10 Effect of Electrolytes on Chemical Equilibria
10A The Effect of Electrolytes on Chemical Equilibria
H3AsO4 + 3I- + 2H+ ↔ H3AsO3 + I3- + H2O
Added electrolyte: Ba(NO3)2, K2SO4, NaClO4; → [I3-]↓ , color ↓
← Fig. 10-1 Effect of electrolyte conc. on
conc.-based equilibrium constants.
K a' × 105
K w' × 1014
CKNO 3 concentration
Concentration NaCl, M
Fig. 10-2 Effect of electrolyte conc. on the
solubility of some salts.
10A-1 The Effect of Ionic Charges on Equilibria
Fig. 10-2
10A-2 The Effect of Ionic Strength
1
2
ionic strength = μ = ([A ]Z A
+ [B]Z B2 + [C]Z C2 + ⋅ ⋅ ⋅)
2
[A], [B], [C], … molar species conc.; ZA, ZB, ZC, … ionic charge
Ex.10-1 Calculate the ionic strength of (a) a 0.1 M soln of KNO3 and (b) a 0.1 M soln
of Na2SO4.
1
(a) For KNO3 soln, μ = (0.1 × 12 + 0.1 × 12 ) = 0.1
2
(b)
For Na2SO4 soln, μ =
1
(0.2 × 12 + 0.1 × 2 2 ) = 0.3
2
Ex. 10-2 What is the ionic strength of a solution that is 0.05 M in KNO3 and 0.1 M in
Na2SO4.
1
μ = (0.05 × 12 + 0.05 × 12 + 0.2 × 12 + 0.1 × 2 2 ) = 0.35 M
2
48
Effect of Charge on Ionic Strength (C = molarity of the salt)
Type
Example
1:1
NaCl
1:2
Ba(NO3)2, Na2SO4
1:3
Al((NO3)3, Na3PO4
2:2
MgSO4
μ
C
3C
6C
4C
10B Activity Coefficients
aX = γ X [ X ]
aX : activity of species X
[X] : molar conc.
γX : activity coefficient
m n
for ppt Xm Yn, thermodynamic equilibrium constant : K sp = aX
⋅ aY
K sp = [X] m [Y] n ⋅ γ Xm ⋅ γ Yn = K sp' ⋅ γ Xm ⋅ γ Yn
'
Concentration solubility product constant: K sp
10B-1 Properties of Activity Coefficients
Activity:
a measure of its effective conc. as
determined by the lowering of the
freezing point of water, by electrical
conductivity and by the mass action
effect.
1. μ↑→ γx↓,
at moderate μ, γx < 1,
As μ → 0, γx → 1,
' → K .
ax →[X] and K sp
sp
Fig.10-3 Effect of ionic strength on activity
coefficient.
2. γx independent of the electrolyte, dependent only μ
3. For a giving μ, species charge↑→ γx↓(departs farther from unity)
4. For a giving μ, γx of ions of the same charge are approxi. equal.
5. γx → effective behavior in all equilibria in which it participates
HCN + H2O ↔ H3O+ + CNAg+ + CN- ↔ AgCN(s)
Ni2+ + 4CN- ↔ Ni(CN)4249
10B-2 The Debye-Hückel Equation
0.51Z x2 μ
− log γ x =
1 + 3.3α x μ
when μ << 0.01
Eq.10-5
1+ μ ≈ 1
− log γ x = 0.51Z x2 μ
1923, P Debye & E. Hückel
γx = activity coefficient of the species X
Zx = charge on the species X
μ = ionic strength of the solution
αx = effective diameter of the hydrated ion
X in nanometers (10-9 m)
Debye-Hückel limiting law (DHLL)
Tab 10-1 Activity Coefficients for Ions at 25℃
Activity Coefficient at Indicated Ionic Strength
Ion
αx, nm 0.001 0.005 0.01 0.05 0.1
+
H3O
0.9
0.967 0.934 0.913 0.85 0.83
+
Li , C6H5COO
0.6
0.966 0.930 0.907 0.83 0.80
+
Na , IO3 , HSO3 , HCO3 , H2PO4 ,
0.4-0.45 0.965 0.927 0.902 0.82 0.77
H2AsO4-, OAcOH-, F-, SCN-, HS-, ClO3-, ClO4-, BrO3-,
0.35 0.965 0.926 0.900 0.81 0.76
IO4-, MnO4K+, Cl-, Br-, I-, CN-, NO2-, NO3-, HCOO0.3
0.965 0.925 0.899 0.81 0.75
+
+
+
+
+
Rb , Cs , Tl , Ag , NH4
0.25 0.965 0.925 0.897 0.80 0.75
2+
2+
Mg , Be
0.8
0.872 0.756 0.690 0.52 0.44
2+
2+
2+
2+
2+
2+
2+
Ca , Cu , Zn , Sn , Mn , Fe , Ni ,
0.6
0.870 0.748 0.676 0.48 0.40
Co2+, Phthalate2Sr2+, Ba2+, Cd2+, Hg2+, S20.5
0.869 0.743 0.668 0.46 0.38
2+
22Pb , CO3, SO3 , C2O4
0.45 0.868 0.741 0.665 0.45 0.36
2+
2222Hg2 , SO4 , S2O3 , CrO4 , HPO4
0.40 0.867 0.738 0.661 0.44 0.35
3+
3+
3+
3+
3+
Al , Fe , Cr , La , Ce
0.9
0.737 0.540 0.443 0.24 0.18
33PO4 , Fe(CN)6
0.4
0.726 0.505 0.394 0.16 0.095
4+
4+
4+
4+
Th , Zr , Ce , Sn
1.1
0.587 0.348 0.252 0.10 0.063
4Fe(CN)6
0.5
0.569 0.305 0.200 0.047 0.020
Ex.10-3 Use Eq. 10-5 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 γHg2+ obtained by interpolation of data
from Table 10-1.
0.51(2) 2 0.085
(1). − log γ Hg =
= 0.4016, γ Hg = antilog(-0.4016) = 0.397 ≈ 0.40
1 + 3.3(0.5) 0.085
2+
2+
(2). Table 10-1: μ = 0.100 → γHg2+ = 0.38; μ = 0.050 → γHg2+ = 0.46
γ Hg 2 + = 0.38 +
0.015
(0.46 − 0.38) = 0.404 ≈ 0.40
0.050
50
10B-3 Equilibrium Calculations using Activity Coefficients
Ex. 10-4 Find the relative error introduced by neglecting activities in calculating the
solubility of Ba(IO3)2 in a 0.033 M solution of Mg(IO3)2. Ksp of Ba(IO3)2 is
1.57 × 10-9.
aBa 2 + ⋅ a 2
IO3-
K sp' =
= K sp = 1.57 × 10−9 = [Ba 2 + ]γ Ba 2 + ⋅ [IO3- ]2 γ 2
IO3-
K sp
(γ Ba 2 + )(γ
2
IO 3-
)
= K sp
= [ Ba 2 + ][ IO 3- ] 2
1
1
μ = ([Mg 2+ ] × 22 + [IO3- ] ×12 ) = (0.033 × 4 + 0.066) = 0.099 ≈ 0.1 M
2
2
Table 10-1: μ = 0.100 → γBa2+ = 0.38; γ - = 0.77
IO3
K sp' =
1.57 × 10 −9
= 6.97 × 10 − 9 = [Ba 2 + ][IO3- ]2
2
0.38 × 0.77
[ IO 3- ] = 2 × 0.033 + 2 [ Ba 2 + ] ≈ 0.066
[Ba2+]×(0.066)2 = 6.97 × 10-9; [Ba2+] = solubility = 1.60 × 10-6
If neglect activities [Ba2+] × (0.066)2 = 1.57 × 10-9; [Ba2+] = solubility = 3.60 × 10-7
3.60 × 10 −7 − 1.60 × 10 −6
relative error =
× 100% = −77%
1.60 × 10 − 6
Ex. 10-5 Use activities to calculate the [H3O+] in a 0.120 M HNO2 solution that is
also 0.050 M in NaCl. What is the relative percent error incurred by neglecting
activity corrections?
1
μ = (0.0500 × 12 + 0.0500 × 12 ) = 0.0500 M
2
μ = 0.050 → γ
γ HNO 2 = 1.0
= 0.85 ; γ NO - = 0.81;
H 3O +
2
K a ⋅ γ HNO2
[H 3 O + ][ NO -2 ]
7.1×10 −4 ×1.0
'
Ka =
=
=
= 1.03 ×10 −3
[HNO 2 ]
0.85 × 0.81
γ H O+ γ NO-
(
3
)( )
2
[ H 3 O + ] = 0.120 × 1.03 × 10 −3 = 1.11 × 10 −2 M
If assuming all γ = 1: [H 3O + ] = 0.120 × 7.1× 10 − 4 = 9.2 × 10-3
relative error =
9.2 × 10 −3 − 1.11 × 10 −2
× 100% = −17%
1.11 × 10 − 2
10B-4 Omitting Activity Coefficients in Equilibrium Calculations
51