EquilibriumClass 11 Chemistry Notes

Here are your comprehensive study notes for the chapter on Equilibrium.

EQUILIBRIUM

Introduction to Equilibrium

Equilibrium is a state in which the rates of opposing processes become equal. This state is not static; it is a dynamic equilibrium, meaning that activity is still occurring at the molecular level, but there is no net change in the overall properties of the system.

Consider a liquid evaporating in a closed container. Initially, molecules escape from the liquid surface to become vapor. As vapor molecules accumulate, some of them return to the liquid phase. Equilibrium is reached when the rate of evaporation equals the rate of condensation. $\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\text{vap})$ The double half-arrows ($\rightleftharpoons$) signify that the process is reversible and occurring in both directions simultaneously. The combination of reactants and products at equilibrium is called an equilibrium mixture.

This principle applies to both physical processes (like phase changes) and chemical reactions. In a chemical reaction, reactants form products (the forward reaction), and products can reform reactants (the reverse reaction). At equilibrium, the concentrations of reactants and products stop changing because the rate of the forward reaction equals the rate of the reverse reaction.

Based on how far they proceed, reactions at equilibrium can be classified into three types:

1. Reactions that go nearly to completion: Almost all reactants are converted to products.
2. Reactions that barely proceed: Only a small amount of product is formed, and most reactants remain.
3. Reactions with comparable concentrations: Significant amounts of both reactants and products are present at equilibrium.

EQUILIBRIUM IN PHYSICAL PROCESSES

Equilibrium can be observed in physical processes, most commonly in phase transformations.

• Solid $\rightleftharpoons$ Liquid
• Liquid $\rightleftharpoons$ Gas
• Solid $\rightleftharpoons$ Gas

Solid-Liquid Equilibrium

Imagine ice and water in a perfectly insulated thermos at 273 K (0°C) and atmospheric pressure. The mass of ice and water remains constant, but at the boundary, water molecules are constantly freezing onto the ice, and ice molecules are melting into the water. At equilibrium: Rate of melting = Rate of freezing.

This specific temperature (at atmospheric pressure) where the solid and liquid phases coexist in equilibrium is called the normal melting point or normal freezing point.

Liquid-Vapour Equilibrium

In a closed container at a constant temperature, a liquid will evaporate, increasing the pressure inside due to the formation of vapor. As vapor concentration increases, the rate of condensation also increases. Equilibrium is reached when the pressure becomes constant. At equilibrium: Rate of evaporation = Rate of condensation.

This constant pressure exerted by the vapor is called the equilibrium vapour pressure.

• Different liquids have different vapor pressures at the same temperature.
• A liquid with a higher vapor pressure is more volatile and has a lower boiling point.
• Equilibrium cannot be reached in an open system because vapor molecules disperse, and the condensation rate remains very low.

The temperature at which the liquid and its vapor are in equilibrium at 1.013 bar (1 atm) pressure is the normal boiling point of the liquid.

Solid - Vapour Equilibrium

Some solids can directly turn into vapor, a process called sublimation. In a closed vessel, solid iodine will sublime to form violet vapor. As the vapor concentration increases, it will start condensing back into solid iodine. At equilibrium: Rate of sublimation = Rate of condensation. $\mathrm{I}_{2} (\text{solid}) \rightleftharpoons \mathrm{I}_{2} (\text{vapour})$ Other examples include camphor and ammonium chloride ($\mathrm{NH}_{4} \mathrm{Cl}$).

Equilibrium Involving Dissolution of Solids or Gases in Liquids

Solids in liquids When you dissolve sugar in water at a given temperature, there's a limit to how much will dissolve. A solution that cannot dissolve any more solute is called a saturated solution. In this state, a dynamic equilibrium exists. At equilibrium: Rate of dissolution = Rate of crystallisation. $\text{Sugar (solution)} \rightleftharpoons \text{Sugar (solid)}$

Gases in liquids The fizzing of a soda bottle when opened is due to an equilibrium between dissolved carbon dioxide and gaseous carbon dioxide above the liquid. $\mathrm{CO}_{2} (\text{gas}) \rightleftharpoons \mathrm{CO}_{2} (\text{in solution})$ This equilibrium is governed by Henry's Law, which states that the mass of a gas dissolved in a solvent is proportional to the pressure of the gas above the solvent. When the bottle is opened, the pressure drops, the equilibrium shifts to the left, and dissolved $CO_2$ escapes as gas.

General Characteristics of Equilibria Involving Physical Processes

• Equilibrium is only possible in a closed system.
• The opposing processes occur at the same rate, resulting in a dynamic but stable state.
• All measurable properties of the system (like pressure, concentration, temperature) remain constant.
• The equilibrium is characterized by a constant value of a specific parameter (e.g., vapor pressure for liquid-vapor equilibrium, solubility for dissolution).

EQUILIBRIUM IN CHEMICAL PROCESSES - DYNAMIC EQUILIBRIUM

Chemical reactions can also be reversible and reach a state of dynamic equilibrium. At this point, the concentrations of reactants and products become constant because the forward and reverse reactions are occurring at the same rate.

For a general reversible reaction: $\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$ As reactants A and B are consumed, the rate of the forward reaction decreases. As products C and D are formed, the rate of the reverse reaction increases. Eventually, the two rates become equal, and the system reaches equilibrium.

The dynamic nature of chemical equilibrium can be proven using isotopic tracers. For example, in the synthesis of ammonia (Haber's process): $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})$ If the reaction is run with deuterium ($D_2$) instead of hydrogen ($H_2$), it reaches equilibrium with the same composition, but with $ND_3$ instead of $NH_3$. If these two equilibrium mixtures are combined, a mass spectrometer reveals the presence of mixed-isotope molecules like $NH_2D$, $NHD_2$, and HD. This "scrambling" of atoms proves that both forward and reverse reactions are continuously happening even after equilibrium is reached.

LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT

In 1864, Cato Guldberg and Peter Waage proposed the Law of Mass Action. This law states that for a reversible reaction at equilibrium and at a constant temperature, the ratio of the product of the concentrations of products to the product of the concentrations of reactants, with each concentration term raised to the power of its stoichiometric coefficient, is a constant. This constant is called the equilibrium constant, denoted by $K_c$.

For a general reaction: $\mathrm{a} \mathrm{~A}+\mathrm{bB} \rightleftharpoons \mathrm{cC}+\mathrm{dD}$ The equilibrium constant expression is: $K_{c}=\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}$ Where [A], [B], [C], and [D] are the molar concentrations of the species at equilibrium. The subscript 'c' indicates that $K_c$ is expressed in terms of concentration.

Properties of the Equilibrium Constant

• Reverse Reaction: The equilibrium constant for the reverse reaction ($K_c'$) is the inverse of the equilibrium constant for the forward reaction ($K_c$). $K_{c}' = \frac{1}{K_{c}}$
• Stoichiometry: If the balanced equation for a reaction is multiplied by a factor 'n', the new equilibrium constant ($K_c''$) is the original equilibrium constant raised to the power of 'n'. $K_{c}'' = (K_{c})^n$ For example, if the reaction $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrmg)$ has an equilibrium constant $K_c$, then the reaction $\frac{1}{2}\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2}\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{HI}(\mathrmg)$ will have an equilibrium constant of $K_c^{1/2}$ or $\sqrt{K_c}$.

HOMOGENEOUS EQUILIBRIA

In a homogeneous equilibrium, all reactants and products are in the same phase (e.g., all gases or all in aqueous solution).

• Example (gas phase): $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})$
• Example (aqueous): $\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SCN}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{Fe}(\mathrm{SCN})^{2+}(\mathrm{aq})$

Equilibrium Constant in Gaseous Systems ($K_p$)

For reactions involving gases, it is often more convenient to express the equilibrium constant in terms of partial pressures. This is called the equilibrium constant $K_p$.

The ideal gas equation, $pV = nRT$, can be rearranged to show that at constant temperature, pressure is directly proportional to concentration ($c = n/V$). $p = \frac{n}{V}RT = cRT$ This relationship allows us to connect $K_p$ and $K_c$.

For the general reaction $aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$: The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n}$ where:

• R is the ideal gas constant ($0.0831$ bar L/mol K).
• T is the absolute temperature in Kelvin (K).
• $\Delta n$ is the change in the number of moles of gas: $\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$ $\Delta n = (c+d) - (a+b)$

• If $\Delta n = 0$, then $K_p = K_c$.
• If $\Delta n > 0$, then $K_p > K_c$.
• If $\Delta n < 0$, then $K_p < K_c$.

HETEROGENEOUS EQUILIBRIA

In a heterogeneous equilibrium, the reactants and products are in two or more different phases (e.g., solid, liquid, gas).

• Example: $\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})$

When writing the equilibrium constant expression for a heterogeneous equilibrium, the concentrations of pure solids and pure liquids are omitted. This is because their concentrations (density/molar mass) are constant and do not change during the reaction. Their constant values are incorporated into the equilibrium constant itself.

For the decomposition of calcium carbonate: $K_c = [\mathrm{CO}_{2}]$ $K_p = p_{\mathrm{CO}_{2}}$ This means that at a given temperature, the pressure of $CO_2$ gas in equilibrium with solid $CaCO_3$ and $CaO$ is constant, regardless of how much of each solid is present.

APPLICATIONS OF EQUILIBRIUM CONSTANTS

The value of the equilibrium constant provides valuable information about a reaction.

Predicting the Extent of a Reaction

The magnitude of $K_c$ or $K_p$ indicates how far a reaction proceeds towards products at equilibrium.

• If $K_c > 10^3$: The reaction proceeds nearly to completion. Products predominate over reactants at equilibrium.
• If $K_c < 10^{-3}$: The reaction hardly proceeds. Reactants predominate over products at equilibrium.
• If $K_c$ is between $10^{-3}$ and $10^3$: Appreciable concentrations of both reactants and products are present at equilibrium.

Predicting the Direction of a Reaction

To predict the direction a reaction will shift to reach equilibrium, we use the reaction quotient, Q. The expression for Q is the same as for K, but it uses the concentrations or partial pressures at any given moment, not necessarily at equilibrium.

For the reaction $aA + bB \rightleftharpoons cC + dD$: $Q_{c}=\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}$ By comparing Q and K, we can determine the direction of the net reaction:

• If $Q_c < K_c$: The ratio of products to reactants is too small. The reaction will proceed in the forward direction (left to right) to reach equilibrium.
• If $Q_c > K_c$: The ratio of products to reactants is too large. The reaction will proceed in the reverse direction (right to left) to reach equilibrium.
• If $Q_c = K_c$: The system is already at equilibrium, and there is no net reaction.

Calculating Equilibrium Concentrations

If the initial concentrations of reactants and the equilibrium constant are known, we can calculate the concentrations of all species at equilibrium.

General Steps:

1. Write the balanced chemical equation.
2. Make a table (often called an ICE table) listing the Initial concentrations, the Change in concentrations (in terms of a variable, x), and the Equilibrium concentrations for each species.
3. Substitute the equilibrium concentration expressions into the equilibrium constant equation.
4. Solve for x. This may involve solving a quadratic equation. Choose the solution that is chemically realistic (e.g., concentrations cannot be negative).
5. Calculate the final equilibrium concentrations using the value of x.

RELATIONSHIP BETWEEN EQUILIBRIUM CONSTANT K, REACTION QUOTIENT Q AND GIBBS ENERGY G

The spontaneity of a reaction is related to the change in Gibbs energy, $\Delta G$. The relationship between $\Delta G$, the standard Gibbs energy change ($\Delta G^\ominus$), and the reaction quotient (Q) is: $\Delta G=\Delta G^{\ominus}+\mathrm{RT} \ln Q$ At equilibrium, there is no free energy to drive the reaction, so $\Delta G = 0$, and the reaction quotient Q becomes equal to the equilibrium constant K. $0 = \Delta G^{\ominus}+\mathrm{RT} \ln K$ This gives us the crucial relationship: $\Delta G^{\ominus}=-\mathrm{RT} \ln K$ This equation connects thermodynamics with equilibrium.

• If $\Delta G^\ominus < 0$ (spontaneous under standard conditions), then $\ln K$ must be positive, which means $K > 1$. The reaction favors the products.
• If $\Delta G^\ominus > 0$ (non-spontaneous under standard conditions), then $\ln K$ must be negative, which means $K < 1$. The reaction favors the reactants.

FACTORS AFFECTING EQUILIBRIA

Le Chatelier's principle provides a way to predict how an equilibrium system will respond to a change in conditions. It states:

When a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will shift its equilibrium position to counteract the effect of the change.

Effect of Concentration Change

• Adding a reactant or product: The equilibrium will shift to consume the added substance. If you add a reactant, the reaction shifts forward (to the right). If you add a product, it shifts in reverse (to the left).
• Removing a reactant or product: The equilibrium will shift to replenish the removed substance. If you remove a product, the reaction shifts forward. If you remove a reactant, it shifts in reverse.

Effect of Pressure Change

A change in pressure (usually by changing the volume) affects gaseous equilibria where the number of moles of gas on the reactant side is different from the product side ($\Delta n \neq 0$).

• Increasing pressure (decreasing volume): The equilibrium will shift to the side with fewer moles of gas to relieve the pressure.
• Decreasing pressure (increasing volume): The equilibrium will shift to the side with more moles of gas.

Effect of Inert Gas Addition

• At constant volume: Adding an inert gas (like argon) increases the total pressure but does not change the partial pressures or concentrations of the reacting gases. Therefore, the equilibrium is undisturbed.
• At constant pressure: Adding an inert gas will increase the total volume, which decreases the partial pressures of all reacting gases. The equilibrium will shift to the side with more moles of gas. (This case is more complex and less commonly discussed at this level).

Effect of Temperature Change

Temperature is the only factor that changes the value of the equilibrium constant (K) itself.

• Exothermic Reactions ($\Delta H$ is negative): Heat is a product. Increasing the temperature will shift the equilibrium to the left (reverse direction) and decrease the value of K. Low temperature favors the products.
• Endothermic Reactions ($\Delta H$ is positive): Heat is a reactant. Increasing the temperature will shift the equilibrium to the right (forward direction) and increase the value of K. High temperature favors the products.

Effect of a Catalyst

A catalyst increases the rate at which equilibrium is reached but does not affect the equilibrium position or the value of the equilibrium constant. It does this by providing an alternative reaction pathway with a lower activation energy, speeding up both the forward and reverse reactions equally.

IONIC EQUILIBRIUM IN SOLUTION

Electrolytes are substances that conduct electricity when dissolved in water because they produce ions. Non-electrolytes do not conduct electricity.

• Strong electrolytes ionize almost completely in solution (e.g., NaCl, HCl).
• Weak electrolytes only partially ionize, establishing an equilibrium between the unionized molecules and their ions (e.g., acetic acid, $CH_3COOH$).

The equilibrium involving ions in an aqueous solution is called ionic equilibrium.

ACIDS, BASES AND SALTS

Arrhenius Concept of Acids and Bases

• An Arrhenius acid is a substance that dissociates in water to produce hydrogen ions ($H^+$).
  • $\mathrm{HX}(\mathrm{aq}) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq})$
• An Arrhenius base is a substance that dissociates in water to produce hydroxide ions ($OH^-$).
  • $\mathrm{MOH}(\mathrm{aq}) \rightarrow \mathrm{M}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$

Limitation: This concept is limited to aqueous solutions and doesn't explain the basicity of substances like ammonia ($NH_3$) that lack an OH group.

The Brönsted-Lowry Acids and Bases

This is a more general definition.

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

When an acid donates a proton, the remaining species is its conjugate base. When a base accepts a proton, the new species formed is its conjugate acid. An acid-base pair that differs by only one proton is called a conjugate acid-base pair.

• $H_2O$ donates a proton, so it's the acid. Its conjugate base is $OH^-$.
• $NH_3$ accepts a proton, so it's the base. Its conjugate acid is $NH_4^+$.
• The conjugate pairs are $H_2O/OH^-$ and $NH_4^+/NH_3$.

A strong acid has a weak conjugate base, and a weak acid has a strong conjugate base.

Lewis Acids and Bases

This is the most general definition.

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

Lewis acids are often electron-deficient species like $BF_3$, $AlCl_3$, and cations like $Mg^{2+}$. Lewis bases are species with lone pairs of electrons, like $NH_3$, $H_2O$, and $OH^-$. $\mathrm{BF}_{3}+: \mathrm{NH}_{3} \rightarrow \mathrm{BF}_{3}: \mathrm{NH}_{3}$ Here, $BF_3$ is the Lewis acid and $NH_3$ is the Lewis base.

IONIZATION OF ACIDS AND BASES

The Ionization Constant of Water and its Ionic Product

Water can act as both an acid and a base (it is amphoteric). It undergoes autoionization: $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$ The equilibrium expression for this reaction gives the ionic product of water, $K_w$. $K_{\mathrm{w}}=[\mathrm{H}_{3} \mathrm{O}^{+}][\mathrm{OH}^{-}]$ At 298 K (25°C), experiments show that in pure water, $[\mathrm{H}_{3} \mathrm{O}^{+}] = [\mathrm{OH}^{-}] = 1.0 \times 10^{-7} \text{ M}$. Therefore, at 298 K: $K_{\mathrm{w}} = (1.0 \times 10^{-7})(1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \text{ M}^2$

• In an acidic solution, $[\mathrm{H}_{3} \mathrm{O}^{+}] > [\mathrm{OH}^{-}]$.
• In a neutral solution, $[\mathrm{H}_{3} \mathrm{O}^{+}] = [\mathrm{OH}^{-}]$.
• In a basic solution, $[\mathrm{H}_{3} \mathrm{O}^{+}] < [\mathrm{OH}^{-}]$.

The pH Scale

The pH scale is a logarithmic scale used to express hydrogen ion concentration. $\mathrm{pH}=-\log[\mathrm{H}^{+}]$

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

Similarly, we can define pOH and pK_w:

• $\mathrm{pOH}=-\log[\mathrm{OH}^{-}]$
• $\mathrm{p}K_{\mathrm{w}}=-\log K_{\mathrm{w}}$

Taking the negative logarithm of the $K_w$ expression gives a useful relationship: $\mathrm{pH}+\mathrm{pOH}= \mathrm{p}K_{\mathrm{w}} = 14 \quad (\text{at 298 K})$

Ionization Constants of Weak Acids and Bases

For a weak acid, HA: $\mathrm{HA}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})$ The acid ionization constant, $K_a$, is: $K_{\mathrm{a}}=\frac{[\mathrm{H}_{3} \mathrm{O}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}$ A larger $K_a$ value indicates a stronger acid. We also use $pK_a = -\log K_a$, where a smaller $pK_a$ indicates a stronger acid.

For a weak base, B: $\mathrm{B}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{BH}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$ The base ionization constant, $K_b$, is: $K_{\mathrm{b}}=\frac{[\mathrm{BH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{B}]}$ A larger $K_b$ value indicates a stronger base. We also use $pK_b = -\log K_b$, where a smaller $pK_b$ indicates a stronger base.

Relation between $K_a$ and $K_b$

For any conjugate acid-base pair, the product of their ionization constants is equal to the ionic product of water. $K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}}$ Taking the negative logarithm of this equation gives: $\mathrm{p} K_{\mathrm{a}}+\mathrm{p} K_{\mathrm{b}}=\mathrm{p} K_{\mathrm{w}}=14 \quad (\text{at 298 K})$ This shows that if an acid is strong (large $K_a$), its conjugate base must be weak (small $K_b$), and vice versa.

Di- and Polyprotic Acids

Acids that can donate more than one proton are called polyprotic acids (e.g., $H_2SO_4$, $H_3PO_4$). They ionize in steps, and each step has its own ionization constant. For a diprotic acid $H_2X$:

1. $\mathrm{H}_{2} \mathrm{X}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{HX}^{-}(\mathrm{aq}) \quad (K_{a1})$
2. $\mathrm{HX}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{2-}(\mathrm{aq}) \quad (K_{a2})$

It is always harder to remove a proton from a negatively charged ion, so the ionization constants decrease with each step: $K_{a1} > K_{a2} > K_{a3}$.

Factors Affecting Acid Strength

The strength of an acid (HA) depends on the strength and polarity of the H-A bond.

• Down a group: The size of atom A increases, making the H-A bond weaker and longer. Bond strength is the dominant factor, so acid strength increases.
  • Example: $HF < HCl < HBr < HI$
• Across a period: The electronegativity of atom A increases, making the H-A bond more polar. Polarity is the dominant factor, so acid strength increases.
  • Example: $CH_4 < NH_3 < H_2O < HF$

Common Ion Effect

The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. This is an application of Le Chatelier's principle.

Hydrolysis of Salts

Hydrolysis is the reaction of an anion or a cation of a salt (or both) with water. This can change the pH of the solution.

• Salt of Strong Acid + Strong Base (e.g., NaCl): Neither ion hydrolyzes. The solution is neutral (pH = 7).
• Salt of Weak Acid + Strong Base (e.g., $CH_3COONa$): The anion hydrolyzes to produce $OH^-$ ions. The solution is basic (pH > 7).
  • $\mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$
• Salt of Strong Acid + Weak Base (e.g., $NH_4Cl$): The cation hydrolyzes to produce $H^+$ ions. The solution is acidic (pH < 7).
  • $\mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) \rightleftharpoons \mathrm{NH}_{4} \mathrm{OH}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})$
• Salt of Weak Acid + Weak Base (e.g., $CH_3COONH_4$): Both ions hydrolyze. The pH depends on the relative strengths of the acid and base.
  • $\mathrm{pH}=7+ \frac{1}{2}(\mathrm{p} K_{\mathrm{a}}-\mathrm{p} K_{\mathrm{b}})$

BUFFER SOLUTIONS

Buffer solutions are solutions that resist a change in pH upon dilution or the addition of small amounts of an acid or an alkali. They are crucial in biological and chemical systems where a stable pH is required.

An acidic buffer consists of a weak acid and its salt with a strong base (e.g., acetic acid and sodium acetate). A basic buffer consists of a weak base and its salt with a strong acid (e.g., ammonium hydroxide and ammonium chloride).

The pH of an acidic buffer is given by the Henderson-Hasselbalch equation: $\mathrm{pH}=\mathrm{p} K_{a}+\log \frac{[\text{Salt}]}{[\text{Acid}]}$ The pOH of a basic buffer is given by: $\mathrm{pOH}=\mathrm{p} K_{b}+\log \frac{[\text{Salt}]}{[\text{Base}]}$ A buffer is most effective when the concentrations of the acid/base and its salt are equal. In this case, $\log(1) = 0$, so pH = pK_a (for an acidic buffer) or pOH = pK_b (for a basic buffer).

SOLUBILITY EQUILIBRIA OF SPARINGLY SOLUBLE SALTS

While some salts are very soluble, others are only slightly or sparingly soluble. For these salts, an equilibrium is established between the undissolved solid and its ions in a saturated solution.

Solubility Product Constant ($K_{sp}$)

For a sparingly soluble salt, the product of the molar concentrations of its ions in a saturated solution, with each concentration raised to the power of its stoichiometric coefficient, is a constant at a given temperature. This constant is the solubility product constant, $K_{sp}$.

For a general salt $M_xX_y$ that dissolves as: $\mathrm{M}_{\mathrm{x}} \mathrm{X}_{\mathrm{y}}(\mathrm{~s}) \rightleftharpoons \mathrm{xM}^{\mathrm{p}^{+}}(\mathrm{aq})+\mathrm{yX}^{\mathrm{q}^{-}}(\mathrm{aq})$ The solubility product expression is: $K_{\mathrm{sp}} = [\mathrm{M}^{\mathrm{p}+}]^{\mathrm{x}}[\mathrm{X}^{\mathrm{q}-}]^{\mathrm{y}}$ If the molar solubility of the salt is S mol/L, then $[M^{p+}] = xS$ and $[X^{q-}] = yS$. So: $K_{\mathrm{sp}} = (xS)^x (yS)^y = x^x y^y S^{(x+y)}$

Common Ion Effect on Solubility

According to Le Chatelier's principle, the solubility of a sparingly soluble salt is decreased by the presence of a common ion.

This effect is used in analytical chemistry to ensure the complete precipitation of an ion from a solution.