Electrolyte Electrolyte SolutionSolution

Muhammad Abbas Ahmad ZainiPhD, CEngCentre of Lipids Engineering & Applied Research, UTM

Week Topic Topic Outcomes6-7 Electrolyte Solution

•The enthalpy, entropy and Gibbs energy of Ion formation in solution

•Activities and activity coefficient

•The Debye-Hückel theory of electrolyte solution

•Chemical equilibrium in electrolyte solution

It is expected that students are able to:

•Define and determine activities and activity coefficient.

•Evaluate mean ionic chemical potential in electrolyte and its characteristic by Debye-Hückellimiting law.

Topic Outcomes

Introduction

Solutions are homogenous mixtures of 2 or more pure substances

In solution, the solute is dispersed uniformly throughout the solvent.

Formation of Solution

Solvent molecules attracted to surface ions.

Each ion is surrounded by solvent molecules.

Enthalpy (∆H) changes with each interaction broken or formed.

Ionic solid dissolving water

Ions are solvated (surrounded by solvent)

If the solvent is water, the ions are hydrated

The intermolecular force here is ion-dipole

Terminology

Electrolyte solutions – solution that can conduct the electricity

Electrolyte – compound that if dissolved in water can to ionized

Ionization process ⇒ Produce +ve and –ve ion.⇒ Charge of the ions that conduct the electricity from

1 electrode to other electrode

Solution

Solution

Electrolyte

Weak electrolyte

Strong electrolyte

NonelectrolyteIdeal & real solutions

Neutral solutes

Strong ElectrolyteProduce ions & conduct electricity; undergo the completely ionized.

NaCl(s) → Na+(aq) + Cl–(aq)

CaBr2(s) → Ca2+(aq) + 2Br–(aq)

H2O

H2O

100% ions

If its test by using electrolyte testerwill produce light lamp and there are gas bubble

Weak Electrolyte

Conduct electricity by weak; undergo half ionized and produce a few ions

HF(g) + H2O ↔ H3O+(aq) + F– (aq)

CH3COOH ↔ H+ + CH3COO –

Non-electrolyte

Solutions that can’t conduct electricity; do not produce ions

Glucose, urea & alcohol

Dissolve as molecules in solution

E.g.

H, S, GH, S, G of ion of ion formation in formation in

solution solution

Thermodynamics of Electrolyte

The thermodynamics of electrolyte solutions is

important for a large number of chemical systems

Acid-base chemistry

Bio-chemical processes

Electrochemical reactions

Materials that dissociate into positively and negatively

charged mobile solvated ions when dissolved in an

appropriate solvent.

Solvation Shell

More energy is gained in the reorientation of the dipolar water molecules around the ions in the solvation shell

Energy flow into the system is needed to dissociate and ionize hydrogen and chlorine

Note: ∆Hof for a pure element in its standard state = 0; A solvation shell is a shell of any

chemical species that acts as a solvent and surrounds a solute species

1/2 H2(g) + 1/2 Cl2(g) H+(aq) + Cl–(aq)

Solvation shell is essential in lowering the energy of the ions thus making the reaction spontaneous

∆HR=-167.2 KJ/mol

Reaction is exothermic

Heat of Reaction

Standard state enthalpy in terms of formation enthalpies,

( ) ( )aq,ClΔHaq,HHH ffreaction−+ +Δ=Δ ooo

Note: ∆Hof for a pure element in its standard state = 0

No contribution of H2(g) & Cl2(g) (pure element) to ∆Ho

f

Cannot be measured directly by calorimetric experiment

How to obtain the information of solvated cations and anions??

Thermodynamics Functions for cations and anions

Can be obtained by making an appropriate choice for the zero of ∆Ho

f, ∆Gof and So

m.

( ) 0aq,HGf =Δ +o

( ) ( ) 0T

aq,HGaq,HSP

f =⎟⎟⎠

⎞⎜⎜⎝

⎛∂

Δ∂−=

++

oo

( ) ( ) ( ) 0aq,HTSaq,HGaq,HH ff =+Δ∂=Δ +++ ooo

and

For all T

∆Horxn, ∆Go

rxn, ∆Sorxn

1/2 H2(g) + 1/2 Cl2(g)→ H+(aq) + Cl– (aq)

From previous reaction,

( )aq,ClHH frxn−Δ=Δ oo

( )aq,ClGG frxn−Δ=Δ oo

( ) ( ) ( )g,ClS1/2g,HS1/2aq,ClSS 2m2mmrxnoooo Δ−Δ−Δ=Δ −

Example: NaCl

NaCl (s) → Na+ (aq) + Cl– (aq)

( ) ( ) ( )( ) ( )

1rxn

111

fffrxn

3.90kJmolH

411.2kJmol240.1kJmol167.2kJmol

sNaCl,Haq,NaHaq,ClHH

−

−−−

+−

+=Δ

−−−+−=

Δ−Δ+Δ=Δ

Values of ∆Gof and So

m can be determined in similar manner

Note: ∆Hof, conventional formation enthalpies; ∆Go

f, conventional Gibbs energies formation; So, conventional formation entropies

Note∆Ho

f, ∆Gof & So

m for ions are defined relative to H+(aq)

∆Hof = –ve; Formation of the solvated ion is more

exothermic than the formation of H+(aq)

Multiply charged ions and smaller ions more exothermic because stronger electrostatic attraction with water in the solvation shell

Entropy decreases as the hydration shell is formed because water molecules are converted to relatively immobile molecules

⇒ Larger charge-size-ratio than H+(aq).⇒ E.g. Mg2+(aq), Zn2+(aq), PO3–

4(aq) ⇒ Solvation shell is more tightly bound

Example 1

Calculate ∆Horeaction, ∆So

reaction and ∆Goreaction for

the reaction AgNO3(aq) + KCl(aq) → AgCl(s) +

KNO3(aq)

Check Your Understanding

Calculate ∆Horeaction, ∆So

reaction and ∆Goreaction for the

reaction Ba(NO3)2(aq) + 2KCl(aq) → BaCl2(s) +

2KNO3(aq)

Thermodynamics of Ion Formation & Solvation

Earlier, ∆Hof, ∆Go

f & Som cannot be determined for

an individual ion in a calorimetric experiment.

Now, the thermodynamic functions associated with

individual ions can be calculated with reasonable

level confidence using a thermodynamic model.

Allows ∆Hof, ∆Go

f & Som values to be converted

to absolute values for individual ions.

Example: Individual Contribution to ∆Gof

( ) ( )( ) ( )

( ) ( )( ) ( )( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( ) 1rxn22

solvationrxn

solvationrxn

1rxn

1rxn

1rxn2

1rxn2

131.2kJmolG aqCl aqHg1/2Clg1/2H

aq,HGG aqHgH

aq,ClGG aqClgCl

349kJmolG gCl egCl

1312kJmolG egH gH

105.7kJmolG gClg1/2Cl

203.3kJmolG gHg1/2H

−−+

+++

−−−

−−−

−−+

−

−

−=Δ+→+

Δ=Δ→

Δ=Δ→

−=Δ→+

=Δ+→

=Δ→

=Δ→

o

oo

oo

o

o

o

o

Dissociation

Formation of ions

Analyze the formation of H+(aq) & Cl– (aq)

The change in the Gibbs energy for overall process,

( ) ( ) 1solvationsolvationrxn 1272kJmolaq,HGaq,ClGG −+− +Δ+Δ=Δ ooo

Pathway

• Play important role in the determination of the Gibbs energies of ion formation

• Can be estimated using Born model

1/2 H2(g) + 1/2 Cl2 → H+(aq) + Cl– (aq)

∆Gorxn

( ) ( )1

solvationsolvation

rxn

1272kJmolaq,HGaq,ClG

G

−

+−

+

Δ+Δ

=−=Δoo

o molKJ /2.131

Determination of ∆Gosolvation

εr4Q'π

φ =

Wnonexp, rev associated with solvation can be calculated, ∆G for the process is known.

Consider, neutral atom A gains the charge q, first in a vacuum and then in a uniform dielectric medium.

∆Gosolvation of an ion with a charge q = Wrev

(A(g)→Aq(aq))solvation – rev. process (A(g) → Aq(g))vacuum

Electrical potential around sphere,Charge Q’

Radius r

Note: Wnon-exp, rev, non-expansion work for a reversible process

Work

∫∫ ===Q

0 0

2

0

Q

0 0 rε8Qdq'Q'

rε41

rε4dq'Q'w

πππ

The work in charging a neutral sphere in vacuum to the charge q’

Permittivity of free space

The work of the same process in a solvent

rεε8Qw

r0

2

π=

Relative permittivity (dielectric const.) of the solvent

Born Model (∆Gosolvation)

⎟⎟⎠

⎞⎜⎜⎝

⎛−=Δ 1

ε1

rε8NezG

r0

A22

solvation πo

For an ion of charge Q = ze, ∆Gosolvation

Note: Values for εr for number of solvents, Table 10.2 (App. A, Data Tables)

Because εr >1, ∆Gosolvation < 0

⇒ Solvation is spontaneous process

Avogadro’s const.Charge number of the ion

radiusRelative permittivity (dielectric const.) of the solvent

Born Model (For Water)

( )14

i

2i

solvation kJmol106.86rzG −××−=Δ o

∆Gosolvation < 0 is strongly negative for small, highly

charged ions in media of high relative permittivity.

Ref.: Atkins, P., Paula, J. D. (2006). Physical Chemistry. 8th ed. W.H Freeman and Company. N. Y.

For water at 25°C,

Example 2

Calculate ∆Gosolvation in an aqueous solution for Cl– (aq)

using the Born model. The radius of the Cl– ion is 1.81 ×

10–10m.

Using the Born Equation

( ) ( ) ( )1

14solvsolv

67kJmol

kJmol106.862201

1811IGClG

−

−−−

−=

××⎟⎠⎞

⎜⎝⎛ −−=Δ−Δ oo

To see how closely the Born equation reproduces the experimental data, we calculate the difference in the values of ∆Go

f for Cl− and I− in water, for which εr= 78.54 at 25°C, given their radii as 181 pm and 220 pm (Table 20.3*), respectively, is

*Ref.: Atkins, P., Paula, J. D. (2006). Physical Chemistry. 8th ed. W.H Freeman and Company. N. Y.

This estimated difference is in good agreement with the experimental difference, which is −61 kJ mol−1.

Activities & Activities & activity activity

coefficient for coefficient for electrolyte electrolyte

solutionsolution

Thermodynamics of Ions in Solutions

Deviations of electrolyte solution from ideal behavior occur at molalities as low as 0.01 mole/kg

Thermodynamic properties of ionic species in solution?

a ln RT μμ += o

Previously, for the H+(aq) ion, we defineo ΔH°f = 0 kJ/mole at all To S°m = 0 J/(K mole) at all To ΔG°f = 0 kJ/mole at all T

occγa =

Activities & Activity Coefficient

Activity & activity coefficient of component of real solution is Not Valid for electrolyte solutions.

Solute-solute interactions are dominated by long range electrostatic forces present between ions in electrolyte solutions

Activity & activity coefficient must be formulated differently for electrolytes to include the Coulomb interactions among ions.

( ) ( ) ( ) ( )aqClaqNalOHsNaCl 2−+ +→+

• NaCl completely dissociated• Solute-solute interactions are electrostatic

in nature

Activities in Electrolyte Solutions

Consider 1 mole of an electrolyte dissociating into ν+

cations & ν- anions, Gibbs energy of the solution,

Note: subscript “+”, cation; “–”, anion

solutesolutesolventsolvent μnμnG +=

In general dissociates completely, −+ vv BA

( )−−++

−−++

++=

++=

μvμvnμn μnμnμnG

solutesolventsolvent

solventsolvent

v+, v- are stoichiometric coefficients of the cations & anions, produced upon dissociation of the electrolyte

Mean Ionic Chemical PotentialSince, v= v+ + v–

−−++ += μvμvμsolutefor a strong electrolyte

Mean ionic chemical potential μ± for the solute

vμvμv

vμμ solute −−++

±+

==

Next task is to relate the chemical potentials of the solute & its individual ions to the activities of these species.

Mean Ionic ActivityDefine the activities, a ln RT μμ += o

For the individuals ions

+++ += a ln RT μμ o−−− += a ln RT μμ o

Note: The standard chemical potentials of the ions (μo+ & μo

–) are based on Henry’s law standard state

vμvμv

vμμ solute −−++

±+

==

±±± += a ln RT μμ o

For the ideal dilute solution

Relationship between a & a±

This gives us the relationship between the electrolyte activity & the mean activity

( )±± +=+= a ln RT μva ln RT μμ oo

aav =±

Mean ionic activity a± is related to the individual ion activities by,

vμμ solute=±

( )1/vvvvvv aaa or aaa −+−+−+±−+± ==

Check Your Understanding

Express a± in terms of a+ and a− for a) Li2CO3, b) CaCl2,

c) Na3PO4 and d) K4Fe(CN)6. Assume complete

dissociation.

Express µ± in terms of µ+ and µ− for a) NaCl, b) MgBr2, c)

Li3PO4, and d) Ca(NO3)2. Assume complete dissociation.

vμvμvμ −−++

±+

=

( )1/vvv aaa −+−+± =

Ionic Activity

If the ionic activities are references to the concentration units of molality,

++

+ = γmma

o −−

− = γmma

o

m+ = v+m m– = v– m

Activity is unitless, the molality must be referenced to a standard state conc. ⇒ m° = 1 mol kg–1

In this standard state, Henry’s law (valid in the limit m→0), is obeyed up to a conc. of m = 1 molal.

occγa =( )1/vvv aaa −+

−+± =

Activities in Electrolyte Solutions

−+−+± = vvv aaa

++

+ = γmma

o −−

− = γmma

o

−+

−+

−+−+

± ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛= vv

vvv γγ

mm

mma

oo

Relationship between a±, m± & γ±

±±

±±±

± ⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛= γ

mmaγ

mma v

vv

ooor

Thus, the mean ionic activity is related to the mean ionic activity coefficient & mean ionic molality,

Simplify,

( ) mvvm

mmm1/vvv

vvv

−+

−+

−+±

−+±

=

=

( )1/vvv

vvv

γγγ

γγγ−+

−+

−+±

−+±

=

=

Mean ionic molality, m± Mean ionic activity coefficient, γ±

occγa =

Chemical Potential Expression

±±± += a ln RT μμ o

v±± += a ln RT vμμsolute

o

( )[ ] ±−+± +⎟⎠⎞

⎜⎝⎛++= −+ γ ln vRTmm ln vRTvv ln RT vμμ vv

solute o

o

“Normal” standard state (usually taken to be Henry’s law standard state)

Obtained from the chemical formula for the solute

vv

v γmma ±

±± ⎟

⎠⎞

⎜⎝⎛=

o

This can be factored into 2 parts

±± +⎟⎠⎞

⎜⎝⎛+= γ ln vRTmm ln vRTμμsolute o

oo

The ideal part (associated with γ± = 1)

Deviations from ideal behavior (γ±can be obtained through exp. or measurement on electrochemical cells or theoretical model)

Check Your Understanding

Express γ± in terms of γ+ and γ− for a) SrSO4, b) MgBr2,

c) K3PO4, and d) Ca(NO3)2. Assume complete

dissociation.

( )1/vvv γγγ −+−+± =

Example 3

Calculate the mean ionic activity of a 0.0150m K2SO4

solution for which the mean activity coefficient is 0.465.

(Ans: 0.0111)

Calculate the mean ionic molality & mean ionic activity of

a 0.150m Ca(NO3)2 solution for which the mean ionic

activity coefficient is 0.165.

Calculate the value of m± in 5.0 x 10–4 molal solutions of

a) KCl, b) Ca(NO3)2, and c) ZnSO4. Assume complete

dissociation.

The The DebyeDebye--HHüückelckel Theory Theory

Estimates of Activity Coefficients

Deviations from ideal solution behavior occur at much lower concentration for electrolytes⇒ Long-range electrostatic Coulomb interaction is

more dominant (interaction between the ions)

Cannot be neglected (even for very dilute solutions of electrolytes)

Allow theoretical estimation of the mean activity coefficients of an electrolyte. ⇒ Each has a limited range of applicability.

±±

± ⎟⎠⎞

⎜⎝⎛= γ

mma

o

The Debye-Hückel Limiting LawThis valid for small concentrations (up to 0.010 molal*)

Izz 0.5092log γ −+± −=

( ) ( )∑∑ −−++−−++ +=+= 2ii

2ii

2ii

2ii zmzm

21zvzv

2mI

Izz 1.173lnγ −+± −=

or

The Debye-Hückel Extended Law (reliably estimate the activity coefficients up to a conc. of 0.10 mole/kg*.

B = 1.00 (kg/mole)1/2IB1

Izz 0.510logγ

+

−= −+

±

*Ref: http://people.stfx.ca/gmarango/chem232

0.1

0.06

The Davies Equation

Can reliably estimate the activity coefficients up to a concentration of 1.00 mole/kg*.

*Ref: http://people.stfx.ca/gmarango/chem232

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛+

⎟⎠⎞

⎜⎝⎛

−= −+± o

o

o

mI0.30

mI1

mI

zz 0.510logγ 1/2

1/2

An empirical modification of Debye-Hückel limiting law for high concentration

Check Your Understanding

Calculate ionic strength, I for 0.05

molal Na2SO4.

( )∑ −−++ += 2ii

2ii zvzv

2mI

Example 4

Using the Debye-Hückel limiting law, calculate the of γ± in

5. 0 ×10–3 m of solutions of

a) KCl (Ans: 0.92)

b) Ca(NO3)2

c) ZnSO4 (Ans: 0.52)

Assume complete dissociation

( )∑ −−++ += 2ii

2ii zvzv

2mI

Izz 1.173lnγ −+± −=

Equilibrium constant in terms of activities

Equilibrium Constant for Electrolyte Solution

( ) jv

i

eqiaK ∏=

occγa i

ii =

The activity of a species relative to its molarity

Activity coefficient of species i

Activities rather than concentration must be taken into account to accurately model chemical equilibrium

Consider the range of ionic strengths for which the Debye-Hückel limiting law is valid

Water auto-ionizes (self-dissociates) to a small extent

The Auto-ionization of Water

These are both equivalent definitions of the autoionization reaction.

2H2O(l) ⇌ H3O+(aq) + OH-(aq)

H2O(l) ⇌ H+(aq) + OH-(aq)

Water is amphoteric.

Note: Amphoteric, can act as either an acid or a base

The activity equilibrium constant,

The Auto-ionization Equilibrium

we know a(H2O) is 1.00,

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+−+

22

3

2 O ︶a ︵H

︶a ︵OH ︶Oa ︵H or O ︶a ︵H

︶a ︵OH ︶a ︵H =K

Kw = a(H+) a(OH-)

Ion product const. for water, Kw, is the product of the activities of the H+ and OH- ions in pure water at a temperature of 298.15 K

Kw = a(H+) a(OH-) = 1.0x10-14 at 298K

Solubility Product Constant

The equilibrium constant in terms of molarities for ionic salts is usually given the symbol Ksp

Note: sp, solubility product

Example: Dissociation MgF2

Consider

( ) ( ) ( )aq2FaqMgsMgF 22

−+ +→

The activity of the pure solid can be set equal to 1,

932

FMg2FMgsp 106.4γ

cc

cc

aaK2

2−

± ×=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛==

−+

−+ oo

From the stoichiometry of the overall equation,

+=− Mg2F 2cc

Fc andγ −± ??

Iteration

Solve for cF–, Giving cMg2+

Calculate ionic strength

Recalculate γ±

Final γ± & cMg2+

1=±γ

( )−−++ += mzmz21I 22

Assume

932

FMg2FMgsp 106.4γ

ca

cc

aaK2

2−

± ×=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛==

−+

−+ oo

Converge

Izz 0.5092logγ −+± −=

TutorialCalculate the solubility of BaSO4 (Ksp = 1.08 × 10–10) (a) in

pure H2O and (b) in an aqueous solution with I = 0.0010 mol

kg–1.

At 25°C, the equilibrium constant for the dissociation of acetic

acid, Ka, is 1.75 ×10–5. Using the Debye-Hückel limiting law,

calculate the degree of dissociation in 0.100m and 1.00m

solutions. Compare these values with what you would obtain if

the ionic interactions had been ignored.