Unit 6: Equilibrium

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The Big Picture: Why Study Chemical and Ionic Equilibrium?#

In Unit 5, we investigated the thermodynamic criteria governing whether a chemical transformation can occur spontaneously. Thermodynamics dictates the ultimate direction of a process through the standard Gibbs energy change:

$\Delta_r G^{\circ} = -RT \ln K$

However, thermodynamics alone does not tell us how fast a system approaches this state, nor does it describe the physical reality of the reaction mixture when the net drive reaches zero. Most chemical reactions do not proceed irreversibly to 100% completion. Instead, as product concentrations accumulate, backward reactions begin to operate, culminating in a state where reactants and products coexist indefinitely in fixed proportions.

This stationary macroscopic state is termed chemical equilibrium. It is of profound importance across natural, biological, and industrial domains:

1. Physiological Oxygen Transport: The binding of oxygen to hemoglobin in pulmonary capillaries and its subsequent release in oxygen-depleted muscular tissues is governed by a reversible chemical equilibrium:

$\mathrm{Hb}\text{(aq)} + 4\mathrm{O}_2\text{(g)} \rightleftharpoons \mathrm{Hb(O}_2)_4\text{(aq)}$

A fatal disruption of this balance occurs in carbon monoxide poisoning, where $\mathrm{CO}$ binds to hemoglobin with an affinity approximately 200 times greater than oxygen, shifting the equilibrium irreversibly and starving cellular respiration. 2. Industrial Synthesis Optimization: The worldwide annual synthesis of ammonia via the Haber process exceeds one hundred million tonnes, providing the primary feedstock for synthetic fertilizers. Because the synthesis reaction is reversible and exothermic:

$\mathrm{N}_2\text{(g)} + 3\mathrm{H}_2\text{(g)} \rightleftharpoons 2\mathrm{NH}_3\text{(g)}, \quad \Delta_r H^{\circ} = -92.38\text{ kJ/mol}$

Chemical engineers must rigorously apply Le Chatelier's principle to select optimum temperatures ($700\text{ K}$ to $750\text{ K}$), elevated pressures ($200\text{ bar}$), and iron catalysts to maximize economic yield without sacrificing reaction rate. 3. Aqueous and Cellular Homeostasis: Human blood maintains a strictly buffered $\text{pH}$ of $7.40 \pm 0.05$. A deviation below $7.35$ (acidosis) or above $7.45$ (alkalosis) causes severe metabolic trauma or death. This stability is maintained by dynamic ionic equilibria involving carbonic acid, bicarbonate ions, and cellular phosphate buffers.

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6.1 Equilibrium in Physical Processes#

The fundamental characteristics of dynamic equilibrium are most clearly recognized by examining phase transformation processes in closed systems. Phase changes involve no rearrangement of chemical bonds, allowing us to isolate the physics of bidirectional rate equality.

The three primary phase equilibria are:

$\text{Solid} \rightleftharpoons \text{Liquid}$ $\text{Liquid} \rightleftharpoons \text{Vapour}$ $\text{Solid} \rightleftharpoons \text{Vapour}$

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6.1.1 Solid-Liquid Equilibrium#

Consider pure water ice and liquid water placed inside a perfectly insulated Dewar flask at standard atmospheric pressure ($1.013\text{ bar}$) and $273.15\text{ K}$ ($0^{\circ}\text{C}$). Because the vessel is insulated, no heat enters or leaves the system.

Experimental observations reveal:

• The mass of ice and the mass of liquid water remain completely invariant over time.
• The temperature remains fixed at $273.15\text{ K}$.

This constancy does not indicate a static, motionless boundary. At the microscopic interface between solid ice and liquid water:

1. Energetic liquid water molecules collide with the ice crystal lattice, lose translational momentum, and freeze into the rigid hydrogen-bonded framework.
2. Simultaneously, surface water molecules within the ice lattice absorb thermal vibrational energy and break away into the liquid phase.
3. At $273.15\text{ K}$ and $1.013\text{ bar}$, the rate of melting equals the rate of freezing:

$\text{Rate of transfer of molecules from ice to water} = \text{Rate of transfer of molecules from water to ice}$

$\mathrm{H}_2\mathrm{O}\text{(s)} \rightleftharpoons \mathrm{H}_2\mathrm{O}\text{(l)}$

For any pure substance at atmospheric pressure, the unique temperature at which solid and liquid phases coexist in dynamic equilibrium is called the normal melting point or normal freezing point.

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6.1.2 Liquid-Vapour Equilibrium#

To understand liquid-vapour equilibrium, consider a transparent evacuated chamber connected to a mercury U-tube manometer:

1. Anhydrous calcium chloride or phosphorus pentoxide is placed inside the chamber for several hours to eliminate all traces of ambient water vapour.
2. The drying agent is tilted away, and a petri dish containing liquid water is quickly introduced into the dry chamber.

$\mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{H}_2\mathrm{O}\text{(vap)}$

The partial pressure exerted by the vapor in equilibrium with the liquid at a specified temperature is defined as the equilibrium vapour pressure (or saturated vapour pressure).

Vapour pressure is an intensive property that depends strictly on temperature and the intermolecular force strength of the liquid:

• Volatile liquids with weak intermolecular forces (such as diethyl ether and acetone) exhibit high equilibrium vapour pressures at room temperature.
• Strongly hydrogen-bonded liquids (such as water and ethylene glycol) exhibit substantially lower equilibrium vapour pressures.
• When the vapour pressure of a liquid equals the external atmospheric pressure, boiling occurs throughout the bulk liquid. At standard atmospheric pressure ($1.013\text{ bar}$), this temperature is defined as the normal boiling point ($100^{\circ}\text{C}$ for pure water).
• At high altitudes where atmospheric pressure is reduced, the boiling point of water decreases proportionally. In an open container, dynamic equilibrium can never be established because vapour molecules continuously diffuse into the atmosphere, meaning $r_{\text{cond}} \ll r_{\text{evap}}$.

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6.1.3 Solid-Vapour Equilibrium#

Substances capable of subliming transition directly between solid and gaseous phases without passing through an intermediate liquid state:

$\mathrm{I}_2\text{(s)} \rightleftharpoons \mathrm{I}_2\text{(vap)}$

When solid crystalline iodine is sealed within a closed glass flask at room temperature:

1. Sublimation produces violet gaseous iodine molecules that diffuse throughout the vessel.
2. The intensity of the purple color increases with time.
3. Eventually, the color intensity reaches a steady, constant value.
4. At this point, the rate of sublimation of solid iodine equals the rate of condensation of gaseous iodine back onto the crystal surfaces.

Similar solid-vapour dynamic equilibria are exhibited by camphor and ammonium chloride:

$\text{Camphor(s)} \rightleftharpoons \text{Camphor(vap)}$ $\mathrm{NH}_4\mathrm{Cl}\text{(s)} \rightleftharpoons \mathrm{NH}_4\mathrm{Cl}\text{(vap)}$

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6.1.4 Equilibrium Involving Dissolution of Solids or Gases in Liquids#

Dissolution of Solids in Liquids

When table sugar (sucrose) is added incrementally to a fixed volume of water at $298\text{ K}$, solute molecules dissolve until reaching a concentration threshold beyond which no further solid dissolves. The solution is now saturated.

In a saturated solution, a dynamic equilibrium operates between the dissolved solute molecules and the solid crystal phase:

$\text{Sugar(solid)} \rightleftharpoons \text{Sugar(solution)}$ $\text{Rate of dissolution} = \text{Rate of crystallisation}$

Dissolution of Gases in Liquids: Henry's Law

When carbonated soda water is sealed under high pressure, gaseous carbon dioxide dissolves into the aqueous phase:

$\mathrm{CO}_2\text{(g)} \rightleftharpoons \mathrm{CO}_2\text{(aq)}$

This equilibrium is governed by Henry's Law, which states that the mass of a gas dissolved in a given volume of liquid at a given temperature is directly proportional to the partial pressure of the gas above the liquid surface:

$m = k_H \cdot p$

Alternatively, in terms of mole fraction $x$ or molar concentration:

$p = K_H \cdot x$

When a soda bottle cap is removed:

1. The headspace pressure drops abruptly from several atmospheres to atmospheric carbon dioxide partial pressure ($p_{\mathrm{CO}_2} \approx 0.0004\text{ bar}$).
2. The equilibrium shifts immediately to relieve the pressure drop: dissolved $\mathrm{CO}_2$ comes out of solution as vigorous effervescence.
3. If left open to ambient air, the beverage eventually turns flat as dissolved carbon dioxide drops to its new, much lower equilibrium concentration.

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6.1.5 General Characteristics of Physical Equilibria#

The unifying physical principles of all equilibria in physical processes are summarized in Table 6.1:

From these systems, five essential rules emerge:

1. Dynamic equilibrium is possible only within a closed system where matter cannot enter or exit.
2. Both opposing processes occur simultaneously at identical rates, maintaining dynamic microscopic turnover despite static macroscopic properties.
3. All measurable macroscopic properties (pressure, concentration, density, color intensity, temperature) remain constant over time.
4. Equilibrium is characterized by a constant numerical value of an intensive thermodynamic parameter at a fixed temperature.
5. The magnitude of this parameter indicates the physical extent to which the process has progressed before achieving equilibrium.

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6.2 Equilibrium in Chemical Processes: Dynamic Equilibrium#

Just as in physical phase transitions, reversible chemical reactions reach a condition where the rates of forward and reverse reactions balance each other.

Consider a generalized reversible elementary chemical transformation:

$\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}$

1. At time $t = 0$, only reactants $\mathrm{A}$ and $\mathrm{B}$ are present at initial concentrations $[\mathrm{A}]_0$ and $[\mathrm{B}]_0$. The forward rate is maximal: $r_f = k_f [\mathrm{A}] [\mathrm{B}]$. The reverse rate is zero: $r_r = k_r [\mathrm{C}] [\mathrm{D}] = 0$.
2. As reactants collide and form products, $[\mathrm{A}]$ and $[\mathrm{B}]$ decrease, causing the forward rate $r_f$ to decline continuously.
3. Simultaneously, product concentrations $[\mathrm{C}]$ and $[\mathrm{D}]$ accumulate, causing the reverse rate $r_r$ to rise from zero.
4. Eventually, a point is reached where:

$r_f = r_r \implies k_f [\mathrm{A}]_{\text{eq}} [\mathrm{B}]_{\text{eq}} = k_r [\mathrm{C}]_{\text{eq}} [\mathrm{D}]_{\text{eq}}$

From this moment onward, the concentrations of all reactants and products remain invariant. This condition is chemical equilibrium.

Classification of Chemical Reactions by Extent of Reaction#

Chemical reactions are grouped into three categories based on where their equilibrium position lies:

1. Reactions proceeding nearly to completion: Equilibrium mixtures contain virtually pure products with barely detectable traces of reactants ($K > 10^3$). Example: $\mathrm{H}_2\text{(g)} + \frac{1}{2}\mathrm{O}_2\text{(g)} \rightleftharpoons \mathrm{H}_2\mathrm{O}\text{(l)}$ with $K_c \approx 2.4 \times 10^{47}$ at $500\text{ K}$.
2. Reactions that barely proceed: Equilibrium mixtures consist almost entirely of unreacted starting materials with only minute traces of products formed ($K < 10^{-3}$). Example: $\mathrm{N}_2\text{(g)} + \mathrm{O}_2\text{(g)} \rightleftharpoons 2\mathrm{NO}\text{(g)}$ with $K_c \approx 4.8 \times 10^{-31}$ at $298\text{ K}$.
3. Reactions with comparable reactant and product concentrations: Significant concentrations of both reactants and products coexist at equilibrium ($10^{-3} \le K \le 10^3$). Example: $\mathrm{H}_2\text{(g)} + \mathrm{I}_2\text{(g)} \rightleftharpoons 2\mathrm{HI}\text{(g)}$ with $K_c = 57.0$ at $700\text{ K}$.

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Demonstrating Dynamic Equilibrium: The Haber Deuterium Experiment#

To rigorously prove that chemical equilibrium is dynamic rather than static, Fritz Haber conducted an elegant isotopic exchange experiment:

1. Dihydrogen and dinitrogen are reacted under high pressure and temperature with an iron catalyst until equilibrium is reached:

$\mathrm{N}_2\text{(g)} + 3\mathrm{H}_2\text{(g)} \rightleftharpoons 2\mathrm{NH}_3\text{(g)}$

1. In an identical separate vessel under identical conditions, deuterium gas ($D_2$, heavy hydrogen) is reacted with dinitrogen until reaching equilibrium:

$\mathrm{N}_2\text{(g)} + 3\mathrm{D}_2\text{(g)} \rightleftharpoons 2\mathrm{ND}_3\text{(g)}$

1. Both equilibrium vessels are connected and their contents allowed to mix. The overall concentrations of total ammonia and total hydrogen remain completely unchanged.
2. However, when the mixture is analyzed using a mass spectrometer, four distinct isotopic forms of ammonia are detected:

$\mathrm{NH}_3, \quad \mathrm{NH}_2\mathrm{D}, \quad \mathrm{NHD}_2, \quad \mathrm{ND}_3$

as well as all three forms of dihydrogen:

$\mathrm{H}_2, \quad \mathrm{HD}, \quad \mathrm{D}_2$

If chemical reactions stopped when reaching equilibrium, zero isotopic scrambling could occur. The formation of mixed species like $\mathrm{NH}_2\mathrm{D}$ and $\mathrm{HD}$ proves that forward and reverse reactions continue to occur at identical rates.

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Reversible Approach from Either Direction#

A fundamental property of chemical equilibrium is that the same equilibrium composition is attained regardless of whether the reaction starts from pure reactants or pure products, provided the total atomic composition and temperature are identical.

For the hydrogen-iodine system:

$\mathrm{H}_2\text{(g)} + \mathrm{I}_2\text{(g)} \rightleftharpoons 2\mathrm{HI}\text{(g)}$

• If we start with $1.0\text{ mol } \mathrm{H}_2$ and $1.0\text{ mol } \mathrm{I}_2$ in a $1\text{ L}$ sealed container at $731\text{ K}$, purple iodine vapour fades as colorless $\mathrm{HI}$ forms until concentrations stabilize.
• If we start with $2.0\text{ mol } \mathrm{HI}$ in an identical $1\text{ L}$ vessel at $731\text{ K}$, colorless gas decomposes to produce purple $\mathrm{I}_2$ vapour and $\mathrm{H}_2$.
• At equilibrium, both vessels reach identical final concentrations of $[\mathrm{H}_2]$, $[\mathrm{I}_2]$, and $[\mathrm{HI}]$.

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6.3 Law of Chemical Equilibrium and Equilibrium Constant#

On the basis of empirical investigations of reversible reactions, Norwegian chemists Cato Maximilian Guldberg and Peter Waage proposed the Law of Mass Action in 1864.

For a general reversible reaction:

$a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$

The Equilibrium Law states that at a given temperature, the product of the equilibrium concentrations of the products raised to their respective stoichiometric coefficients, divided by the product of the equilibrium concentrations of the reactants raised to their respective stoichiometric coefficients, is a constant:

$K_c = \frac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b}$

where:

• $[\mathrm{A}], [\mathrm{B}], [\mathrm{C}], [\mathrm{D}]$ denote molar concentrations at equilibrium (expressed in $\text{mol/L}$ or $\text{M}$).
• $a, b, c, d$ are the stoichiometric coefficients from the balanced chemical equation.
• $K_c$ is the equilibrium constant expressed in concentration units.

Kinetic Derivation of the Equilibrium Law#

For an elementary bidirectional reaction:

$\mathrm{A} + \mathrm{B} \underset{k_r}{\overset{k_f}{\rightleftharpoons}} \mathrm{C} + \mathrm{D}$

The rate of the forward reaction is:

$r_f = k_f [\mathrm{A}][\mathrm{B}]$

The rate of the reverse reaction is:

$r_r = k_r [\mathrm{C}][\mathrm{D}]$

At dynamic equilibrium, the forward and reverse rates are equal:

$r_f = r_r \implies k_f [\mathrm{A}]_{\text{eq}} [\mathrm{B}]_{\text{eq}} = k_r [\mathrm{C}]_{\text{eq}} [\mathrm{D}]_{\text{eq}}$

Rearranging gives the ratio of rate constants:

$\frac{k_f}{k_r} = \frac{[\mathrm{C}]_{\text{eq}} [\mathrm{D}]_{\text{eq}}}{[\mathrm{A}]_{\text{eq}} [\mathrm{B}]_{\text{eq}}} = K_c$

Because the forward and reverse rate constants $k_f$ and $k_r$ are intrinsic properties that depend solely on temperature, their quotient $K_c$ is also constant at a specified temperature.

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Rules for Writing Equilibrium Expressions#

When manipulating chemical equations, the form of the equilibrium constant changes systematically:

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6.4 Homogeneous Equilibria#

In a homogeneous equilibrium, all reacting species and products exist within a single phase (gas phase, pure liquid phase, or uniform liquid solution).

Examples:

• Gaseous homogeneous equilibrium:

$\mathrm{N}_2\text{(g)} + 3\mathrm{H}_2\text{(g)} \rightleftharpoons 2\mathrm{NH}_3\text{(g)}$

• Liquid esterification homogeneous equilibrium:

$\mathrm{CH}_3\mathrm{COOH}\text{(l)} + \mathrm{C}_2\mathrm{H}_5\mathrm{OH}\text{(l)} \rightleftharpoons \mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5\text{(l)} + \mathrm{H}_2\mathrm{O}\text{(l)}$

• Aqueous ionic homogeneous equilibrium:

$\mathrm{Fe}^{3+}\text{(aq)} + \mathrm{SCN}^-\text{(aq)} \rightleftharpoons [\mathrm{Fe(SCN)}]^{2+}\text{(aq)}$

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6.4.1 Equilibrium Constant in Gaseous Systems ($K_p$)#

For reactions involving gases, measuring partial pressures in bar or atmospheres is often more practical than determining molar concentrations directly.

From the Ideal Gas Equation:

$pV = nRT \implies p = \left(\frac{n}{V}\right)RT = cRT$

where:

• $c = \frac{n}{V}$ is the molar concentration in $\text{mol/L}$.
• $p$ is the partial pressure in bar.
• $R = 0.0831\text{ bar L mol}^{-1}\text{ K}^{-1}$.
• $T$ is the absolute temperature in Kelvin.

Therefore, the partial pressure of any ideal gas component is directly proportional to its molar concentration at constant temperature:

$p_i = [i]RT$

Consider the general gaseous reaction:

$a\mathrm{A}\text{(g)} + b\mathrm{B}\text{(g)} \rightleftharpoons c\mathrm{C}\text{(g)} + d\mathrm{D}\text{(g)}$

The equilibrium constant expressed in partial pressures is:

$K_p = \frac{(p_{\mathrm{C}})^c (p_{\mathrm{D}})^d}{(p_{\mathrm{A}})^a (p_{\mathrm{B}})^b}$

Substituting $p_i = [i]RT$:

$K_p = \frac{([\mathrm{C}]RT)^c ([\mathrm{D}]RT)^d}{([\mathrm{A}]RT)^a ([\mathrm{B}]RT)^b} = \frac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b} \cdot \frac{(RT)^{c+d}}{(RT)^{a+b}}$

$K_p = K_c (RT)^{(c+d) - (a+b)}$

$K_p = K_c (RT)^{\Delta n_g}$

where:

$\Delta n_g = (c + d) - (a + b) = \sum n_{\text{gaseous products}} - \sum n_{\text{gaseous reactants}}$

Three Scenarios for the Relation between $K_p$ and $K_c$

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6.5 Heterogeneous Equilibria#

An equilibrium system containing substances in more than one phase is designated a heterogeneous equilibrium.

Classic examples include:

• Liquid-vapour water in a closed flask: $\mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{H}_2\mathrm{O}\text{(g)}$
• Saturated slaked lime solution: $\mathrm{Ca(OH)}_2\text{(s)} \rightleftharpoons \mathrm{Ca}^{2+}\text{(aq)} + 2\mathrm{OH}^-\text{(aq)}$
• Thermal decomposition of limestone in a lime kiln:

$\mathrm{CaCO}_3\text{(s)} \overset{\Delta}{\rightleftharpoons} \mathrm{CaO}\text{(s)} + \mathrm{CO}_2\text{(g)}$

Why Pure Solids and Pure Liquids are Omitted from Equilibrium Constant Expressions#

Writing the equilibrium expression for limestone decomposition according to mass action:

$K_c = \frac{[\mathrm{CaO(s)}][\mathrm{CO}_2\text{(g)}]}{[\mathrm{CaCO}_3\text{(s)}]}$

By definition, the molar concentration of any pure solid or pure liquid is the number of moles per unit volume:

$[\text{Pure Solid or Liquid}] = \frac{n}{V} = \frac{m / M}{V} = \frac{m}{V} \cdot \frac{1}{M} = \frac{\rho}{M}$

Because density $\rho$ and molar mass $M$ of a pure crystalline solid or pure liquid are intensive constants independent of the total quantity present, their molar concentrations are constants at a given temperature:

• $[\mathrm{CaCO}_3\text{(s)}] = \text{constant}$
• $[\mathrm{CaO(s)}] = \text{constant}$

Multiplying these constants into $K_c$ yields the modified equilibrium constant:

$K'_c = K_c \cdot \frac{[\mathrm{CaCO}_3\text{(s)}]}{[\mathrm{CaO(s)}]} = [\mathrm{CO}_2\text{(g)}]$

or in terms of partial pressure:

$K_p = p_{\mathrm{CO}_2}$

This demonstrates that at a given temperature, there is a single, fixed equilibrium pressure of carbon dioxide in contact with calcium oxide and calcium carbonate, regardless of whether you have ten grams or ten tonnes of solid limestone.

Experimentally, at $1100\text{ K}$, the equilibrium pressure of $\mathrm{CO}_2$ is $2.0 \times 10^5\text{ Pa} = 2.0\text{ bar}$. Therefore:

$K_p = \frac{2.0 \times 10^5\text{ Pa}}{10^5\text{ Pa}} = 2.00$

Other heterogeneous equilibrium expressions:

• Mond process for nickel refining:

$\mathrm{Ni}\text{(s)} + 4\mathrm{CO}\text{(g)} \rightleftharpoons \mathrm{Ni(CO)}_4\text{(g)} \implies K_c = \frac{[\mathrm{Ni(CO)}_4]}{[\mathrm{CO}]^4}$

• Silver oxide in nitric acid:

$\mathrm{Ag}_2\mathrm{O}\text{(s)} + 2\mathrm{HNO}_3\text{(aq)} \rightleftharpoons 2\mathrm{AgNO}_3\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \implies K_c = \frac{[\mathrm{AgNO}_3]^2}{[\mathrm{HNO}_3]^2}$

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6.6 Applications of Equilibrium Constants#

The equilibrium constant provides three vital predictive tools for chemical processes:

1. Predicting the extent of reaction from the magnitude of $K$.
2. Predicting the direction of reaction by evaluating the Reaction Quotient $Q$.
3. Calculating equilibrium concentrations from initial quantities.

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6.6.1 Predicting the Extent of a Reaction#

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6.6.2 Predicting the Direction of Reaction: The Reaction Quotient ($Q$)#

For any arbitrary mixture of reactants and products (not necessarily at equilibrium), we define the Reaction Quotient $Q_c$:

$Q_c = \frac{[\mathrm{C}]_t^c [\mathrm{D}]_t^d}{[\mathrm{A}]_t^a [\mathrm{B}]_t^b}$

where subscript $t$ designates instantaneous concentrations measured at non-equilibrium time $t$.

Comparing $Q_c$ with the equilibrium constant $K_c$ determines the spontaneous direction of the reaction:

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6.7 Relationship between Equilibrium Constant $K$, Reaction Quotient $Q$, and Gibbs Energy $G$#

Thermodynamics provides the fundamental energetic foundation for chemical equilibrium. From classical thermodynamics, the change in Gibbs energy for a reaction mixture at any arbitrary composition is given by:

$\Delta_r G = \Delta_r G^{\circ} + RT \ln Q$

where:

• $\Delta_r G$ is the instantaneous Gibbs free energy gradient (driving force).
• $\Delta_r G^{\circ}$ is the standard Gibbs energy change when all reactants and products are in their standard states ($1\text{ bar}$ partial pressure for gases, $1\text{ M}$ for solutes).
• $Q$ is the reaction quotient.

The Physical Condition of Equilibrium#

1. If $\Delta_r G < 0$, the forward reaction is spontaneous and proceeds until free energy is minimized.
2. If $\Delta_r G > 0$, the reverse reaction is spontaneous.
3. When the system achieves dynamic chemical equilibrium:

$\Delta_r G = 0 \quad \text{and} \quad Q = K$

Substituting these into the thermodynamic relation:

$0 = \Delta_r G^{\circ} + RT \ln K$

$\Delta_r G^{\circ} = -RT \ln K$

Converting natural logarithms to base-10 logarithms:

$\Delta_r G^{\circ} = -2.303 RT \log_{10} K$

Solving for the equilibrium constant $K$:

$K = e^{-\Delta_r G^{\circ} / RT}$

Thermodynamic Interpretation of Reaction Spontaneity#

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6.8 Factors Affecting Equilibria: Le Chatelier's Principle#

When a chemical system reaches dynamic equilibrium, it will maintain its composition indefinitely provided external conditions are kept constant. However, industrial chemical synthesis requires shifting equilibria to maximize product yield.

The qualitative response of an equilibrium system to external perturbation is governed by Le Chatelier's Principle (formulated by French chemist Henri Le Chatelier in 1884):

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6.8.1 Effect of Concentration Change#

When the concentration of any participating species is altered:

• Adding a substance: The equilibrium shifts in the direction that consumes the added substance.
• Removing a substance: The equilibrium shifts in the direction that replenishes the removed substance.

Consider the equilibrium:

$\mathrm{H}_2\text{(g)} + \mathrm{I}_2\text{(g)} \rightleftharpoons 2\mathrm{HI}\text{(g)}$

1. If extra $\mathrm{H}_2$ gas is injected at equilibrium, $[\mathrm{H}_2]$ increases, making the reaction quotient:

$Q_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} < K_c$

1. Because $Q_c < K_c$, a net forward reaction occurs. Additional $\mathrm{H}_2$ and $\mathrm{I}_2$ react to produce $\mathrm{HI}$.
2. When new equilibrium is restored, the final concentration of $\mathrm{H}_2$ is less than immediately after injection, but greater than in the original mixture, while $[\mathrm{I}_2]$ has decreased and $[\mathrm{HI}]$ has increased.

Industrial Utility of Product Removal

In the manufacture of quicklime ($\mathrm{CaO}$) from limestone:

$\mathrm{CaCO}_3\text{(s)} \rightleftharpoons \mathrm{CaO}\text{(s)} + \mathrm{CO}_2\text{(g)}$

If the reaction were carried out in a closed container, equilibrium would be reached when $p_{\mathrm{CO}_2} = K_p$, stopping further decomposition. In industrial rotary kilns, a continuous draft of air sweeps carbon dioxide out of the kiln flue ($p_{\mathrm{CO}_2} \ll K_p$), driving decomposition of $\mathrm{CaCO}_3$ to 100% completion.

Laboratory Demonstration: Iron(III) Thiocyanate Complex

The reversible formation of the deep blood-red complex ion:

$\underset{\text{Pale Yellow}}{\mathrm{Fe}^{3+}\text{(aq)}} + \underset{\text{Colourless}}{\mathrm{SCN}^-\text{(aq)}} \rightleftharpoons \underset{\text{Deep Blood-Red}}{[\mathrm{Fe(SCN)}]^{2+}\text{(aq)}}$

$K_c = \frac{[[\mathrm{Fe(SCN)}]^{2+}]}{[\mathrm{Fe}^{3+}][\mathrm{SCN}^-]}$

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6.8.2 Effect of Pressure and Volume Change#

Because liquids and solids are virtually incompressible, changes in pressure have negligible effects on heterogeneous systems containing only condensed phases. For gaseous equilibria, pressure changes obtained by altering the container volume have substantial effects whenever $\Delta n_g \neq 0$.

Consider the industrial methanation reaction:

$\mathrm{CO}\text{(g)} + 3\mathrm{H}_2\text{(g)} \rightleftharpoons \mathrm{CH}_4\text{(g)} + \mathrm{H}_2\mathrm{O}\text{(g)}$

• Reactant gas moles: $1 + 3 = 4\text{ moles}$
• Product gas moles: $1 + 1 = 2\text{ moles}$
• Change in gas moles: $\Delta n_g = 2 - 4 = -2$

If the volume of the reaction container is halved at constant temperature, Boyle's Law dictates that total pressure doubles, and every individual concentration doubles ($c' = 2c$).

The new reaction quotient is:

$Q_c = \frac{[2\mathrm{CH}_4][2\mathrm{H}_2\mathrm{O}]}{[2\mathrm{CO}][2\mathrm{H}_2]^3} = \frac{4}{16} \cdot \frac{[\mathrm{CH}_4][\mathrm{H}_2\mathrm{O}]}{[\mathrm{CO}][\mathrm{H}_2]^3} = \frac{1}{4} K_c$

Because $Q_c = \frac{1}{4}K_c < K_c$, the forward reaction proceeds preferentially. The equilibrium shifts in the forward direction, which contains fewer gaseous molecules ($2$ vs $4$), thereby decreasing internal pressure and counteracting the applied compression.

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6.8.3 Effect of Inert Gas Addition#

When an inert gas (such as argon, helium, or neon) that does not participate chemically is introduced into an equilibrium vessel:

1. Addition at Constant Volume ($V = \text{constant}$):
   Total pressure increases due to additional non-reactive atoms. However, because volume and temperature are fixed, the partial pressure and molar concentration of each reacting gas ($p_i = \frac{n_i RT}{V}$, $[i] = \frac{n_i}{V}$) remain strictly unchanged. Therefore, $Q_c$ remains equal to $K_c$, and the equilibrium position is completely undisturbed.
2. Addition at Constant Pressure ($p = \text{constant}$):
   To accommodate the inert gas while maintaining constant pressure, the container volume must expand. This expansion decreases the partial pressures of all reacting gases. The system responds as if subjected to an overall pressure decrease, shifting toward the side with the larger number of gaseous moles (if $\Delta n_g > 0$, shifts forward; if $\Delta n_g < 0$, shifts reverse).

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6.8.4 Effect of Temperature Change#

Perturbations of concentration, pressure, or volume alter the reaction quotient $Q_c$, causing the system to shift until $Q_c$ matches an invariant $K_c$. In contrast, a change in temperature alters the numerical value of the equilibrium constant $K$ itself.

The temperature dependence of $K$ is governed by the sign of the standard enthalpy of reaction $\Delta_r H^{\circ}$, as formalized by the van 't Hoff equation:

$\ln \left(\frac{K_2}{K_1}\right) = \frac{\Delta_r H^{\circ}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Experimental Demonstration: Nitrogen Dioxide Dimerization

Consider the dimerization of brown nitrogen dioxide into colorless dinitrogen tetroxide:

$2\mathrm{NO}_2\text{(g)} \rightleftharpoons \mathrm{N}_2\mathrm{O}_4\text{(g)}, \quad \Delta_r H^{\circ} = -57.2\text{ kJ/mol (Exothermic)}$

Two sealed test tubes containing identical amounts of brown $\mathrm{NO}_2$ gas are prepared:

• Tube 1 is placed in a freezing bath at $270\text{ K}$ ($0^{\circ}\text{C}$). Cooling favors the forward exothermic reaction. The deep brown color fades, turning nearly colourless as $\mathrm{N}_2\mathrm{O}_4$ forms.
• Tube 2 is placed in a boiling water bath at $363\text{ K}$ ($90^{\circ}\text{C}$). Heating drives the reverse endothermic reaction. The gas darkens into an intense reddish-brown color as $\mathrm{N}_2\mathrm{O}_4$ dissociates into $\mathrm{NO}_2$.

Endothermic Demonstration: Cobalt Chloride Hexahydrate

$\underset{\text{Pink}}{[\mathrm{Co(H}_2\mathrm{O})_6]^{2+}\text{(aq)}} + 4\mathrm{Cl}^-\text{(aq)} \rightleftharpoons \underset{\text{Deep Blue}}{[\mathrm{CoCl}_4]^{2-}\text{(aq)}} + 6\mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta_r H^{\circ} > 0$

• At room temperature, the solution is blue-violet.
• When placed in an ice bath, it shifts in the reverse exothermic direction, turning distinctly pink.
• When heated over a burner, it shifts forward in the endothermic direction, turning vivid deep blue.

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6.8.5 Effect of a Catalyst#

A catalyst accelerates a chemical transformation by providing an alternative reaction pathway with a lower activation energy ($E_a$).

Key features of catalysts in equilibrium systems:

1. A catalyst lowers the activation energy of the forward reaction ($E_{a,f}$) and reverse reaction ($E_{a,r}$) by the exact same numerical amount:

$\Delta E_a = E_{a,f}^{\text{cat}} - E_{a,f}^{\text{uncat}} = E_{a,r}^{\text{cat}} - E_{a,r}^{\text{uncat}}$

1. Consequently, the forward and reverse rate constants are accelerated by the exact same multiplier:

$\frac{k_f^{\text{cat}}}{k_f^{\text{uncat}}} = \frac{k_r^{\text{cat}}}{k_r^{\text{uncat}}}$

1. The equilibrium constant is the quotient of these rate constants:

$K_c = \frac{k_f}{k_r} = \frac{k_f^{\text{cat}}}{k_r^{\text{cat}}}$

Therefore, a catalyst does not alter the numerical value of $K_c$, nor does it shift the equilibrium position or change product yield.

A catalyst merely accelerates the rate at which the dynamic equilibrium state is achieved. In industrial processes where reactions have high activation barriers (such as the contact process for sulphuric acid, $2\mathrm{SO}_2 + \mathrm{O}_2 \rightleftharpoons 2\mathrm{SO}_3$ with a $\mathrm{V}_2\mathrm{O}_5$ catalyst, or the Haber process with iron), catalysts allow commercial production rates at moderate operating temperatures.

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6.9 Ionic Equilibrium in Solution#

In 1834, Michael Faraday discovered that chemical substances can be divided into two classes based on their electrical conductivity in aqueous solution:

1. Electrolytes: Substances whose aqueous solutions conduct an electric current (acids, bases, salts).
2. Non-electrolytes: Substances whose aqueous solutions do not conduct electricity (aqueous sucrose, urea, glycerol).

Faraday further classified electrolytes into two categories:

• Strong Electrolytes: Dissociate almost completely ($100\%$ ionization) into ions upon dissolution in water. Examples: $\mathrm{HCl}, \mathrm{HNO}_3, \mathrm{H}_2\mathrm{SO}_4, \mathrm{NaOH}, \mathrm{KOH}, \mathrm{NaCl}$.
• Weak Electrolytes: Ionize only partially in aqueous solution, leaving most solute molecules un-ionized. Examples: $\mathrm{CH}_3\mathrm{COOH}, \mathrm{HF}, \mathrm{HCN}, \mathrm{NH}_3, \mathrm{NH}_4\mathrm{OH}$.

In weak electrolytes, a dynamic equilibrium operates between the un-ionized neutral molecules and the generated ions:

$\mathrm{CH}_3\mathrm{COOH}\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{CH}_3\mathrm{COO}^-\text{(aq)} + \mathrm{H}_3\mathrm{O}^+\text{(aq)}$

This equilibrium involving ions in aqueous solution is defined as ionic equilibrium.

Water facilitates ionic dissociation because of its high dielectric constant ($\epsilon_r \approx 80$ at $298\text{ K}$). According to Coulomb's law:

$F = \frac{q_1 q_2}{4\pi \varepsilon_0 \varepsilon_r r^2}$

The electrostatic force holding oppositely charged ions together within a crystal lattice is reduced by a factor of 80 when immersed in water, while exothermic ion hydration stabilizes the separated ions.

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6.10 Acids, Bases, and Salts#

Three successive theoretical frameworks describe acid-base behavior:

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6.10.1 The Hydronium Ion ($mathrm{H}_3mathrm{O}^+$)#

A bare proton ($\mathrm{H}^+$) is an unshielded atomic nucleus with a radius of only $\approx 10^{-15}\text{ m}$, creating an intense electric field ($10^{10}\text{ V/m}$). A bare proton cannot exist independently in aqueous solution.

It immediately binds to a lone pair on a water molecule to form the trigonal pyramidal hydronium ion:

$\mathrm{H}^+ + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{H}_3\mathrm{O}^+$

In liquid water, hydronium is hydrated further into extended hydrogen-bonded clusters:

$\mathrm{H}_5\mathrm{O}_2^+, \quad \mathrm{H}_7\mathrm{O}_3^+, \quad \mathrm{H}_9\mathrm{O}_4^+$

Similarly, hydroxide is hydrated into species such as $\mathrm{H}_3\mathrm{O}_2^-$ and $\mathrm{H}_7\mathrm{O}_4^-$. In this text, $\mathrm{H}^+\text{(aq)}$ and $\mathrm{H}_3\mathrm{O}^+\text{(aq)}$ are used interchangeably to represent hydrated protons.

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6.10.2 Brönsted-Lowry Conjugate Acid-Base Pairs#

According to the Brönsted-Lowry definition, an acid donates a proton to a base. In the reverse direction, the resulting species can transfer a proton back:

$\underset{\text{Base 1}}{\mathrm{NH}_3\text{(aq)}} + \underset{\text{Acid 2}}{\mathrm{H}_2\mathrm{O}\text{(l)}} \rightleftharpoons \underset{\text{Conjugate Acid 1}}{\mathrm{NH}_4^+\text{(aq)}} + \underset{\text{Conjugate Base 2}}{\mathrm{OH}^-\text{(aq)}}$

$\underset{\text{Acid 1}}{\mathrm{HCl}\text{(aq)}} + \underset{\text{Base 2}}{\mathrm{H}_2\mathrm{O}\text{(l)}} \rightleftharpoons \underset{\text{Conjugate Acid 2}}{\mathrm{H}_3\mathrm{O}^+\text{(aq)}} + \underset{\text{Conjugate Base 1}}{\mathrm{Cl}^-\text{(aq)}}$

Amphiprotic / Amphoteric Nature of Water

Water can act as both a Brönsted acid and a Brönsted base depending on the reaction partner:

• When reacting with $\mathrm{HCl}$, water accepts a proton, acting as a base: conjugate acid is $\mathrm{H}_3\mathrm{O}^+$.
• When reacting with $\mathrm{NH}_3$, water donates a proton, acting as an acid: conjugate base is $\mathrm{OH}^-$.

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6.11 Ionization of Acids and Bases#

6.11.1 The Autoionization of Water and Its Ionic Product ($K_w$)#

Pure water is an extremely weak electrolyte that undergoes self-ionization (autoprotolysis):

$\mathrm{H}_2\mathrm{O}\text{(l)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+\text{(aq)} + \mathrm{OH}^-\text{(aq)}$

The equilibrium constant is:

$K = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-]}{[\mathrm{H}_2\mathrm{O}]^2}$

Because water is present in vast excess, its molar concentration is effectively constant. For pure water at $298\text{ K}$ with density $1000\text{ g/L}$ and molar mass $18.02\text{ g/mol}$:

$[\mathrm{H}_2\mathrm{O}] = \frac{1000\text{ g/L}}{18.02\text{ g/mol}} = 55.55\text{ mol/L}$

Incorporating this constant concentration into $K$ yields the ionic product of water ($K_w$):

$K_w = [\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-] = [\mathrm{H}^+][\mathrm{OH}^-]$

At $298\text{ K}$ ($25^{\circ}\text{C}$), conductivity measurements establish:

$[\mathrm{H}^+] = [\mathrm{OH}^-] = 1.0 \times 10^{-7}\text{ M}$

$K_w = (1.0 \times 10^{-7}) \times (1.0 \times 10^{-7}) = 1.0 \times 10^{-14}\text{ M}^2$

The fraction of ionized water molecules is:

$\frac{1.0 \times 10^{-7}\text{ M}}{55.55\text{ M}} \approx 1.8 \times 10^{-9} \quad (\text{roughly } 2 \text{ molecules out of every billion})$

Temperature Dependence of $K_w$

Because autoionization is an endothermic bond-cleaving process:

$2\mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+\text{(aq)} + \mathrm{OH}^-\text{(aq)}, \quad \Delta H > 0$

Increasing the temperature shifts the equilibrium forward, increasing $K_w$:

• At $298\text{ K}$ ($25^{\circ}\text{C}$): $K_w = 1.0 \times 10^{-14} \implies [\mathrm{H}^+] = 1.0 \times 10^{-7}\text{ M} \implies \text{pH} = 7.00$
• At $310\text{ K}$ ($37^{\circ}\text{C}$, body temperature): $K_w = 2.7 \times 10^{-14} \implies [\mathrm{H}^+] = \sqrt{2.7 \times 10^{-14}} = 1.64 \times 10^{-7}\text{ M} \implies \text{pH} = 6.78$
• At $373\text{ K}$ ($100^{\circ}\text{C}$): $K_w \approx 5.13 \times 10^{-13} \implies [\mathrm{H}^+] = 7.16 \times 10^{-7}\text{ M} \implies \text{pH} = 6.14$

Even though the $\text{pH}$ of neutral water at $100^{\circ}\text{C}$ is $6.14$, the water is not acidic because $[\mathrm{H}^+] = [\mathrm{OH}^-]$ remains strictly satisfied. The neutral point shifts with temperature:

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6.11.2 The pH Scale#

In 1909, Danish biochemist Søren P. L. Sørensen introduced the logarithmic pH scale to represent hydrogen ion concentrations:

$\text{pH} = -\log_{10} a_{\mathrm{H}^+} \approx -\log_{10} \left(\frac{[\mathrm{H}^+]}{1\text{ mol/L}}\right)$

Similarly, for hydroxide ions:

$\text{pOH} = -\log_{10} [\mathrm{OH}^-]$

Taking the negative logarithm of the water autoionization expression:

$K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14}$

$-\log K_w = -\log [\mathrm{H}^+] - \log [\mathrm{OH}^-] = -\log(10^{-14})$

$\text{pK}_w = \text{pH} + \text{pOH} = 14.00 \quad (\text{at } 298\text{ K})$

Because the scale is logarithmic:

• A change of $1\text{ pH unit}$ corresponds to a $10\text{-fold}$ change in $[\mathrm{H}^+]$.
• A change of $2\text{ pH units}$ corresponds to a $100\text{-fold}$ change in $[\mathrm{H}^+]$.

Common Solution pH Values (NCERT Table 6.5)

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6.11.3 Ionization Constants of Weak Acids ($K_a$) and Ostwald's Dilution Law#

Consider a monoprotic weak acid $\mathrm{HX}$ dissolving in water at initial concentration $c\text{ (mol/L)}$. Let $\alpha$ represent the degree of ionization (the fraction of acid molecules that dissociate):

$\mathrm{HX}\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+\text{(aq)} + \mathrm{X}^-\text{(aq)}$

The acid ionization constant is:

$K_a = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{X}^-]}{[\mathrm{HX}]} = \frac{(c\alpha)(c\alpha)}{c(1 - \alpha)} = \frac{c\alpha^2}{1 - \alpha}$

For very weak electrolytes where $\alpha \ll 1$ (typically $\alpha < 0.05$ or $5\%$), the approximation $1 - \alpha \approx 1$ holds:

$K_a \approx c\alpha^2 \implies \alpha = \sqrt{\frac{K_a}{c}}$

This relation is Ostwald's Dilution Law:

• The degree of ionization $\alpha$ is inversely proportional to the square root of concentration: $\alpha \propto \frac{1}{\sqrt{c}}$.
• As an acidic solution is diluted with water ($c \rightarrow 0$), $\alpha$ increases, approaching $1.0$ ($100\%$ ionization) at infinite dilution.

The hydronium concentration in terms of $K_a$:

$[\mathrm{H}^+] = c\alpha = c \sqrt{\frac{K_a}{c}} = \sqrt{K_a \cdot c}$

$\text{pH} = \frac{1}{2}(\text{pK}_a - \log_{10} c)$

where $\text{pK}_a = -\log_{10} K_a$. A larger $K_a$ (or smaller $\text{pK}_a$) corresponds to a stronger acid.

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6.11.4 Ionization Constants of Weak Bases ($K_b$)#

For a weak molecular base $\mathrm{B}$ or metal hydroxide $\mathrm{MOH}$:

$\mathrm{B}\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{BH}^+\text{(aq)} + \mathrm{OH}^-\text{(aq)}$

$K_b = \frac{[\mathrm{BH}^+][\mathrm{OH}^-]}{[\mathrm{B}]} = \frac{c\alpha^2}{1 - \alpha}$

When $\alpha \le 0.05$:

$\alpha \approx \sqrt{\frac{K_b}{c}}, \quad [\mathrm{OH}^-] = c\alpha = \sqrt{K_b \cdot c}$

$\text{pOH} = \frac{1}{2}(\text{pK}_b - \log_{10} c), \quad \text{pH} = 14 - \text{pOH}$

where $\text{pK}_b = -\log_{10} K_b$.

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6.11.5 Relation between $K_a$ and $K_b$ of a Conjugate Acid-Base Pair#

Consider the weak base ammonia $\mathrm{NH}_3$ and its conjugate acid, the ammonium ion $\mathrm{NH}_4^+$:

1. Base ionization of $\mathrm{NH}_3$:

$\mathrm{NH}_3\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{NH}_4^+\text{(aq)} + \mathrm{OH}^-\text{(aq)}, \quad K_b = \frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]}$

1. Acid ionization of conjugate acid $\mathrm{NH}_4^+$:

$\mathrm{NH}_4^+\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+\text{(aq)} + \mathrm{NH}_3\text{(aq)}, \quad K_a = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{NH}_3]}{[\mathrm{NH}_4^+]}$

Multiplying the two equilibrium expressions:

$K_a \times K_b = \left(\frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{NH}_3]}{[\mathrm{NH}_4^+]}\right) \times \left(\frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]}\right) = [\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-] = K_w$

$K_a \times K_b = K_w$

Taking negative logarithms:

$\text{pK}_a + \text{pK}_b = \text{pK}_w = 14.00 \quad (\text{at } 298\text{ K})$

This relationship confirms that knowing either $K_a$ or $K_b$ for one partner in a conjugate pair allows immediate calculation of the other.

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6.11.6 Di- and Polybasic Acids and Polyacidic Bases#

Acids capable of donating more than one ionizable proton per molecule are termed polybasic or polyprotic acids.

For a diprotic acid $\mathrm{H}_2\mathrm{X}$ (such as carbonic acid $\mathrm{H}_2\mathrm{CO}_3$ or oxalic acid $\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4$), ionization proceeds in successive stages:

$\mathrm{H}_2\mathrm{X}\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+\text{(aq)} + \mathrm{HX}^-\text{(aq)}, \quad K_{a1} = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{HX}^-]}{[\mathrm{H}_2\mathrm{X}]}$

$\mathrm{HX}^-\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+\text{(aq)} + \mathrm{X}^{2-}\text{(aq)}, \quad K_{a2} = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{X}^{2-}]}{[\mathrm{HX}^-]}$

Why $K_{a1} \gg K_{a2} \gg K_{a3}$

For every polyprotic acid, the successive ionization constants decrease by several orders of magnitude:

$K_{a1} \gg K_{a2} \gg K_{a3}$

Physical Origin: Electrostatic attraction dictates that removing a positively charged proton from a neutral molecule ($\mathrm{H}_2\mathrm{X}$) requires significantly less energy than pulling a proton away from a negatively charged monoanion ($\mathrm{HX}^-$), which in turn is vastly easier than pulling a proton from a doubly negative dianion ($\mathrm{X}^{2-}$).

Because $K_{a1} \gg K_{a2}$, the hydronium concentration in polyprotic acid solutions is governed almost exclusively by the first dissociation step.

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6.11.7 Factors Affecting Acid Strength#

The extent to which an acid $\mathrm{H}-\mathrm{A}$ ionizes in water depends primarily on two atomic parameters:

1. Bond Strength (Bond Dissociation Enthalpy): The energy required to break the covalent $\mathrm{H}-\mathrm{A}$ bond.
2. Bond Polarity: The electronegativity difference between $\mathrm{H}$ and $\mathrm{A}$, which induces partial ionic charge separation ($\mathrm{H}^{\delta+} - \mathrm{A}^{\delta-}$).

Trend Down a Group: Bond Strength Dominates

When comparing binary acids formed by elements in the same group of the Periodic Table, atomic radius increases down the column, lengthening the bond and weakening bond dissociation enthalpy.

For Group 17 halogen acids:

$\mathrm{HF} \ll \mathrm{HCl} \ll \mathrm{HBr} \ll \mathrm{HI}$

Even though fluorine is the most electronegative element (making the $\mathrm{H}-\mathrm{F}$ bond the most polar), the tiny fluoride radius produces a very short, strong bond ($574\text{ kJ/mol}$). Consequently, $\mathrm{HF}$ is a weak acid in water, while $\mathrm{HI}$, with its long, weak bond ($299\text{ kJ/mol}$), is a powerful strong acid.

Trend Across a Period: Electronegativity / Polarity Dominates

When comparing binary hydrides of elements in the same row of the Periodic Table, atomic radii are comparable, so bond polarity becomes the deciding factor.

For Period 2 hydrides:

$\mathrm{CH}_4 < \mathrm{NH}_3 < \mathrm{H}_2\mathrm{O} < \mathrm{HF}$

As electronegativity increases from carbon ($2.5$) to fluorine ($4.0$), electron density is pulled away from hydrogen, facilitating proton transfer to solvent water molecules.

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6.11.8 The Common-Ion Effect#

The Common-Ion Effect is the shift in equilibrium caused by the addition of an ionic compound that provides an ion already present in the equilibrium mixture. It is a direct manifestation of Le Chatelier's principle.

Consider the weak acid acetic acid:

$\mathrm{CH}_3\mathrm{COOH}\text{(aq)} \rightleftharpoons \mathrm{CH}_3\mathrm{COO}^-\text{(aq)} + \mathrm{H}^+\text{(aq)}$

If sodium acetate ($\mathrm{CH}_3\mathrm{COONa}$, a strong electrolyte) is dissolved into this solution, it dissociates completely, introducing a high concentration of common acetate ions ($\mathrm{CH}_3\mathrm{COO}^-$).

In response to this added product ion, the equilibrium shifts in the reverse direction, consuming acetate ions and hydronium ions to regenerate un-ionized acetic acid molecules. As a result:

• The degree of ionization $\alpha$ drops sharply.
• The free $[\mathrm{H}^+]$ concentration decreases.
• The solution $\text{pH}$ rises.

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6.11.9 Hydrolysis of Salts and the pH of Their Solutions#

When an ionic salt dissolves in water, it dissociates completely into its constituent cations and anions:

$\mathrm{BA}\text{(s)} \rightarrow \mathrm{B}^+\text{(aq)} + \mathrm{A}^-\text{(aq)}$

These ions can interact with water molecules. Hydrolysis is defined as the chemical interaction between water and the cation, the anion, or both ions of a salt, producing acidic or alkaline solutions.

1. Salt of a Strong Acid and a Strong Base (e.g. $mathrm{NaCl}, mathrm{KNO}_3, mathrm{BaCl}_2$)

• Cations of strong bases ($\mathrm{Na}^+, \mathrm{K}^+, \mathrm{Ca}^{2+}$) and anions of strong acids ($\mathrm{Cl}^-, \mathrm{NO}_3^-, \mathrm{SO}_4^{2-}$) have negligible affinity for protons or hydroxide ions.
• They undergo hydration but do not hydrolyze.
• The neutral autoionization balance of water ($[\mathrm{H}^+] = [\mathrm{OH}^-] = 10^{-7}\text{ M}$) remains undisturbed.
• Solution $\text{pH} = 7.00$ (strictly neutral).

2. Salt of a Weak Acid and a Strong Base (e.g. $mathrm{CH}_3mathrm{COONa}, mathrm{NaCN}, mathrm{K}_2mathrm{CO}_3$)

• The cation ($\mathrm{Na}^+$) does not hydrolyze.
• The anion ($\mathrm{CH}_3\mathrm{COO}^-$), being the conjugate base of a weak acid, reacts with water:

$\mathrm{CH}_3\mathrm{COO}^-\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH}\text{(aq)} + \mathrm{OH}^-\text{(aq)}$

• This anionic hydrolysis produces excess $\mathrm{OH}^-$ ions, making the solution alkaline:

$\text{pH} = 7 + \frac{1}{2}(\text{pK}_a + \log_{10} c) > 7$

3. Salt of a Strong Acid and a Weak Base (e.g. $mathrm{NH}_4mathrm{Cl}, mathrm{AlCl}_3, mathrm{CuSO}_4$)

• The anion ($\mathrm{Cl}^-$) does not hydrolyze.
• The cation ($\mathrm{NH}_4^+$), being the conjugate acid of a weak base, donates a proton to water:

$\mathrm{NH}_4^+\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{NH}_4\mathrm{OH}\text{(aq)} + \mathrm{H}^+\text{(aq)}$

• This cationic hydrolysis produces excess $\mathrm{H}^+$ ions, making the solution acidic:

$\text{pH} = 7 - \frac{1}{2}(\text{pK}_b + \log_{10} c) < 7$

4. Salt of a Weak Acid and a Weak Base (e.g. $mathrm{CH}_3mathrm{COONH}_4, mathrm{NH}_4mathrm{CN}$)

• Both cation and anion undergo simultaneous hydrolysis:

$\mathrm{NH}_4^+ + \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4\mathrm{OH} + \mathrm{CH}_3\mathrm{COOH}$

• Remarkably, the degree of hydrolysis and the resulting $\text{pH}$ are independent of solution concentration:

$\text{pH} = 7 + \frac{1}{2}(\text{pK}_a - \text{pK}_b)$

• If $\text{pK}_a < \text{pK}_b$: acid is stronger than base $\implies \text{pH} < 7$ (slightly acidic).
• If $\text{pK}_a > \text{pK}_b$: base is stronger than acid $\implies \text{pH} > 7$ (slightly basic).
• If $\text{pK}_a = \text{pK}_b$: $\text{pH} = 7.00$ (neutral).

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6.12 Buffer Solutions#

Solutions that resist drastic changes in $\text{pH}$ upon the addition of small quantities of strong acid or base, or upon dilution, are termed Buffer Solutions.

Buffers are categorized into two major classes:

1. Acidic Buffers: Composed of a weak acid and its salt with a strong base (e.g. $\mathrm{CH}_3\mathrm{COOH} + \mathrm{CH}_3\mathrm{COONa}$). Operates effectively in the acidic range ($\text{pH } 4 - 6$).
2. Basic Buffers: Composed of a weak base and its salt with a strong acid (e.g. $\mathrm{NH}_4\mathrm{OH} + \mathrm{NH}_4\mathrm{Cl}$). Operates effectively in the basic range ($\text{pH } 8 - 10$).

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6.12.1 Mechanism of Buffer Action (Acidic Buffer Example)#

Consider an equimolar buffer containing acetic acid and sodium acetate:

• Sodium acetate dissociates completely: $\mathrm{CH}_3\mathrm{COONa} \rightarrow \mathrm{CH}_3\mathrm{COO}^- + \mathrm{Na}^+$
• Acetic acid exists in dynamic equilibrium: $\mathrm{CH}_3\mathrm{COOH} \rightleftharpoons \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+$

1. When strong acid ($\mathrm{H}^+$) is added:
   The added protons are captured by the reservoir of conjugate base acetate ions:

$\mathrm{CH}_3\mathrm{COO}^-\text{(aq)} + \mathrm{H}^+\text{(aq)} \rightarrow \mathrm{CH}_3\mathrm{COOH}\text{(aq)}$

Because strong $\mathrm{H}^+$ ions are converted into weakly dissociated neutral acetic acid molecules, the free hydronium concentration remains virtually constant. 2. When strong base ($\mathrm{OH}^-$) is added:
The added hydroxide ions react with un-ionized acetic acid molecules:

$\mathrm{CH}_3\mathrm{COOH}\text{(aq)} + \mathrm{OH}^-\text{(aq)} \rightarrow \mathrm{CH}_3\mathrm{COO}^-\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)}$

Because strong $\mathrm{OH}^-$ ions are neutralized to form water and acetate, the $\text{pH}$ change is negligible.

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6.12.2 The Henderson-Hasselbalch Equation#

For the weak acid equilibrium:

$\mathrm{HA}\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+\text{(aq)} + \mathrm{A}^-\text{(aq)}$

$K_a = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{A}^-]}{[\mathrm{HA}]}$

Rearranging for $[\mathrm{H}_3\mathrm{O}^+]$:

$[\mathrm{H}_3\mathrm{O}^+] = K_a \cdot \frac{[\mathrm{HA}]}{[\mathrm{A}^-]}$

Taking negative base-10 logarithms of both sides:

$-\log_{10}[\mathrm{H}_3\mathrm{O}^+] = -\log_{10} K_a - \log_{10}\left(\frac{[\mathrm{HA}]}{[\mathrm{A}^-]}\right)$

$\text{pH} = \text{pK}_a + \log_{10}\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)$

Because the weak acid ionizes only minimally, $[\mathrm{HA}] \approx [\text{Acid}]_{\text{initial}}$. Because sodium salt dissociates completely, $[\mathrm{A}^-] \approx [\text{Salt}]$. This gives the celebrated Henderson-Hasselbalch Equation for Acidic Buffers:

$\text{pH} = \text{pK}_a + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right)$

Similarly, for basic buffers:

$\text{pOH} = \text{pK}_b + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Base}]}\right)$

$\text{pH} = 14.00 - \text{pOH}$

Two Essential Properties of Buffers

1. Equal Concentration Equivalence: When $[\text{Salt}] = [\text{Acid}]$, the log term becomes $\log_{10}(1) = 0$, so $\text{pH} = \text{pK}_a$. To prepare a buffer of target $\text{pH}$, select an acid whose $\text{pK}_a$ is as close as possible to the target $\text{pH}$.
2. Dilution Invariance: Diluting a buffer with pure water decreases $[\text{Salt}]$ and $[\text{Acid}]$ by the exact same ratio. Because their quotient remains unchanged inside the logarithm, the $\text{pH}$ of a buffer is unaffected by dilution.

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6.13 Solubility Equilibria of Sparingly Soluble Salts#

Solubility varies dramatically across ionic compounds:

• Soluble Salts: Molar solubility $> 0.1\text{ M}$ (e.g. $\mathrm{NaCl}, \mathrm{KNO}_3$).
• Slightly Soluble Salts: Molar solubility between $0.01\text{ M}$ and $0.1\text{ M}$.
• Sparingly Soluble Salts: Molar solubility $< 0.01\text{ M}$ (e.g. $\mathrm{AgCl}, \mathrm{BaSO}_4, \mathrm{CaF}_2$).

Dissolution of an ionic solid involves an energetic competition between:

1. Lattice Enthalpy ($\Delta_{\text{lattice}}H^{\circ}$): Positive energetic penalty to separate cations and anions from the crystal lattice.
2. Hydration Enthalpy ($\Delta_{\text{hyd}}H^{\circ}$): Negative enthalpy released as polar water molecules hydrate the freed ions.

For a salt to dissolve appreciably, its negative hydration enthalpy must overcome its lattice enthalpy. When a sparingly soluble salt is stirred in water, dissolution proceeds until an equilibrium is established between undissolved solid and dissolved hydrated ions.

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6.13.1 The Solubility Product Constant ($K_{sp}$)#

Consider barium sulphate in dynamic equilibrium with its saturated solution:

$\mathrm{BaSO}_4\text{(s)} \overset{\text{water}}{\rightleftharpoons} \mathrm{Ba}^{2+}\text{(aq)} + \mathrm{SO}_4^{2-}\text{(aq)}$

Because $[\mathrm{BaSO}_4\text{(s)}]$ is a pure solid with constant molar density, it is incorporated into the equilibrium constant:

$K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{SO}_4^{2-}]$

The constant $K_{sp}$ is defined as the Solubility Product Constant.

If $S$ represents the molar solubility of barium sulphate in pure water (in $\text{mol/L}$):

$[\mathrm{Ba}^{2+}] = S, \quad [\mathrm{SO}_4^{2-}] = S$

$K_{sp} = S \cdot S = S^2 \implies S = \sqrt{K_{sp}}$

At $298\text{ K}$, $K_{sp}(\mathrm{BaSO}_4) = 1.1 \times 10^{-10}$:

$S = \sqrt{1.1 \times 10^{-10}} = 1.05 \times 10^{-5}\text{ mol/L}$

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General Stoichiometric Formula for $K_{sp}$ and Solubility $S$#

For any general sparingly soluble ionic salt of formula $\mathrm{M}_x\mathrm{X}_y$:

$\mathrm{M}_x\mathrm{X}_y\text{(s)} \rightleftharpoons x\mathrm{M}^{p+}\text{(aq)} + y\mathrm{X}^{q-}\text{(aq)}$

Let $S$ be the molar solubility:

• $[\mathrm{M}^{p+}] = xS$
• $[\mathrm{X}^{q-}] = yS$

$K_{sp} = [\mathrm{M}^{p+}]^x [\mathrm{X}^{q-}]^y = (xS)^x (yS)^y = x^x y^y S^{x+y}$

$S = \left(\frac{K_{sp}}{x^x y^y}\right)^{\frac{1}{x+y}}$

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Criteria for Precipitation: Ionic Product ($Q_{sp}$) vs $K_{sp}$#

The product of actual instantaneous ion concentrations in a solution (each raised to their stoichiometric power) is defined as the Ionic Product ($Q_{sp}$):

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Common-Ion Applications in Chemical Analysis and Industry#

1. Purification of Common Salt ($mathrm{NaCl}$):
   Crude rock salt contains impurities like sodium sulphate, calcium chloride, and magnesium chloride. When a saturated solution of crude salt is treated by bubbling dry hydrogen chloride gas ($\mathrm{HCl}$) through it, the high concentration of chloride ions causes $Q_{sp}(\mathrm{NaCl}) > K_{sp}(\mathrm{NaCl})$. High-purity crystalline $\mathrm{NaCl}$ precipitates out, while more soluble calcium and magnesium chlorides remain dissolved.
2. Qualitative Cation Analysis (Salt Analysis):
   • Group II basic radicals ($\mathrm{Cu}^{2+}, \mathrm{Cd}^{2+}, \mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}$) are precipitated as sulphides in the presence of dilute $\mathrm{HCl}$. The $\mathrm{H}^+$ from $\mathrm{HCl}$ suppresses the ionization of weak diprotic acid $\mathrm{H}_2\mathrm{S}$, providing just enough $[\mathrm{S}^{2-}]$ to exceed the very low $K_{sp}$ of Group II sulphides without precipitating the higher $K_{sp}$ sulphides of Group IV ($\mathrm{Zn}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Ni}^{2+}$).
   • Group III basic radicals ($\mathrm{Fe}^{3+}, \mathrm{Al}^{3+}, \mathrm{Cr}^{3+}$) are precipitated as hydroxides using $\mathrm{NH}_4\mathrm{OH}$ in the presence of $\mathrm{NH}_4\mathrm{Cl}$. The common ammonium ion suppresses hydroxide concentration, precipitating only the least soluble Group III hydroxides while keeping Group V cations in solution.

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Effect of pH on Solubility of Salts of Weak Acids#

The solubility of sparingly soluble salts containing basic anions (such as carbonates $\mathrm{CO}_3^{2-}$, phosphates $\mathrm{PO}_4^{3-}$, fluorides $\mathrm{F}^-$, or sulphides $\mathrm{S}^{2-}$) increases dramatically in acidic solutions (lower $\text{pH}$).

Consider calcium fluoride:

$\mathrm{CaF}_2\text{(s)} \rightleftharpoons \mathrm{Ca}^{2+}\text{(aq)} + 2\mathrm{F}^-\text{(aq)}$

When acid is added to the solution, excess hydronium ions react with fluoride anions to form weakly dissociated hydrofluoric acid:

$\mathrm{F}^-\text{(aq)} + \mathrm{H}^+\text{(aq)} \rightleftharpoons \mathrm{HF}\text{(aq)}$

This consumption depletes free $[\mathrm{F}^-]$ in solution. According to Le Chatelier's principle, the solid dissolution equilibrium shifts forward to replenish $[\mathrm{F}^-]$, dramatically increasing the solubility of $\mathrm{CaF}_2$.

Quantitatively, if $K_a$ is the acid dissociation constant of the conjugate acid $\mathrm{HX}$, the effective solubility $S$ at a given $[\mathrm{H}^+]$ is:

$S = \left[\frac{K_{sp}([\mathrm{H}^+] + K_a)}{K_a}\right]^{1/2}$

As $[\mathrm{H}^+]$ increases (decreasing $\text{pH}$), $S$ increases without bound until all solid dissolves. This explains why teeth enamel (primarily hydroxyapatite / calcium phosphate) dissolves rapidly in acidic oral environments, leading to tooth decay.

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6.14 Chapter Summary: Essential Takeaways#

1. Dynamic Equilibrium: Attained when opposing processes occur simultaneously at identical rates ($r_f = r_r$). Macro-properties remain static while microscopic exchange continues.
2. Equilibrium Constant $K_c$ and $K_p$: Governed by the Law of Mass Action. Related for gaseous reactions by $K_p = K_c(RT)^{\Delta n_g}$.
3. Pure Condensed Phases: The concentrations of pure solids and pure liquids are constant and omitted from equilibrium constant expressions.
4. Reaction Quotient $Q$: Determines spontaneous direction: $Q < K$ moves forward, $Q > K$ moves backward, $Q = K$ is at equilibrium.
5. Thermodynamic Link: $\Delta_r G^{\circ} = -RT \ln K = -2.303 RT \log_{10} K$. Spontaneous reactions have $\Delta_r G^{\circ} < 0$ and $K > 1$.
6. Le Chatelier's Principle: External perturbations (concentration, pressure, volume) shift equilibria to counteract the disturbance. Temperature change alters $K$ itself (van 't Hoff equation). Catalysts accelerate rate of attainment without altering equilibrium position or $K$.
7. Acids and Bases: Arrhenius ($\mathrm{H}^+/\mathrm{OH}^-$ in water), Brönsted-Lowry (proton donor/acceptor, conjugate pairs), Lewis (electron pair acceptor/donor).
8. Water Autoionization and pH: $K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14}$ at $298\text{ K}$. $\text{pH} = -\log_{10}[\mathrm{H}^+]$, and $\text{pH} + \text{pOH} = 14.00$.
9. Weak Electrolytes: Governed by Ostwald's dilution law $\alpha \approx \sqrt{K / c}$. Polyprotic acids have $K_{a1} \gg K_{a2} \gg K_{a3}$ due to electrostatic barriers.
10. Salt Hydrolysis: Salts of weak acids and strong bases form alkaline solutions ($\text{pH} > 7$); salts of strong acids and weak bases form acidic solutions ($\text{pH} < 7$); salts of weak acids and weak bases have $\text{pH} = 7 + \frac{1}{2}(\text{pK}_a - \text{pK}_b)$, independent of concentration.
11. Buffer Solutions: Resist $\text{pH}$ changes; formulated via Henderson-Hasselbalch equations: $\text{pH} = \text{pK}_a + \log_{10}([\text{Salt}]/[\text{Acid}])$. $\text{pH}$ is invariant to dilution.
12. Solubility Product $K_{sp}$: Governing constant for sparingly soluble salts. Precipitation occurs when $Q_{sp} > K_{sp}$. Common-ion addition suppresses solubility. Salts of weak acids dissolve extensively at lower $\text{pH}$.