Unit 3: Comprehensive Study Guide: Chemical and Ionic Equilibria

Chemical Equilibrium

Free energy change in a chemical reaction

In chemical reactions, the change in Gibbs free energy (ΔG) determines the spontaneity and direction of a reaction under constant temperature and pressure. A reaction proceeds spontaneously in the direction that decreases its free energy until equilibrium is reached.

Gibbs Free Energy Change (ΔG): A thermodynamic potential that measures the "useful" or process-initiating work obtainable from an isothermal, isobaric thermodynamic system. For a chemical reaction, ΔG indicates spontaneity.

• If ΔG < 0: The reaction is spontaneous in the forward direction.
• If ΔG > 0: The reaction is non-spontaneous in the forward direction (spontaneous in the reverse direction).
• If ΔG = 0: The system is at equilibrium.

Thermodynamic derivation of the law of chemical equilibrium

For a general reversible reaction: aA + bB ⇌ cC + dD

The change in Gibbs free energy (ΔG) at any given moment is related to the standard Gibbs free energy change (ΔG°) and the reaction quotient (Q) by the equation:

ΔG = ΔG° + RT ln Q

Where:

• ΔG = Gibbs free energy change under non-standard conditions
• ΔG° = Standard Gibbs free energy change (when all reactants and products are in their standard states)
• R = Ideal gas constant (8.314 J mol^-1 K^-1)
• T = Absolute temperature in Kelvin
• Q = Reaction quotient, which has the same form as the equilibrium constant but uses current concentrations/pressures.

At equilibrium, the system's free energy is at a minimum, meaning there is no net change in free energy, so ΔG = 0. Also, at equilibrium, the reaction quotient Q becomes the equilibrium constant K.

Substituting ΔG = 0 and Q = K into the equation:

0 = ΔG° + RT ln K
ΔG° = -RT ln K

This equation provides the thermodynamic derivation of the law of chemical equilibrium, relating the standard free energy change to the equilibrium constant.

Distinction between ΔG and ΔG°

Le Chatelier’s principle

Le Chatelier’s Principle: If a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress or counteracts the applied change.

This principle helps predict the direction of shift in an equilibrium in response to various stresses:

• Change in Concentration:
  If a reactant is added, the equilibrium shifts to the right (products) to consume the added reactant. If a product is added, the equilibrium shifts to the left (reactants). Removing a reactant or product has the opposite effect.
• Change in Pressure or Volume (for gaseous reactions):
  Pressure Increase (Volume Decrease): The equilibrium shifts towards the side with fewer moles of gas to reduce the pressure.
  Pressure Decrease (Volume Increase): The equilibrium shifts towards the side with more moles of gas to increase the pressure.
  Note: If the number of moles of gas is the same on both sides, pressure change has no effect on the equilibrium position.
• Change in Temperature:
  Increase in Temperature: The equilibrium shifts in the endothermic direction (absorbs heat) to counteract the added heat.
  Decrease in Temperature: The equilibrium shifts in the exothermic direction (releases heat) to compensate for the lost heat.
• Addition of an Inert Gas:
  If an inert gas is added at constant volume, it increases the total pressure but does not change the partial pressures of reactants and products, so the equilibrium position remains unaffected.
  If an inert gas is added at constant pressure (volume increases), it reduces the partial pressures of all reactants and products, causing the equilibrium to shift towards the side with more moles of gas.
• Addition of a Catalyst:
  A catalyst increases the rates of both forward and reverse reactions equally. It helps the system reach equilibrium faster but does not affect the position of equilibrium (the values of ΔG or K).

Relationships between Kp, Kc and Kx for reactions involving ideal gases

For reactions involving ideal gases, equilibrium can be expressed in terms of concentrations (Kc), partial pressures (Kp), or mole fractions (Kx).

Consider a general reversible reaction involving ideal gases: aA(g) + bB(g) ⇌ cC(g) + dD(g)

1. Relationship between Kp and Kc:

Kc is the equilibrium constant in terms of molar concentrations:

Kc = ([C]^c[D]^d) / ([A]^a[B]^b)

Kp is the equilibrium constant in terms of partial pressures:

Kp = (P_C^c P_D^d) / (P_A^a P_B^b)

Using the ideal gas law, PV = nRT, we know P = (n/V)RT = CRT (where C is molar concentration).

• P_A = [A]RT
• P_B = [B]RT
• P_C = [C]RT
• P_D = [D]RT

Substituting these into the Kp expression:

Kp = (([C]RT)^c ([D]RT)^d) / (([A]RT)^a ([B]RT)^b)
Kp = ([C]^c[D]^d / [A]^a[B]^b) × (RT)^{(c+d) - (a+b)}
Kp = Kc (RT)^Δn

Where Δn = (sum of stoichiometric coefficients of gaseous products) - (sum of stoichiometric coefficients of gaseous reactants).

2. Relationship between Kp and Kx:

Kx is the equilibrium constant in terms of mole fractions:

Kx = (x_C^c x_D^d) / (x_A^a x_B^b)

The partial pressure of a gas is related to its mole fraction (x) and the total pressure (P_total):

P_i = x_i P_total

Substituting this into the Kp expression:

Kp = ((x_C P_total)^c (x_D P_total)^d) / ((x_A P_total)^a (x_B P_total)^b)
Kp = (x_C^c x_D^d / x_A^a x_B^b) × (P_total)^{(c+d) - (a+b)}
Kp = Kx (P_total)^Δn

Where Δn is the same as defined above.

Ionic Equilibria

Strong, moderate and weak electrolytes

Electrolyte: A substance that produces ions when dissolved in a solvent (usually water) and thereby conducts electricity.

• Strong Electrolytes: Substances that dissociate or ionize completely (or nearly completely) in solution. They are excellent conductors of electricity. Examples: strong acids (HCl, H₂SO₄), strong bases (NaOH, KOH), and most soluble salts (NaCl, KNO₃).
• Moderate Electrolytes: Substances that show significant but incomplete ionization in solution, lying between strong and weak electrolytes. Their degree of ionization is higher than weak electrolytes but less than strong ones. Some organic acids or bases can fall into this category under certain conditions.
• Weak Electrolytes: Substances that dissociate or ionize only partially (a small fraction) in solution. They are poor conductors of electricity compared to strong electrolytes. Examples: weak acids (CH₃COOH, H₂CO₃), weak bases (NH₃), and sparingly soluble salts.

Degree of ionization

Degree of Ionization (α): The fraction of the total number of molecules of an electrolyte that ionize or dissociate into ions in a solution. It ranges from 0 to 1 (or 0% to 100%).

α = (Number of moles ionized) / (Total number of moles initially dissolved)

Factors affecting degree of ionization

• Nature of the Electrolyte: Strong electrolytes have α ≈ 1, while weak electrolytes have α << 1.
• Concentration: For weak electrolytes, the degree of ionization increases as the concentration of the solution decreases (i.e., with dilution). This is because the ions are further apart, reducing the chance of recombination.
• Temperature: Ionization is typically an endothermic process. According to Le Chatelier's principle, increasing temperature generally increases the degree of ionization for most electrolytes.
• Presence of Common Ion: The addition of a common ion (an ion already present in the equilibrium) suppresses the ionization of a weak electrolyte. This is known as the common ion effect.

Ionization constant and ionic product of water

Ionization Constant (Ka for acids, Kb for bases):

For a weak acid (HA): HA(aq) ⇌ H⁺(aq) + A^-(aq)
Ka = ([H⁺][A^-]) / [HA]

For a weak base (BOH): BOH(aq) ⇌ B⁺(aq) + OH^-(aq)
Kb = ([B⁺][OH^-]) / [BOH]

Ka and Kb are equilibrium constants that quantify the strength of a weak acid or base, respectively. A larger Ka or Kb value indicates a stronger weak acid or base.

Ionic Product of Water (Kw):

Water undergoes a slight self-ionization (autoionization):

H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH^-(aq) or H₂O(l) ⇌ H⁺(aq) + OH^-(aq)

Kw = [H⁺][OH^-]

At 25°C, Kw = 1.0 × 10^-14. In pure water at 25°C, [H⁺] = [OH^-] = 1.0 × 10^-7 M.

Ionization of weak acids and bases

Weak acids and bases ionize partially in water, establishing an equilibrium. Their ionization constants (Ka and Kb) are used to calculate the concentrations of ions and subsequently the pH of their solutions.

For a weak acid HA:

Initial: HA(aq) ⇌ H⁺(aq) + A^-(aq)
Initial conc: C 0 0
Change: -Cα +Cα +Cα
Equil conc: C(1-α) Cα Cα

Ka = (Cα × Cα) / C(1-α) = Cα² / (1-α)

If α is very small (typically < 0.05 or 5%), we can approximate 1-α ≈ 1. Then,

Ka ≈ Cα² → α = √(Ka/C)

And [H⁺] = Cα = C × √(Ka/C) = √(Ka × C)

For a weak base BOH (or B + H₂O):

Similar derivation leads to:

Kb ≈ Cα² → α = √(Kb/C)

And [OH^-] = Cα = √(Kb × C)

pH scale

pH Scale: A logarithmic scale used to specify the acidity or basicity of an aqueous solution. It ranges typically from 0 to 14.

Definitions:

pH = -log₁₀[H⁺]
pOH = -log₁₀[OH^-]
pKw = -log₁₀Kw

Relationship between pH and pOH:

pH + pOH = pKw

At 25°C, pKw = 14, so pH + pOH = 14.

• pH < 7: Acidic solution
• pH = 7: Neutral solution
• pH > 7: Basic (alkaline) solution

Common ion effect

Common Ion Effect: The phenomenon where the ionization of a weak electrolyte is suppressed by the addition of a strong electrolyte containing an ion common to the weak electrolyte. This is a direct consequence of Le Chatelier's principle.

Example: Acetic acid (weak acid) and sodium acetate (its salt).

CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO^-(aq) (Weak acid equilibrium)

CH₃COONa(aq) → Na⁺(aq) + CH₃COO^-(aq) (Strong electrolyte, complete dissociation)

When sodium acetate is added to acetic acid, the concentration of the common ion (CH₃COO^-) increases. According to Le Chatelier's principle, the equilibrium for acetic acid shifts to the left, consuming H⁺ ions and decreasing the ionization of the weak acid. This results in a higher pH (less acidic solution) than if only acetic acid were present.

Salt hydrolysis-calculation of hydrolysis constant, degree of hydrolysis and pH for different salts

Salt Hydrolysis: The reaction of a salt's cation or anion (or both) with water, leading to the formation of an acidic or basic solution, or sometimes a neutral one. Essentially, it's the reverse of neutralization.

The type of hydrolysis depends on the nature of the parent acid and base from which the salt is formed.

1. Salt of Strong Acid and Strong Base (e.g., NaCl, KNO₃):
Neither cation nor anion hydrolyzes. The solution remains neutral (pH = 7 at 25°C).

Example: Na⁺(aq) + Cl^-(aq) + H₂O(l) → No reaction.

2. Salt of Strong Acid and Weak Base (e.g., NH₄Cl, FeCl₃):
The cation (from the weak base) hydrolyzes, producing H⁺ ions, making the solution acidic.

Example: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)

Hydrolysis Constant (Kh): For NH₄⁺, Kh = ([NH₃][H⁺]) / [NH₄⁺]
Relationship with Kw and Kb: Kh = Kw / Kb

Degree of hydrolysis (h): h = √(Kh / C)

pH calculation: [H⁺] = √(Kh × C) = √(Kw × C / Kb)
pH = -log[H⁺]

3. Salt of Weak Acid and Strong Base (e.g., CH₃COONa, KCN):
The anion (from the weak acid) hydrolyzes, producing OH^- ions, making the solution basic.

Example: CH₃COO^-(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH^-(aq)

Hydrolysis Constant (Kh): For CH₃COO^-, Kh = ([CH₃COOH][OH^-]) / [CH₃COO^-]
Relationship with Kw and Ka: Kh = Kw / Ka

Degree of hydrolysis (h): h = √(Kh / C)

pH calculation: [OH^-] = √(Kh × C) = √(Kw × C / Ka)
pOH = -log[OH^-]
pH = 14 - pOH

4. Salt of Weak Acid and Weak Base (e.g., CH₃COONH₄):
Both cation and anion hydrolyze. The pH of the solution depends on the relative strengths of the weak acid and weak base (i.e., the relative values of Ka and Kb).

Example: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)
CH₃COO^-(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH^-(aq)

Hydrolysis Constant (Kh): Kh = Kw / (Ka × Kb)

Degree of hydrolysis (h): h = √(Kh)

pH calculation (approximate):

• If Ka ≈ Kb, the solution is nearly neutral.
• If Ka > Kb, the solution is acidic.
• If Ka < Kb, the solution is basic.

pH = 7 + (1/2)(pKa - pKb)

Where pKa = -logKa and pKb = -logKb.

Buffer solutions

Buffer Solution: An aqueous solution that resists significant changes in pH upon the addition of small amounts of acid or base, or upon dilution.

Types of Buffer Solutions:

• Acidic Buffer: Consists of a weak acid and its salt with a strong base (e.g., CH₃COOH and CH₃COONa). They maintain an acidic pH (< 7).
• Basic Buffer: Consists of a weak base and its salt with a strong acid (e.g., NH₃ and NH₄Cl). They maintain a basic pH (> 7).

Mechanism of Buffer Action (Example: Acidic Buffer CH₃COOH/CH₃COO^-):

• Addition of a strong acid (H⁺): The H⁺ ions are consumed by the conjugate base from the salt:
  CH₃COO^-(aq) + H⁺(aq) → CH₃COOH(aq)
  Since CH₃COOH is a weak acid, it remains largely undissociated, and the pH change is minimal.
• Addition of a strong base (OH^-): The OH^- ions are consumed by the weak acid:
  CH₃COOH(aq) + OH^-(aq) → CH₃COO^-(aq) + H₂O(l)
  The weak base (CH₃COO^-) produced does not significantly affect pH, and the pH change is minimal.

Henderson-Hasselbalch Equation (for calculating pH of buffer solutions):

For acidic buffer: pH = pKa + log([Salt] / [Acid])
For basic buffer: pOH = pKb + log([Salt] / [Base]) or pH = 14 - (pKb + log([Salt] / [Base]))

Solubility and solubility product of sparingly soluble salts – applications of solubility product principle

Solubility (s):

Solubility (s): The maximum amount of a substance (solute) that can dissolve in a given amount of solvent at a specific temperature to form a saturated solution. For sparingly soluble salts, it is often expressed in mol/L or g/L.

Solubility Product (Ksp):

Solubility Product Constant (Ksp): The equilibrium constant for the dissolution of a sparingly soluble ionic compound in water. It represents the product of the concentrations of its ions in a saturated solution, each raised to the power of its stoichiometric coefficient in the balanced dissolution equation.

For a sparingly soluble salt A_xB_y(s) that dissolves as:

A_xB_y(s) ⇌ xA^y+(aq) + yB^x-(aq)

Ksp = [A^y+]^x[B^x-]^y

If 's' is the molar solubility of A_xB_y, then at equilibrium:

• [A^y+] = xs
• [B^x-] = ys

Ksp = (xs)^x (ys)^y = x^x y^y s^(x+y)

Calculations:

• Calculating Ksp from solubility (s): Given 's' (in mol/L), substitute it into the Ksp expression.
• Calculating solubility (s) from Ksp: Given Ksp, rearrange the expression to solve for 's'.

Applications of Solubility Product Principle:

• Predicting Precipitation: The ion product (Qsp) is calculated in the same way as Ksp, but with non-equilibrium concentrations.
  • If Qsp < Ksp: The solution is unsaturated; no precipitation occurs.
  • If Qsp = Ksp: The solution is saturated; equilibrium exists.
  • If Qsp > Ksp: The solution is supersaturated; precipitation will occur until Qsp = Ksp.
• Common Ion Effect on Solubility: The solubility of a sparingly soluble salt decreases significantly when a strong electrolyte containing a common ion is added to the solution. This is a direct application of Le Chatelier's principle. For example, adding NaCl to a saturated solution of AgCl (AgCl(s) ⇌ Ag⁺(aq) + Cl^-(aq)) will increase [Cl^-], shifting the equilibrium to the left and decreasing [Ag⁺], thus reducing the solubility of AgCl.
• Selective Precipitation: By carefully controlling the concentration of a common ion, different ions in a mixture can be selectively precipitated. For instance, if a solution contains both Ba²⁺ and Sr²⁺, and their sulfates (BaSO₄ and SrSO₄) have different Ksp values, addition of SO₄^2- can be controlled to precipitate the less soluble sulfate first.