Ionic Equilibrium

Acid-base equilibrium Solubility equilibrium

The Arrhenius Theory

1. Acid-substances that dissociate in water to produce H+

eg: HA (aq) H+ (aq) + A- (aq)

2. Base-substance that dissociate in water to produce OH-

eg: M(OH) (aq) M+ (aq) + OH- (aq)

Limitations

a) Cannot account for the basic properties of certain compounds that do not contain –OH groups/ ions in their molecules such as amines,RNH2,NH3,NA2CO3 and C2H5ONa2

NH3(aq) + HCl( aq) NH4Cl(g)

NH3 reacts with HCl by receiving proton or H+ but not by donating OH-

H2O

H2O

Acid – base Equilibrium

BRONSTED-LOWRY THEORY1.Bronsted Acid: Acid substance( molecule or ion) that can donate a proton, H+ to another substance/to a baseEg: HCl, HNO3,HF2. Anionic and cationic salts that have proton that can be transferred/donated such as

NH4 +, HSO4- ,HCO3-

3. Base-substance that can accept a proton from an acid/another substance.Eg: Negative ion or neutral molecule with a lone pair of electrons which can form a dative

covalent bond with the proton such as H2O,F-, NH3

and OH-

HA + B BH+ + A-

Acid Base Conjugate Conjugate Acid BaseProton Proton Donor Acceptor- loses - accepts - species - speciesIts proton proton formed when formed whento form to form when a H+ a proton isbase an acid is added to removed from

a base an acid

Acid-base conjugate pair1. Species differ by a proton

HA & A-

Acid Conjugate Base B & BH+

Base Conjugate Acid

2. Water is an amphoteric solventActs as a base

CH3COOH(aq) + H20(l) CH3COO-(aq) + H30+(aq) Acid Base Conjugate Conjugate Proton Donor Proton Acceptor Base Acid coz has lone

hydroxonium ion/ pair of ē oxonium ion/

hydronium ionActs as an acid

NH3(aq) + H20(l) NH4+ (aq) + OH-(aq) Base Acid Conjugate Conjugate Proton Acceptor Proton Donor Acid Base

ADVANTAGES- can be used to define acids & bases in gaseous phase.

HCl (g) + NH3 (g) → NH4Cl (g)acid base(proton (proton

donor) acceptor)

- Can be used to discuss the strength of an acid or a base.HA + H2O A– + H3O+

is increased, eq. position lies to the right, when HA is a stronger acid.B + H2O BH+ + OH–

B is a stronger base( = degree of dissociation)

hexaaquo complexes, [M(H2O)6]n+ which have small highly charged cations (such as Al3+, Fe2+, Cr3+, Be2+) act as Bronsted–Lowry acids in water.

eg. [Al(H2O)6]3+ + H2O [Al(H2O)5OH]2+ + H3O+

acid base(proton (proton donor) acceptor)

Mn+

Central metal ion- empty valence

orbitals

H2OH2O

H2O

H2O

H2O

H2O

EXERCISE 1

Classify the underlined substances in the following equations as eitheracids or base according to Bronsted-Lowry’s definition.

a) HCO3 – + HCl H2CO3 + Cl–

b) NH4+ + NH2

– 2NH3

c) CH3CONH2 + H20 CH3CONH3+ + OH–

d) NaH + H20 NaOH + H2

EXERCISE 2

Give the formula of the conjugate acids for the following base.

a) Cl – –b) C2H5O – c) NH2OH –d) C6H5NH2 – e) HCO3

– –f) OH – –

LIMITATIONChemical species that do not contain H+ are not defined as acids .eg: BF3

THE BASICITY OF AN ACID

ACID

Monoprotic acid Polyprotic acid/Polybasic acid

- Produces only one single -Donates more than one proton

proton permolecule on permolecule of the acid on dissociation.

dissociation eg: Diprotic acid are – H2SO4,H2CO3

OR Polyprotic acids dissociate in stages

Monobasic acid and have more than one Ka.

- Forms only one conjugate base Eg: H3PO4 H+ + H2PO4 –(K1)

permolecule of the acid H2PO4 - H+ + HPO4 2- (K2)

Eg: HCl , CH3COOH, HNO3 HPO4 2- H+ + PO4 3- (K3)

H3PO4 3H+ + PO4 3-

Ka for H3PO4 = K1 X K2 X K3

= [H+]3 [PO4 2-]

[H3PO4]

Lewis Theory

Acid - species( an atom, ion or molecule ) which can form a dative covalent bond by accepting one pair of electron from a base. - electron pair acceptor. - eg : 1. all the positive ions 2. oxidising agents 3. molecules with an incomplete octet of electrons such as BF3, BeCl2, BCl3, AlCl3 Base – species that has an unshared electron pair which can form a dative bond with an atom, molecule or ion. - electron pair donor - eg:- negative ions - reducing agents - molecules with lone pair of electrons such as H2O:, :NH3,

R-CH ..

Neutralisation

The formation of a dative covalent bond between a species that donates an electron pair (Lewis base) and the species that accepts an electron pair( Lewis Acid)

An acid- base reaction( Bronsted- Lowry Theory)

A reaction where an electron pair of a base is accepted by an acid through the formation of

a dative covalent bond. H

Eg. :NH3 + BCl3 H N B Cl

Cl

Cl

H

H+ + H– :

: : O

:

:

O

H H

H+ + : NH3

H H

H

H

N ↓

+

Lewis base

Lewis acid

eg:

HO B + H2O HO B OH + H+

OH

OH

OH

OH

–

Lewis acidLewis base

eg:

Cu2+ + 4NH3 Cu

:

–H2N :

H3N :

NH

3

:

NH3

: 2+

Lewisacid

Lewisbase

(Central metal ions)

(liquids) Complex ion

LIMITATION

• HCl(g), Bronsted acid, is not an acid from the point of Lewis Theory- because it cannot accept an electron pair.

ADVANTAGES

1. Acid-base reaction can be extended to include reactions in which protons are not involved.

eg. BF3 + NH3 BF3 NH3

Acid Base (No transfer of proton)

2. Lewis bases are also Bronsted – Lowry bases as all electron pair donors can accept a proton.

3. Lewis acids( eg. Metal ions) need not be Bronsted – Lowry acids ( proton donors).

So, Lewis Theory broadens the concept of an acid. Lewis acids includes not only H+ ions but also cations and molecules

with empty valence orbitals that can accept electron pairs from a Lewis base.

1. Electrolyte- A chemical compound that will conduct electricity in molten state or in aqueous solution.( because of the presence of free/ mobile ions)

2. Non- electrolyte- A chemical compound that cannot conduct electricity both in molten state or in aqueous solution.( because they do not have mobile ions)

Electrolytes

Strong electrolytes Weak Electrolytes

Example

- Mineral acids (HCl, HNO3, H2SO4) - Organic acids (CH3COOH)

- Alkalis ( NaOH, KOH, Ca(OH)2, Ba(OH)2) - Organic bases (amines, NH3 )

- ionic salts.

- Degree of dissociation ~ 100% - Low < 0.1, <0 ( partially ionised)

- Electrical conductivity - Good Poor

- [ions]- high - Low ( most of them remain as

undissociated molecules)

pH

The strength of acids & bases

Degree of dissociation the dissociation constant

Can be compared byObj. questions

•pH = - log [ H+]

•0 7 14

•Acid neutral base

•pH value changes with [ ]

* Not a good way because [ ] must be the same in order to compare.

• At constant T,Ka Strength of acid or Kb Strength of base

• Dissociation constant will not change with [ ]

• It only depend on the T

• Commonly used to compare the strength of acid/ base

•The [ ] must be the same

• Strength

• Degree of dissociation varies with [ ].

% dissociation

[HA] dissociate x 100%

[HA] initial

• [ ] ↓ , ↑

[ ]

1 [ ]

1.0

Exercise

1.Arrange the acids in ascending order of the strength of the acid.

Ka/M pKa

HCOOH 2.09X10-4 3.68

CH3COOH 1.80X10-5 4.75

HCN 4.90X10-10 9.30

H2CO3 7.47X10-7 6.35

Solution: HCN < H2CO3 < CH3COOH < HCOOH

acid strength increase

(Ka or pKa ↓ )↑

↑

Acids

Weak AcidsStrong Acids

-Most mineral acids

Eg. : HCl , H2SO4 , HNO3

HCl + H2O → H3O+ + Cl–

(almost complete dissociation)

α : 100%

[H+} :

pH :

Ka :

pKa :

-Organic acids

Eg. : CH3 COOH

CH3 COOH + H2O CH3COO– + H3O+

(partially dissociates)

< 100%

Way to determine

pH : determine [H+] determine [H+] by

–[H+] is obtained directly from [acid] [H+] = KaCeg: [H+] = [monoprotic acid] = c

2[H+] = [dibasic acid] c = concentration of weak acid

eg: H2SO4 2H+ + SO42– pH = –log [H+]

0.1 M 2 x 0.1 M

pH = – log [H+]

Calculate the pH ofi) 0.01 mol dm–3 sulphuric acidii) 0.01 mol dm–3 CH3COOH (Ka = 1.80 x 10–5 M)

Bases

Weak BasesStrong Bases

Eg. : NaOH, KOH, Ba(OH)2

NaOH → Na+ + OH–

α : 100%

(almost completely dissociated)

[OH –] :

pOH :

Kb :

pKb :

Amines, NH3

NH3 + H2O NH4 + + OH–

10%

(partially dissociated)

Way to determine

pH : determine [OH –] determine [OH+] by

–[OH –] is obtained directly from [base] [OH –] = KbCeg: NaOH Na+ + OH – = c 0.1 M 0.1 M pOH = –log [OH –]eg: Ba(OH)2 Ba2+ + 2OH–

0.1 M 2 x 0.1 M pH = 14 – pOH

pOH = – log [OH] –

pH = 14 – pOH

Ostwald Dilution Law

― shows relationship between the concentrations of weak acids / bases and / Ka; Kb

For weak Acids

HA + H2O H3O+ + A–

Initialmoles in c o o1dm3 solution

No. of molesdissociated/ c c cformed

Eq. moles c - c c c

Exercise

1.Calculate the pH of i. 50 cm3 of 0.5M of Ba(OH)2

ii. 50 cm3 of 0.5M of NH3

(Kb NH3 = 1.75 x 10–5 mol dm–3)

2.Calculate the pH of50 cm3 of 0.5M of NH3 + 50 cm3 of 1M of H2SO4

[A–] [H3O+]

[HA]

[OH–] [BH+] [B]

c cc (1 - )

c2

1 -

Kb

c

[A–] [H3O+]

[HA][BH+] [OH–] [B]

c cc (1 - )

Ostwald Dilution Law (continued)

Ka =

For weak acids,

is very small,

1 - 1

Ka = c2

=

[H+] = c

=

= Ka C

[H+] = Ka C

c2

1 -

Ka

c

Ka

cFor weak bases

B + H2O BH + + OH– Initial moles in 1dm3 solution c

o o

No. of molesdissociated/formed c c

c

Eq. moles c - c c c