Class 11 Chemistry

# Equilibrium

## pH Scale

The pH of a solution is the negative logarithm to base 10 of the activity (aH+) of hydrogen ion. In dilute solutions (< 0.01 M) activity of hydrogen ion is equal in magnitude to molarity represented by [H+].

aH+ = [H+] /mol L-1

So, pH can be represented as follows:

pH = - log aH+ = - log ([H+] /mol L-1)

Thus, an acidic solution of HCL (10-2 M) will have a pH = 2.

• Acidic solution has pH < 7
• Basic solution has pH > 7
• Neutral solution has pH = 7

### Ionization Constants of Weak Acid

Let us consider a weak acid HX which is partially ionized in aqueous solution. The equilibrium can be given by following equation.

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

Initial concentration and change in concentration can be illustrated as follows:

HXH3O+X-
Initial concentration (M)c00
Extent of ionization change (M) = α-cα+cα+cα
Equilibrium concentrationc – cα

So, equilibrium constant for this acid-base dissociation equilibrium can be given as follows:

K_a=(c^2α^2)/(c(1-α)=(cα^2)/(1-α)

Ka is called dissociation or ionization constant of acid HX. Alternately, it can also be represented in terms of molar concentration as follows:

K_a=([H^+][X])/([HX])

At a given temperature T, the value of Ka shows the strength of the acid. It means larger the value of Ka, the stronger is the acid.

### Ionization of Weak Bases

Let us consider the ionization of a base MOH; as given by following equation.

MOH (aq) ⇋ M+ (aq) + OH- (aq)

Ionization constant of a base can be given by following equation:

K_b=(c^2α^2)/(c(1-α)=(cα^2)/(1-α)

Ionization constant of base can also be expresses in terms of molar concentration, as follows:

K_b=([M^+][OH^-])/([MOH])

#### Relation Between Ka and Kb

Let us consider following reactions

NH4+ (aq) + H2O ⇋ H3O+ (aq) + NH3 (aq)

K_a=([H_3O^+][NH_3])/([NH_4^+]) = 5.6 × 10-10

NH3 (aq) + H2O (l) ⇋ NH4+ (aq) + OH- (aq)

K_b=([NH_4^+][OH^-])/([NH_3]) = 1.8 × 10-5

Net: 2H2O (l) ⇋ H3O+ (aq) + OH- (aq)

K_w= [H_3O^+][OH^-] = 1.0 × 10-14 M

Now, K_a\xx\K_b=([H_3O^+][NH_3])/([NH_4^+])xx([NH_4^+][OH^-])/([NH_3])

=[H_3O^+][OH^-]=K_w

Or, (5.6xx10^(-10))xx(1.8xx10^(-10))=1.0xx10^(-14) M

It can be said that K_a\xx\K_b=K_w

#### Di- and Polybasic Acids and Di- and Polybasic Bases

Some acids contain more than one ionizable proton per molecule of the acid. Such acids are called polybasic or polyprotic acids.

Example: Ionization reaction of a diabasic acid H2X can be given by following equations:

H2X (aq) ⇋ H+ (aq) + HX- (aq)

HX- (aq) ⇋ H+ (aq) + X2- (aq)

Corresponding equilibrium constants for first and second ionization are as follows:

K_(a1)=([H^+][HX^-])/([H_2X])

K_(a2)=([H^+][X^(2+)])/([HX^-])

Similarly, for tribasic acids like H3PO4 you get three ionization constants.

It can be observed that higher order ionization constants are smaller than lower order ionization constants. The reason for this is that it is more difficult to remove a positively charged proton from a negatively charged ion due to electrostatic forces.

## Factors Affecting Acid Strength

It is a complex phenomenon, but in broad terms it can be said that the extent of dissociation of an acid depends on the strength and polarity of H-A bond. When strength of H-A bond decreases, the acid becomes stronger. When H-A bond becomes more polar, the acid becomes stronger.

Trend in Periodic Table: When we go down a group, H-A bond strength decreases and so acid strength increases. When we move across a period, H-A bond polarity increases and so acid strength increases. H-A bond strength is important factor while discussing acid strength down a group. But H-A bond polarity becomes important when we go across a period.

Common Ion Effect in Ionization of Acid and Base

When a substance is added to a reaction at equilibrium and the substance provides more of an ionic species already present in the dissociation equilibrium then the equilibrium shifts in a direction to negate the effect of added ions. This effect is called common ion effect.

Let us consider the following example of dissociation of acetic acid at equilibrium.

CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

This equation can be written in short as follows:

HAc (aq) ⇌ H+ (aq) + Ac- (aq)

K_a=([H^+][Ac^-])/([HAc])

If acetate ions are added to acetic acid solution, it results in reduction in concentration of hydrogen ions. If H+ ions are added from external source, it results in reduction in concentration of hydrogen ions. So, addition of either HAc- or H+ results in reduction in concentration of hydrogen ions.

## Hydrolysis of Salts

A salt is formed after neutralization reaction between acids and bases. They undergo ionization when dissolved in water. Cations and anions of salt (in aqueous solution) either exist in hydrated form or become hydrolysed. Cations of strong bases and anions of strong acids become hydrated. Hence, solution of salts formed from strong acids and bases are neutral. On the other hand, salts of weak acid and/or weak base get hydrolysed.

Hydrolysis of salts of weak acid and strong base: Cation becomes hydrated, while anion undergoes hydrolysis. Following is an example:

CH3COONa (aq) → CH3COO- (aq) + Na+

CH3COO- (aq) + H2O (l) ⇌ CH3COOH + OH-

Hydrolysis of salts of weak base and strong acid: Anion becomes hydrated while cation gets hyrolysed. Following is an example:

NH4Cl (aq) → NH4+ (aq) + Cl- (aq)

NH4+ (aq) + H2O (l) ⇌ NH4OH

Hydrolysis of salts of weak acid and weak base: Both cation and anion get hydrolysed. Following is an example:

CH3COONH4 + H2O ⇌ CH3COOH + NH4OH

Degree of hydrolysis is independent of concentration of solution, and pH of such solution is determined by their pK and pH values.

pH = 7 + ½ (pKa - pKb)

If the difference is positive then pH can be greater than 7. If the difference is negative then pH will be less than 7.

Buffer Solutions: The solutions which resist change in pH are called Buffer solution. Buffer solution of known pH can be prepared from the knowledge of pKa or acid and pKb of base and by controlling the ratio of salt and acid or salt and base. A mixture of acetic acid and sodium acetate acts as buffer solution at pH around 4.75.

### Solubility Equilibria of Sparingly Soluble Salts

Solubility of a salt depends on many factors but important among them are lattice enthalpy of salt and salvation enthalpy of ions in solution. If ion-solvent interaction overcomes the lattice enthalpy then the salt is soluble in given solvent. In case of non-polar solvent, salvation enthalpy is small and hence is not enough to overcome the lattice enthalpy of the salt. So, salts are insoluble in non-polar solvent but soluble in polar solvent.

#### Solubility Product Constant

Following equation shows the equilibrium between un-dissolved solid and the ions in a saturated solution of barium sulphate.

BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq)

Equilibrium constant can be given as follows:

K=([Ba^(2+)][SO_4^(2-)])/([BaSO_4])

Concentration of pure solid substance remains constant and can be given as follows:

Ksp = K[BaSO4 = [Ba2+][SO42-]

Here, Ksp is called the solubility product constant or solubility product.