a small solvent cage, an effect
that is ignored in the electrostatic treatment.
Note The simple model of Born accounts
satisfactorily for the main effects. The
deviation are primarily for small ions.
64
Problem 7.30
Estimate the change in Gibbs energy ?G when 1 mol
of K ions (radius 0.133 nm) is transported from
aqueous solution (?78) to the lipid environment
of a cell membrane (?4) at 25 ?C.
Solution
In membrane,
In water,
65
More Advanced Theories
The effective dielectric constant of a solvent
varies in the neighborhood of an ion. The
dielectric behavior of a liquid is related to the
tendency of ions to orient themselves in an
electric field. At very high field strengths the
molecules become fully aligned in the field and
there can be no further orientation. The
dielectric constant then falls to a very low
value of about 2, an effect that is known as
dielectric saturation.
Figure 7.16
Figure 7.16 shows the effective dielectric
constant in the neighborhood of ions of various
types. For example, for a ferric (Fe3) ion the
dielectric constant has a value of about 1.78 up
to a distance of about 0.3 nm from the center of
the ion, owing to the high field produced by the
ion. At greater distances, where the field is
less, the effective dielectric constant rises
toward its limiting value of 78.
Note When the Born treatment of Gibbs energies
and entropies of hydration of ions are modified
by taking these dielectric saturation effects
into account, the agreement with experiment is
considerably better.
66
Qualitative Treatments
The main attractive force between an ion and a
surrounding water molecule is the ion-dipole
(? ?-???) attraction. Figure 7.17 shows that
the way in which a water molecule is expected
to orient itself in the neighborhood of a
positive or negative ion. There are two
possible orientations for a negative ion and
it appears that both may play a role. Quite
strong binding can result from ion-dipole
forces the binding energy for a Na ion and a
neighboring water molecule, for example, is about
80 kJ mol-1.
Figure 7.17
67
Figure 7.18
The smaller the ion, the greater the binding
energy between the ion and a water molecule,
because the water molecule can approach the ion
more closely. Figure 7.18 shows that a Li ion
in water is surrounded tetrahedrally by four
water molecules. Li has a hydration number of
4. The four water molecules are sufficiently
strongly attached to the Li ion that during
electrolysis they are dragged along with it,
therefore, Li has a lower mobility than Na
and K the latter ions being larger, have a
smaller tendency to drag water molecules. A
moving Li ion drags not only its four
neighboring water molecules but also some of
the water molecules that are further away from
it.
68
Figure 7.19
Fank et al. have concluded that for ions in
aqueous solution it is possible to
distinguish three different zones in the
neighborhood of the solute molecule, as shown
in Figure 7.19. In the immediate vicinity of the
ion there is a shell of water molecules that
are more or less immobilized by the very high
field due to the ion. These water molecules can
be described as constituting the inner
hydration shell and are sometimes described as
an iceberg, since their structure has some
icelike characteristics. (ice is less dense
than liquid water, whereas the iceberg around
the ion is more compressed than normal liquid
water).
Surrounding this iceberg there is a second
region referred to as a region of structure
breaking. Here the water molecules are oriented
more randomly than in ordinary water, where
there is considerable ordering due to hydrogen
bonding. The occurrence of this
structure-breaking region results from two
competing orienting influences on each water
molecule. One of these is the normal structure
effect of the neighboring water molecules the
other is the orienting influence upon the dipole
of the spherically symmetrical ionic
field. The third region around the ion comprise
all the water sufficiently far from the ion that
its effect is not felt.
69
The behavior of the proton H in water deserves
some special discussion. Because of its small
size, the proton attaches itself very strongly
to a water molecule,
The equilibrium for this process, which is
strongly exothermic, lies very far to the right.
Thus, the hydrogen ion is frequently regarded as
existing as the H3O ion which is known as the
oxonium ion.
oxonium ion (???)
Figure 7.21
Because of ion-dipole attractions, other
molecules are held quite closely to the H3O
species. Figure 7.21 shows that three water
molecules are held particularly strongly by
hydrogen bonds involving three hydrogen atoms of
the H3O species. The hydrated proton may be
written as H3O.3H2O, or as H9O4. The species
H9O4 has the same kind of structure as the
hydrated Li ion, in which the proton is
surrounded tetrahedrally by four water
molecules. An additional H2O molecule held by
ion-dipole force but not by hydrogen bonding,
and it therefore is not held so strongly as
the three other water molecules.
hydronium ion (?????)
70
7-10 Activity Coefficients
Debye-Huckel Limiting Law
The behavior of a single ion of type i is ideal
ci is the concentration of the ion.
Note we have used kB, the Boltzmann constant,
instead of the gas constant R, since we are
concerned with single ions instead of a mole of
ions.
due to the presence of the ionic atmosphere
The behavior of a single ion of type i is
nonideal
? is the activity coefficient
work of charging an isolated ion
The reciprocal 1/? plays the same role for the
atmosphere as does r for the ion, and the change
of sign is required because the net charge on the
atmosphere is opposite to that on the ion.
work of charging the ionic atmosphere
71
The expression for the activity coefficient ?i is
? is proportional to the square root of the
quantity
Ni is the number of ions of the ith type per unit
volume and zi is its valency.
The ionic strength I of a solution is defined as
ci is the molar concentration of the ions of the
type i.
for a 1M solution of a uni-univalent electrolyte
such as NaCl,
for a 1M solution of a uni-bivalent electrolyte
such as K2SO4,
72
Problem 7.32
What concentration of the following have the same
ionic strength of 0.1 M NaCl? CuSO4, Ni(NO3)2,
Al2(SO4)3, Na3PO4 Assume complete dissociation
and neglect hydrolysis.
Solution
x0.025
x M CuSO4
x0.033
x M Ni(NO3)2
x0.00667
x M Al2(SO4)3
x0.0167
x M Na3PO4
73
B is a quantity that depends on properties such
as ? and T. When water is the solvent at 25 ?C,
the value of B is 0.51 mol-1/2 dm3/2.
Define the mean activity coefficient ?? in terms
of the individual values for ? and ?- by the
relationship
? and ?- are the number of ions of the two kinds
produced by the electrolyte.
For ZnCl2, ?1 and ?-2. For an uni-univalent
electrolyte (??-1) the mean activity
coefficient is the geometric mean (??-)1/2 of
the individual values.
74
Problem 7.36
Assuming the Born equation (Eq. 7.86) to apply,
make an estimate of the reversible work of
charging 1 mol of NaCl- in aqueous solution at
25 ?C (?78), under the following conditions a.
The electrolyte is present at infinite
dilution. b. The electrolyte is present at such a
concentration that the mean activity coefficient
is 0.70. The ionic radii are 95 pm for Na
and 181 pm for Cl-.
Solution
(a)
For 1 mol of NaCl- at infinite dilution,
75
(b)
For 1 mol of Na ions, of activity coefficient
?, the work of charging the ionic atmosphere is
Similarly for the chloride ion, the work per mole
is
For 1 mol of NaCl-,
If ??0.70
The net work of charging is thus
76
for aqueous solutions at 25 ?C,
Debye-Huckel Limiting Law (DHLL)
Figure 7.22
The term is known as the salting out term, it
accounts for the lowered solubilities of salts at
high ionic strengths.
a??????????????????
??????????
77
Problem 7.40
Make use of the Debye-Huckel limiting law to
estimate the activity coefficients of the ions in
an aqueous 0.004 M solution of sodium sulfate at
298 K. Estimate also the mean activity
coefficient.
Solution
for Na
for SO4-2
78
7-11 Ionic Equilibria
Activity Coefficients from Equilibrium Constant
Measurements
The practical equilibrium constant is
? and ?- are the activity coefficients of the
ions and ?u is the activity coefficient of the
undissociated acid
Figure 7.23
??????????????? ??????-??????????
?????????????0
79
Solubility Products
For a sparingly soluble uni-univalent salt AB,
The solubility product is
Figure 7.24
???????, ??????????????,??s???,?????salting-in. ??
?????, ??????????????,??s???,?????salting-out.
?????????????0
80
Problem 7.34
Calculate the solubility of silver acetate in
water at 25 ?C, assuming the DHLL to apply the
solubility product is 4.0?10-3 mol2dm-6.
Solution
81
Example 7.6
The solubility product of BaSO4 is 9.2?10-11 mol2
dm-6. Calculate the mean activity coefficient of
the Ba2 and SO42- ions in a solution that is
0.05 M in KNO3 and 0.05 M in KCl, assuming the
Debye-Huckel limiting law to apply. What is the
solubility of BaSO4 in that solution, and in pure
water?
Solution
The ionic strength of the solution is
According to the DHLL
If the solubility in the solution is s,
In pure water,
Note the solubility is higher in the salt
solution than in water (salting in), because
the activity coefficients have been lowered.
82
Problem 7.38
Silver chloride, AgCl, is found to have a
solubility 1.561?10-5 M in a solution that is
0.01 M in K2SO4. Assume the DHLL to apply and
calculate the solubility in pure water.
Solution
In pure water,
83
Problem 7.33
The solubility product of PbF2 at 25 ?C is
4.0?10-9 mol3 dm-9. Assuming the Debye-Huckel
limiting law to apply, calculate the solubility
of PbF2 in (a) pure water (b) 0.01 M NaF.
Solution
(a)
First, neglect the effect of activity
coefficient if s is the solubility
The ionic strength is
According to the DHLL
If now the true solubility is s, the activities
of the ions are
84
(b)
In 0.01 M NaF, the ionic strength is essentially
0.01 mol dm-3 and
If s is the solubility,
85
7.12 Ionization of Water
Because the concentration of water hardly varies
from solution to solution, it is usual to
incorporate this concentration into the
equilibrium constant.
Ionic product of water is
At 25 ?C the value of the ionic product is almost
exactly 10-14 mol2 dm-6.
In chemically pure water the concentrations of
hydrogen and hydroxyl ions are equal and