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Lecture 18 Electrolyte Solutions - Debye-Huckel Theory. Charge neutrality Electrostatics effects Charge distribution Activity coefficient Thermodynamics functions. Charge neutrality. When ionic material dissociates in the solvent, it has to stay charge neutral - PowerPoint PPT Presentation
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Lecture 18 Electrolyte Solutions - Debye-Huckel Theory
Charge neutrality Electrostatics effects Charge distribution Activity coefficient Thermodynamics functions
Charge neutralityWhen ionic material dissociates in the solvent, it has to stay charge neutral
Where zi is the charge of an ion and and ni is concentration
Considering equilibrium between ionic solid and solution, e.g., for NaCl
Which means that due to charge neutrality only the sum of the chemical potentials for anions and cations is defined, and not independent chemical potentials
€
zini = 0i
∑
€
μNasd + μCl
sd ⇔ μNasol + μCl
sol
Chemical potentialWe express chemical potential as
This definition is slightly different than before
vs
But the difference can be absorbed into the definition of
For electrostatic problem one can split the activity coefficient into one due to short range interactions and one due to electrostatic (long range) interactions
€
μ jsol = μ j
0( )
sol+ kT lnn j + kT lnγ
€
kT ln n j
€
kT ln X j
€
μ j0
( )sol
€
μ jsol = μ j
0( )
sol+ kT lnn j + kT lnγ + kT lnγ E
Electrostatics
Electrostatic potential around a single charge in a medium with a dielectric constant
With many ions the potential will be modified by the time averaged potential due to other ions. It has to satisfy the Poisson equation
Where is the charge density around ion J
€
ϕ J (r) =qJ
εr
€
∇2ϕ J (r) =−4π
ερ e (r)
€
ρe (r)
Charge distribution In spherical coordinates
The charge distribution is connected with potential via Boltzmann distribution
Where the sum is over all species. Expanding exponential and inserting to the top equation
Note that the first term in the expansion is zero due to charge neutrality
€
1
r
d2 rϕ J (r)( )dr2
=−4π
ερ e (r)
€
ρe (r) = e zini
i
∑ e−zi eϕ J (r) / kT
€
1
r
d2 rϕ J (r)( )dr2
=−4πe
εzini
i
∑ (1− zieϕ J (r) /kT)
Charge distribution - II Using charge neutrality condition
Or
Where
Is so called ionic strength
€
1
r
d2 rϕ J (r)( )dr2
=4πe2
εkTnizi
2
i
∑ ϕ J (r)
€
1
r
d2 rϕ J (r)( )dr2
= κ 2ϕ J (r)
€
κ 2 =4πe2
εkTnizi
2
i
∑
Charge distribution - II Equation
Has a physical solution
Where A is constant that can be evaluated requiring that total charge in the cloud around ion J is negative zJ leading to
Where a is the distance of minimum approach between cation and anion
€
1
r
d2 rϕ J (r)( )dr2
= κ 2ϕ J (r)
€
ϕ J (r) =Ae−κr
r
€
ϕ J (r) =zJe
ε
eκa
1+ κa
e−κr
r
Screening length
The inverse of is the Debye screening length over which the ion is neutralized by the cloud of other ions. Remembering that
One can see that the Debye length is long at high T and low ion concentrations. Large dielectric constant also promotes long Debye length as the interactions between ions are weaker. When the Debye length is ~ ion size the theory does not apply. Why?
€
κ 2 =4πe2
εkTnizi
2
i
∑€
κ
Excess chemical potential
To find our excess chemical potential we can use so called charging process where initially neutral ion is charged from 0 to its final charge qJ=ezJ. We can calculate the work done during the charging process as
With
Upon integration
€
Wel =0
1
∫ dr4πr2
a
∞
∫d ξzi( )eρ J (r)
r
€
ρJ (r) =−eκ 2ξzJe
κa
4πrε 1+ κa( )e−κr
€
μJex = Wel = −
zJ2e2
2ε
κ
1+ κa
Activity coefficientWith
The activity coefficient is
Which for dilute solutions becomes
€
μJex = Wel = −
zJ2e2
2ε
κ
1+ κa
€
lnγ JE =
μ Jex
kT= −
zJ2e2
2εkT
κ
1+ κa
€
lnγ JE = −
zJ2e2κ
2εkT
€
κa → 0
Excess Gibbs free energy
Where the last equality comes from
€
Gex = μ iex
i
∑ N i = −e2κ
2εzi
2
i
∑ N i = −Ve 2κ
2εzi
2
i
∑ ni = −Vκ 3
8πkT
€
κ 2 =4πe2
εkTnizi
2
i
∑
Osmotic pressureFrom math and thermodynamics
Thus
Integrating both sides over dV from infinite volume to V
€
∂G
∂V
⎛
⎝ ⎜
⎞
⎠ ⎟N ,T
=∂G
∂P
⎛
⎝ ⎜
⎞
⎠ ⎟N ,T
∂P
∂V
⎛
⎝ ⎜
⎞
⎠ ⎟N ,T
= V∂P
∂V
⎛
⎝ ⎜
⎞
⎠ ⎟N ,T
€
∂P
∂V
⎛
⎝ ⎜
⎞
⎠ ⎟N ,T
=1
V
∂G
∂V
⎛
⎝ ⎜
⎞
⎠ ⎟N ,T
€
∂P
∂V ⎛ ⎝
⎞ ⎠N ,T
=∞
V
∫1
V
∂G
∂V ⎛ ⎝
⎞ ⎠N ,T∞
V
∫ = P(V ) − P(∞) = P(V )
Osmotic pressure -2