1

CHAPTER 6 / HARVEY

EQUILIBRIUM CHEMISTRY

TOPICS• CHEMICAL EQUILIBRIUM• THERMODYNAMICS AND EQUILIBRIUM• MANUPULATING EQUILIBRIUM CONSTANTS• EQUILIBRIUM CONSTANTS FOR CHEMICAL REACTIONS

– Precipitation Reactions– Acid-Base Reactions– Complexation Reactions– Oxidation-Reduction Reactions

• LE CHÂTELIER'S PRINCIPLE• LADDER DIAGRAMS• SOLVING EQUILIBRIUM PROBLEMS• BUFFER SOLUTIONS• ACTIVITY EFFECTS• SUGGESTED PROBLEMS• 6.1, 6.2, 6.4, 6.5, 6.7, 6.12, 6.14, 6.19• APPENDIX 3 AND 4

2

A. EQUILIBRIUM Gulberg and Waage (1867) A reaction is at equilibrium when the rates of the forward and reverse reactions are equal. It is a dynamic state.

baff BAkrate ][][=

dcrr DCkrate ][][=

aA bB cC dD

[ ] [ ][ ] [ ] r

fba

dc

eq kk

BADCK ==

rf raterate =

Method of derivation ok for elementary stepreaction only!

The expression obtained is the correct expression for the equilibrium constant, but the method of derivation has no general validity. Because, reaction rates depend on the mechanism of the reaction! Reaction rates depend on the number of colliding species (molecularity) whereas the equilibrium constant expression depends only on the stoichiometry of the reaction. e.g. S2O8

2− + 3I− → 2 SO42− + I3

− Rate = kf [S2O8

2−] [I−]

3

Change in concentrations of reactants and products as the reaction proceeds What can be predicted from knowledge of the equilibrium constant? 1) the tendency of the reaction to occur and in what direction 2) Not whether it is fast enough to be feasible in practice ** Even with a large K, a reaction may proceed from right to left if sufficiently large concentrations of products are initially present

Na2CO3 + CaCl2 2 NaCl + CaCO3

B. THERMODYNAMICS AND EQUILIBRIUM

A chemical system evolves spontaneously towards its lowest energy state. Energy of a chemical system changes during a chemical reaction. What determines the final position of a reaction? 1) Enthalpy 2) Entropy and temperature

TSHG −= (1) At constant pressure and temperature

STHG ∆−∆=∆ (2)

G∆ change in Gibbs free energy H∆ change in enthalpy (net flow of energy as heat) S∆ change in entropy

T temperature in Kelvins

aA bB cC dD

4

H∆

ST∆

G∆

Forward reaction

Reverse reaction

< 0

> 0

< 0

Spontaneous

< 0

< 0 , ST∆ < H∆

< 0

Spontaneous

< 0

< 0 , ST∆ > H∆

> 0

Spontaneous

< 0

< 0, ST∆ = H∆

0

EQUILIBRIUM

EQUILIBRIUM

> 0

< 0

>0

Spontaneous

> 0

> 0, ST∆ > H∆

< 0

Spontaneous

> 0

> 0, ST∆ < H∆

> 0

Spontaneous

> 0

> 0, ST∆ = H∆

0

EQUILIBRIUM EQUILIBRIUM

Gibbs free energy is a function of concentrations

QRTGG o ln+∆=∆ (3)

[ ] [ ][ ] [ ]ba

dc

BA

DCQ = (4)

oG∆ change in Gibb's free energy under standard -state conditions

Standard-state conditions 298 K 1M solute Pure solid Pure liquid 1 atm partial pressure At equilibrium 0=∆G Therefore KRTGo ln−=∆ (5)

5

Thermodynamic Equilibrium Constant versus Concentration Equilibrium Constant For very dilute/ ideal solutions (mM), in the absence of interactions between reactants, the equilibrium constant can be calculated using concentrations. Equilibrium constant based on concentration is sometimes referred to as concentration equilibrium constant.

[ ] [ ][ ] [ ]beq

aeq

deq

ceq

CBA

DCK = (6)

Real thermodynamic equilibrium constant

bB

aA

dD

cC

aa

aaK = (7)

[ ]AaA γ= (8)

a activity γ activity coefficient

+−= HapH log Molecular solutes have activity coefficients very near unity up to an ionic strength of 0.1. Activity coefficients of pure liquids and solids are equal to one.

Activity coefficients of ionic solutes For ionic solutes the extended Debye-Huckel equation is used to calculate the activity coefficient

µα

µγ

××+

××=−

A

AA

z

3.31

51.0log

2 (9) at 25°C, 1.0≤µ

∑=i

ii zc 221µ (10)

Az charge of the ion Aα effective diameter of the hydrated ion in nanometers

µ ionic strength of the solution It is handy to recognize the following Type of ionic compound

Ionic strength

−+ BA C ++2

2 BA C3

−+ 22 BA C4 ++3

3 BA C6

6

Relationship between concentration equilibrium constant and thermodynamic equilibrium constant

bB

aA

dD

cC

CKKγγ

γγ=

bB

aA

dD

cC

CKKγγ

γγlogloglog −−=−

bB

aA

dD

cC

C pKpKγγ

γγlog+=

BADCC badcpKpK γγγγ loglogloglog +−++=

)3.31

51.0)((

2222

µµ

αααα +++−−+=

B

B

A

A

D

D

C

CC

zb

za

zd

zcpKpK

Measure CK at different ionic strength and extrapolate to zero to obtain K .

µα

µγ

××+

××=−

A

AA

z

3.31

51.0log

2

µα

µγ

××+

××=−

A

AA

z

3.31

51.0log

2

As the ionic strength approaches zero..,.

When charge increase …

When effective diameter of the hydrated ions decreases …

When the mean distance of approach decreases …

C. MANUPULATING EQUILIBRIUM CONSTANTS

C.1 Reverse reaction's equilibrium constant

12

1K

K =

C.2 Overall equilibrium constant

4321 KKKKK overall = If reactions (A + B) = reaction C, the equilibrium constant of reaction C (Kc) is equal to the product of the equilibrium constants for reaction A and B.

C.3 Equilibrium Constants are used for: •Calculation of concentrations of chemical species at equilibrium under conditions where the same information might be difficult to measure or is unobtainable by direct experiment. •Prediction of conditions that will lead to desired results

N i+ 2 N H 3 N i(N H 3)+2

N i(N H 3)+ 2 N H 3 N i(N H 3)2+ 2

N i(N H 3)2+ 2 N H 3 N i(N H 3)3

+ 2

N i(N H 3)3+ 2 N H 3 N i(N H 3)4

+ 2

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D. EQUILIBRIUM CONSTANTS FOR CHEMICAL REACTIONS

D.1 PRECIPITATION REACTIONS Metathesis reaction

33 )( NaNOsAgClNaClAgNO +→+

()()()()()()( 33 aNOaqNasAgClaqClaqNaaqNOaqAg −+−+−+ ++→+++ Net ionic equation

)()()( sAgClaqClaqAg →+ −+ Solubility product : spK

)()()( aqClaqAgsAgCl −+ +⎯→←

== −+ ]][[ ClAgK sp 10108.1 −× 74.9=sppK

D.2 ACID-BASE REACTIONS

D.2.1 Brønsted and Lowry definition Acid: proton donor Base: proton acceptor

D.2.2 Strong and weak acids and bases Strong acids and bases are completely dissociated in water HCl, HNO3, HCLO4, HI, First proton of H2SO4 NaOH, KOH Weak acids and weak bases Weak acids and weak bases are weak electrolytes, they are partially dissociated in water.

Conjugate base

Conjugate acid

Conjugate acid

Conjugate base

C H 3 C O O H H 2 O C H 3 C O O - H 3 O +

N H 3 H 2 O N H 4+ O H -

C H 3 C O O - H 2 O C H 3 C O O H O H -

N H 4+

H 2 O N H 3 H 3 O +

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D.2.3 Acid and base dissociation constants

][]][[

3

33COOHCH

OHCOOCHKa

+−=

][

]][[

3

3−

−=

COOCH

OHCOOHCHKb

wba KOHOHKK ==× −+ ]][[ 3 ion-product constant

D.2.4 Dissociation of water

Water is amphiprotic

14100000.1 −×=wK at 24 °C 1410008.1 −×=wK at 25°C

][][

3+

− =OH

KOH w

14==+ wpKpOHpH

14==+ wba pKpKpK

H2O(l) H2O(l) H3O+(aq) OH-(aq)Autoprotolysis/ self ionization

The tendency of a solvent to accept or donate protons determines the strength of the solute Acid or base

9

• Acidity of HCl and HClO4 in Anhydrous Acetic Acid, a weaker proton acceptor

• Anhydrous Acetic acid is a differentiating solvent for HCl and HClO4.

• Water is a levelling solvent for HCl, HClO4, HNO3, etc…

D.2.5 Other amphiprotic species

D.2.6 Polyprotic acids H2CO3 H3PO4 H2C2O4

HCO3- H2O H2CO3(aq) OH-(aq)

HCO3-(aq) H2O(l) CO3

-2(aq) H3O+(aq)

10

7

32

331 1045.4

][]][[ −

+−×==

COHOHHCO

Ka

11

3

323

2 1069.4][

]][[ −−

+−×==

HCO

OHCOKa

321 aaa KKK ff

H2CO3(aq) H2O(l) HCO3-(aq) H3O+(aq)

HCO3-(aq) H2O(l) CO3

-2(aq) H3O+(aq)

329.102 =apK

352.61 =apK

STRENGTHS OF ACIDS AND BASES (addendum)

• Weak acid (chloroacetic, 1.36E-3) Conjugate base• Weaker acid (acetic, 1.75E-5) Conjugate base• Weakest acid (phenol, 1.0E-11) Conjugate base

• Weak base (Ammonia,1.75E-5) Conjugate acid• Weaker base (Acetate, 5.75E-10) Conjugate acid• Weakest base (oxalate, 1.79E-13) Conjugate acid

11

D.4 OXIDATION-REDUCTION REACTIONS

D.4.1 Definitions Reaction in which electrons are transferred from one reactant to another. The reducing reagent is oxidized (oxidation state increases) The oxidizing reagent is reduced (oxidation state decreases)

Oxalate is oxidized [oxidation state of C changes from +3 to +4] Permanganate is reduced (oxidation state of manganese changes from +7 to +2) Oxalate is the reducing agent Permaganate is the oxidizing agent

5 C2O4-2(aq) 2 MnO4

-(aq) H+(aq) 10 CO2(g) 2 Mn+2 8H2O(l)16

Balancing a Redox Equations1) Split the equation into two half-equations 2) Balance the half-equations with respect to both atoms and charge3) Combine the two half-reactions in such a way as to eliminate electrons_______________________________________________________________a) Balance the atoms of the element being oxidized or reducedb) Balance oxidation number by adding electrons (to the left for the reduction

reaction and to the right for the oxidation reaction)c) Balance charge by adding H+ ions in acidic solution, OH- in basic solutiond) Balance hydrogen by adding H2O moleculese) Check that oxygen is balancedf) Multiply the two half reaction by factors such as the number of electrons

are equal in the two half reactions (i.e. “cross multiply” the number of electrons)

g) Combine the two half reactions and eliminate electrons

Fe2+(aq) + MnO4-(aq) Fe3+(aq) + Mn2+(aq) acidic solution

Cl2 + Cr(OH)3 (s) Cl-(aq) + CrO42- (aq) basic solution

12

D.4.2 Equilibrium Constant for Redox reactions? Because electrochemical potentials are easily measured for redox reactions, they are conveniently used to express equilibrium and to calculate equilibrium constants. Free energy for moving a charge (Q) under the influence of a potential E is given by:

QEG ×=∆ (1)

nFQ = (2)

nFEG −=∆ (3) (J/mol)

n number of moles of electrons per mole of reatant F Faraday's constant (96,485 C.mol–1) Negative sign????

oo nFEG −=∆ (4)

QRTGG o ln+∆=∆ (5)

E° is the standard state potential

Substituting (3) and (4) in (5), yields (6)

QRTnFEnFE o ln+−=− (6) Divide (6) by (−nF)

QnFRTEE o ln−= (7)

1131451.8 −−= molJKR

15.298=T QQ log30258.2ln =

mVVF

RT 16.5905916.030258.2 ==×

Qn

EE o log05916.0−= (8)

At equlibrium E = 0, therefore

KnFRTE o log= (9)

Therefore, if the standard-state potential of a reaction is known, the equilibrium constant of the reaction can be calculated.

.ln not log

Nernst equation

13

Standard-State Electrochemical Potential for the Reaction ( oE ) At equlibrium E = 0, therefore

KnFRTEo log= (9)

Therefore, if the standard-state potential of a reaction is known, the equilibrium constant of the reaction can be calculated. Standard-state potentials for chemical reaction can be calculated using available standard state-potentials of the oxidation and reduction half-reactions.

oox

ored

oreac EEE −=

oredE standard-state reduction potential of the reduced reactant

ooxE standard-state reduction potential of the oxidized reactant

.ln not log

What are Standard-State Reduction Potentials? (Appendix 3D) Standard-state reduction potential of chemical species relative to the reduction potential of the hydronium ion, which by convention is set to 0.000 V.

They express the tendency of chemical species to be reduced relative to the hydronium ion.

VE o

HOH 000.023 / =+

If oE > 0.000, species has a greater tendency to be reduced than H+

Strong oxidizing agents have large positive oE (dioxygen, permanganate)

If oE < 0, species has lesser tendency to be reduced than hydrogen and a greater tendency to be oxidized.

Strong reducing agents have large negative oE (Zinc, Sodium)

)(2)(2 2 gHeaqH ⇔+ −+ VE o 000.0=

)(22 sZneZn ⇔+ −+ VE o 763.0−=

OHeHgO 22 244)( ⇔++ −+ VE o 229.1=

+−+ ⇔+ 23 FeeFe VE o 771.0=

−−− ⇔+ IeI 323 VE o 536.0=

2 H3O+(aq) 2 e- 2 H2O(l) H2(g)

14

Reference Electrodes: SHE and SCE)

ESCE = 0.244 V at 25°C

Hg2Cl2(s) + 2e- ↔2Hg(l) + 2 Cl-

[ ] [ ]22/ log

205916.02682.0log

205916.0

22

−− −=−= ClClEE oHgClHg

Calculate the equilibrium constant of the following reaction

−+−+ +⇔+ 323 232 IFeIFe

1) Standard-State reaction potential 2) Equilibrium constant

15

E. LE CHÂTELIER PRINCIPLE Predicting qualitatively the effects of changes in temperature, pressure, total concentration of reactant or total concentration of products (stresses) on the equilibrium Le Chatelier's principle

Equilibrium shifts in the direction that tends to relieve the 'stress'.

The mass-action effect

An equilibrium shift caused by changing the amount of one of the participating species is called the mass-action effect.

Examples 1) Adding reactant 2) Adding product 3) Common ion effect 4) Effect of complexation on solubility 5) Dilution 6) Concentrating

F. LADDER DIAGRAM Systematic quantitative approach involves elaborate calculations

F.1 ACID-BASE EQUILIBRIA Species versus pH

][]][[ 3

HAOHA

Ka+−

=

][][log

HAApKapH−

+=

][

][−

+=A

HApKpOH b

HA H2O A- H3O+

A- H2O HA OH-

][

]][[−

−=

A

OHHAKb

16

Amino AcidR group

zwitterion

F.2 COMPLEXATION EQUILIBRIA Species versus p(ligand) Mn+ (aq) + L (aq) ↔ M(L)n+

]][[

])([1

LM

LMKn

n

+

+=

])([

][loglog 1 +

++=

n

n

LM

MKpL

17

F.3 OXIDATION-REDUCTION EQUILIBRIA Species versus E

][][Relog05916.0

Oxd

nEE o −=

Ox n e Red

18

Indicators [ ][ ]+

+

−= 4

2

log2

05916.0SnSnEE o

G. SOLVING EQUILIBRIUM PROBLEMS

G.1 SIMPLE CASES Solubility of ions in absence of pH effect Solubility of AgCl in distilled water Ksp = 1.8 x 10 -10 AgCl(s)

Ag+ Cl-

Initial concentrations

solid 0 0

Change in concentration

solid x x

Ksp = [Ag+][Cl-]=x2

Equilibrium solid x x

[ ] MKAg sp51034.1 −+ ×==

19

Solubility of calcium fluoride

CaF2 Ca2+ F-

Initial

Solid

0

0

Change

Solid

x

2 x

Equilibium

solid

0 + x = x

0 + 2x = 2x

113222 109.34)2)((]][[ −−+ ×==== xxxFCaK sp

Mx 41014.2 −×= the solubility of CaF2 is M41014.2 −× .

G.2 USING APPROXIMATIONS

G.2.1 Calculate the hydronium and hydroxide concentrationsin 0.200 M aqueous NaOH.

20

G.2.2 Calculate the pH of a 0.100 M acetic acid solution

02468

1012141618

0 1 2 3 4 5

Log (CHA/Ka)

%er

ror

Series1

[ ][ ]+

+

−=

OHCOHK

HAa

3

23

[ ] HAaCKOH =+3

G.2.3 Common ion effect• Calculate the molar solubility of Ba(IO3)2 in a solution that is

0.0200 M in Ba(NO3)2.

21

G.3 SYSTEMATIC APPROACH TO SOLVING EQUILIBRIUM PROBLEMS

Use• Equilibrium Constants expressions for all

relevant equilibria• Mass balance equations (conservation of

mass)• Charge balance equations (solution

electroneutrality)• Reasonable assumptions.

G.4 pH of Monoprotic Acids and Bases

1.00 M NH3

22

G.5 pH of Polyprotic Acids and Bases Phosphoric acid

148.21 =apK , 31 1011.7 −×=aK

199.72 =apK , 8

2 1032.6 −×=aK

35.123 =apK , 133 105.4 −×=aK

][][][][ 3

4244243

−−− +++= POHPOPOHPOHC T

1321 =+++ αααα o

TCPOH ][ 43

0 =α

TCPOH ][ 42

1−

=α

TCHPO ][ 2

42

−=α

TCPO ][ 3

43

−=α

[ ][ ] [ ] [ ] 321321

231

33

33

aaaaaa

oKKKOHKKOHKOH

OH+++

=++

α

[ ][ ] [ ] [ ] 321321

231

33

231

1

aaaaaa

a

KKKOHKKOHKOHOHK

+++=

++α

[ ][ ] [ ] [ ] 321321

231

33

3212

aaaaaa

aa

KKKOHKKOHKOHOHKK

+++=

++

+

α

[ ] [ ] [ ] 3213212

313

3

3213

aaaaaa

aaa

KKKOHKKOHKOHKKK

+++=

++α

8 18E 011 82E 012 88E 074 05E 181 00E 134 50E 136 32E 087 11E 0313 00

5.87E-014.13E-012.07E-069.18E-173.16E-134.50E-136.32E-087.11E-0312.50

3.10E-016.90E-011.09E-051.53E-151.00E-124.50E-136.32E-087.11E-0312.00

1.25E-018.75E-014.38E-051.95E-143.16E-124.50E-136.32E-087.11E-0311.50

4.31E-029.57E-011.51E-042.13E-131.00E-114.50E-136.32E-087.11E-0311.00

1.40E-029.85E-014.93E-042.19E-123.16E-114.50E-136.32E-087.11E-0310.50

4.47E-039.94E-011.57E-032.21E-111.00E-104.50E-136.32E-087.11E-0310.00

1.41E-039.94E-014.97E-032.21E-103.16E-104.50E-136.32E-087.11E-039.50

4.43E-049.84E-011.56E-022.19E-091.00E-094.50E-136.32E-087.11E-039.00

1.36E-049.52E-014.76E-022.12E-083.16E-094.50E-136.32E-087.11E-038.50

3.89E-058.63E-011.37E-011.92E-071.00E-084.50E-136.32E-087.11E-038.00

9.48E-066.66E-013.33E-011.48E-063.16E-084.50E-136.32E-087.11E-037.50

1.74E-063.87E-016.13E-018.62E-061.00E-074.50E-136.32E-087.11E-037.00

2.37E-071.67E-018.33E-013.71E-053.16E-074.50E-136.32E-087.11E-036.50

2.67E-085.94E-029.40E-011.32E-041.00E-064.50E-136.32E-087.11E-036.00

2.79E-091.96E-029.80E-014.36E-043.16E-064.50E-136.32E-087.11E-035.50

2.82E-106.27E-039.92E-011.40E-031.00E-054.50E-136.32E-087.11E-035.00

2.83E-111.99E-039.94E-014.42E-033.16E-054.50E-136.32E-087.11E-034.50

2.80E-126.23E-049.86E-011.39E-021.00E-044.50E-136.32E-087.11E-034.00

2.72E-131.91E-049.57E-014.26E-023.16E-044.50E-136.32E-087.11E-033.50

2.49E-145.54E-058.77E-011.23E-011.00E-034.50E-136.32E-087.11E-033.00

1.97E-151.38E-056.92E-013.08E-013.16E-034.50E-136.32E-087.11E-032.50

1.18E-162.63E-064.16E-015.84E-011.00E-024.50E-136.32E-087.11E-032.00

5.22E-183.67E-071.84E-018.16E-013.16E-024.50E-136.32E-087.11E-031.50

1.89E-194.20E-086.64E-029.34E-011.00E-014.50E-136.32E-087.11E-031.00

alpha3alpha2alpha1alpha0[H+]ka3Ka2Ka1pH

23

Mole fractions vs pH

0.00E+00

1.00E-01

2.00E-01

3.00E-01

4.00E-01

5.00E-01

6.00E-01

7.00E-01

8.00E-01

9.00E-01

1.00E+00

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00

pH

G.6 EFFECTS OF COMPLEXATION ON SOLUBILITY The solubility of a precipitate can increase dramatically in the presence of reagents that form complexes with the anion or the cation of the precipitate. Calculate the molar solubility of BaSO4 in a solution in which [H3O+] is 0.10 M, 0.50 M, 1.0 M, 2.0 M. The Ksp of BaSO4 is 1.1 x 10-10 The Ka of HSO4

− is 1.02 x 10-2

24

F. BUFFER SOLUTIONS

H.1 Definition A buffer is a solution of a conjugate acid/base pair that resists changes in pH when a strong acid or base is added, or when the solution is diluted.

H.2 Henderson-Hasselbach equation

HA

Aa C

CpKpH log+=

H.3 Buffer Capacity

Number of moles of a strong acid or a strong base that causes a 1.00 L of the buffer to undergo a 1.00-unit change in pH.

Depends not only on the concentrations of the conjugate acid and base, but also on their ratio.

Falls off rapidly as the concentrations ratio departs from unity. Therefore, the pKa of the acid chosen for an application should be within one unit of the desired pH for the buffer to have reasonable buffer capacity.

Buffer Capacity

• Calculate the buffer capacity for a 0.10M TRIS buffer of pH = 9.075. The pKa of TRIS is 8.075. Ka= 8.41x10-9.

• TRIS: (HOCH2)3NH2.• Buffer capacity at pH 8.075?

0.00811-0.008111.000Change

0.0990.0009910.075Final

0.09090.009099.075Initial

(M)(M)

TRISTRISH+pH

25

Buffers systems

12.137.1992.148Phosphoric acid

5.408KHP

5.4082.95Phtalic acid

8.075TRIS

9.244Ammonia

pKa3pKa2pKa1Substances