Class 11 Chemistry Chapter 7 Equilibrium Notes

Equilibrium:

When rate of forward process is equal to rate of reverse process then this condition is known as (dynamic) equilibrium.

Equilibrium can be established for both physical processes and chemical reactions.

Solid−Liquid Equilibria

A perfectly insulated thermos flask is taken in which ice and water are kept at 273 K and atmospheric pressure.

Observation: There is no change of mass of ice and water. (Despite being at the boundary between ice and water, molecules from the liquid water collide against ice and adhere to it. Similarly, some molecules of ice also escape into the liquid phase.)

Reason: At 273 K and atmospheric pressure, the rate of transfer of molecules from ice to water is equal to that of transfer of molecules from water to ice.

This state is known as equilibrium state.

H₂O_(s) ⇌ H₂O_(l)

Ice and water are in equilibrium only at particular temperature and pressure.

Normal melting point or normal freezing point − Temperature at which the solid and liquid phases of the substance are at equilibrium at atmospheric pressure

Conclusion:

Both the opposing processes occur simultaneously.

Both processes occur at the same rate hence, the amount of ice and water remains constant.

Liquid-Vapour Equilibrium

Experimental set up

Anhydrous CaCl₂ is kept inside for few hours to remove moisture.

Anhydrous CaCl₂ is removed and a dish of water is quickly placed.

Observation: Mercury level in the right limb of the manometer slowly increases and attains a constant value. That is, pressure inside the box increases and reaches a constant value.

Volume of water in the watch glass decreases.

Reason: In the beginning, there was no water vapour inside the box. After addition of water, evaporation starts and water molecules escape into the gaseous phase. As a result, pressure goes on increasing. However, the rate of increase in pressure decreases with time due to condensation of vapour into water and after sometime, rate of evaporation becomes equal to the rate of condensation.

H₂O_(l)  ⇌  H₂O_(g)

Therefore, the pressure becomes constant. This state is called equilibrium state.

Vapour Pressure

Equilibrium vapour pressure: Constant pressure exerted by water vapour molecules at equilibrium at a given temperature.

Increases with increase in temperature.

Equilibrium vapour pressure is different for different liquids at the same temperature.

Higher the vapour pressure, more volatile is the liquid and lower is the boiling point.

Time taken for evaporation depends upon the

·       nature of the liquid

·       amount of the liquid

·       temperature

It is not possible to reach equilibrium in open system.

Reason: The rate of condensation is much less than the rate of evaporation in open system.

Boiling Point

Normal boiling point: Temperature at which the liquid and the vapours are at equilibrium at atmospheric pressure (normally 1 atm / 1.013 bar)

It depends upon the atmospheric pressure, which depends upon the altitude of the place. Higher the altitude, lower is the boiling point.

Exists when solids sublime to vapour

Example: When some solid iodine is warmed in a closed vessel, iodine sublimes to iodine vapours (violet). The intensity of colour increases with time but after sometime, the intensity becomes constant. This state represents equilibrium state.

I_2(solid)   ⇌  I_2(vapour)

Some other examples:

Camphor_(solid)  ⇌  Camphor_(vapour)

NH₄Cl_(solid)   ⇌  NH₄Cl_(vapour)

Equilibrium Involving Dissolution of Solids or Gases in Liquids

Solids in liquids

In saturated solution, a dynamic equilibrium exists between the solute molecules in the solid state and in the solution.

Example:

Sugar_(solution)  ⇌   Sugar_(solid)

At equilibrium,

Rate of dissolution of sugar = Rate of crystallisation of sugar

Confirmation of equilibrium with the help of radioactive sugar: If some radioactive sugar is dropped into a saturated solution of non-radioactive sugar, then after sometime, radioactivity is observed both in the solid sugar and in the solution. The ratio of the radioactive to non-radioactive molecules in the solution increases till it attains a constant value.

 

Gases in Liquids

Equilibrium exists between the molecules in the gaseous state and the molecules dissolved in the liquid.

Example: The following equilibrium exists when carbon dioxide is dissolved in soda water.

CO_2(gas)   ⇌   CO_{2(in solution)}

This equilibrium is governed by Henry’s Law.

According to Henry’s law, the mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the pressure of the gas above the solvent.

When soda water bottle is opened, the carbon dioxide dissolved in it escapes out rapidly with fizz. Carbon dioxide escapes to reach a new equilibrium condition required for the lower pressure (partial pressure in the atmosphere).

General Characteristics of Equilibria Involving Physical Processes

Equilibrium can be established only in a closed system at a given temperature.

Both opposing processes occur at the same rate and there is a dynamic equilibrium.

All the measurable properties of the system are constant at equilibrium.

Liquid ⇌ Vapour H2O(l) ⇌ H2O (g)	pH2O (vapour pressure) is constant at given temperature.
Solid ⇌ Liquid H2O(S) ⇌ H2O (l)	Melting point is fixed at constant pressure.
Solute(s) ⇌ Solute (solution) Sugar(s) ⇌ Sugar (solution)	Concentration of solute in solution is constant at a given temperature.
Gas(g) ⇌ Gas(aq) CO2(g) ⇌ CO2(aq)	[gas(aq)]/[gas(g)] (ratio of concentrations) is constant at a given temperature.

Dynamic Equilibrium:

Rate of forward reaction = Rate of reverse reaction. No net change in composition.

Plot of concentration vs. time for a reversible reaction: A  +  B  ⇌  C  +  D

Depiction of equilibrium for the reaction: N₂(g)  +  3H₂(g)  ⇌  2NH₃(g)

Depiction of equilibrium in the reaction, H₂(g)  +  I₂(g)  ⇌  2HI(g) from either direction:

Law of Chemical Equilibrium or Equilibrium Law

At a given temperature, the product of the concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation, divided by the product of the concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the law of chemical equilibrium or equilibrium law.

For a general reaction,

aA  +  bB  ⇌  cC  +  dD

Equilibrium equation is,

 

Where, K_c = Equilibrium constant

 = Equilibrium constant expression

Example:

H₂(g)  +  I₂(g)  ⇌  2HI(g)

 

The subscript ‘C’ indicates concentrations in terms of mol L⁻¹

Chemical Equilibrium	Equilibrium Constant
aA + bB ⇌ cC + dD	K
cC + dD ⇌ aA + bB
naA + nbB ⇌ ncC + ndD
aA + bB ⇌ cC + dD

Types of Chemical Equilibria:

Two types of chemical equilibria –

Homogeneous equilibria

Heterogeneous Equilibria

 

Homogeneous Equilibria

All the reactants and products are in the same phase.

Example:

In gaseous phase,

N₂(g)  +  3H₂(g)  ⇌  2NH₃(g)

In solution phase,

Fe³⁺(aq)  +  SCN^–(aq) ⇌  Fe(SCN)²⁺(aq)

CH₃COOC₂H₅(aq)  +  H₂O(l)  ⇌  CH₃COOH(aq)  +  C₂H₅OH(aq)

Equilibrium constant for homogeneous reaction in gaseous systems:

Ideal gas equation is given by,

pV = nRT

 

If concentration C is in mol L⁻¹ or mol dm⁻³ and p is in bar, then we can write

p = c RT

Or, p = [gas] RT                                                            …..(i)

Where, R = 0.0831 bar L mol⁻¹ K⁻¹

For a general reaction,

aA + bB ⇌ cC + dD

Equilibrium constant in terms of pressure,

 

 

 

 

 

where R = Gas constant

T = Absolute temperature

∆n_g = (Number of moles of gaseous products – Number of moles of gaseous reactants) in balanced chemical equation.

= (c + d) – (a + b)

While calculating K_p, pressure should be expressed in bar.

1 bar = 10⁵ Pa = 10⁵ Nm⁻²

Heterogeneous Equilibria

Have more than one phase.

Examples: H₂O(l)  ⇌  H₂O(g)

   Ca(OH)₂(s)  +  (aq)  ⇌  Ca²⁺(aq)  +  2OH^–(aq)

   CaCO₃(s)  ⇌  CaO(s)  +  CO₂(g)

Equilibrium constant for CaCO₃(s)  ⇌  CaO(s)  +  CO₂(g)

 

As CaO and CaCO₃ are pure solids, [CaO] and [CaCO₃] are constants.

Hence, equilibrium constant,

Or 

Applications of Equilibrium Constant

Predicting the extent of a reaction:

If K_C > 10³, then the products predominate over the reactants.

If K_C < 10⁻³, then the reactants predominate over the products.

If 10⁻³ < K_C< 10³, then appreciable concentrations of both reactants and products are present.

Predicting the direction of a reaction:

Reaction quotient, Q (Q_C with molar concentration and Q_p with partial pressure)

For a general reaction,

aA  +  bB  ⇌  cC  +  dD

 

The concentrations are not necessarily equilibrium values.

If Q_C > K_C, then the reaction will proceed in the reverse direction.

If Q_C< K_C, then the reaction will proceed in the forward direction.

If Q_C = K_C , then the reaction is at equilibrium.

Relationship between Equilibrium Constant (K), Reaction Quotient (Q) and Gibbs Energy (G)

If ΔG = − ve, then the reaction is spontaneous and proceeds in the forward direction.

If ΔG = + ve, then the reaction is non-spontaneous and proceeds in the backward direction.

If ΔG = 0, then the reaction is at equilibrium.

Relationship between K, Q and G

∆G = ∆G^o + RT ln Q

where, ∆G^o = Standard Gibbs Energy

At equilibrium, ΔG = 0, Q = K_C

Hence,   

Interpretation of Spontaneity, Using the Equation  in Terms Of 

If 

 

 

Hence, the reaction is spontaneous.

If 

 

Hence, the reaction is non-spontaneous.

Factors Affecting Equilibria:

Le Chatelier’s principle

According to this principle, if a system is in equilibrium and it is subjected to any change in any of the factors that determine the equilibrium conditions of the system, then it will shift the equilibrium in such a way so as to reduce or to counteract the effect of the change.

Effect of Concentration Change

The concentration stress of an added reactant/ product is relieved by the net reaction in the direction that consumes the added substance.

The concentration stress of a removed reactant/product is relieved by the net reaction in the direction that replenishes the removed substance.

Example:

H₂(g)  +  I₂(g)  ⇌  2HI(g)

Effect of addition of H₂: Equilibrium shifts in the forward direction

In terms of reaction quotient Q_C,

When H₂ is added at equilibrium, Q_C becomes less than K_C. Therefore, equilibrium shifts in the forward direction until Q_C = K_C

Effect of Pressure Change

In case of solids and liquids, the effect of pressure change is neglected.

Reason: The volume of solid or liquid is independent of pressure.

For gases, if pressure is increased, then the equilibrium shifts in the direction in which the number of moles of gas decreases.

Example:

When pressure in increased, the reaction goes in the reverse direction.

Effect of Addition of Inert Gas

If an inert gas is added at constant volume, then the equilibrium remains undisturbed.

Reason: Partial pressures or molar concentrations of the substance do not change with addition of inert gas at constant volume.

Effect of Change in Temperature

We know that when concentration, pressure, or volume is changed, equilibrium is disturbed.

Reason: Q_C changes and no longer equals to K_C

When temperature is changed, equilibrium is disturbed.

Reason: K_C changes

Change in equilibrium constant with temperature depends upon the sign of ΔH for the reaction.

For exothermic reaction (negative ΔH), the equilibrium constant decreases with the increase in temperature.

For endothermic reaction (positive ΔH), the equilibrium constant increases with the increase in temperature.

Example:

With increase in temperature, the equilibrium shifts in the backward direction.

Effect of a Catalyst

Catalyst increases the rate of reaction by lowering the activation energy for the forward and reverse reactions by exactly the same amount.

Catalyst does not affect the equilibrium.

Ionic Equilibrium

Equilibrium that involves ions in aqueous solution.

Acids, Bases and Salts

Arrhenius concept of acids and bases:

Acids − substances that dissociate in water to give hydrogen ions  or

Or

H⁺ exists as hydronium ion H₃O⁺ {[H(H₂O)]⁺}

Reason: H⁺ is very reactive and cannot exist freely, so it binds to the oxygen atom in H₂O.

Bases − substances that produce hydroxyl ions

Hydroxyl ion:

Exists in hydrated forms in aqueous solution such as  , etc.

The Bronsted-Lowry acids and bases

Acids − substances that can donate a hydrogen ion H⁺

Bases − substances that can accept a hydrogen ion H⁺

Representation of dissolution of NH₃ in H₂O is shown in the figure.

Representation of dissolution of HCl in H₂O is shown in the figure.

Conjugate acid-base pair:

Acid-base pair that differs only by one proton

Some species that can act both as Bronsted acids and bases are listed in the following table with their respective conjugate acids and bases.

Species	Conjugate acid	Conjugate base
H2O	H3O+	OH−
HCO3–	H2CO3	CO32–
HSO4–	H2SO4	SO42–
NH3	NH4+	NH2–

Lewis acids and bases:

Acids − species that accept an electron pair

Bases − species that donate an electron pair

Though BF₃ does not have a proton, it acts as an acid, because it accepts a lone pair of electrons.

Examples of Lewis acids − AlCl₃, BF₃, BCl₃ CO³⁺, Mg²⁺, H⁺

Examples of Lewis bases − H₂O, NH₃, OH⁻, F⁻

Ionisation of Acids and Bases:

Arrhenius Concept

Strong acids are those which completely dissociate in aqueous solutions to give H₃O⁺ ions.

Examples: Perchloric acid (HClO₄), hydrochloric acid (HCl), hydrobromic acid (HBr), hydroiodic acid (HI), sulphuric acid (H₂SO₄), nitric acid (HNO₃)

Strong bases are those which completely dissociate in aqueous solutions to give OH⁻ ions.

Examples: Lithium hydroxide (LiOH), sodium hydroxide (NaOH), potassium hydroxide (KOH), barium hydroxide (Ba(OH)₂)

 

Bronsted-Lowry Concept

The conjugate base of a strong acid is a weak base.

      HCl_(aq)  +   H₂O_(l)  ⇌  H₃O⁺_(aq) + Cl⁻_(aq)

Strong acid      Weak base

The conjugate base of a weak acid is a strong base.

CH₃COOH_(aq) +   H₂O_(l)  ⇌  H₃O⁺_(aq) + CH₃COO⁻_(aq)

  Weak acid           Strong base

The conjugate acid of a strong base is a weak acid.

CH₃COOH_(aq) +   OH⁻_(aq)  ⇌  H₂O_(l) + CH₃COO⁻_(aq)

   Strong base         Weak acid

The conjugate acid of a weak base is a strong acid.

     HCl_(aq) +    H₂O_(l)  ⇌  H₃O⁺_(aq) + Cl⁻_(aq)

   Weak base     Strong acid

Weaker the conjugate base, stronger is the acid. Weaker the conjugate acid, stronger is the base.

Ionisation constant of water and its ionic product

Water acts as acid as well as base.

H₂O_(l) + H₂O_(l)    ⇌    H₃O⁺_(aq)   +   OH⁻_(aq)

 Acid        Base               Conjugate       Conjugate

                                             base                acid

Dissociation constant,

[H₂O] is omitted as H₂O is a pure liquid and [H₂O] is constant.

K_w = [H⁺] [OH⁻]

Where, K_w = Ionic product of water

At 298 K, [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M

So,    K_w = [H⁺] [OH⁻] = (1.0 × 10⁻⁷ M)² = 1.0 × 10⁻¹⁴ M²

The value of K_w depends upon temperature.

Density of pure water = 1000 g L⁻¹

Molar mass of water = 18 g mol⁻¹

Therefore, molarity of pure water,

[H₂O] =

Therefore, the ratio of dissociated water to un-dissociated water

 

Thus, equilibrium lies mainly towards un-dissociated water.

Comparison between [H₃O⁺] and [OH⁻]

For acidic solution, [H₃O⁺] > [OH⁻]

For neutral solution, [H₃O⁺] = [OH⁻]

For basic solution, [H₃O⁺] < [OH⁻]

Strong and Weak Electrolyte

1. ​Strong electrolyte: Electrolytes which ionise in water completely are termed as strong electrolytes such as NaCl, MgCl₂ etc. Strong acids, strong bases and soluble salts are strong electrolytes.

2. Weak electrolyte: Electrolytes which do not ionise completely in water are termed as weak electrolytes such as acetic acid. Weak acids, weak bases and sparingly soluble salts are weak electrolytes.

Degree of Dissociation

When an electrolyte is dissolved in water or any solvent it may either completely dissociate or sometimes partially. Their solubility in solvent such as water is expressed by the degree of dissociation.
It is defined as the extent to which an electrolyte dissociates into ions in a solvent. It is represented by the symbol α. It is calculated as follows
                                          α =
The value of α is 1 for strong electrolytes as they are completely dissociated and less than 1 for weak electrolytes since they are not completely dissociated.

The value of α for an electrolyte depends on

(i)      Nature of solvent: Solvent having high dielectric constant will favour the dissociation.

(ii)    Nature of electrolytes: Strong electrolytes dissociate completely while weak electrolytes dissociate partially.

(iii)  Dilution: Increasing dilution will increase the degree of dissociation of weak electrolytes.

(iv)  Temperature: Increase in temperature generally favours the dissociation.

 The pH Scale

Logarithmic scale in which the concentration of hydronium ion (in molarity) is expressed.

 

For pure water: pH = −log (10⁻⁷) = 7

Value of pH:

For acidic solution: pH < 7

For basic solution: pH > 7

For neutral solution: pH = 7

K_w = [H₃O⁺] [OH⁻] = 10⁻¹⁴

⇒ −log K_w = −log {[H₃O⁺] [OH⁻]} = −log 10⁻¹⁴

⇒ pK_w = −log [H₃O⁺] − log [OH⁻] = 14

⇒ pK_w = pH + pOH = 14

Determination of pH:

Can be roughly determined with a pH paper. The given figure is that of a pH paper with four strips that have different colours at the same pH. Can be accurately determined by a pH meter.

Ionization of monobasic weak acids (HX):

                                HX_(aq)+ H₂O_(l)     ⇌   H₃O⁺_(aq) + X⁻_(aq)

Initial  Conc. (M)         c                                0             0

Let α be the extent of ionization.

Change                    −cα                              +cα         +cα

Conc. at eq.            c− cα                              cα            cα

Where, c = Initial concentration of the undissociated acid HX

Equilibrium constant of above acid-base equilibrium,

K_a is called the dissociation or ionization constant of acid HX.

Significance of K_a

Measure of strength of acid

Stronger the acid, larger is the value of K_a.

Dimensionless quantity

pK_a = −log(K_a)

Ionization of monoacidic Weak Bases (MOH)

Ionization of weak base MOH:

MOH_(aq)   ⇌  M⁺_(aq)+ OH⁻_(aq)

Equilibrium constant is given by,

K_b is called base ionization constant.

If c = Initial concentration of base

α= Degree of ionization of base

Then, 

pK_b = −log (K_b)

Relation between K_a and K_b

For weak acid HA,

HA_(aq)+ H₂O  ⇌  H₃O⁺_(aq)+ A⁻_(aq)

The conjugate base A⁻behaves as a weak base in water.

A⁻+ H₂O  ⇌  HA + OH⁻

That is, K_a(acid) × K_b(conjugate base) = K_w

In general, at 298 K,

K_a × K_b= K_w= 10⁻¹⁴ mol² L⁻²

pK_a + pK_b= pK_w= 14

 

Polybasic or polyprotic acids

Donate more than one proton per molecule

Examples: Oxalic acid (H₂C₂O₄), sulphuric acid (H₂SO₄), phosphoric acid (H₃PO₄)

Ionisation reaction for a dibasic acid, H₂X

= First ionisation constant of H₂X

 = Second ionisation constant of H₂X

Higher order ionisation constants  are smaller than the lower order ionisation constant .

Reason: Difficult to remove a positively charged proton from a negatively charged ion due to increased electrostatic force

Polyacidic base

Accept more than one proton

Examples: carbonate  ion, oxalate  ion.

 are first, second and third constants respectively for a tri-acidic base.

Factors Affecting Acid Strength

For an acid, HA

Stronger the H − A bond, weaker is the acid.

More the electronegativity difference between H and A, stronger is the acid

Reason: Easier to break the H − A bond due to charge separation

Acidic strength of HA increases down the group

Size increases

––––––––––––––––––––→

HF << HCl << HBr << HI

––––––––––––––––––––→

  Acid strength increases

Reason: Down the group, the size of A increases, so, the strength of the H − A bond decreases.

Acid strength of HA increases from left to right across a period

Electronegativity of ‘A’ increases

–––––––––––––––––––––––––––––––→

CH₄ < NH₃ < H₂O < HF

–––––––––––––––––––––––––––––––→

Acid strength increases

Reason: Across a period, electronegativity increases

Common Ion Effect in the Ionisation of Acids and Bases

Shift in the equilibrium by the addition of a substance which dissociates to give more of an ionic species already present in the dissociation equilibrium

Based on Le Chatelier’s principle, Ionisation suppresses, i.e., the equilibrium shifts towards the un-dissociated substance when any one of the products is added to the solution.

Dissociation of acetic acid:

If any one of H⁺ ions or CH₃COO⁻ is added from an external source, then the equilibrium will shift in the direction of the un-dissociated acetic acid (i.e., in the backward direction).

Let us illustrate this effect by considering the effect on the pH of the solution by adding 0.05 M acetate ion to 0.05 M acetic acid solution.

Dissociation equilibrium of acetic acid:

Initial concentration (M)       0.05                       0                   0.05

Let x be the extent of the dissociation of acetic acid.

Then, concentration at equilibrium:

[CH₃COOH] = 0.05 − x

[H⁺] = x

[CH₃COO^-] =0.05 + x

Since  is very small for a weak acid, x << 0.05

Therefore, 0.05 + x ≈ 0.05 ≈ 0.05 − x

Thus, 

By experiments, we know that for acetic acid,  = 1.8 × 10⁻⁵

Therefore, x = 1.8 × 10⁻⁵

⇒ [H⁺] = 1.8 × 10⁻⁵

Hence, pH = − log [H⁺]

= − log (1.8 × 10⁻⁵)

= 4.74

Hydrolysis of Salts and pH of Their Solutions:

Hydrolysis: Interaction of the anion or the cation (or both) of a salt with water to produce an acidic or a basic solution.

Salts of strong acids and strong bases are neutral (pH = 7)

Reason: The cations of strong bases and the anions of strong bases do not undergo hydrolysis; they only get hydrated.

Hydrolysis of the salts of a weak acid and a strong base:

For example, CH₃COONa

Completely ionised

Acetate ion undergoes hydrolysis in water

pH is more than 7.

Reason:

Acetic acid is a weak acid

↓

Remains mainly un-ionised in solution

↓

Increase in concentration of OH− ion in solution

↓

Solution becomes alkaline (pH > 7)

Hydrolysis of the salts of a strong acid and a weak base:

For example, NH₄Cl

Completely ionised

Ammonium ion undergoes hydrolysis in water

pH is less than 7

Reason:

Ammonium hydroxide is a weak base

↓

Remains mainly un-ionised in solution

↓

Increase in concentration of H+ ion in solution

↓

Solution becomes acidic (pH < 7)

Hydrolysis of the salts of a weak acid and a weak base

For example, CH₃COONH₄

Is not completely ionised

The ions undergo hydrolysis in water

 and  remain partially ionised in solution

Buffer Solution

Solution which resists change in hydrogen ion concentration (and, thereby pH) on dilution, or with the addition of a small amount of an acid or a base

Acidic buffer: Equimolar mixture of a weak acid and its salt with a strong base. E.g., mixture of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa); pH = 4.75

Basic buffer: Equimolar mixture of a weak base and its salt with a strong acid. E.g., mixture of ammonium hydroxide (NH₄OH) and ammonium chloride (NH₄Cl); pH = 9.25

Henderson Equation

Buffers are required to maintain the pH level of the solution. So they must be chosen wisely according to the pH range. The pH range of buffer solution is determined by the Henderson's equation.

Dissociation of an acid is given as

HA ⇌ H⁺  +  A^–

Equilibrium constant for the above reaction is represented as

 

Taking negative logarithm on both sides

 

 

 

The above equation represents the Henderson equation for acids.

Solubility Equilibria of Salts

Factors on which solubility depends:

Lattice enthalpy of the salt

Solvation enthalpy of the ions in a solution

For a salt to be dissolved, lattice enthalpy should be less than solvation enthalpy.

Category I	Soluble salt	Solubility > 0.1 M
Category II	Slightly soluble salt	0.01 M < Solubility < 0.1 M
Category III	Sparingly soluble salt	Solubility < 0.01 M

Solubility product constant:

Equilibrium between the un-dissolved solid and the ions in a saturated solution

Equilibrium constant,

 

BaSO₄ is a pure solid, hence, [BaSO₄] = Constant

We can write,

Where,  = Solubility product constant or Solubility product

It is found experimentally that at 298 K,

 = 1.1 × 10⁻¹⁰

If molar solubility of BaSO₄ is S, then

General formula:

Where,

Solubility product

 

 

Common ion effect on the solubility of ionic salts:

According to Le Chatelier’s principle, if the concentration of any one of the ions is increased, then it would combine with the ion of its opposite charge, and some of the salt will be precipitated, till once again .

Similarly, if the concentration of one of the ions is decreased, then more salt will need to be dissolved to increase the concentration of both the ions, till once again .

Effect of pH on solubility:

The solubility of the salts of weak acids increases with a decrease in pH.

Reason: With the decrease in pH, the concentration of the anion decreases due to its protonation. This increases the solubility of the salt, till .

If considered quantitatively, two equilibria have to be satisfied simultaneously.

 

 

On taking the reciprocal of both sides and then adding 1, we get

 

 

On taking the reciprocal again, we get

 

It can be observed that f decreases with increase in [H⁺] that is, with the decrease in pH.

Let S be the solubility of the salt at the given pH

Now,

 

 

 

Hence, it can be concluded that S increases with an increase in  , i.e., with a decrease in pH.

 

****** End of Chapter ******

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