The ionic equilibrium

The ionic equilibrium

The law of mass action and equilibrium

Chemical compounds

Ionic compounds:

Positive and negative ions are bonded by electrostatic forces

Example: Sodium chloride NaCl

-It dissociates completely in water into Na and Cl- ions

Covalent compounds

The atoms are bonded by covalent bonds

Example:

-Hydrogen chloride gas (HCl) ionizes completely in water

– Acetic acid

CH₃COOH incompletely ionized in water

IONIZATION:

Process in which unionized molecules are changed into ions

Types of ionization

a) Complete ionization                   

b) incomplete ionization

1)Complete ionization:–

All unionized molecules  are changed into ions(strong electrolyte)

HCl_(aq)  ⇔ H⁺_(aq) + Cl^–_(aq)

2)incomplete ionization:

Small part of unionized molecules are changed into ions (weak electrolyte)

CH₃COOH ⇔ H⁺   + CH₃COO^–

Question: compare between complete and incomplete ionization with example

Experiment: testing the electric conductivity of hydrogen chloride gas dissolved in benzene and pure acetic acid

Observation: the light bulb don’t light up in both cases

Conclusion:

The solution of HCl in benzene and pure acetic acid both don’t contains ions so they are nonelectrolytes

Expermint2:

The effect dilution of both hydrochloric acid and acetic acid 

HCl acid (1M concentration) acetic acid(1 M concentration)

Then increases dilution to (0.01 M then 0.001M)

Observation:

In case of HCl : the illumination of the light bulb intense and not affected by dilution

In case of acetic acid:

Increases the light intense by increasing dilution

Conclusion:

HCl is strong acid ionizes completely in water not affected by dilution

HCl ⇔H⁺   +     Cl^–

-Acetic acid is weak acid

-it ionizes incompletely in water

-Its ionization increases by increasing dilution

CH₃COOH ⇔CH₃COO^– + H⁺

GR: the ability of hydrochloric acid to conduct the electricity is not affected by dilution with water while acetic acid ability to conduct the electricity is affected by dilution

Hydronium ion (Hydroxonium ion)

(Hydrated proton)

Hydronium ion:

It is a positive ion produced from combination of water molecules with H+(proton)of ionized acid by coordinate bond

HCL_(s) +H₂O_(l) ⇔H₃O⁺_(aq)   +Cl^–_(aq)

Why hydronium ion is formed ?

-Due to the attraction of H⁺ ( of acid) with the lone pair electrons of oxygen atom in water molecules

GR: Hydronium ion is called hydrated proton?

IONIC equilibrium in weak electrolytes

There are two opposite process

– A weak (incomplete ) ionization takes place in weak electrolytes

Ionization of small amount  and  Recombination between ions to form molecules

CH₃COOH_(aq)+H₂O_(L)  ⇔ CH₃COO^–_(aq) + H₃O⁺_(aq)

A dynamic equilibrium state occurs:

CH₃COO^–_(aq) + H₃O⁺_(aq) ⇔ CH₃COOH_(aq)+H₂O_(L)

Ionic equilibrium: the equilibrium state between molecules of a weak electrolytes and their ions

Question : compare between : chemical equilibrium and ionic equilibrium with example

GR: The law of mass action can  not be applied to the strong electrolyte

Ostwald s law of dilution

He discovered (1888) the relation between :

The degree of ionization (dissociation(alpha  ) and the concentration (C) mol/L in the weak electrolyte

Ostwald s law:

At a constant temperature the degree of ionization (ᾳ) of the weak electrolytes increases dilution

On dissolving of a weak monoprotic acid (HA) in water it ionizes incompletely.

   HA_{(aq) ⇔}⁺ _(aq) + A^– _(aq)

By substitution in the law of mass action:

Equilibrium constant of weak acid(Ka)LIonization constant):

= the produced ions concentration / molecules concentration before ionization

Ka=[H⁺] [A^–] / [HA]

Ka= Cᾳ .  C ᾳ / C(1-ᾳ)

Ka= C² ᾳ² / C(1-ᾳ)

Ka=Cᾳ /(1-ᾳ)

The degree of ionization of weak acid (ᾳ) is negligible so

(1-ᾳ)=1

To calculate the concentration of  weak base :

Example:

calculate the degree of ionization of 0.1 M hydrocyanic acid knowing that its equilibrium constant (ionization constant ) = 7.2 x 10^-10  

Answer:

 ᾳ²=( Ka/Ca )² 

ᾳ²=( 7.2 x10^-10 /0.1 )2   

then    ᾳ = 8.5 x 10^-5

Example: calculate the concentration of weak acid acetic acid CH₃COOH  if its ionization percentage = 0.42 % and its equilibrium constant  K_a=1.8 x 10^-5

Answer:

ᾳ= 0.42/100 =0.0042

C_a= K_a/ ᾳ² =1.8 x 10^-5 /(0.0042)²

Example:

a weak acid its degree of ionization = 2 x 10^-2

in a 1L volume which contains 0.25 mol of it calculation the equilibrium constant of this acid.

Answer:

Concentration = No. of moles / volume (L)

C_a= 0.25 /1 = 0.25 M

K_a=C_a  X ᾳ² =

0.25 x (2×10^-2)²  

= 1 x 10^-4

Example:

a weak acid its degree of ionization = 0.008 (at 25 c) in 0.15 M solution

Calculate its degree of ionization in 0.1 M solution at the same temperature what would you deduce

Answer:

Ka value is constant at constant temperature

C1 (ᾳ1)² = C2 (ᾳ2)² 

0.15 x (0.008)² =0.1 (ᾳ2)² 

(ᾳ2  =   0.15 x (0.008)² /0.1

= 0.00098

deduction: the degree of ionization increases by increasing the dilution

To calculate of hydronium ion(H3O+) concentration in weak acid:

Acetic acid ionizes in water as:

CH₃COOH + H₂O ⇔ CH₃COO^– + H₃O⁺

So    K_a= [CH₃COO] [H₃O⁺]  /[CH3COOH]

The no. of moles of H₃O⁺= the No. of moles of CH₃COO^–

[H3O⁺] = [CH₃COO^–]

So K_a = [H3O⁺]² / [CH3COOH]

ᾳ of weak acid is negligible value

so : acid concentration before ionization = acid concentration at equilibrium

To calculate Hydronium ion concentration [H₃O⁺] in a weak acids by knowing its degree of ionization (ᾳ):

Example:

calculate hydration ion concentration in a 0.1M solution of acetic if its ionization percentage =1.34 %

Answer:  ᾳ =1.34/100

= 1.34 x 10^-2 

[H₃O⁺] = ᾳ C_a = 1.34 x 10^-2 x 0.1

= 1.34 x 10^-3 M

To calculate of hydroxyl ion con.[OH^–] in weak base solution:

Example:

NH₃ gas dissolves in water giving a weakly ionized basic

NH_3(g) + H₂O_{(L) ⇔}NH₄^+(aq)    + OH^– _(aq)

K_b=[NH₄⁺] [OH^–] / [NH₃]

No. of HO^– ions = No. of NH₄⁺ ions

  [NH₄⁺] = [OH^–]

K_b = [OH^–]²  / [NH₃]

Example:

calculate  hydroxyl ionization in 0.2 M solution of methyl amine CH₃NH₂ (at 25C) if equilibrium constant =3.6 x 10^-4