Equilibrium is a fundamental concept in chemistry and is vital in many biological and environmental processes, such as oxygen transport in the body by hemoglobin. When a process reaches a state where the forward and reverse rates are equal, it is said to be in equilibrium. This state is dynamic, meaning both opposing processes are still occurring simultaneously at equal rates, resulting in no net change in the macroscopic properties of the system (like concentration, pressure, etc.). The composition of the system at equilibrium is called the equilibrium mixture.

Equilibrium can be established in both physical processes (like phase changes or dissolution) and chemical reactions. Chemical reactions can reach equilibrium to varying extents:

• Some reactions proceed almost to completion, with very little reactant remaining at equilibrium.
• Some reactions form only a small amount of product, with most reactants remaining.
• Some reactions reach equilibrium with comparable concentrations of both reactants and products.

The extent of a reaction at equilibrium depends on experimental conditions. Understanding equilibrium allows for optimising conditions to favour desired products in industrial processes.

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Equilibrium In Physical Processes

Equilibrium is not limited to chemical reactions; it also occurs in physical processes, especially phase transformations or dissolution in closed systems at constant temperature.

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Solid-Liquid Equilibrium

Consider ice and water in a perfectly insulated container at 273 K and atmospheric pressure. At this temperature, the system is at equilibrium. The amounts of ice and water remain constant, and the temperature stays at the melting/freezing point. However, molecules are continuously exchanging between the solid (ice) and liquid (water) phases. Molecules from liquid water freeze onto the ice surface, while molecules from ice melt into the liquid. At equilibrium, the rate of melting equals the rate of freezing.

Solid $\rightleftharpoons$ Liquid

For a pure substance, the normal melting point (or freezing point) is the specific temperature at 1 atm pressure where solid and liquid phases coexist in equilibrium.

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Liquid-Vapour Equilibrium

When a liquid is placed in a closed container, molecules evaporate into the vapour phase, and molecules from the vapour condense back into the liquid. Initially, evaporation is faster than condensation, causing vapour pressure to increase. As vapour concentration rises, the rate of condensation increases. Eventually, the rates become equal, and dynamic equilibrium is established:

Liquid $\rightleftharpoons$ Vapour

At equilibrium, the vapour pressure above the liquid remains constant at a given temperature; this is the equilibrium vapour pressure or saturated vapour pressure. Vapour pressure increases with temperature. Different liquids have different vapour pressures at the same temperature; liquids with higher vapour pressure are more volatile and have lower boiling points.

Boiling occurs when the vapour pressure of the liquid equals the external atmospheric pressure. The normal boiling point is the temperature at 1 atm where liquid and vapour are in equilibrium. Boiling point depends on pressure.

Equilibrium between liquid and vapour is not possible in an open system where vapour can escape into the surroundings, preventing condensation from balancing evaporation.

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Solid – Vapour Equilibrium

Some solids can directly transform into the vapour phase (sublimation), and the vapour can condense back into the solid. In a closed container, this process reaches equilibrium:

Solid $\rightleftharpoons$ Vapour

Examples include iodine (I$_2$), camphor, and ammonium chloride (NH$_4$Cl).

I$_2$(solid) $\rightleftharpoons$ I$_2$(vapour)

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Equilibrium Involving Dissolution Of Solid Or Gases In Liquids

Solids in liquids: When a solid solute is added to a solvent, it dissolves until a saturated solution is formed, where no more solute can dissolve at that temperature. In a saturated solution, a dynamic equilibrium exists between the solid solute and the dissolved solute in the solution:

Solute(solid) $\rightleftharpoons$ Solute(solution)

At equilibrium, the rate of dissolution equals the rate of crystallisation. This is demonstrated by adding radioactive solute to a saturated solution, where radioactivity is observed transferring between the solid and dissolved phases.

Gases in liquids: Gases dissolve in liquids, and an equilibrium is established between the gas in the gaseous state and the gas dissolved in the liquid. The solubility of a gas in a liquid depends on pressure (Henry's Law) and temperature.

CO$_2$(gas) $\rightleftharpoons$ CO$_2$(in solution)

When a soda bottle is opened, the CO$_2$ pressure above the liquid decreases to atmospheric pressure, reducing the solubility and causing dissolved CO$_2$ to effervesce out until a new equilibrium is reached at the lower pressure.

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General Characteristics Of Equilibria Involving Physical Processes

Several characteristics are common to equilibrium in physical processes:

1. Equilibrium can only be attained in a closed system at a given temperature.
2. Equilibrium is dynamic; opposing processes continue at equal rates.
3. All measurable macroscopic properties of the system (like temperature, pressure, concentration, mass of phases) remain constant at equilibrium.
4. At a given temperature, equilibrium for a physical process is characterized by a constant value of a specific parameter (e.g., melting point for solid-liquid, vapour pressure for liquid-vapour, solubility for solid-solution, gas concentration proportional to pressure for gas-solution).
5. The magnitude of this characteristic parameter indicates the extent of the process at equilibrium.

Process	Equilibrium	Conclusion
Liquid Vapour	H$_2$O (l) $\rightleftharpoons$ H$_2$O (g)	pH$_2$O constant at given temperature
Solid Liquid	H$_2$O (s) $\rightleftharpoons$ H$_2$O (l)	Melting point is fixed at constant pressure
Solute(s) Solute (solution)	Sugar(s) $\rightleftharpoons$ Sugar (solution)	Concentration of solute in solution is constant at a given temperature
Gas(g) Gas (aq)	CO$_2$(g) $\rightleftharpoons$ CO$_2$(aq)	[gas(aq)]/[gas(g)] is constant at a given temperature

Equilibrium In Chemical Processes – Dynamic Equilibrium

Similar to physical processes, chemical reactions can also reach a state of equilibrium. In a reversible chemical reaction, reactants form products (forward reaction), and products form reactants (reverse reaction). Initially, the forward reaction rate is high, and the reverse reaction rate is zero or low. As reactants are consumed and products are formed, the forward rate decreases, and the reverse rate increases. Eventually, the rates of the forward and reverse reactions become equal, and the system reaches a state of chemical equilibrium.

Reactants $\rightleftharpoons$ Products

At chemical equilibrium, the concentrations of reactants and products remain constant over time. This equilibrium is dynamic; the forward and reverse reactions continue to occur at equal rates, resulting in no net change in concentrations.

The dynamic nature can be shown by using isotopes. In the Haber process, $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$, if D$_2$ is used instead of H$_2$, at equilibrium, a mixture of $\text{NH}_3$, $\text{NH}_2\text{D}$, $\text{NHD}_2$, $\text{ND}_3$ and $\text{H}_2$, HD, D$_2$ will be present, demonstrating the continuous forward and reverse reactions swapping H and D atoms.

Chemical equilibrium can be reached from either direction, starting with reactants or products.

The extent to which a reaction proceeds towards products at equilibrium varies. Some go almost to completion, some hardly proceed, and some have significant amounts of both reactants and products at equilibrium. This extent depends on experimental conditions.

Law Of Chemical Equilibrium And Equilibrium Constant

The relationship between the concentrations of reactants and products at equilibrium is described by the Law of Chemical Equilibrium, also known as the Law of Mass Action. Based on experimental observations, Norwegian chemists Cato Maximillian Guldberg and Peter Waage proposed in 1864 that for a general reversible reaction:

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

at equilibrium, the ratio of the product of the concentrations of the products raised to their stoichiometric coefficients to the product of the concentrations of the reactants raised to their stoichiometric coefficients is a constant at a given temperature.

This constant is called the equilibrium constant ($K_c$), where the subscript 'c' indicates that concentrations are expressed in molarity (mol/L).

$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ (at equilibrium)

The concentrations in the equilibrium constant expression are always the equilibrium concentrations, although the 'eq' subscript is often omitted for simplicity. Square brackets [ ] indicate molar concentration.

Example: H$_2$(g) + I$_2$(g) $\rightleftharpoons$ 2HI(g)

$K_c = \frac{[HI]^2}{[H_2][I_2]}$ (at equilibrium)

The value of $K_c$ is unique for a particular reaction at a specific temperature and is independent of the initial concentrations of reactants or products.

The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction. If $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$ has $K_c$, then $2\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)}$ has $K'_c = 1/K_c$.

If a reaction is multiplied by a factor 'n', the new equilibrium constant is the original constant raised to the power of 'n'. If $\text{a A} + \text{b B} \rightleftharpoons \text{c C} + \text{d D}$ has $K_c$, then $\text{na A} + \text{nb B} \rightleftharpoons \text{nc C} + \text{nd D}$ has $K''_c = (K_c)^n$.

It is important to specify the balanced chemical equation when quoting the value of an equilibrium constant.

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Units of Equilibrium Constant: The unit of $K_c$ is $(\text{mol/L})^{\Delta n_{gas}}$, where $\Delta n_{gas}$ is the change in the number of moles of gas in the balanced equation. For the reaction $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$, $\Delta n_{gas} = 2 - (1+1) = 0$, so $K_c$ is dimensionless. For $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$, $\Delta n_{gas} = 2 - (1+3) = -2$, so $K_c$ has units of $(\text{mol/L})^{-2}$ or $\text{L}^2/\text{mol}^2$ or $\text{M}^{-2}$. Similarly, the unit of $K_p$ is $(\text{pressure})^{\Delta n_{gas}}$. To make equilibrium constants dimensionless, standard state concentrations (1 M) or pressures (1 bar or 1 atm) are used in the expression.

Chemical equation	Equilibrium constant
a A + b B $\rightleftharpoons$ c C + d D	$K_c$
c C + d D $\rightleftharpoons$ a A + b B	$K'_c = (1/K_c )$
na A + nb B $\rightleftharpoons$ ncC + ndD	$K''_c = (K_c^n)$

Homogeneous Equilibria

In a homogeneous equilibrium system, all reactants and products are in the same phase. Examples:

• Gas phase: $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$
• Solution phase: $\text{CH}_3\text{COOH(aq)} + \text{C}_2\text{H}_5\text{OH(aq)} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5\text{(aq)} + \text{H}_2\text{O(aq)}$ (if water is the solvent, it's often treated as pure liquid, making it heterogeneous, but if reaction is in another solvent, it's homogeneous).
• Aqueous ion phase: $\text{Fe}^{3+}\text{(aq)} + \text{SCN}^-\text{(aq)} \rightleftharpoons \text{Fe(SCN)}^{2+}\text{(aq)}$

Equilibrium constant expressions for homogeneous equilibria follow the general form using concentrations ($K_c$).

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Equilibrium Constant In Gaseous Systems

For reactions involving gases, the amount of substance is often measured by partial pressure instead of molar concentration. The equilibrium constant can be expressed in terms of partial pressures, denoted as $K_p$.

From the ideal gas law, $pV = nRT$, the pressure $p$ of a gas is proportional to its molar concentration ($n/V$) at constant temperature: $p = (n/V) RT = cRT$.

For a reaction like $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$, $K_c = \frac{[HI]^2}{[H_2][I_2]}$.

The equilibrium constant in terms of partial pressures is $K_p = \frac{(p_{HI})^2}{(p_{H_2})(p_{I_2})}$.

Substituting $p = cRT$ (or $[gas] = p/RT$) into the $K_c$ expression:

$K_c = \frac{(p_{HI}/RT)^2}{(p_{H_2}/RT)(p_{I_2}/RT)} = \frac{(p_{HI})^2/(RT)^2}{(p_{H_2})(p_{I_2})/(RT)^2} = \frac{(p_{HI})^2}{(p_{H_2})(p_{I_2})} = K_p$.

In this specific reaction, $K_p = K_c$. This occurs because the total number of moles of gas on the product side equals the total number of moles of gas on the reactant side ($\Delta n_{gas} = 0$).

For a general gaseous reaction $\text{a A} + \text{b B} \rightleftharpoons \text{c C} + \text{d D}$, the relationship between $K_p$ and $K_c$ is:

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

where $\Delta n_{gas} = (c + d) - (a + b)$, the change in the total number of moles of gas from reactants to products. $R$ must be in units consistent with pressure and volume (e.g., L bar K$^{-1}$ mol$^{-1}$ for pressure in bar and volume in L, or J K$^{-1}$ mol$^{-1}$ for pressure in Pa and volume in m$^3$). The standard state for pressure is typically 1 bar or 1 atm when using $K_p$.

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Heterogeneous Equilibria

A heterogeneous equilibrium system involves reactants and products in more than one phase. Examples include reactions between solids and gases, or solids/liquids and species in solution.

Example: Thermal decomposition of calcium carbonate:

CaCO$_3$(s) $\rightleftharpoons$ CaO(s) + CO$_2$(g)

This involves a solid phase (CaCO$_3$, CaO) and a gas phase (CO$_2$).

The equilibrium constant expression includes only the species that are in variable concentrations. The molar concentration of a pure solid or a pure liquid at a given temperature is constant, regardless of the amount present. Thus, their concentration terms are incorporated into the equilibrium constant itself and do not explicitly appear in the expression.

For $\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)}$, the general $K_c$ expression is $K_c = \frac{[CaO(s)][CO_2(g)]}{[CaCO_3(s)]}$.

Since $[CaO(s)]$ and $[CaCO_3(s)]$ are constant, we define a modified equilibrium constant $K'_c = K_c \frac{[CaCO_3(s)]}{[CaO(s)]} = [CO_2(g)]$.

Or, more commonly in terms of pressure for the gas phase:

$K_p = p_{CO_2}$

This means that at equilibrium, at a fixed temperature, the partial pressure of CO$_2$ above the solid is constant, as long as both CaCO$_3$(s) and CaO(s) are present.

Example: $\text{Ni (s)} + 4 \text{ CO (g)} \rightleftharpoons \text{Ni(CO)}_4\text{ (g)}$

$K_c = \frac{[Ni(CO)_4(g)]}{[CO(g)]^4}$ (Concentration of solid Ni is omitted)

$K_p = \frac{p_{Ni(CO)_4}}{(p_{CO})^4}$ (Partial pressure of solid Ni is omitted)

For heterogeneous equilibria to exist, all pure solid or liquid phases involved must be present at equilibrium, even in small amounts. Their presence is implicit in the equilibrium constant expression, even though their concentrations don't appear.

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Applications Of Equilibrium Constants

Equilibrium constants are powerful tools in chemistry, allowing us to predict and understand the behavior of reversible reactions.

Important features of equilibrium constants:

1. Applicable only at equilibrium.
2. Independent of initial concentrations.
3. Constant value at a given temperature.
4. Value for reverse reaction is the inverse of the forward reaction's value.
5. Value changes if the balanced equation is multiplied or divided by a coefficient.

Applications of equilibrium constants:

• Predicting the extent of a reaction.
• Predicting the direction of a reaction.
• Calculating equilibrium concentrations.

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Predicting The Extent Of A Reaction

The magnitude of $K_c$ or $K_p$ indicates how far a reaction proceeds towards products at equilibrium. A large $K$ means a high concentration of products relative to reactants, suggesting the reaction goes almost to completion. A small $K$ means a high concentration of reactants relative to products, suggesting the reaction does not proceed significantly towards products.

• If $K_c > 10^3$ (large K): Products highly predominate. The reaction proceeds nearly to completion.
• If $K_c < 10^{-3}$ (small K): Reactants highly predominate. The reaction proceeds only to a small extent.
• If $10^{-3} \le K_c \le 10^3$: Appreciable concentrations of both reactants and products are present at equilibrium.

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Predicting The Direction Of The Reaction

The reaction quotient ($Q_c$) is defined in the same way as the equilibrium constant $K_c$, but the concentrations are the current concentrations at any given time, not necessarily at equilibrium.

For a reaction a A + b B $\rightleftharpoons$ c C + d D, $Q_c = \frac{[C]_t^c[D]_t^d}{[A]_t^a[B]_t^b}$, where $[ ]_t$ are concentrations at time $t$.

By comparing $Q_c$ with $K_c$, we can predict the direction of the net reaction needed to reach equilibrium:

• If $Q_c < K_c$: The ratio of products to reactants is too low. The net reaction proceeds in the forward direction (towards products) to increase products and decrease reactants until $Q_c = K_c$.
• If $Q_c > K_c$: The ratio of products to reactants is too high. The net reaction proceeds in the reverse direction (towards reactants) to decrease products and increase reactants until $Q_c = K_c$.
• If $Q_c = K_c$: The system is already at equilibrium, and there is no net reaction.

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Calculating Equilibrium Concentrations

Given the initial concentrations of reactants and the value of $K_c$ (or $K_p$), we can calculate the concentrations of all species at equilibrium by following a systematic approach.

Steps:

1. Write the balanced chemical equation.
2. Set up an "ICE" table (Initial, Change, Equilibrium). List the initial concentrations of all species. Use stoichiometry to define the change in concentration in terms of a variable (e.g., $x$), assuming reaction proceeds in one direction (usually forward). Write the equilibrium concentrations by adding initial and change.
3. Substitute the equilibrium concentrations into the equilibrium constant expression ($K_c$ or $K_p$).
4. Solve the resulting equation for the variable ($x$). If it's a quadratic or higher-order equation, choose the physically meaningful root (concentrations must be non-negative). Simplifications (like neglecting $x$ if $K$ is very small) can sometimes be made.
5. Calculate the equilibrium concentrations of all species using the determined value of $x$.
6. Verify the result by plugging the equilibrium concentrations back into the $K$ expression to see if it matches the given $K$ value.

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Relationship Between Equilibrium Constant K, Reaction Quotient Q And Gibbs Energy G

Thermodynamics provides a link between the equilibrium constant ($K$) and the change in Gibbs energy ($\Delta G$) for a reaction. The change in Gibbs energy for a reaction ($\Delta G$) indicates the spontaneity of the reaction under given conditions.

• If $\Delta G < 0$, the reaction is spontaneous in the forward direction.
• If $\Delta G > 0$, the reaction is non-spontaneous in the forward direction (spontaneous in the reverse).
• If $\Delta G = 0$, the reaction is at equilibrium.

The Gibbs energy change ($\Delta G$) under non-standard conditions is related to the standard Gibbs energy change ($\Delta G^0$) and the reaction quotient ($Q$) by the equation:

$\Delta G = \Delta G^0 + RT \ln Q$

where $R$ is the gas constant and $T$ is the absolute temperature.

At equilibrium, $\Delta G = 0$ and $Q = K$. Substituting these into the equation:

$0 = \Delta G^0 + RT \ln K$

Rearranging gives the fundamental relationship between standard Gibbs energy change and the equilibrium constant:

$\Delta G^0 = -RT \ln K$

or $\Delta G^0 = -2.303 RT \log K$

This equation shows that the equilibrium constant is directly related to the standard Gibbs energy change of the reaction. A negative $\Delta G^0$ corresponds to $K > 1$, favouring products at equilibrium. A positive $\Delta G^0$ corresponds to $K < 1$, favouring reactants at equilibrium.

We also know that $\Delta G^0 = \Delta H^0 - T\Delta S^0$. Combining these gives:

$-RT \ln K = \Delta H^0 - T\Delta S^0$

$\ln K = \frac{-(\Delta H^0 - T\Delta S^0)}{RT} = \frac{-\Delta H^0}{RT} + \frac{\Delta S^0}{R}$

This equation shows how temperature, standard enthalpy change, and standard entropy change determine the equilibrium constant.

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Factors Affecting Equilibria

Once a system reaches equilibrium, its composition can be altered by changing conditions like concentration, pressure, or temperature. The response of a system at equilibrium to such changes is described by Le Chatelier's Principle: If a system at equilibrium is subjected to a change in any of the factors that determine the equilibrium conditions, the system will shift in a direction that counteracts or reduces the effect of that change.

Le Chatelier's principle allows qualitative prediction of how changing conditions affects the equilibrium composition and yield of products.

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Effect Of Concentration Change

If the concentration of a reactant or product is changed in a system at equilibrium, the system is no longer at equilibrium ($Q_c \ne K_c$). The net reaction will shift in the direction that consumes the added substance or replenishes the removed substance to restore equilibrium.

• Adding a reactant shifts the equilibrium towards the product side (forward reaction).
• Removing a reactant shifts the equilibrium towards the reactant side (reverse reaction).
• Adding a product shifts the equilibrium towards the reactant side (reverse reaction).
• Removing a product shifts the equilibrium towards the product side (forward reaction).

In industrial processes, products are often continuously removed (e.g., condensing ammonia gas out of the reaction mixture) to drive the reaction forward and maximise yield.

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Effect Of Pressure Change

Pressure changes primarily affect gaseous equilibria where the total number of moles of gas on the reactant side is different from the product side ($\Delta n_{gas} \ne 0$). Changes in pressure are usually achieved by changing the volume of the container.

Increasing the total pressure (by decreasing volume) shifts the equilibrium towards the side with fewer moles of gas to reduce the pressure. Decreasing the total pressure (by increasing volume) shifts the equilibrium towards the side with more moles of gas to increase the pressure.

Solids and liquids are nearly incompressible, so pressure changes have negligible effect on their concentrations and thus on heterogeneous equilibria involving only solids and liquids (unless the pressure change is enormous).

Example: $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ ($\Delta n_{gas} = 2 - 4 = -2$). Increasing pressure shifts equilibrium towards the product side (2 moles of gas), decreasing the number of moles of gas and relieving the pressure stress. High pressure favours ammonia synthesis.

Example: $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$ ($\Delta n_{gas} = 2 - 1 = +1$). Increasing pressure shifts equilibrium towards the reactant side (1 mole of gas), decreasing the number of moles of gas.

Example: $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$ ($\Delta n_{gas} = 2 - 2 = 0$). Pressure changes (by volume change) have no effect on the equilibrium position as the number of moles of gas is the same on both sides.

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Effect Of Inert Gas Addition

Adding an inert gas (a gas that does not participate in the reaction) affects the equilibrium differently depending on whether the volume or the total pressure is kept constant.

• Adding an inert gas at constant volume: The partial pressures (and molar concentrations) of the reacting gases remain unchanged. The reaction quotient $Q_c$ (or $Q_p$) is unaffected, and the equilibrium position remains undisturbed.
• Adding an inert gas at constant total pressure: To keep the total pressure constant, the volume of the container must increase. This increase in volume decreases the partial pressures and molar concentrations of all reacting gases. The equilibrium will shift in the direction of the side with more moles of gas to counteract the decrease in total moles per unit volume, thus increasing the total number of moles and restoring the partial pressures towards their original relative values.

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Effect Of Temperature Change

Unlike concentration and pressure changes which alter $Q$ relative to $K$, changing the temperature actually changes the value of the equilibrium constant ($K$) itself. The effect of temperature on $K$ depends on whether the reaction is exothermic or endothermic.

Le Chatelier's principle applied to temperature changes considers heat as a reactant or product:

• For an exothermic reaction ($\Delta H < 0$, heat is a product), increasing the temperature (adding heat) shifts the equilibrium towards the reactant side (consuming heat), decreasing the value of $K$. Decreasing temperature shifts the equilibrium towards the product side, increasing $K$.
• For an endothermic reaction ($\Delta H > 0$, heat is a reactant), increasing the temperature (adding heat) shifts the equilibrium towards the product side (consuming heat), increasing the value of $K$. Decreasing temperature shifts the equilibrium towards the reactant side, decreasing $K$.

High temperatures favor endothermic reactions; low temperatures favor exothermic reactions.

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Effect Of A Catalyst

A catalyst is a substance that increases the rate of a chemical reaction without being consumed. Catalysts provide an alternative reaction pathway with a lower activation energy. In a reversible reaction at equilibrium, a catalyst increases the rates of both the forward and reverse reactions equally.

Therefore, a catalyst helps the system reach equilibrium faster but does not change the equilibrium constant or the composition of the equilibrium mixture. Catalysts are useful in industry to speed up reactions that are slow at temperatures favourable for high product yield (e.g., using iron catalyst in Haber process at moderate temperatures to achieve a reasonable rate).

A catalyst is ineffective if the equilibrium constant is exceedingly small; it cannot make a non-spontaneous reaction spontaneous or shift the equilibrium position.

Ionic Equilibrium In Solution

While studying the effect of concentration, we saw reactions involving ions in solution, like $\text{Fe}^{3+}\text{(aq)} + \text{SCN}^-\text{(aq)} \rightleftharpoons \text{Fe(SCN)}^{2+}\text{(aq)}$. Many equilibria occur in aqueous solutions and involve ions. These are called ionic equilibria.

Substances that conduct electricity when dissolved in water are called electrolytes because they produce ions in solution. Acids, bases, and salts are electrolytes. Some electrolytes (strong electrolytes) ionise almost completely in water (e.g., NaCl). Others (weak electrolytes) ionise only partially, establishing an equilibrium between the unionised molecules and the ions produced (e.g., acetic acid, $\text{CH}_3\text{COOH}$).

Ionic equilibrium specifically refers to the equilibrium established between the ions and the unionised molecules (or solids) in the aqueous solution of a weak electrolyte or a sparingly soluble salt.

The conductivity of electrolyte solutions is due to the mobility of these ions. Strong electrolytes are highly conductive due to high ion concentration, while weak electrolytes are less conductive due to low ion concentration.

Acids, Bases And Salts

Acids, bases, and salts are common classes of electrolytes with distinct properties. Acids often taste sour, turn blue litmus red, and react with active metals to produce hydrogen gas. Bases often taste bitter, feel soapy, and turn red litmus blue. Acids and bases react in neutralisation reactions to form salts and often water.

The behavior of acids and bases can be described by different concepts:

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Arrhenius Concept Of Acids And Bases

According to Svante Arrhenius (1884):

• An acid is a substance that dissociates in water to produce hydrogen ions, $\text{H}^+\text{(aq)}$. These hydrated protons exist as hydronium ions, $\text{H}_3\text{O}^+\text{(aq)}$, formed by bonding with water molecules ($\text{H}^+ + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+$). We often use H$^+$ and $\text{H}_3\text{O}^+$ interchangeably in aqueous solutions.
• A base is a substance that dissociates in water to produce hydroxyl ions, $\text{OH}^-\text{(aq)}$.

Examples: HCl(aq) $\rightarrow$ H$^+$(aq) + Cl$^-$ (aq), NaOH(aq) $\rightarrow$ Na$^+$(aq) + OH$^-$ (aq).

Limitation: Arrhenius concept is limited to aqueous solutions and does not explain the basicity of substances without OH group (like NH$_3$) or the acidity of substances without H$^+$ (like CO$_2$).

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The Brönsted-Lowry Acids And Bases

Johannes Brönsted and Thomas M. Lowry (1923) provided a more general definition:

• A Brönsted-Lowry acid is a proton donor (donates H$^+$).
• A Brönsted-Lowry base is a proton acceptor (accepts H$^+$).

In the reaction $\text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)}$, NH$_3$ accepts a proton from H$_2$O, so NH$_3$ is a Brönsted base, and H$_2$O is a Brönsted acid. In the reverse reaction, $\text{NH}_4^+$ donates a proton to $\text{OH}^-$, so $\text{NH}_4^+$ is a Brönsted acid, and $\text{OH}^-$ is a Brönsted base.

An acid-base pair that differs by only one proton is called a conjugate acid-base pair. When an acid donates a proton, it forms its conjugate base. When a base accepts a proton, it forms its conjugate acid.

Acid $\rightleftharpoons$ Proton + Conjugate base

Base + Proton $\rightleftharpoons$ Conjugate acid

Example: HCl + H$_2$O $\rightleftharpoons$ H$_3$O$^+$ + Cl$^-$

HCl (acid) and Cl$^-$ (conjugate base) form a pair. H$_2$O (base) and H$_3$O$^+$ (conjugate acid) form a pair.

Stronger acids have weaker conjugate bases, and vice-versa. Water is amphoteric; it can act as an acid (donating H$^+$ to NH$_3$) or a base (accepting H$^+$ from HCl).

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Lewis Acids And Bases

G.N. Lewis (1923) proposed a broader definition based on electron pair transfer:

• A Lewis acid is an electron pair acceptor.
• A Lewis base is an electron pair donor.

This concept includes species that don't involve proton transfer. Lewis bases are similar to Brönsted-Lowry bases (they must have a lone pair to donate). However, Lewis acids include proton-containing species (H$^+$ accepts an electron pair from a base) and electron-deficient species like metal ions (e.g., $\text{Al}^{3+}, \text{Mg}^{2+}$) or molecules with incomplete octets (e.g., BF$_3$, AlCl$_3$).

Example: $\text{BF}_3 + :\text{NH}_3 \rightarrow \text{F}_3\text{B} \leftarrow :\text{NH}_3$

BF$_3$ accepts an electron pair from NH$_3$, so BF$_3$ is a Lewis acid. NH$_3$ donates a lone pair, so NH$_3$ is a Lewis base. The bond formed is a coordinate covalent bond (dative bond).

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Ionization Of Acids And Bases

The extent to which acids and bases dissociate or ionise in water determines their strength. Strong acids and bases dissociate almost completely, while weak acids and bases only partially dissociate, establishing an ionic equilibrium.

Strong acids: HClO$_4$, HCl, HBr, HI, HNO$_3$, H$_2$SO$_4$. They are strong proton donors in water. $\text{HCl(aq)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_3\text{O}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$ (almost complete dissociation)

Strong bases: LiOH, NaOH, KOH, CsOH, Ba(OH)$_2$. They are strong producers of $\text{OH}^-$ ions in water. $\text{NaOH(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)}$ (almost complete dissociation)

Weak acids and bases: They only partially ionise in water. An equilibrium is established between the unionised species and the ions.

Example: Weak acid HA in water: $\text{HA(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{A}^-\text{(aq)}$

Strength of acids/bases can be compared using their tendency to donate/accept protons (Brönsted-Lowry concept). A stronger acid will donate a proton to a stronger base, favouring the formation of weaker acid and weaker base in the equilibrium.

Strong acids have very weak conjugate bases (e.g., Cl$^-$ is a very weak base). Weak acids have strong conjugate bases (e.g., CN$^-$ is a relatively strong base).

Weak acids: HNO$_2$, HF, CH$_3$COOH. Weak bases: NH$_3$, most organic amines.

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The Ionization Constant Of Water And Its Ionic Product

Pure water undergoes a slight autoionization (or autoprotolysis), acting as both an acid and a base:

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

The equilibrium constant for this reaction is $K = \frac{[\text{H}_3\text{O}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}$. Since water is a pure liquid and its concentration is essentially constant, it is incorporated into the equilibrium constant, yielding the ionic product of water ($K_w$):

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

At 298 K (25$^\circ$C), experimental measurements show $[\text{H}^+] = [\text{OH}^-] = 1.0 \times 10^{-7}$ M in pure water. Thus, $K_w = (1.0 \times 10^{-7})(1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \text{ M}^2$ at 298 K.

$K_w$ is temperature-dependent. In any aqueous solution, the product of $[\text{H}^+]$ and $[\text{OH}^-]$ is always equal to $K_w$ at that temperature.

• Acidic solution: $[\text{H}^+] > [\text{OH}^-]$.
• Neutral solution: $[\text{H}^+] = [\text{OH}^-]$.
• Basic solution: $[\text{H}^+] < [\text{OH}^-]$.

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The pH Scale

The concentration of $\text{H}^+$ (or $\text{H}_3\text{O}^+$) in aqueous solutions is often expressed using the pH scale, a logarithmic scale defined as the negative logarithm (base 10) of the hydrogen ion activity. For dilute solutions, activity is approximated by molarity:

$\text{pH} = -\log_{10}[\text{H}^+]$ (more precisely, $\text{pH} = -\log_{10} a_{H^+}$)

Similarly, $\text{pOH} = -\log_{10}[\text{OH}^-]$.

From $K_w = [\text{H}^+][\text{OH}^-]$, taking the negative logarithm of both sides gives:

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

$\text{p}K_w = \text{pH} + \text{pOH}$

At 298 K, $\text{p}K_w = -\log(10^{-14}) = 14$. So, $\text{pH} + \text{pOH} = 14$ at 298 K.

At 25$^\circ$C:

• Acidic solutions: pH < 7 (and pOH > 7, $[\text{H}^+] > 10^{-7}$ M).
• Neutral solutions: pH = 7 (and pOH = 7, $[\text{H}^+] = 10^{-7}$ M).
• Basic solutions: pH > 7 (and pOH < 7, $[\text{H}^+] < 10^{-7}$ M).

The pH scale is logarithmic; a change of 1 pH unit means a tenfold change in $[\text{H}^+]$. pH is measured using pH paper (rough estimation) or pH meters (more accurate).

Name of the Fluid	pH	Name of the Fluid	pH
Saturated solution of NaOH	~15	Black Coffee	5.0
0.1 M NaOH solution	13	Tomato juice	~4.2
Lime water	10.5	Soft drinks and vinegar	~3.0
Milk of magnesia	10	Lemon juice	~2.2
Egg white, sea water	7.8	Gastric juice	~1.2
Human blood	7.4	1M HCl solution	~0
Milk	6.8	Concentrated HCl	~–1.0
Human Saliva	6.4

The p scale can be applied to other constants like ionization constants: $\text{p}K_a = -\log K_a$, $\text{p}K_b = -\log K_b$.

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Ionization Constants Of Weak Acids

Weak acids (HA) partially ionise in water, establishing an equilibrium: $\text{HA(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{A}^-\text{(aq)}$. The equilibrium constant for this reaction is the acid ionization constant ($K_a$):

$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$ (at equilibrium, water concentration is constant and omitted)

Alternatively, using $c$ as initial concentration and $\alpha$ as the degree of ionization (fraction of acid that ionizes), $K_a$ can be expressed as $K_a = \frac{c\alpha^2}{1-\alpha}$.

$K_a$ is a measure of acid strength; a larger $K_a$ indicates a stronger weak acid. $K_a$ is temperature-dependent.

Acid	Ionization Constant, $K_a$
Hydrofluoric Acid (HF)	3.5 $\times$ 10$^{-4}$
Nitrous Acid (HNO$_2$)	4.5 $\times$ 10$^{-4}$
Formic Acid (HCOOH)	1.8 $\times$ 10$^{-4}$
Niacin (C$_5$H$_4$NCOOH)	1.5 $\times$ 10$^{-5}$
Acetic Acid (CH$_3$COOH)	1.74 $\times$ 10$^{-5}$
Benzoic Acid (C$_6$H$_5$COOH)	6.5 $\times$ 10$^{-5}$
Hypochlorous Acid (HClO)	3.0 $\times$ 10$^{-8}$
Hydrocyanic Acid (HCN)	4.9 $\times$ 10$^{-10}$
Phenol (C$_6$H$_5$OH)	1.3 $\times$ 10$^{-10}$

Similar to pH and pOH, $\text{p}K_a = -\log_{10} K_a$. A smaller $\text{p}K_a$ means a larger $K_a$ and a stronger weak acid.

The degree of ionization ($\alpha$) is the fraction of the weak electrolyte that has ionized. It can be calculated from initial concentration and $K_a$ using the equilibrium expression. Percent dissociation is $\alpha \times 100\%$.

Calculating pH of weak acid solutions typically involves setting up an ICE table and solving for the equilibrium concentration of H$^+$. If the acid is very weak or dilute, the autoionization of water may need to be considered.

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Ionization Of Weak Bases

Weak bases (B or MOH) partially ionise in water, establishing an equilibrium: $\text{B(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{BH}^+\text{(aq)} + \text{OH}^-\text{(aq)}$ or $\text{MOH(aq)} \rightleftharpoons \text{M}^+\text{(aq)} + \text{OH}^-\text{(aq)}$. The equilibrium constant is the base ionization constant ($K_b$):

$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$ or $K_b = \frac{[\text{M}^+][\text{OH}^-]}{[\text{MOH}]}$ (at equilibrium, water concentration is constant and omitted)

Using initial concentration $c$ and degree of ionization $\alpha$, $K_b = \frac{c\alpha^2}{1-\alpha}$.

$K_b$ measures base strength; a larger $K_b$ means a stronger weak base. $\text{p}K_b = -\log_{10} K_b$. A smaller $\text{p}K_b$ means a larger $K_b$ and a stronger weak base.

Base	$K_b$
Dimethylamine, (CH$_3$)$_2$NH	5.4 $\times$ 10$^{-4}$
Triethylamine, (C$_2$H$_5$)$_3$N	6.45 $\times$ 10$^{-5}$
Ammonia, NH$_3$ or NH$_4$OH	1.77 $\times$ 10$^{-5}$
Quinine, (A plant product)	1.10 $\times$ 10$^{-6}$
Pyridine, C$_5$H$_5$N	1.77 $\times$ 10$^{-9}$
Aniline, C$_6$H$_5$NH$_2$	4.27 $\times$ 10$^{-10}$
Urea, CO (NH$_2$)$_2$	1.3 $\times$ 10$^{-14}$

Calculating pH of weak base solutions is similar to weak acids but involves calculating $[\text{OH}^-]$ first, then pOH, then pH using $\text{pH} = 14 - \text{pOH}$ (at 298 K).

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Relation Between Ka And Kb

For a conjugate acid-base pair (e.g., weak acid HA and its conjugate base A$^-$, or weak base B and its conjugate acid BH$^+$), their ionization constants are related through the ionic product of water, $K_w$.

Consider the ionization of a weak acid HA: $\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$; $K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$.

Consider the ionization of its conjugate base A$^-$: $\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$; $K_b = \frac{[\text{HA}][\text{OH}^-]}{[\text{A}^-]}$.

Multiplying $K_a$ and $K_b$:

$K_a \times K_b = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \times \frac{[\text{HA}][\text{OH}^-]}{[\text{A}^-]} = [\text{H}_3\text{O}^+][\text{OH}^-] = K_w$.

So, for any conjugate acid-base pair:

$K_a \times K_b = K_w$ (at a given temperature)

Taking the negative logarithm of both sides:

$\text{p}K_a + \text{p}K_b = \text{p}K_w$.

At 298 K, $\text{p}K_w = 14$, so $\text{p}K_a + \text{p}K_b = 14$.

This relationship shows that a strong acid (large $K_a$, small p$K_a$) has a weak conjugate base (small $K_b$, large p$K_b$), and a weak acid (small $K_a$, large p$K_a$) has a strong conjugate base (large $K_b$, small p$K_b$). The same applies to base-conjugate acid pairs.

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Di- And Polybasic Acids And Di- And Polyacidic Bases

Some acids have more than one ionizable proton per molecule and are called polybasic or polyprotic acids (e.g., H$_2$SO$_4$, H$_3$PO$_4$). They ionize in successive steps, each with its own ionization constant ($K_{a1}, K_{a2}, K_{a3}, \dots$).

Example: Dibasic acid H$_2$X

Step 1: H$_2$X $\rightleftharpoons$ H$^+$ + HX$^-$ ; $K_{a1} = \frac{[\text{H}^+][\text{HX}^-]}{[\text{H}_2\text{X}]}$

Step 2: HX$^-$ $\rightleftharpoons$ H$^+$ + X$^{2-}$ ; $K_{a2} = \frac{[\text{H}^+][\text{X}^{2-}]}{[\text{HX}^-]}$

Successive ionization constants decrease ($K_{a1} > K_{a2} > K_{a3} > \dots$) because it becomes increasingly difficult to remove a positively charged proton from an ion that is already negatively charged.

Similarly, polyacidic bases can accept more than one proton (e.g., CO$_3^{2-}$ can accept two protons to form H$_2$CO$_3$).

Acid	$K_{a1}$	$K_{a2}$	$K_{a3}$
Oxalic acid (H$_2$C$_2$O$_4$)	5.9 $\times$ 10$^{-2}$	6.4 $\times$ 10$^{-5}$	–
Ascorbic acid (Vitamin C)	7.9 $\times$ 10$^{-5}$	1.6 $\times$ 10$^{-12}$	–
Sulfurous acid (H$_2$SO$_3$)	1.7 $\times$ 10$^{-2}$	6.4 $\times$ 10$^{-8}$	–
Sulfuric acid (H$_2$SO$_4$)	Very large (>10$^2$)	1.2 $\times$ 10$^{-2}$	–
Carbonic acid (H$_2$CO$_3$)	4.3 $\times$ 10$^{-7}$	5.6 $\times$ 10$^{-11}$	–
Phosphoric acid (H$_3$PO$_4$)	7.5 $\times$ 10$^{-3}$	6.2 $\times$ 10$^{-8}$	4.2 $\times$ 10$^{-13}$
Citric acid (C$_6$H$_8$O$_7$)	7.4 $\times$ 10$^{-4}$	1.7 $\times$ 10$^{-5}$	4.0 $\times$ 10$^{-6}$

In solutions of polyprotic acids, the first ionization step is usually the most significant source of H$^+$.

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Factors Affecting Acid Strength

The strength of an acid (HA) is determined by the ease with which it donates a proton, which depends primarily on the strength and polarity of the H-A bond.

• Bond Strength: Weaker H-A bonds are easier to break, leading to stronger acids. Down a group in the periodic table, as the size of atom A increases, the H-A bond length increases, and bond strength decreases, making the acid stronger. Example: HF << HCl << HBr << HI (increasing acid strength).
• Bond Polarity: More polar H-A bonds (where A is more electronegative) facilitate the release of H$^+$. Across a period, as the electronegativity of atom A increases, the H-A bond polarity increases, making the acid stronger. Example: CH$_4$ < NH$_3$ < H$_2$O < HF (increasing acid strength).

When comparing acids in the same group, bond strength is usually the dominant factor. When comparing acids in the same period, bond polarity is usually the dominant factor.

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Common Ion Effect In The Ionization Of Acids And Bases

The common ion effect is a specific case of Le Chatelier's principle. It describes the decrease in the degree of ionization of a weak electrolyte (acid or base) when a strong electrolyte containing a common ion is added to the solution.

Example: Acetic acid dissociation: $\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{CH}_3\text{COO}^-\text{(aq)}$.

If sodium acetate (a strong electrolyte) is added to this solution, it dissociates completely to produce acetate ions ($\text{CH}_3\text{COO}^-$): $\text{CH}_3\text{COONa(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{CH}_3\text{COO}^-\text{(aq)}$. The concentration of acetate ions, a common ion, increases. According to Le Chatelier's principle, the equilibrium for acetic acid dissociation shifts to the left, consuming H$^+$ and $\text{CH}_3\text{COO}^-$ ions and forming more unionised $\text{CH}_3\text{COOH}$. This reduces the equilibrium concentration of H$^+$ and increases the pH compared to a solution of acetic acid alone.

Similarly, adding a strong acid (source of H$^+$) to a weak acid solution suppresses the weak acid's ionization. Adding a strong base (source of $\text{OH}^-$) to a weak base solution suppresses the weak base's ionization (as OH$^-$ is the common ion). The common ion effect is particularly important in regulating pH in buffer solutions.

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Hydrolysis Of Salts And The pH Of Their Solutions

When salts dissolve in water, they dissociate into cations and anions. These ions can interact with water molecules, potentially affecting the solution's pH. The reaction of a salt's cation or anion (or both) with water to form the corresponding acid or base is called hydrolysis.

Salts derived from strong acids and strong bases (e.g., NaCl, KBr) dissociate into cations (like Na$^+$) of strong bases and anions (like Cl$^-$) of strong acids. These ions are very weak conjugate acid/bases and do not react significantly with water (hydrolyse). Their solutions are neutral, with pH = 7.

Salts derived from weak acids and strong bases (e.g., $\text{CH}_3\text{COONa}$, KCN) dissociate into cations of strong bases (like Na$^+$) and anions of weak acids (like $\text{CH}_3\text{COO}^-$). The cation of the strong base does not hydrolyse. However, the anion of the weak acid acts as a relatively strong conjugate base and hydrolyses water, producing $\text{OH}^-$ ions:

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

The production of $\text{OH}^-$ makes the solution basic (pH > 7).

Salts derived from strong acids and weak bases (e.g., NH$_4$Cl, Fe(NO$_3$)$_3$) dissociate into cations of weak bases (like $\text{NH}_4^+$) and anions of strong acids (like Cl$^-$). The anion of the strong acid does not hydrolyse. However, the cation of the weak base acts as a relatively strong conjugate acid and hydrolyses water, producing $\text{H}^+$ ions:

$\text{BH}^+\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{B(aq)} + \text{H}_3\text{O}^+\text{(aq)}$

The production of $\text{H}^+$ makes the solution acidic (pH < 7).

Salts derived from weak acids and weak bases (e.g., $\text{CH}_3\text{COONH}_4$) dissociate into both a cation of a weak base and an anion of a weak acid. Both ions undergo hydrolysis:

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

$\text{BH}^+\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{B(aq)} + \text{H}_3\text{O}^+\text{(aq)}$

The pH of the solution depends on the relative strengths of the weak acid (HA) and the weak base (B). If $K_a(\text{HA}) > K_b(\text{B})$, the solution is acidic (pH < 7). If $K_a(\text{HA}) < K_b(\text{B})$, the solution is basic (pH > 7). If $K_a(\text{HA}) \approx K_b(\text{B})$, the solution is nearly neutral (pH $\approx$ 7). More precisely, for salts of weak acid and weak base:

$\text{pH} = 7 + \frac{1}{2} (\text{p}K_a - \text{p}K_b)$

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Buffer Solutions

Buffer solutions are solutions that resist significant changes in pH upon dilution or addition of small amounts of acids or bases. They are typically composed of a mixture of a weak acid and its conjugate base (e.g., acetic acid and sodium acetate) or a weak base and its conjugate acid (e.g., ammonia and ammonium chloride).

When a small amount of acid is added, the weak base component reacts with it. When a small amount of base is added, the weak acid component reacts with it. This neutralises the added acid or base, minimising the change in pH.

Buffer solutions are essential in many chemical and biological systems where maintaining a stable pH is critical (e.g., blood, biological experiments, chemical synthesis, pharmaceuticals).

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Designing Buffer Solution

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation.

For an acidic buffer (weak acid HA and its conjugate base A$^-$):

$\text{pH} = \text{p}K_a + \log_{10} \frac{[\text{Conjugate Base}]}{[\text{Weak Acid}]} = \text{p}K_a + \log_{10} \frac{[\text{A}^-]}{[\text{HA}]}$.

Often, the concentration of the conjugate base comes primarily from the salt, and the concentration of the weak acid is approximately its initial concentration. So:

$\text{pH} = \text{p}K_a + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$

The pH of an acidic buffer is close to the p$K_a$ of the weak acid. If the concentrations of the weak acid and its salt are equal, $\log_{10}(1) = 0$, so pH = p$K_a$. To prepare a buffer of a desired pH, select a weak acid whose p$K_a$ is close to that pH and adjust the ratio of [Salt]/[Acid].

For a basic buffer (weak base B and its conjugate acid BH$^+$):

$\text{pOH} = \text{p}K_b + \log_{10} \frac{[\text{Conjugate Acid}]}{[\text{Weak Base}]} = \text{p}K_b + \log_{10} \frac{[\text{BH}^+]}{[\text{B}]}$.

Then, pH can be calculated using $\text{pH} = \text{p}K_w - \text{pOH}$. Alternatively, using $K_a \times K_b = K_w$ and $\text{p}K_a + \text{p}K_b = \text{p}K_w$:

$\text{pH} = \text{p}K_a(\text{of conjugate acid}) + \log_{10} \frac{[\text{Base}]}{[\text{Conjugate Acid}]}$.

Or $\text{pH} = \text{p}K_w - \text{p}K_b + \log_{10} \frac{[\text{Base}]}{[\text{Conjugate Acid}]}$.

For a basic buffer, the pH is close to $\text{p}K_a$ of the conjugate acid (or $14 - \text{p}K_b$ of the base). If [Base] = [Conjugate Acid], then pH = p$K_a$.

Buffer pH is relatively unaffected by dilution because the ratio of concentrations [Salt]/[Acid] or [Conjugate Acid]/[Base] remains constant upon dilution.

Solubility Equilibria Of Sparingly Soluble Salts

Even "insoluble" ionic salts dissolve to a very small extent in water. When a sparingly soluble ionic salt is in contact with its saturated aqueous solution, a heterogeneous equilibrium is established between the undissolved solid salt and the ions in solution.

Example: Barium sulfate (BaSO$_4$)

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

The equilibrium constant expression is $K = \frac{[\text{Ba}^{2+}][\text{SO}_4^{2-}]}{[\text{BaSO}_4\text{(s)}]}$. Since BaSO$_4$(s) is a pure solid, its concentration is constant. This constant is incorporated into the equilibrium constant, defining the solubility product constant ($K_{sp}$).

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Solubility Product Constant

$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$ (at equilibrium in a saturated solution)

$K_{sp}$ is the product of the molar concentrations of the constituent ions raised to their stoichiometric coefficients in the balanced dissolution equation, for a saturated solution at a given temperature.

If $S$ is the molar solubility of BaSO$_4$ (moles of BaSO$_4$ that dissolve per liter), then in a saturated solution, $[\text{Ba}^{2+}] = S$ and $[\text{SO}_4^{2-}] = S$.

$K_{sp} = S \times S = S^2$

For a general sparingly soluble salt $M_xX_y(s)$ that dissociates into $xM^{p+}(aq)$ and $yX^{q-}(aq)$, where $(xp = yq)$: $M_xX_y(s) \rightleftharpoons xM^{p+}(aq) + yX^{q-}(aq)$.

If $S$ is the molar solubility, then at equilibrium, $[M^{p+}] = xS$ and $[X^{q-}] = yS$.

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

The solubility $S$ can be calculated from $K_{sp}$: $S = \left(\frac{K_{sp}}{x^x y^y}\right)^{1/(x+y)}$.

The value of $K_{sp}$ indicates the solubility of the salt; a larger $K_{sp}$ means higher solubility for salts of the same stoichiometric type (same $x$ and $y$). $K_{sp}$ is temperature-dependent.

The ion product $Q_{sp}$ is defined similarly to $K_{sp}$ but uses concentrations at any given time (not necessarily equilibrium). Comparison of $Q_{sp}$ with $K_{sp}$ predicts precipitation or dissolution:

• If $Q_{sp} < K_{sp}$: The solution is unsaturated. More salt can dissolve.
• If $Q_{sp} > K_{sp}$: The solution is supersaturated. Precipitation occurs until $Q_{sp} = K_{sp}$.
• If $Q_{sp} = K_{sp}$: The solution is saturated. Equilibrium exists.

Salt	$K_{sp}$	Salt	$K_{sp}$
AgCl	1.8 $\times$ 10$^{-10}$	Mg(OH)$_2$	8.9 $\times$ 10$^{-12}$
AgBr	5.0 $\times$ 10$^{-13}$	CaSO$_4$	9.1 $\times$ 10$^{-6}$
AgI	8.5 $\times$ 10$^{-17}$	SrSO$_4$	3.2 $\times$ 10$^{-7}$
Ag$_2$CrO$_4$	1.1 $\times$ 10$^{-12}$	BaSO$_4$	1.1 $\times$ 10$^{-10}$
BaF$_2$	1.7 $\times$ 10$^{-6}$	MgCO$_3$	6.8 $\times$ 10$^{-6}$
CaF$_2$	3.4 $\times$ 10$^{-11}$	CaCO$_3$	3.36 $\times$ 10$^{-9}$
PbCl$_2$	1.6 $\times$ 10$^{-5}$	BaCO$_3$	5.1 $\times$ 10$^{-9}$
Mn(OH)$_2$	1.9 $\times$ 10$^{-13}$	AgCN	6.0 $\times$ 10$^{-17}$
Fe(OH)$_2$	7.9 $\times$ 10$^{-16}$	Ca$_3$(PO$_4$)$_2$	1.2 $\times$ 10$^{-26}$
Hg$_2$Cl$_2$	1.1 $\times$ 10$^{-18}$	ZnS	1.2 $\times$ 10$^{-23}$

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Common Ion Effect On Solubility Of Ionic Salts

The solubility of a sparingly soluble salt is decreased when a soluble salt containing a common ion is added to the solution. This is a specific application of the common ion effect to solubility equilibria.

Example: Dissolution of BaSO$_4$: BaSO$_4$(s) $\rightleftharpoons$ Ba$^{2+}$(aq) + SO$_4^{2-}$(aq).

If sodium sulfate (Na$_2$SO$_4$), a soluble salt, is added to a saturated solution of BaSO$_4$, the concentration of sulfate ions ($\text{SO}_4^{2-}$), the common ion, increases. According to Le Chatelier's principle, the equilibrium shifts to the left, causing some BaSO$_4$ to precipitate out of the solution until $Q_{sp} = K_{sp}$ again. This reduces the solubility of BaSO$_4$ compared to its solubility in pure water.

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