Jessica Sproul
Chapter 9 Summary
Dr. Rahni
Analytical Chemistry
Electrolyte Effects: Activity or Concentration
9A) Effects of Electrolytes on Chemical Equilibria
1) The position of most solution equilibria depends on the electrolyte
concentration of the medium, even when the electrolyte contains no ion in
common with those involved in the equilibrium.
2) If an electrolyte is added to a reaction involving the oxidation of iodide ion
by arsenic acid, the color of the triiodide ion becomes less, indicating that
the concentration of I3- has decreased and that the electrolyte has shifted
the equilibrium to the left.
3) As the electrolyte concentration becomes very small, concentration-based
equilibrium constants approach their thermodynamic values: K w, Ksp, Ka.
4) Since the electrostatic forces of all singly charged ions are approximately
the same, different salts exhibit almost the exact same effects on
equilibria.
9A-1) How Do Ionic Charges Affect Equilibria?
1) The magnitude of the effect that the electrolyte has is highly dependent on
the charge of the ions participating in the equilibrium.
2) When only neutral species are involved in the equilibrium, the electrolyte
has no real effect, no matter how concentrated.
i) In the above figure, the molar solubility of the doubly charged ions
in BaSO4 increases by a factor of 2 in a solution of 0.02 M KNO3
over a solution of pure water. On the other hand, the molar
solubility of a solution of Ba(IO3)2 with one singly and one doubly
charged ion only increases by a factor of 1.25 in a solution of 0.02
M KNO3 over a solution of pure water. And finally, the molar
solubility of AgCl with two singly charged ions only increases by a
factor of 1.2 in a solution of 0.02 M KNO3 over a solution of pure
water.
ii) The above chart shows that with ionic participants, the magnitude
of the electrolyte effect increases with charge.
9A-2) What Is the Effect of Ionic Strength on Equilibria?
1) The effect of added electrolyte on equilibria is independent of the chemical
nature of the electrolyte but depends on a property of the solution called
ionic strength:
i) Ionic strength = μ = 0.5([A] ZA2 + [B] ZB2 + [C] ZC2 +…)
ii) Where [A], [B], [C],… represent the molar species concentrations of
ions A, B, C,… and ZA, ZB, ZC,… are their ionic charges.
Example 9-1
Calculate the ionic strength of (a) a 0.1 M solution of KNO3 and (b) a 0.1 M
solution of Na2SO4.
(a) For the KNO3 solution, [K+] and [NO3-] are 0.1 M and
μ = 0.5([K+] ZA2 + [NO3-] ZB2)
μ = 0.5([0.1]12 + [0.1]12)
μ = 0.1
(b) For the Na2SO4 solution, [Na+] = 0.2 and [SO42-] = 0.1. Therefore,
μ = 0.5([Na+] ZA2 + [SO42-] ZB2)
μ = 0.5([0.2]12 + [0.1]22)
μ = 0.3
Example 9-2
What is the ionic strength of a solution that is 0.05 M in KNO3 and 0.1 M in
Na2SO4?
μ = 0.5([K+] ZA2 + [NO3-] ZB2 + [Na+] ZC2 + [SO42-] ZD2)
μ = 0.5([0.05]12 + [0.05]12 + [0.2]12 + [0.1]22)
μ = 0.35 M
2) The ionic strength of a solution of a strong electrolyte consisting solely of
singly charged ions is identical with its total molar salt concentration. But
when the ions are no longer singly charged, the ionic strength increases.
Effect of Charge on Ionic Strength
Type
Electrolyte
Example
Ionic
Strength*
1:1
NaCl
c
1:2
1:3
2:2
Ba(NO3)2,
Na2SO4
Al(NO3)3,
Na3PO4
MgSO4
3c
6c
4c
*c = molarity of the salt
3) For a solution with an ionic strength of 0.1 or less, the electrolyte effect is
independent of the kind of ions and dependent only on the ionic strength.
9A-3) The Salt Effect
1) The electrolyte effect results from the electrostatic attractive and repulsive
forces that exist between the ions of an electrolyte and the ions involved in
an equilibrium. These forces cause each ion from the dissociated reactant
to be surrounded by a sheath of solution that contains a slight excess of
electrolyte ions of opposite charge.
These charged atmospheres lessen the charges on the ions in
equilibrium, the Ba2+ and SO42- in the figure, thus decreasing their overall
attraction to each other and increasing solubility.
2) The effective concentration of barium ions and of sulfate ions becomes
less as the ionic strength of the medium becomes greater.
9B) Activity Coefficients
1) The term activity, a, is used to account for the effects of electrolytes on
chemical equilibria.
2) The activity, or effective concentration, of species X depends on the ionic
strength of the medium and is defined as
ax =
x
[X]
where ax is the activity of X, [X] is its molar concentration, and
x
is a
dimensionless quantity called the activity coefficient.
3) The activity coefficient and the activity of X very with ionic strength.
4) If XmYn is a precipitate, the thermodynamic solubility product expression is
defined by the equation
Ksp = (axm)(ayn)
But if the equation for activity is applied, a new equation arises:
Ksp =
m
n
m
n
X
Y [X] [Y]
=
m
n
X
Y
• K`sp
where K`sp is the concentration solubility product constant and Ksp is the
thermodynamic equilibrium constant. The activity coefficients
X
and
Y
vary with the ionic strength in such a way as to keep Ksp numerically
constant and independent of ionic strength.
9B-1) Properties of Activity Coefficients
1) In very dilute solutions, where the ionic strength is minimal, the
effectiveness with which a species influences an equilibrium in which it is
a participant becomes constant and the activity coefficient becomes unity.
As the ionic strength increases, however, an ion loses some of its
effectiveness and its activity coefficient decreases. At high ionic
strengths, activity coefficients often increase and may even become
greater than unity.
2) In dilute solutions, the activity coefficient for a given species is
independent of the nature of the electrolyte and dependent only on ionic
strength.
3) For a given ionic strength, the activity coefficient of an ion departs farther
from unity as the charge carried by the species increases.
4) At any given ionic strength, the activity coefficients of ions of the same
charge are approximately equal.
5) The activity coefficient of a given ion describes its effective behavior in all
equilibria in which it participates.
9B-2) The Debye-Hückel Equation
1) The Debye-Hückel equation takes the ionic atmosphere model and
derives a theoretical expression that permits the calculation of activity
coefficients of ions from their charge and their average size.
2) J. Kielland calculated various values of άx for numerous ions from a variety
of experimental data.
Table 9-1
Activity Coefficient at Indicated Ionic Strength
Ion
άx, nm
0.005
0.01
0.05
0.1
H+
0.9
0.934
0.913
0.854
0.825
(C3H7)4N+
0.8
0.932
0.911
0.847
0.816
(C3H7)3NH+, {OC6H2(NO3)3}
0.7
0.931
0.909
0.841
0.806
Li+, C6H5COO, (C2H5)4N+
0.6
0.930
0.906
0.833
0.795
CHCl2COO, (C2H5)3NH+
0.5
0.928
0.904
0.825
0.783
0.4–0.45
0.927
0.902
0.818
0.773
OH, F, HS, MnO4, IO4, HCOO
0.35
0.926
0.900
0.811
0.761
K+, Cl, Br, I, CN, NO2, NO3
0.3
0.925
0.898
0.806
0.753
Rb+, Cs+, NH4+, Tl+, Ag+
0.25
0.924
0.897
0.801
0.744
Na+, H2PO4, HCO3, HSO3,
CH COO, (CH ) NH+, C H NH +
3
3 3
2 5
3
Mg2+, Be2+
0.8
0.755
0.689
0.516
0.444
Ca2+, Cu2+, Zn2+, Mn2+, Sn2+,
Ni2+, Co2+, Fe2+
0.6
0.747
0.675
0.482
0.400
Sr2+, Ba2+, Cd2+, Hg2+, S2
0.5
0.742
0.668
0.463
0.376
Pb2+, CO32, SO32, MnO42
0.45
0.740
0.664
0.454
0.363
Hg22+, SO42, CrO42, HPO42
0.4
0.738
0.660
0.443
0.350
Al3+, Fe3+, Cr3+, La3+
0.9
0.538
0.442
0.241
0.177
{Co(H3NCH2CH2NH3)3}3
0.6
0.518
0.413
0.193
0.127
{OOCCOH(CH2COO)2}3
0.5
0.511
0.403
0.177
0.110
PO43, Fe(CN)63, {Co(NH3)6}3+
0.4
0.504
0.392
0.160
0.094
Th4+, Zr4+, Ce4+, Sn4
Fe(CN) 4
0.11
0.346
0.251
0.098
0.062
0.5
0.303
0.199
0.046
0.020
6
(www.chemistry.msu.edu/courses/ cem834/Handouts/Handout4-ActivityCoef.doc)
Example 9-3
Use the Debye-Hückel equation to calculate the activity coefficient for
Hg2+ in a solution that has an ionic strength of 0.085. Use 0.5 nm for the
effective diameter of the ion. Compare the calculated value with
obtained by interpolation of data from Table 9-1.
= antilog (-0.0416)
2+
Hg
= 0.397 = 0.4
Table 9-1 indicates that
that
= 0.38 when the ionic strength is 0.1, and
= 0.46 when the ionic strength is 0.085,
= 0.38 + (0.015/0.050) (0.46 – 0.38) = 0.404 = 0.4
9B-3) Equilibrium Calculations with Activity Coefficients
1) Equilibrium calculations with activities yield values that are better in
agreement with experimental results than those obtained with molar
concentrations.
Example 9-4
Find the relative error introduced by neglecting activities in
calculating the solubility of Ba(IO3)3 in a 0.033 M solution of Mg (IO3)2.
The thermodynamic solubility product for Ba(IO3)3 is 1.57 X 10-9.
At the outset, we write the solubility-product expression in terms of
activities:
aBa2+ • (aIO3-)2 = Ksp = 1.57 X 10-9
where aBa2+ and aIO3- are the activities of barium and iodate ions.
Replacing activities in this equation by activity coefficients and
concentrations yields
[Ba2+]
• [IO3-]2
= Ksp
where
and
are the activity coefficients for the two ions.
Rearranging this expression gives
K’sp = Ksp/(
+
) = [Ba2+] [IO3-]2
where K’sp is the concentration-based solubility product.
The ionic strength of the solution is found by plugging values into
the equation:
μ = 0.5([Mg2+] 22 + [IO3-] 12)
μ = 0.5([0.033]4 + [0.066]1) = 0.099 = 0.1
Turning now to Table 9-1, we find that at an ionic strength of 0.1,
= 0.38
= 0.77
Substituting into the thermodynamic solubility-product expression give
K’sp = (1.57 X 10-9) / (0.38 • 0.772) = 6.8 X 10-9
[Ba2+] [IO3-]2 = 6.8 X 10-9
Proceeding now as in earlier solubility calculations yields
Solubility = [Ba2+]
[IO3-] = 2 • 0.033 + 2[Ba2+] = 0.066
[Ba2+](0.066)2 = 1.56 X 10-6 M
If we neglect activities, the solubility is
[Ba2+](0.066)2 =1.57 X 10-9
[Ba2+] = solubility = 3.60 X 10-7 M
relative error = [(3.60 X 10-7 – 1.56 X 10-6) / (1.56 X 10-6)] • 100%
relative error = -77%
9B-4) Omitting Activity Coefficients in Equilibrium Calculations
1) We shall ordinarily neglect activity coefficients and simply use molar
concentrations in applications of the equilibrium law. This simplifies
calculations and decreases necessary data. For most purposes, the error
introduced by this method is minimal.
i) Significant discrepancies occur when the ionic strength is larger
than 0.01 or when the ions involved have multiple charges.
ii) With dilute solutions of nonelectrolytes or of simply charged ions,
the use of concentrations in a mass-law calculation often provides
reasonably accurate results.
End of Chapter Problems
(http://webpage.pace.edu/dnabirahni/rahnidocs/che111/Chapter9.doc)
Problem 9-8
-log γX =
0.51 Z2X √μ
1 + 0.33 αX √μ
a) Fe3+ at μ = 0.075
-log γX =
0.51 (3)2 √0.075
= 0.20
1 + 0.33 (0.9) √0.075
b) Pb2+ at μ = 0.012
-log γX =
0.51 (2)2 √0.012
1 + 0.33 0.45 √0.012
= 0.64
c) Ce4+ at μ = 0.080
-log γX =
0.51 (4)2 √0.080
= 0.073
1 + 0.33 1.1 √0.080
d) Sn4+ at μ 0.060
-log γX =
0.51 (4)2 √0.060
= 0.088
1 + 0.33 1.1 √0.060
Problem 9-13
Calculate the solubilities of compounds (a) – (d) in a 0.0167M solution of
Ba(NO3)2 using activites and molar concentration:
μ = ½ [0.0167(2)2 + 2(0.0167)(1)2] = 0.05
(a) AgIO3
1)
γAg+ = 0.80 & γIO3- = 0.82
K′sp = 3.1x10-8/[(0.80)(0.82)] = 4.73x10-4
S = √4.73x10-8 = 2.2x10-4M
1)
S = √3.1x10-8 = 1.8x10-4M
(b) Mh(OH)2
1)
γMg = 0.52 & γOH = 0.81
K′sp = 7.1x10-12/[(0.52)(0.81)2 = 2.08x10-11
S = [Mg] = ½[OH]
S = (2S)2 = 2.08x10-11
S = 1.7x10-4M
2)
S = (7.1x10-12/4)1/3 = 1.2x10-4M
(c) BaSO4
1)
γBa = 0.46 & γSO4 = 0.44
K′sp = 1.1x10-10/[(.46)(0.44)] = 5.43x10-10
S = 5.43x10-10/0.0167 = 3.3x10-8M
2)
S = 1.1x10-10/0.0167 = 6.6x10-9M
(d) La(IO3)2
1)
γLa = 0.24 & γIO3 = 0.82
K′sp = 1.0x10-11/[(0.24)(0.82)3] = 7.56x10-11
S = [La] = 1/3 [IO3]
S(3S)3 = 7.56x10-11
S = (7.56x10-11/27)1/4 = 1.3x10-3M
2)
S = (1.0x10-11/27)1/4 = 8x10-4M