Acids, bases and solution equilibria

A four lecture course for the 1st year

Jose M. Goicoechea

http://course.chem.ox.ac.uk/acids-bases-and- solutions-equilibria-year-1-2014.aspx http://goicoechea.chem.ox.ac.uk/teaching.html Acid-base reactions

+ + NH3 + H3O NH4 + H2O

+ - H2O + HI H3O + I

- 2- + HSO4 + H2O SO4 + H3O

NH3 + BF3 NH3:BF3

C5H5N + I2 C5H5N:I2

Species highlighted act as acids Redox reactions

A redox reaction is a reaction in which there is a change in oxidation state

Fe3+(aq) + Cr2+(aq) Fe2+(aq) + Cr3+(aq)

+ 2+ Zn(s) + 2H3O Zn + H2(g)

2PCl3 + O2 2OPCl3

Ca(s) + H2 CaH2(s)

+ - O2 + Pt + 3F2 [O2] [PtF6]

One or two electrons are transferred entirely

Species highlighted act as oxidants Definitions of Acid/Base

Arrhenius/Ostwald

Brønsted/Lowry

Lux/Flood

‘Solvent system’

Lewis

Usanovich Arrhenius/Ostwald

Acids and bases

dissociate in H2O, + + releasing H (H3O ) and OH-.

Arrhenius Ostwald

+ - H2O H (aq) + OH (aq)

+ H (aq) is an acid - OH (aq) is a base Brønsted/Lowry

Proton theory retained but the definition is now independent of solvent

Brønsted Lowry An acid is a proton donor and a base is a proton acceptor.

+ - Other solvents are also NH4 + NH2 2NH3 capable of self-ionisation

+ - Proton HCl + NH3 [NH ] Cl transferred from 4 acid to base Donor acid Acceptor base Brønsted/Lowry

Every acid has a conjugate base and every base has a conjugate acid

HA + B- A- + HB

The conjugate base of a weak acid is a strong base, and the conjugate base of a strong acid is a weak base. TRUE FACT! Lux/Flood

A definition for anhydrous/dry systems: used in solid state chemistry

The concept focuses on the oxide ion (O2-)

A base is a oxide donor and an acid is an oxide acceptor.

CaO + SiO2 CaSiO3

The Lux-Flood base is a basic anhydride CaO + H O Ca2+ + 2OH- 2 The Lux-Flood acid is an acid anhydride.

SiO2 + H2O H2SiO3 Solvent system

+ - This definition 2H2O H3O + OH is based on 2NH NH + + NH - solvent 3 4 2 autoionisation + - 2H2SO4 H3SO4 + HSO4 cationsolv anionsolv

An acid is a species that increases the concentration of cationsolv

A base is a species that increases the concentration of anionsolv Lewis

An acid is an electron pair acceptor A base is a an electron pair donor

NH3 + BF3 NH3BF3

H H H N + H N B H B H H H H H H H

Has ‘lone’ pair Has six valence electrons More examples of Lewis acid/base interactions

OH2 3+ 3+ H2O OH2 Al + 6H2O Al H2O OH2 OH2

All metal cations in donor solvents are Lewis acids interacting with solvent molecules which act as Lewis bases

+ + Ag + 2NH3 H3N Ag NH3 Usanovich

A broad definition which encompasses all other acid/base concepts

Usanovich

An acid is a species that Lewis reacts with bases. It gives up cations or Ionotropic accepts anions or electrons

A base is a species that Brønsted Solvent Lowry Arrhenius system reacts with acids. It gives up anions, combines with cations,

or donates electrons. Lux Flood Proton equilibria in water

+ - HX(aq) + H2O H3O + X (aq)

[H O+][X-] K = 3 a [HX]

pKa = –log10Ka

+ pH = –log10[H3O ] Autoprotolysis

- + (aq.) (aq.) 2H2O H3O + OH + - Kw = [H3O ][OH ]

–14 For H2O, Kw = 10 at 25°C

Easy way to treat pKa [H O+][X-] Since K = 3 a [HX]

+ - - Ka = [H (aq.)] when [HX] = [X ] i.e. when HX is 50% converted to X

pKa is pH at which HX is 50% dissociated – Acid HA A Ka pKa Hydriodic HI I– 1011 –11

- 10 Perchloric HCIO4 CIO4 10 –10 Hydrobromic HBr Br– 109 –9 Hydrochloric HCl Cl– 107 –7

- 2 Sulfuric H2SO4 HSO4 10 –2 – 2 Nitric HNO3 NO3 10 –2 + Hydronium ion H3O H2O 1 0.0 – –1 Chloric HCIO3 CIO3 10 1 – –2 Sulfurous H2SO3 HSO3 1.5 × 10 1.81 - 2- –2 Hydrogensulfate ion HSO4 SO4 1.2 × 10 1.92 - –3 Phosphoric H3PO4 H2PO4 7.5 × 10 2.12 Hydrofluoric HF F– 3.5 × 10–4 3.45

– –4 Formic HCOOH HCO2 1.8 × 10 3.75 – –5 Ethanoic CH3COOH CH3CO2 1.74 × 10 4.76 + –6 Pyridinium ion HC5H5N C5H5N 5.6 × 10 5.25 – –7 Carbonic H2CO3 HCO3 4.3 × 10 6.37 – –8 Hydrogen sulfide H2S HS 9.1 × 10 7.04 - 2- –8 Dihydrogenphosphate ion H2PO4 HPO4 6.2 × 10 7.21 + –10 Ammonium ion NH4 NH3 5.6 × 10 9.25 Hydrocyanic HCN CN– 4.9 × 10–10 9.31

– 2– –11 Hydrogencarbonate ion HCO3 CO3 4.8 × 10 10.32 2- 3- –13 Hydrogenphosphate ion HPO4 PO4 2.2 × 10 12.67 Hydrogensulfide ion HS– S2– 1.1 × 10–19 19 Polyprotic acids

A polyprotic acid loses protons in succession, with each deprotonation becoming progressively less favourable.

The behaviour of such species in solution be represented by a distribution diagram

pKa1 = 2.21 pKa2 = 7.21 pKa3 = 12.68

[H3PO4] a(H3PO4) = - 2- 3- [H3PO4] + [H2PO4 ] + [HPO4 ] + [PO4 ] Hammet acidity function H+ + X- HX

For any base B: H+ + B HB+

[BH+] H = pK + – log 0 BH [B]

[BH+] can be measured with a dye [B] Very concentrated solutions: Hammet acidity function HX + Ind IndH+ + X-

NH2

Ind is a coloured indicator e.g.

[Ind] NO2 H = pK + + log 4-Nitroaniline 0 IndH [IndH+]

H0 is the Hammett function of HX

Pure H2SO4, H0 = –11.9 Superacid (HSO3F), H0 = –15.1 Strength of Brønsted acids and bases

Two principal factors control acid/base strength:

1) Inherent properties of the species itself

To quantify this value we must measure proton transfer reactions in the gas phase. Chemistry controlled by ΔH

+ + H (g) + X(g) HX (g) Ap = –ΔH A = proton affinity of X p

2) Environment (e.g. solvation) WORTH REMEMBERING! Proton-Transfer reactions

- + HA(g) + B(g) A (g) + HB (g)

Can be considered as two “half reactions”

ΔHB + + H (g) + B(g) HB (g)

ΔHA + - H (g) + A (g) HA(g)

Then ΔHrxn = ΔHB – ΔHA ΔHrxn = Ap(A) – Ap(B)

The reaction is favourable is Ap(B) > Ap(A) + - H(g) + e + X(g)

EA –ΔHion + - H(g) + X(g)

H(g) + X(g)

–Ap

–BHX

HX(g)

Ap = BHX + ΔHion – EA Ap = BHX + ΔHion - EA

ROW

Decreasing EA

GROUP

Increasing BHX

Decreasing Ap; i.e., increasing acidity Proton affinity (Ap; KJ/mol)

- - - - - H CH3 NH2 OH F 1675 1745 1689 1635 1554 - - - - SiH3 PH2 SH Cl

1554 1552 1469 1395 Ap for - - - - H2O: 723 GeH3 AsH2 SeH Br 1509 1515 1466 1354 I- 1315

- - PH2 and F have almost identical Ap values but their pKa values differ (27 to 3.45, respectively) making F- a much stronger acid in water. Predictions ignoring solvation

- + HI + H2O I + H3O 723 1315

- + NH3 + H2O OH + NH4 872 1635

This can’t be right, other factors must be at play Taking into account solvation effects

B(aq) + H+(aq) BH+(aq)

Large –ΔHhyd. or alternatively… - B (aq) + H+(aq) BH(aq)

The Born formula tells us 2 Large –ΔH –z hyd that AHsolv is proportional to: r Effective proton affinity (Ap′; KJ/mol) and pKa

- - - - - H CH3 NH2 OH F 1380 1351 1188 1150 49 39 15.7 3.2 - - - - SiH3 PH2 SH Cl A ′ for 1283 1090 p H2O: 1130 35 27 7.05 –6.1 - - - - GeH3 AsH2 SeH Br 1079 25 23 3.8 –8 I- 1068 –9 Predictions taking into account solvation

- + HI + H2O I + H3O 1130 1068

- + NH3 + H2O OH + NH4 1195 1188 Range of discrimination

+ - HY + HY H2Y + Y

+ - Kauto = [H2Y ][Y ] pKauto = –logKauto

Acid-base discrimination windows

NH3

H2O

CH3COOH

HF

1400 1200 1000 800

Ap′/KJ/mol Ap vs. Ap′ 1900 - CH3 - NH2 1700 OH- F- - PH2 CN- 1500 Cl- Br- I- 1300

1100

900 py

NH3

700 H2O

500 Solvent levelling

+ + H2F (aq) + H2O HF + H3O (aq)

- - NH2 (aq) + H2O OH (aq) + NH3

+ Strongest acid that can exist is H2Y

Strongest base is Y- Metal aqua ion acidity

(n-1)+ OH2 n+ OH2

H2O OH2 H2O OH2 M M

H2O OH2 H2O OH

OH2 OH2

n+ (n–1)+ + Maq. MOH + Haq.

Arrange in order of increasing acidity:

+ 3+ 2+ 2+ [Na(OH2)6] , [Sc(OH2)6] , [Mn(OH2)6] , [Ni(OH2)6]

+ 2+ 2+ 3+ [Na(OH2)6] < [Mn(OH2)6] < [Ni(OH2)6] < [Sc(OH2)6] Hydrolysis constants (pKhyd) for cations at 25°C

Li+ Be2+ 13.9 6.2 Na+ Mg2+ Al3+ 14.7 11.4 5.0 K+ Ca2+ Sc3+ Ti3+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+ Ga3+ - 12.6 4.7 2.3 6.5 8.7 10.6 10.1 9.6 10.0 7.6 9.5 2.6 V3+ Cr3+ Fe3+ Co3+ 2.6 3.9 2.0 3.2 Rb+ Sr2+ Y3+ Rh3+ Pd2+ Ag+ Cd2+ In3+ - 13.1 8.0 3.2 1.6 11.8 7.9 3.2 Cs+ Ba2+ La3+ Ir3+ Pt2+ Hg2+ Tl+ - 13.3 9.5 4.8 >2.5 2.5 13.3

Burgess, J. Metal Ions in Solution; Ellis Horwood: Chichester, 1978, pp. 264-265 Pauling’s rules for oxoacids

OpE(OH)q

Rule 1: pKa1 = 8 – 5p WORTH REMEMBERING!

The higher the number of oxo- groups bound to the central element, the greater the acidity of the acid. HClO4; O3Cl(OH); pKa = –10

O O

Cl Cl The higher acidity O O O O is due to the O O greater number of resonance structures of the conjugate base

O O

Cl Cl O O O O O O Pauling’s rules to the test

pKa1 = 8 – 5p p = 0 1 2 3 O

Cl Cl HO Cl Cl HO O HO O HO O Hypochlorous Chlorous O Chloric Perchloric O pKa= 7.2 2.0 -1.0 -10 Pauling’s rules for oxoacids

The successive pKa values of polyprotic acids (those with q > 1), increase by five for each successive proton transfer

Rule 2: pKa2 = pKa1 + 5 pKa3 = pKa2 + 5 pKa4 = pKa3 + 5 WORTH REMEMBERING! O pKa1 = 2.1 P OH pKa2 = 7.4 HO OH pKa3 = 12.7

O pK = 1.8 P a1 OH HO pKa2 = 6.6 H

O pKa1 = 2.3 As pKa2 = 6.9 OH HO pKa3 = 11.5 OH Compare the acidities of the following acids: O pK = 8 – 5p = 3 H3PO4 a P OH Real value: 2.12 HO OH

O H SO 2 4 pK = 8 – 5p = –2 S a OH O Real value: –2 OH O

HClO4 Cl pKa = 8 – 5p = –7 HO O O Real value: –10 So what’s up with boric acid?

OH According to Pauling’s first rule for oxo- HO B acids it should have a pK of 8. OH a Tabulated values indicate pKa = 9.2

However, it does not act as a conventional Brønsted acid! Not deprotonated at the hydroxide H

O H Lewis acidity at OH boron results in the formation of OH HO B B an adduct OH OH weakening O–H HO OH bond in water + H+ Poloxo- compounds As the pH of a solution is increased the aqua ions of metals with basic or amphoteric oxides undergo polmerisation and precipitation

4+ OH2 3+ OH2 2+ OH2 OH2 H H2O OH2 H2O OH2 H O O 2 OH2 Fe Fe Fe Fe

H2O OH2 H2O OH H2O O OH H 2 OH2 OH2 OH2 OH2

precipitate

3+ + [Fe(OH2)6] (aq) + (3 + n)H2O Fe(OH)3•nH2O + 3H3O

3+ + [Al(OH2)6] (aq) + (3 + n)H2O Al(OH)3•nH2O + 3H3O Syntheses using acid/base reactions in alternative solvents

PH3 PH3 + NaNH2 NaPH2 + NH3

CH3Cl NaPH2 + CH3Cl CH3PH2 + NaCl

C2H2 C2H2 + NaNH2 NaC2H + NH3

C4H9Cl NaC2H + C4H9Cl C4H9C2H + NaCl From 2012 Prelims: Acids, bases and solution equilibria

LECTURE 2 Lewis acid/base theory A + :B A:B acid base adduct e- pair e- pair complex acceptor donor How do these interactions take place?

In addition to a base with a lone pair of electrons the acid must also have an empty orbital capable of interacting with the lone pair on the base. Electronically saturated complexes will not form an adduct. Examples of Lewis acids (1) H+

H+ + :B [H:B]+

A Brønsted acid exhibits Lewis acidity because it donates H+, which in turn can accept a lone pair of electrons from a Lewis base. Examples of Lewis acids (2) Mn+ Metal cations

n+ n+ M + x:B [M(B)x] acid base adduct metal ligand complex

3+ e.g. aqua ions [Fe(OH2)6] Examples of Lewis acids

Main-group compounds (3) with incomplete ‘octet’

CH3 CH3

B Ga

H3C CH3 H3C CH3

H3C Be CH3 Examples of Lewis acids

Molecules that can arrange valence e- to (4) create a vacant orbital

O O

C + O H C OH

O O Examples of Lewis acids

Compounds of heavy (5) p-block elements

- - PF5 + F [PF6] - - SbCl5 + Cl [SbCl6]

- 2- SiF4 + 2F [SiF6] How Lewis acid/base chemistry influences structure (self acid-base interactions)

H-bonding Lewis base

Lewis acid/base interactions have a strong influence on the physical Lewis acid properties of compounds How Lewis acid/base chemistry influences structure (self acid-base interactions)

Structure of BeCl2

Cl Be Cl Gas phase High temperature

Cl Cl Cl Cl Be Be Be Be Be Cl Cl Cl Cl Solid state One dimensional chain structure How Lewis acid/base chemistry influences structure (self acid-base interactions)

Oligomerisation/Polymerisation

Structure of AlBr3? Br Br Br Al Al Br Br Br

NOT monomeric AlBr3! How Lewis acid/base chemistry influences structure (self acid-base interactions)

Oligomerisation/Polymerisation

+ 3+ - H 2+ Al (aq) [Al(OH2)5(OH)] dimer Brønsted acid Lewis acid

OH2 OH2 H H2O O OH2 Al Al

H2O O OH2 H OH2 OH2 Strength of Lewis acids and bases

Factors: 1) Electronic properties 2) Structure

Techniques: 1) Spectroscopy 2) Electrochemistry 3) Calorimetry Equilibrium constants A + :B A:B [AB] K = f [A][B]

A + nB ABn

[AB ] b = n n [A][B]n Experimental observations in WATER

Mg2+ - - - - Kf: F > Cl > Br > I

Hg2+ - - - - Kf: I > Br > Cl > F

Why??? HARD/SOFT interactions

Concept introduced by R. G. Pearson (1963) Class a compounds = HARD Small, Highly charged, Low polarizability

Class b compounds = SOFT Large, Small or no charge, High polarizability

Observations by Ahrland, Chatt and Davies (1958) Metal + Halide - - TRUE FACT! Class a: F > I Well not really, but Class b: I- > F- close enough Examples Metal + Halide Class a: F- > I- Class b: I- > F-

3+ + 2+ 2+ Al (aq) + HgF (aq) AlF (aq) + Hg (aq)

However,

AlI2+(aq) + Hg2+(aq) Al3+(aq) + HgI+(aq) Everyday examples

Goldschmidt classification of elements Lithophiles: H-H Chalcophiles: S-S

The tarnish on Al is OXIDE The tarnish on Ag is SULPHIDE

- Gold extracted as [Au(CN)2] HARD/SOFT

Hard likes hard Soft likes soft

WHY??? Combination of Intrinsic and Extrinsic effects HARD ACIDS

H+, Li+, Ca2+, Al3+, Ti4+ Non-polarisable, often small and highly charged

HARD BASES

- 2- - 2- 2- F , O , OH , CO3 , SO4 , NH3 Non-polarisable, ions based on 2p elements SOFT ACIDS

Cu+, Ag+, Cd2+, Hg2+, Tl+ Large, small charge, polarisable

SOFT BASES

- - - I , CN , RS , PR3, CO Typically involve elements from 3p and lower rows Irving-Williams series

2+ 2+ [M(H2O)n] + L [M(L)(H2O)n–1] + H2O

Ordering of Kf

Ba2+Zn2+

Approximately in order of decreasing r+ mercaptoamine

Logarithms of the stability constants for the 1:1 complexes between Ba2+ 2+ through Zn and the mercaptoacetic bidentate ligands acid oxalic acid, glycine, ethylenediamine, oxalic acid mercaptoacetic acid, and mercaptoethylamine.

ethylenediamine glycine

Sigel, H.; McCormick, D. B. Acc. Chem. Res. 1970, 3, 201 M2+ Steric effects

Me Me Me B +

N N N Me

Me -71 -74 -42

-1 Hrxn./kJmol More steric effects: Frustrated Lewis Pairs

Borane: Steric bulk prevents association of Lewis acidic Lewis acid and Lewis basic sites:

Phosphine: Unquenched reactivity! Lewis base

Borate

Figure form Science 2006, 314, 1124 Phosphonium Zwitterion Acidity of BX3

The order of thermodynamic stability of complexes

of :N(CH3)3 with BX3 is BF3

Opposite of what one would expect using halogen electronegativites as an argument. F F F

B B B F F F F F F p-orbital overlap between full halogen p orbitals and empty B 2p orbitals must be broken in order to form acid/base adduct Trends in Lewis acidity

SiI4 < SiBr4 < SiCl4 < SiF4

but

BF3 < BCl3 < BBr3 < BI4 ‘Frontier’ orbitals

LUMO Empty acceptor orbital

HOMO Donor orbital. Contains e- pair HARD-HARD

Ion-ion, ion-dipole Non-polarlisable

SOFT-SOFT

Covalent bonding Polarisable HARD ACIDS

H+, Li+, Ca2+, Al3+, Ti4+ Non-polarisable, often small and highly charged

HARD BASES

- 2- - 2- 2- F , O , OH , CO3 , SO4 , NH3 Non-polarisable, ions based on 2p elements SOFT ACIDS

Cu+, Ag+, Cd2+, Hg2+, Tl+ Large, small charge, polarisable

SOFT BASES

- - - I , CN , RS , PR3, CO Typically involve elements from 3p and lower rows HARD SOFT LUMO Hard base has Soft base has high-lying LUMO low-lying LUMO LUMO

HOMO Soft base has high-lying HOMO HOMO

Hard base has low-lying HOMO HARD-HARD

Ion-ion, ion-dipole Non-polarlisable

SOFT-SOFT

Covalent bonding Polarisable Interpretation of hardness and softness

Hard acid-base interactions are predominantly electrostatic.

Soft acid-base interactions are predominantly covalent. π-acidity; π-basicity

Stabilise soft Lewis acids (soft bases)

Stabilise hard Lewis acids (hard bases) What reactions are we measuring?

2+ - 2- Hg (aq) + 4X (aq) HgX4 (aq)

X logK Cl 16 Br 22 I 30 Gas phase

HgF2 + BeI2 BeF2 + HgI2

H (KJ/mol) for MX2(g) → M(g) + 2X(g)

BeF2 1264 HgF2 536

BeI2 578 HgI2 291 Survivor HARD-SOFT H2O H2O H2O H2O

H2O Competition! H2O H O 2 H2O

H O - H2O 2 I - HgF (aq) + I (aq) H2O 2 H2O H2O H2O H2O H2O Very favourable

H2O H2O H2O

H2O - - F H2O HgI (aq) + F (aq) H O 2 2 H2O H2O

Electron density is much higher Stronger enthalpy of solvation Effect of solvent

The solvent is not an innocent bystander… it is involved in the chemical process.

- - HgCl2 + 2Br HgBr2 + 2Cl

Ωm GAS PHASE

In H2O Solvents The solvent dielectric constant is a good guide for the ε importance of the solvent effect with charged species

However this is an oversimplification Need to consider the solvent effect in terrms of Acid/Base interactions

Solvents are characterised by their DONOR numbers and ACCEPTOR numbers DONOR NUMBER

Measured by taking a base :B and determining

Hrxn with SbCl5 in CH2Cl-CH2Cl

SbCl5 + :B SbCl5–B

DONOR NUMBER = –H (kcal/mol) ACCEPTOR NUMBER Measured NMR chemical shifts of 31P in

Et3P=O (triethylphosphine oxide) in the pure solvent

Et3P=O Arbitrary scale:

Hexane = 0

SbCl5 = 100 SOLVENT DN AN ε Pyridine 33.1 14.2 12.3 DMSO 29.8 19.3 45 Diethylether 19.2 3.9 4.3

H2O 18 54.8 81.7 Ethanoic acid - 52.9 6.2 Acetone 17 12.5 20.7 Acetonitrile 14.1 19.3 36 Benzene 0.1 8.2 2.3

CCl4 - 8.6 2.2

SbCl5 - 100 -

CF3COOH - 105.3 - SOLVENT DONOR NUMBER ACCEPTOR NUMBER Acetic acid - 52.9 Acetic anhydride 10.5 - Acetone 17.0 12.5 Acetonitrile 14.1 19.3 Acetyl chloride 0.7 - Antimony pentachloride - 100 Benzene 0.1 8.2 Carbon tetrachloride - 8.6 Chloroform - 23.1 Dichloromethane - 20.4 Diethylether 19.2 3.9 Dimetylsulfoxide (DMSO) 29.8 19.3 Dioxane 14.8 10.8 Ethanol 19.0 37.1 Ethyl acetate 17.1 - Formamide - 39.8 Hexamethylphosphorotriamide (HMPA) 38.8 10.6 Methanol 20.0 41.3 Methyl acetate 16.5 - Methylsulfonic acid - 126.1 Nitrobenzene 4.4 14.8 Nitromethane 2.7 20.5 Phosphorus oxychloride 11.7 - Propanol 18.0 33.5 Pyridine 33.1 14.2 Tetrahydrofuran (THF) 20.0 8.0 Drago + Wayland For: A (g) + B (g) AB (g)

-H = EAEB + CACB

EA, B = ionic susceptibility

CA, B = covalent susceptibility Drago – Wayland parametres E C acids

SbCl5 15.1 10.5

BMe3 12.6 3.48

SO2 1.88 1.65 Iodine 2.05 2.05 bases

NH3 2.78 7.08 Benzene 0.23 2.9 Pyridine 1.17 6.40 Methylamine 1.30 5.88 Drago-Weyland parametres reflect electrostatic (E) and covalent (C) factors, but not solvation Using the Drago – Wayland equation

Me3B + NH3 Me3B-NH3

H = -(EAEB + CACB)

H = -[(12.6 × 2.78) + (3.48 × 7.08)] H = -59.7 kJ/mol

Experimentally: H = -57.5 kJ/mol Superacids A Lewis acid can be used to ‘boost’ the acidity of a Brønsted acid

+ - 2HF + SbF5 [H2F] + [SbF6]

and…

2HSO3F + SbF5

+ - [H2SO3F] + [FSO3-SbF5] Magic acid Generation of reactive carbocations

CH3 CH3

- C + SbF5 C + [SbF6] H H H C H C 3 F 3

Di-Xenon Xe + + - XeF + HF/SbF5 [Xe2] + [Sb4F21] ACID-BASE THEORY: CONCEPTS TO

REMEMBER

Brønsted-Lowry acids/bases: proton transfer; pKa

Proton affinity; Relative proton affinity. Solvent effect.

Pauling’s rules for oxo-acids.

Hard-soft interactions. Effect of solvent.

Measures of solvent polarity and/or

donor/acceptor ability. From 2012 Prelims: Acids, bases and solution equilibria

LECTURE 3 Electron transfer The simplest chemical reaction: Xm + Yn Xm–1 + Yn+1 X = oxidant Y = reductant m, n are formal oxidation states before reaction e.g.:

3+ 2+ 2+ 3+ Fe (aq) + Cr (aq) Fe (aq) + Cr (aq) Atom transfer XO + Y X + YO

X + YH2 XH2 + Y e.g.: V IV III VI - 2- - 2- NO3 + SO3 NO2 + SO4

II I IV 0 CO + H2O CO2 + H2 Redox reactions occur entirely in solution… –I VI 2- + 3RCH2OH + 2CrO4 +10H I III 3RCHO + 2Cr3+(aq)

… or at the interface; e.g. an electrode

2+ - Cu (aq) + 2 e Cu(s) + Aox + H2 Ared + 2H (aq)

1 atm H2 “1 M” H+ (i.e. pH = 0) Standard Hydrogen electron

[A ][H+]2 K = red [Aox][H2] p(H2)

Van’t Hoff equation (–Gө/2.3RT) R = universal gas cte. K = 10 8.314472(15) JK−1mol−1 Write as Half-reactions Electron is “common currency”…

- 1 Aox + 2e Ared + - 2 2H aq + 2e 2H2

overall reaction is: 1 – 2

o ө ө Grxn = GA – GH ө By definition GH = 0 ө ө Grxn = GA

ө ө Grxn = - n F E so... 2.3RT Eө = logK n F KAB Aox + Bred Ared + Box

The two half-reactions are: - ө Aox + ne Ared E A - ө Box + ne Bred E B

n F ө ө logK = (E A – E B) AB 2.3RT Generalise for any system… G = Gө + 2.3RTlogQ Q = reaction quotient 2.3RT [A ] E = Eө - red = log n F [Aox]

2.3RT [A ] = Eө + ox log n F [Ared]

Nernst equation At 298K:

0.059 [Aox] Eө + log n [Ared]

We can now calculate equilibrium constants and direction of reaction for ANY redox reaction

But first consider ‘simple’ electron-transfer reactions To note: ө Gө nE = - 1) ( F ) VOLT EQUIVALENT

2) Convention of direction of half-reaction - Aox + ne Ared

Always REDUCTION POTENTIAL Observe:

I.E.1 < I.E.2 < I.E.3 < … However trends in Eө vary

OXIDATION STATES CONTROLLED BY CHEMISTRY

Hatm, Hcomplexation e.g. solvation FROM PRELIMS 2006: M2+(g) 2+ M2+(solv) Hsolv (M )

I.E.2 ≈ Eө (M2+/+) -Gө -Hө = ≈ nF nF + + + ignoring S M (g) Hsolv (M ) M (solv)

I.E.1 ≈ Eө (M+/0) M(g)

Hatm M(s) Lattimer diagrams

0.95V 1.51V 3+ 2+ MnO2 Mn aq Mn aq

2+ MnO2 Mn aq 0.95 + 1.51 = 1.23V 2 E ө = 0.56 2.26 0.95 1.51 -1.18 - 2- 3+ 2+ 0 MnO4 MnO4 MnO2 Mn Mn Mn

1.70 1.23

Σ(nEө) E = Σn ө - 2+ E MnO4 /Mn

0.56 + (2×2.26) + 0.95 + 1.51 5

= 1.51 V Observe:

Redox dispropotionation:

There are numerous examples in which a particular oxidation state is inherently unstable. Let’s consider a generic case, Mn+

K n+ disp (n+1)+ (n–1)+ 2M M + M

The two half reactions are:

n+ - (n–1)+ ө M + e M E 1 (n+1)+ - n+ ө M + e M E 2

n F ө ө logK = (E 1 – E 2) disp 2.3RT Mn+ is unstable ө ө if E 1 > E 2 Examples

2Cu+(aq) Cu2+(aq) + Cu(s)

2Mn3+(aq) Mn2+(aq) + MnO (s) 2 III I V - - - 2ClO2 (aq) ClO (aq) + ClO3 (aq)

These also include main-group species with an odd number of electrons AlI, PIV, SV, SIII, PbIII etc. and T.M. compounds in +1 oxidation state Let’s consider Cu+

K + disp + 2Cu (aq) 2Cu (aq) + Cu(s)

The two half reactions are:

Cu+(aq) + e- Cu(s) Eө = 0.52V 2+ - + ө Cu (aq) + e Cu (aq) E = 0.15V

logKdisp = 16.9 (0.52 – 0.15) 6.3 Kdisp = 10 Stability of Mn3+(aq)

3+ 2+ + 2Mn (aq) Mn (aq) + MnO2(aq) + 4H

The two half reactions are:

3+ 2+ ө Mn (aq) Mn (aq) E = 1.51V

+ 3+ Eө = 0.95V MnO2(s) + 4H Mn (aq)

Gө = - (1.51-0.95) F = -27 kJ/mol The Frost diagram Interpreting Frost diagrams Frost diagram Manganese (acidic)

6 HMnO4 5 H2MnO4 4 (H3MnO3)

3

/ V / 2 ө ө

NE 1 Mn 0 Mn3+ MnO2 -1

-2 Mn2+ -3 0 1 2 3 4 5 6 7 Oxidation number, N Latimer diagrams

0.15V 0.52V 2+ + 0 Cu Cu Cu

0.34V

0.8 N N Eө 0.7 0.6

0 0 0.5 / V / +1 0 + 0.52 ө 0.4

NE 0.3

+2 0.52 + 0.15 0.2

0.1

0 0 1 2 Oxidation number, N +7 +6 +5 +4 +3 +2 0 - - 3+ 2+ 0 MnO4 HMnO4 (H3MnO4) MnO2 Mn Mn Mn 0.90V 1.28V 2.9V 0.95V 1.5V –1.18V

N Species Calculation nEө 0 Mn 0 0.0 +2 Mn2+ 2 × –1.18 –2.36 +3 Mn3+ 1 × 1.5 + –2.36 –0.86

+4 MnO2 1 × 0.95 + –0.86 0.09

+5 H3MnO4 1 × 2.9 + 0.09 2.99 - +6 HMnO4 1 × 1.28 + 2.99 4.27 - +7 MnO4 1 × 0.90 + 4.27 5.17 Frost diagram Manganese (acidic)

6 HMnO4 5 H2MnO4 4 (H3MnO3)

3

/ V / 2 ө ө

NE 1 Mn 0 Mn3+ MnO2 -1

-2 Mn2+ -3 0 1 2 3 4 5 6 7 Oxidation number, N Acids, bases and solution equilibria

LECTURE 4 +7 +6 +5 +4 +3 +2 0 - - 3+ 2+ 0 MnO4 HMnO4 (H3MnO4) MnO2 Mn Mn Mn 0.90V 1.28V 2.9V 0.95V 1.5V -1.18V

N Species Calculation nEө 0 Mn 0 0.0 +2 Mn2+ 2 × –1.18 –2.36 +3 Mn3+ 1 × 1.5 + –2.36 –0.86

+4 MnO2 1 × 0.95 + –0.86 0.09

+5 H3MnO4 1 × 2.9 + 0.09 2.99 - +6 HMnO4 1 × 1.28 + 2.99 4.27 - +7 MnO4 1 × 0.90 + 4.27 5.17 Frost diagram Manganese (acidic)

6 HMnO4 5 H2MnO4 4 (H3MnO3)

3

/ V / 2 ө ө

NE 1 Mn 0 Mn3+ MnO2 -1

-2 Mn2+ -3 0 1 2 3 4 5 6 7 Oxidation number, N +7 +6 +5 +4 +3 +2 0 - 2- 3- 0 MnO4 MnO4 MnO4 MnO2 Mn2O3 Mn(OH)2 Mn 0.56V 0.27V 0.93V 0.15V -0.23V -1.56V

N Species Calculation nEө 0 Mn 0 0.0 +2 Mn2+ 2 × –1.56 –3.12 +3 Mn3+ 1 × –0.23 + –3.12 –3.35

+4 MnO2 1 × 0.15 + –3.35 –3.20

+5 H3MnO4 1 × 0.93 + –3.20 –2.27 - +6 HMnO4 1 × 0.27 + –2.27 –2.0 - +7 MnO4 1 × 0.56 + –2.0 –1.44 Frost diagram Manganese (basic)

0 Mn0

-0.5

-1 - MnO4

-1.5

/ V / ө ө -2 3-

NE MnO4 2- MnO4 -2.5

-3 Mn(OH) -3.5 2 MnO2 Mn2O3

-4 0 1 2 3 4 5 6 7 Oxidation number, N Frost diagram Manganese

6 HMnO4

5 H2MnO4

4 (H3MnO3) 3

2 / V / ө ө 1 MnO NE Mn 2 0 0 1 2 3 4 5 6 7 -1 Mn3+ Mn2+ MnO 3- 4 - -2 MnO4 MnO 2- -3 4

Mn(OH)2 Mn O MnO -4 2 3 2 Oxidation number, N

Stabilisation of oxidation states by complexation

III 3- - II 4- [Fe (CN)6] + e [Fe (CN)6] Eө = 0.41V

However, Fe3+ + e- Fe2+ Eө = 0.77V M2+(g) 2+ M2+(solv) Hsolv (M )

I.E.2 ≈ Eө (M2+/+) -Gө -Hө = ≈ nF nF + + + ignoring S M (g) Hsolv (M ) M (solv)

I.E.1 ≈ Eө (M+/0) M(g)

Hatm M(s) ө E 1 - Mox + e Mred

+L Kox +L Kred ө E 2 - MoxL + e MredL ө ΣG = 0 ө ө –nE 1F – RTlnKred + nE 2F + RTlnKox= 0

ө ө 2.3RT Kred E 2 – E 1 = log n F Kox ө E 1 3+ - 2+ Fe + e Fe

ox red +6CN b6 +6CN b6 Eө III 3- - 2 II 4- [Fe (CN)6] + e [Fe (CN)6]

o o red ox E 2 – E 1 = (2.3RT/F)log(b6 /b6 ) pH dependence of redox equilibria

Nernst equation

2.3RT [A ] E = Eө + ox log n F [Ared] + - Aox + 2H + 2e AredH2

+ 2 2.3RT [Aox][H ] E = Eө + = log 2 F [AredH2]

2.3RT [Aox] 2.3RT = Eө + - 2 × log pH 2 F [AredH2] 2 F

2.3RT [Aox] 2.3RT = Eө + - log pH 2 F [AredH2] F pH independent pH dependent 0.059 [A ] = Eө + ox - 0.059pH log 2 [AredH2]

In general, for reaction: + - Aox + mH + ne AredHm

ө 0.059 [Aox] m = E + log - 0.059pH n n [AredHm] pH dependence “Trimmed down” equation:

Assuming concentrations of Aox, AredHm are 1M (unit activities):

m E = Eө - 0.059 pH pH n Oxo-transfer reactions

- - - ClO4 + 2e ClO3 + ‘O’ ‘O’ = O2-, a very strong base write as: - + - - ClO4 + 2H + 2e ClO3 + H2O

m = 1 pH dependence = n -0.059 V/pHunit Pourbaix diagrams

Show how E varies with pH

gradient = m - 0.059 n Pourbaix diagrams

Include proton transfer within Nernst framework

+ + Aox + H AoxH

m = 1, n=0 gradient = ∞ H pH < pK ox + - AoxH + e AredH m = 0, n = 1 gradient = 0

H H pK ox < pH < pK red + - Aox + H + e AredH m = 1, n = 1 gradient = –0.059

H pH > pK red - - Aox + e Ared m = 0, n = 1 gradient = 0

Constructing a Pourbaix diagram

3+ - 2+ Eө = 0.77V 1 Fe (aq) + e Fe (aq) m = 0, n = 1 slope = 0

3+ + 2 Fe (aq) + 3H2O Fe(OH)3(s) + 6H n = 0 slope = ∞ (pK ≈ 3)

+ - 2+ 3 Fe(OH)3(s) + 3H + e Fe (aq) + 3H2O n = 1, m=3 slope = –0.177 2 1

3 Constructing a Pourbaix diagram (continued)

2+ + 4 Fe (aq) + 2H2O Fe(OH)2(s) +2H m = 0, n = 0 slope = ∞ (pK ≈ 9)

+ - 5 Fe(OH)3(s) +H +e Fe(OH)2(s) + H2O m = 1, n = 1 slope = –0.059 4 5 For water the limits are:

+ - ө O2 + 2H + 2e 2H2O E = 1.23V m = 2, n = 2 slope = –0.059

+ - ө 2H aq + 2e H2 E = 0V m = 2, n = 2 slope = –0.059