Module 5: Equilibrium and Acid Reactions

Master dynamic equilibrium, Le Chatelier's principle, equilibrium calculations, and solubility equilibria with clear theory and exam-ready problem-solving strategies.

Pillar 1 of 3

Equilibrium Concepts

Explore the fundamental concepts of chemical equilibrium, including static versus dynamic equilibrium, thermodynamic spontaneity, Le Chatelier's principle, and collision theory explanations for equilibrium shifts.

Static vs Dynamic Equilibrium

Understanding the difference between static and dynamic equilibrium is fundamental to chemical equilibrium. These two types of equilibrium describe very different situations despite both involving unchanging macroscopic properties.

Static Equilibrium

Static equilibrium occurs when a system is at rest with no processes occurring. Both forward and reverse reaction rates are zero. The system is unchanging because nothing is happening at the molecular level.

Example: A sealed container of solid sodium chloride sitting on a shelf represents static equilibrium. No dissolution or crystallization is occurring. The amount of solid remains constant because the rate of dissolution is zero and the rate of crystallization is also zero. This is not a true chemical equilibrium - it's simply an inactive system.

Key characteristic: In static equilibrium, if you could observe individual molecules, you would see no movement between reactants and products. The system is frozen in its current state.

Dynamic Equilibrium

Dynamic equilibrium occurs when forward and reverse processes occur continuously at equal rates. The system appears unchanging macroscopically, but at the molecular level, particles constantly convert between reactants and products.

Mathematical definition: At dynamic equilibrium, the rate of the forward reaction equals the rate of the reverse reaction:

\text{rate}_{forward} = \text{rate}_{reverse}

Because the rates are equal, the concentrations of all species remain constant over time, even though individual molecules continue to react.

Example: A saturated solution of sodium chloride in water at constant temperature represents dynamic equilibrium. Sodium and chloride ions continuously dissolve from the solid into solution, while other ions crystallize from solution onto the solid. The rates of dissolution and crystallization are equal, so the amount of dissolved salt remains constant.

Representation: Dynamic equilibrium reactions are always written with a double equilibrium arrow ( $\rightleftharpoons$ ) to indicate both forward and reverse processes occur simultaneously. For example:

\mathrm{NaCl}{\text{(s)}} \rightleftharpoons \text{Na}^{+}{\text{(aq)}} + \text{Cl}^{-}{\text{(aq)}}

All true chemical equilibria are dynamic, not static. When we say a reaction "reaches equilibrium," we mean it reaches a state where opposing processes balance each other, not where all processes stop. This distinction is crucial for understanding Le Chatelier's principle and how equilibrium systems respond to disturbances.

Worked Example

Consider the equilibrium: $\mathrm{N_2O_4}{\text{(g)}} \rightleftharpoons 2\mathrm{NO_2}{\text{(g)}}$ . At equilibrium at 25°C, the forward reaction rate is 0.050 mol/(L·s). What is the reverse reaction rate? Explain what is happening at the molecular level.

Given:

Forward rate at equilibrium = 0.050 mol/(L·s)

Solution:

At dynamic equilibrium, the forward rate equals the reverse rate by definition.

Therefore, reverse rate = 0.050 mol/(L·s)

Molecular explanation:

At the molecular level, N₂O₄ molecules are continuously dissociating into NO₂ molecules at a rate of 0.050 mol/(L·s). Simultaneously, NO₂ molecules are combining to form N₂O₄ at exactly the same rate of 0.050 mol/(L·s).

Because these two opposing processes occur at equal rates, the concentrations of N₂O₄ and NO₂ remain constant, even though individual molecules constantly interconvert. If you could tag specific molecules, you would see them continuously changing between N₂O₄ and NO₂ forms, but the overall population of each species stays constant.

Practice Question

Explain the difference between static and dynamic equilibrium. Give an example of each and explain why chemical equilibria are always dynamic, never static.

Show Answer

Static equilibrium occurs when both forward and reverse processes have stopped completely (both rates equal zero). The system is unchanging because nothing is happening. An example is a sealed container of pure solid NaCl at room temperature where no dissolution occurs. The amount of solid remains constant simply because no reactions are taking place.

Dynamic equilibrium occurs when forward and reverse processes occur continuously at equal rates. The system appears unchanging macroscopically because the rate of formation equals the rate of consumption for each species, but at the molecular level, particles constantly interconvert. An example is a saturated NaCl solution where ions continuously dissolve from the solid and crystallize back onto it at equal rates, maintaining constant concentrations.

Chemical equilibria are always dynamic because molecules have kinetic energy and are in constant motion. Even when macroscopic properties like concentration appear constant, individual molecules continue to collide, react, and interconvert. The equilibrium state results from balanced opposing processes, not from molecular processes stopping. If processes stopped, we would have static equilibrium, which doesn't occur in chemical systems at temperatures above absolute zero where molecules have thermal energy.

Entropy & Gibbs Free Energy

Not all reactions proceed to completion. Whether a reaction is spontaneous (able to occur without continuous external input of energy) depends on the balance between enthalpy and entropy changes, quantified by Gibbs free energy.

Enthalpy (ΔH)

Enthalpy is the heat content of a system. The enthalpy change (ΔH) represents heat absorbed or released during a reaction at constant pressure.

Exothermic reactions (ΔH negative): Release heat to the surroundings. Bonds formed in products are stronger than bonds broken in reactants, releasing excess energy. Examples include combustion and many oxidation reactions. Exothermic reactions are thermodynamically favored from an enthalpy perspective.

Endothermic reactions (ΔH positive): Absorb heat from the surroundings. Bonds broken in reactants require more energy than is released when product bonds form. Examples include photosynthesis and thermal decomposition. Endothermic reactions are thermodynamically unfavored from an enthalpy perspective, though they can still be spontaneous if entropy increases sufficiently.

Entropy (ΔS)

Entropy is a measure of disorder or the number of ways energy can be distributed in a system. The second law of thermodynamics states that the total entropy of the universe always increases in spontaneous processes.

Positive entropy change (ΔS positive): Disorder increases. This occurs when solids dissolve, liquids vaporize, or reactions produce more gas molecules than they consume. Example: $\mathrm{CaCO_3}{\text{(s)}} \rightarrow \mathrm{CaO}{\text{(s)}} + \mathrm{CO_2}{\text{(g)}}$ (one gas molecule produced from solid reactants increases disorder).

Negative entropy change (ΔS negative): Disorder decreases. This occurs when gases condense, solutions crystallize, or reactions produce fewer gas molecules than they consume. Example: $3\mathrm{H_2}{\text{(g)}} + \mathrm{N_2}{\text{(g)}} \rightarrow 2\mathrm{NH_3}{\text{(g)}}$ (4 moles of gas become 2 moles, decreasing disorder).

Predicting entropy changes: Count gas molecules on each side. More gas molecules on the product side usually means ΔS is positive. Dissolution of solids and melting also increase entropy. Entropy generally increases with temperature.

Gibbs Free Energy (ΔG)

Gibbs free energy combines enthalpy and entropy to determine spontaneity:

\Delta G = \Delta H - T\Delta S

Where T is temperature in Kelvin. This equation reveals how temperature affects spontaneity.

ΔG negative (spontaneous): The reaction can occur without continuous external energy input. The system releases free energy that can do useful work. Reactions proceed forward until equilibrium is reached.

ΔG = 0 (at equilibrium): Forward and reverse reactions are equally favorable. No net change occurs. The system is at dynamic equilibrium.

ΔG positive (non-spontaneous): The reaction requires continuous energy input to proceed. The reverse reaction is spontaneous. If forced to occur, the system will revert to reactants when energy input stops.

Temperature dependence: The -TΔS term shows temperature's effect. For endothermic reactions with positive ΔS (disorder increases), raising temperature can make ΔG negative and the reaction spontaneous. This explains why some reactions only occur at high temperatures.

Important exception: Combustion reactions always go to completion (never reach equilibrium) because they are highly exothermic with large negative ΔG values, and products typically escape as gases. They should always be written with a single arrow (→), never an equilibrium arrow.

Worked Example

For the reaction $\mathrm{CaCO_3}{\text{(s)}} \rightarrow \mathrm{CaO}{\text{(s)}} + \mathrm{CO_2}{\text{(g)}}$ , ΔH = +178 kJ/mol and ΔS = +160 J/(mol·K). Calculate ΔG at: (a) 298 K (25°C), (b) 1200 K. Is the reaction spontaneous at each temperature?

Given:

$\Delta H = +178 \text{ kJ/mol} = +178,000 \text{ J/mol}$

$\Delta S = +160 \text{ J/(mol K)}$

(a) At 298 K:

$\Delta G = \Delta H - T\Delta S$

$\Delta G = 178,000 - (298)(160)$

$\Delta G = 178,000 - 47,680$

$\Delta G = +130,320 \text{ J/mol} = \mathbf{+130.3 \text{ kJ/mol}}$

ΔG is positive, so the reaction is non-spontaneous at room temperature. Limestone doesn't decompose at 25°C.

(b) At 1200 K:

$\Delta G = 178,000 - (1200)(160)$

$\Delta G = 178,000 - 192,000$

$\Delta G = \mathbf{-14,000 \text{ J/mol} = -14 \text{ kJ/mol}}$

ΔG is negative, so the reaction is spontaneous at high temperature. This is how limestone is thermally decomposed to make quicklime (CaO) industrially.

This endothermic reaction with positive entropy change becomes spontaneous at high temperature because the -TΔS term eventually outweighs the positive ΔH.

Practice Question

Using the Gibbs free energy equation, explain why some endothermic reactions can be spontaneous and why increasing temperature favors reactions with positive entropy changes. Why are combustion reactions always written with a single arrow instead of an equilibrium arrow?

Show Answer

The Gibbs free energy equation ΔG = ΔH - TΔS shows that spontaneity depends on both enthalpy and entropy. Endothermic reactions have positive ΔH, which alone would make ΔG positive (non-spontaneous). However, if the reaction also has a large positive ΔS (significant increase in disorder), the -TΔS term becomes increasingly negative as temperature increases. At sufficiently high temperatures, the -TΔS term can outweigh the positive ΔH, making ΔG negative overall and the reaction spontaneous.

Increasing temperature favors reactions with positive ΔS because the entropy term is multiplied by temperature. For a reaction with ΔS greater than 0, the -TΔS contribution becomes more negative as T increases, making ΔG more negative and the reaction more thermodynamically favorable. This explains why dissolving some salts (endothermic with positive ΔS) occurs more readily at higher temperatures, and why thermal decomposition reactions require heating.

Combustion reactions are always written with a single arrow (→) because they go to completion rather than reaching equilibrium. They are highly exothermic with very large negative ΔG values, meaning they are overwhelmingly thermodynamically favorable in the forward direction. Additionally, combustion products (CO₂ and H₂O) typically escape as gases, continuously removing products from the system. Without products present to drive the reverse reaction, the system cannot establish equilibrium. The reaction proceeds essentially 100% to completion, so an equilibrium arrow would be inappropriate.

Le Chatelier's Principle

Le Chatelier's principle states that when a system at equilibrium experiences a disturbance, the equilibrium position shifts to minimize that disturbance. This principle allows us to predict how equilibrium systems respond to changes in concentration, pressure, or temperature.

Effect of Concentration Changes

Adding or removing reactants or products disturbs the equilibrium, causing it to shift to minimize the change.

Adding a reactant: The equilibrium shifts to the right (toward products) to consume the added reactant and minimize the disturbance. Example: For $\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$ , adding N₂ shifts equilibrium right, producing more NH₃.

Removing a product: The equilibrium shifts to the right to replace the removed product. This is why continuous removal of products in industrial processes (like distilling ammonia away from the reaction mixture) drives the reaction toward completion.

Adding a product or removing a reactant: The equilibrium shifts left (toward reactants) to minimize the disturbance by consuming the added product or regenerating the removed reactant.

Important note: Adding or removing pure solids or liquids does not affect equilibrium position because their concentrations are constants incorporated into the equilibrium constant. Only changes to species in solution or gas phase affect equilibrium.

Effect of Pressure/Volume Changes

Pressure and volume changes only affect equilibria involving gases, and only when there are different numbers of gas moles on each side of the equation.

Increasing pressure (decreasing volume): The equilibrium shifts toward the side with fewer gas moles to minimize the pressure increase. Example: For $\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$ , the left side has 4 moles of gas while the right has 2 moles. Increasing pressure shifts the equilibrium right (toward fewer moles) to reduce total gas pressure.

Decreasing pressure (increasing volume): The equilibrium shifts toward the side with more gas moles. For the same reaction, decreasing pressure would shift equilibrium left, favoring the side with 4 moles of gas over 2 moles.

Equal gas moles: If both sides have the same number of gas moles, pressure changes do not affect equilibrium position. Example: $\mathrm{H_2} + \mathrm{I_2} \rightleftharpoons 2\mathrm{HI}$ has 2 gas moles on each side, so pressure changes have no effect on equilibrium position.

Critical analysis tip: Always explicitly state the number of gas moles on each side when analyzing pressure effects. For example: "The left side has 4 moles of gas (1 N₂ + 3 H₂) while the right side has 2 moles (2 NH₃), so increasing pressure shifts the equilibrium right toward fewer moles."

Effect of Temperature Changes

Temperature changes affect equilibrium by favoring either the endothermic or exothermic direction. Think of heat as a reactant in endothermic reactions and a product in exothermic reactions.

Increasing temperature: The equilibrium shifts to absorb the added heat, favoring the endothermic direction. For an exothermic reaction (ΔH negative), increasing temperature shifts equilibrium left (toward reactants, the endothermic direction). For an endothermic reaction (ΔH positive), increasing temperature shifts equilibrium right (toward products, which absorbs heat).

Example - exothermic reaction: For $\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$ , ΔH = -92 kJ/mol (exothermic). Increasing temperature shifts the equilibrium left to absorb heat, producing less ammonia. This is why the Haber process operates at moderate temperatures despite higher temperatures increasing reaction rate.

Example - endothermic reaction: For $\mathrm{N_2O_4} \rightleftharpoons 2\mathrm{NO_2}$ , ΔH = +58 kJ/mol (endothermic). Increasing temperature shifts equilibrium right, producing more brown NO₂ gas. This is observable: a sealed tube of N₂O₄/NO₂ mixture becomes darker brown when heated.

Le Chatelier phrasing: Always use the phrase "shifts to minimize the disturbance" rather than anthropomorphizing the system. Never say the equilibrium "tries to" or "wants to" do something - equilibrium systems are not conscious entities. The shift is a consequence of how rates change, not intentional behavior.

Worked Example

Consider the equilibrium: $2\mathrm{SO_2}{\text{(g)}} + \mathrm{O_2}{\text{(g)}} \rightleftharpoons 2\mathrm{SO_3}{\text{(g)}}$ , ΔH = -198 kJ/mol. Predict the effect on SO₃ concentration of: (a) adding O₂, (b) increasing pressure, (c) increasing temperature.

(a) Adding O₂:

Adding O₂ increases the concentration of a reactant. The equilibrium shifts to the right to minimize this disturbance by consuming the added O₂.

Effect: [SO₃] increases

(b) Increasing pressure:

Count gas moles: Left side has 3 moles (2 SO₂ + 1 O₂), right side has 2 moles (2 SO₃).

Increasing pressure shifts equilibrium toward the side with fewer gas moles to minimize the pressure increase.

The equilibrium shifts right (toward 2 moles).

Effect: [SO₃] increases

ΔH = -198 kJ/mol, so the forward reaction is exothermic (releases heat).

Increasing temperature shifts equilibrium to absorb the added heat, favoring the endothermic direction (reverse reaction, toward reactants).

The equilibrium shifts left.

Effect: [SO₃] decreases

Practice Question

For the equilibrium $\mathrm{PCl_5}{\text{(g)}} \rightleftharpoons \mathrm{PCl_3}{\text{(g)}} + \mathrm{Cl_2}{\text{(g)}}$ , ΔH = +93 kJ/mol, predict the direction of shift for: (a) adding Cl₂, (b) decreasing volume, (c) decreasing temperature. Explain each answer using Le Chatelier's principle.

Show Answer

(a) Adding Cl₂ increases the concentration of a product. The equilibrium shifts to minimize this disturbance by consuming the added Cl₂. The equilibrium shifts left (toward reactants), producing more PCl₅.

(b) Decreasing volume increases pressure. The system shifts to minimize this pressure increase by moving toward the side with fewer gas moles. The left side has 1 mole of gas (PCl₅) while the right side has 2 moles (PCl₃ + Cl₂). The equilibrium shifts left (toward fewer moles), favoring PCl₅ formation.

(c) The reaction is endothermic (ΔH positive), meaning heat is absorbed in the forward direction. Decreasing temperature removes heat from the system. The equilibrium shifts to minimize this disturbance by releasing heat, favoring the exothermic direction (reverse reaction). The equilibrium shifts left (toward reactants), producing more PCl₅ which releases the absorbed heat.

Collision Theory & Equilibrium

Collision theory explains equilibrium shifts at the molecular level. By understanding how concentration, temperature, and other factors affect collision frequency and activation energy, we can explain why Le Chatelier's principle works.

Fundamentals of Collision Theory

For a reaction to occur, reactant molecules must collide with sufficient energy and proper orientation. The rate of reaction depends on collision frequency and the fraction of collisions that have enough energy to overcome the activation energy barrier.

Collision frequency: The number of molecular collisions per unit time. Collision frequency increases with concentration (more particles in a given volume means more collisions) and temperature (particles move faster at higher temperatures, colliding more often).

Activation energy (Ea): The minimum energy required for a collision to result in a reaction. Only collisions with energy greater than or equal to Ea can break bonds and form new ones. The fraction of molecules with energy greater than Ea increases exponentially with temperature.

Orientation factor: Molecules must collide with appropriate geometry for reaction to occur. Even energetic collisions fail if molecules approach from the wrong angle, preventing proper orbital overlap for bond formation.

Explaining Concentration Effects with Collision Theory

When reactant concentration increases, more reactant molecules occupy the same volume, increasing collision frequency between reactants. This increases the forward reaction rate.

Example: For $\text{A} + \text{B} \rightleftharpoons \text{C}$ , adding more A increases [A]. With more A molecules present, B molecules collide with A more frequently. The forward rate increases while the reverse rate (which depends only on [C]) initially remains constant. The system is no longer at equilibrium - forward rate exceeds reverse rate.

As the forward reaction proceeds faster, [C] increases. This increases the reverse reaction rate because more C molecules are available to convert back to A and B. Eventually, the forward and reverse rates balance at new values (both higher than before), establishing a new equilibrium with higher [C] and lower [B]. The equilibrium has shifted right, consistent with Le Chatelier's principle.

Explaining Temperature Effects with Collision Theory

Temperature affects both collision frequency and the fraction of collisions with sufficient energy. However, the energy effect dominates - even a small temperature increase significantly increases the fraction of molecules with E greater than or equal to Ea.

Effect on both reactions: Increasing temperature increases both forward and reverse reaction rates because more molecules in both directions have sufficient energy to overcome their respective activation energy barriers. However, the two rates don't increase equally.

Endothermic vs exothermic: The reaction direction with higher activation energy is more sensitive to temperature changes. For an exothermic forward reaction, the reverse (endothermic) direction has higher Ea. Increasing temperature increases the reverse rate more than the forward rate, shifting equilibrium left. This explains why increasing temperature favors the endothermic direction.

Example: For $\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$ (exothermic forward), raising temperature increases the decomposition of NH₃ (endothermic, high Ea) more than the formation of NH₃ (exothermic, lower Ea). The reverse rate increases more than the forward rate, establishing new equilibrium with less NH₃. The equilibrium shifts left, favoring the endothermic direction.

Explaining Pressure Effects with Collision Theory

Increasing pressure by decreasing volume compresses gas molecules into a smaller space, increasing concentration of all gaseous species. This increases collision frequency for all reactions involving gases.

For the forward reaction: $\mathrm{N_2} + 3\mathrm{H_2} \rightarrow 2\mathrm{NH_3}$ , the rate depends on [N₂][H₂]³. When volume decreases, both [N₂] and [H₂] increase. Because the forward rate depends on the product of multiple concentrations, it increases dramatically.

For the reverse reaction: $2\mathrm{NH_3} \rightarrow \mathrm{N_2} + 3\mathrm{H_2}$ , the rate depends on [NH₃]². When volume decreases, [NH₃] also increases, but the rate increase is less dramatic because only one concentration is involved (though squared).

The reaction involving more gas molecules (4 on the left vs 2 on the right) experiences a larger rate increase when pressure rises. Initially, the forward rate exceeds the reverse rate. As the system shifts right and produces more NH₃, the reverse rate increases until a new equilibrium is established with more products. This explains why increasing pressure shifts equilibrium toward fewer moles - the side with fewer molecules is less affected by the pressure increase.

Worked Example

For $2\mathrm{NO_2} \rightleftharpoons \mathrm{N_2O_4}$ , explain using collision theory why increasing [NO₂] shifts the equilibrium right. Describe what happens to forward and reverse rates immediately after adding NO₂, and how equilibrium is reestablished.

Immediately after adding NO₂:

Adding NO₂ increases [NO₂], putting more NO₂ molecules in the container. This increases collision frequency between NO₂ molecules.

The forward rate (2 NO₂ → N₂O₄) depends on [NO₂]². Doubling [NO₂] quadruples the forward rate because collision frequency between NO₂ molecules increases by a factor of four (each molecule has twice as many collision partners).

The reverse rate (N₂O₄ → 2 NO₂) depends on [N₂O₄], which hasn't changed yet. The reverse rate initially remains constant.

Result: Forward rate greater than reverse rate. The system is no longer at equilibrium.

Reestablishing equilibrium:

Because forward rate exceeds reverse rate, NO₂ is consumed and N₂O₄ is produced. As [N₂O₄] increases, the reverse reaction rate increases. As [NO₂] decreases (though it remains higher than the original concentration), the forward rate decreases from its initial spike.

This continues until forward and reverse rates become equal again at new values. The new equilibrium has higher [N₂O₄] and lower [NO₂] than before adding NO₂, though [NO₂] is still higher than in the original equilibrium.

Conclusion: The equilibrium has shifted right, producing more N₂O₄, exactly as Le Chatelier's principle predicts. Collision theory explains the mechanism: increased [NO₂] increases forward reaction collision frequency, temporarily disrupting equilibrium until product accumulation rebalances the rates.

Practice Question

For the exothermic reaction $\mathrm{H_2} + \mathrm{I_2} \rightleftharpoons 2\mathrm{HI}$ , use collision theory to explain why: (a) increasing temperature shifts equilibrium left, (b) the shift occurs even though both forward and reverse rates increase.

Show Answer

(a) The forward reaction (H₂ + I₂ → 2 HI) is exothermic, releasing energy, so it has a lower activation energy. The reverse reaction (2 HI → H₂ + I₂) is endothermic, absorbing energy, so it has a higher activation energy. When temperature increases, both forward and reverse rates increase because more molecules have energy exceeding their respective activation energies. However, the reverse reaction, with its higher Ea, is more sensitive to temperature - the fraction of molecules with E ≥ Ea(reverse) increases more dramatically than the fraction with E ≥ Ea(forward). Therefore, the reverse rate increases more than the forward rate.

(b) After increasing temperature, the reverse rate temporarily exceeds the forward rate, disturbing equilibrium. HI decomposes into H₂ and I₂ faster than it forms. As [HI] decreases, the reverse rate decreases (fewer HI collisions). As [H₂] and [I₂] increase, the forward rate increases (more H₂-I₂ collisions). Eventually, the two rates equalize again at new, higher values. The new equilibrium has lower [HI] and higher [H₂] and [I₂] compared to the original equilibrium - the equilibrium has shifted left. Both rates are higher than initially, but they're balanced at a position favoring reactants, explaining why increasing temperature shifts exothermic equilibria toward reactants even though all processes speed up.

Pillar 2 of 3

Equilibrium Calculations

Master quantitative equilibrium analysis including constructing equilibrium constant expressions, solving ICE table problems, and using the reaction quotient to predict reaction direction.

The Equilibrium Constant (K_eq)

The equilibrium constant (K_eq) quantifies the position of equilibrium, indicating the relative amounts of products and reactants present when the system reaches equilibrium. Every equilibrium has a unique K value at a given temperature.

Constructing the Equilibrium Expression

For a general equilibrium reaction:

aA + bB \rightleftharpoons cC + dD

The equilibrium constant expression is:

K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}

Products in the numerator, reactants in the denominator. Each concentration is raised to the power of its stoichiometric coefficient.

Critical rule: Pure solids (s) and pure liquids (l) are never included in the equilibrium expression. Their concentrations are constants incorporated into K_eq. Only aqueous (aq) and gaseous (g) species appear in the expression.

Example: For $\mathrm{CaCO_3}{\text{(s)}} \rightleftharpoons \mathrm{CaO}{\text{(s)}} + \mathrm{CO_2}{\text{(g)}}$ , the equilibrium expression is simply $K_{eq} = [\mathrm{CO_2}]$ . The solid CaCO₃ and CaO do not appear.

Interpreting K_eq Values

The magnitude of K_eq indicates which side of the equilibrium is favored at equilibrium.

K_eq much greater than 1 (K >> 1): Products are strongly favored. At equilibrium, [products] >> [reactants]. The numerator (product concentrations) is much larger than the denominator. The reaction proceeds nearly to completion. Example: K_eq = 1 × 10⁸ means products dominate.

K_eq approximately 1 (0.1 < K < 10): Neither side is strongly favored. Significant amounts of both reactants and products exist at equilibrium. The system is balanced between forward and reverse directions.

K_eq much less than 1 (K << 1): Reactants are strongly favored. At equilibrium, [reactants] >> [products]. The denominator is much larger than the numerator. Very little product forms. Example: K_eq = 1 × 10⁻⁸ means reactants dominate.

Temperature dependence: K_eq changes with temperature. For exothermic reactions, K_eq decreases as temperature increases (less product at higher T). For endothermic reactions, K_eq increases with temperature (more product at higher T). K_eq is constant at a given temperature regardless of initial concentrations.

Worked Example

For the equilibrium $2\mathrm{NO_2}{\text{(g)}} \rightleftharpoons \mathrm{N_2O_4}{\text{(g)}}$ , equilibrium concentrations are: [NO₂] = 0.0172 M and [N₂O₄] = 0.00140 M. Calculate K_eq.

Given:

[NO₂] = 0.0172 M, [N₂O₄] = 0.00140 M

Solution:

Step 1: Write the equilibrium expression (products over reactants):

$K_{eq} = \frac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}$

Note: NO₂ has coefficient 2, so it's squared in the denominator.

Step 2: Substitute equilibrium concentrations:

$K_{eq} = \frac{0.00140}{(0.0172)^2}$

Step 3: Calculate:

$K_{eq} = \frac{0.00140}{2.96 \times 10^{-4}}$

$K_{eq} = \mathbf{4.73}$

Since K_eq ≈ 5, neither reactants nor products are strongly favored. Both NO₂ and N₂O₄ exist in significant amounts at equilibrium.

Practice Question

Write the equilibrium constant expression for: (a) $\mathrm{N_2}{\text{(g)}} + 3\mathrm{H_2}{\text{(g)}} \rightleftharpoons 2\mathrm{NH_3}{\text{(g)}}$ , (b) $\text{AgCl}{\text{(s)}} \rightleftharpoons \text{Ag}^{+}{\text{(aq)}} + \text{Cl}^{-}{\text{(aq)}}$ . Explain why solids don't appear in K_eq expressions.

Show Answer

(a) For N₂ + 3H₂ ⇌ 2NH₃, the equilibrium expression is:

$K_{eq} = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$

Products (NH₃) in the numerator raised to power 2. Reactants (N₂ and H₂) in the denominator, with H₂ raised to power 3 due to its coefficient.

(b) For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), the equilibrium expression is:

$K_{eq} = [\text{Ag}^{+}][\text{Cl}^{-}]$

The solid AgCl does not appear in the expression. Only the aqueous ions appear.

Pure solids and pure liquids don't appear in equilibrium expressions because their "concentrations" are constant. The concentration of a pure solid or liquid is determined by its density, which doesn't change during the reaction. For example, solid NaCl always has the same concentration of Na⁺ and Cl⁻ in its crystal structure regardless of how much solid is present. Since these are constants, they're incorporated into the numerical value of K_eq itself rather than appearing as variables in the expression. Only species whose concentrations can vary (aqueous solutions and gases) appear in the K_eq expression.

ICE Tables

ICE tables (Initial, Change, Equilibrium) are systematic tools for solving equilibrium problems. They organize initial concentrations, changes that occur as the system reaches equilibrium, and final equilibrium concentrations.

ICE Table Structure

An ICE table has three rows and one column for each species in the balanced equation:

Initial (I): Starting concentrations before any reaction occurs. These are given in the problem or can be calculated from given information.

Change (C): How much each concentration changes as the system moves toward equilibrium. Express changes in terms of a variable (typically x) based on stoichiometry. If reactant A decreases by x, and the balanced equation shows A reacts with B in a 1:2 ratio, then B decreases by 2x.

Equilibrium (E): Final concentrations at equilibrium. Each equilibrium concentration equals the initial concentration plus the change: [A]_eq = [A]_initial + change. For reactants, changes are negative (consumed). For products, changes are positive (formed).

Stoichiometric ratios: Changes must respect the balanced equation's stoichiometry. For $aA + bB \rightleftharpoons cC + dD$ , if A changes by -ax, then B changes by -bx, C by +cx, and D by +dx. The coefficients scale the change variable.

Solving ICE Table Problems

The general approach to solving equilibrium problems with ICE tables:

Step 1: Write the balanced equation with an equilibrium arrow (⇌).

Step 2: Construct the ICE table with columns for each species and rows for I, C, E.

Step 3: Fill in initial concentrations from the problem. If a species starts at zero, write 0.

Step 4: Define the change variable (usually x) for one species, then use stoichiometry to express all other changes in terms of x. Reactants decrease (-x scaled by coefficient), products increase (+x scaled by coefficient).

Step 5: Write equilibrium concentrations as initial ± change for each species.

Step 6: Substitute equilibrium expressions into the K_eq expression and solve for x. This may require algebra or the quadratic formula.

Step 7: Calculate final equilibrium concentrations by substituting x back into the equilibrium row expressions. Check that your answer makes chemical sense (all concentrations positive, K_eq value matches).

Worked Example

For $\mathrm{H_2}{\text{(g)}} + \mathrm{I_2}{\text{(g)}} \rightleftharpoons 2\mathrm{HI}{\text{(g)}}$ , K_eq = 54.3 at 430°C. If initial concentrations are [H₂] = 0.200 M, [I₂] = 0.200 M, and [HI] = 0, find the equilibrium concentrations.

Step 1 & 2: Set up ICE table

	H₂	I₂	2HI
I	0.200	0.200	0
C	-x	-x	+2x
E	0.200 - x	0.200 - x	2x

Step 3: Write K_eq expression and substitute:

$K_{eq} = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$

$54.3 = \frac{(2x)^2}{(0.200-x)(0.200-x)}$

$54.3 = \frac{4x^2}{(0.200-x)^2}$

Step 4: Solve for x (take square root of both sides):

$\sqrt{54.3} = \frac{2x}{0.200-x}$

$7.37 = \frac{2x}{0.200-x}$

$7.37(0.200-x) = 2x$

$1.474 - 7.37x = 2x$

$1.474 = 9.37x$

$x = 0.157 \text{ M}$

Step 5: Calculate equilibrium concentrations:

[H₂] = 0.200 - 0.157 = 0.043 M

[I₂] = 0.200 - 0.157 = 0.043 M

[HI] = 2(0.157) = 0.314 M

Check: K = (0.314)²/[(0.043)(0.043)] = 0.0986/0.00185 = 53.3 ✓ (close to 54.3, difference due to rounding)

Practice Question

For $\mathrm{N_2O_4}{\text{(g)}} \rightleftharpoons 2\mathrm{NO_2}{\text{(g)}}$ , K_eq = 0.36 at 100°C. If [N₂O₄] initially = 0.100 M and [NO₂] initially = 0, set up an ICE table and determine the equilibrium concentrations.

Show Answer

ICE Table:

	N₂O₄	2NO₂
I	0.100	0
C	-x	+2x
E	0.100 - x	2x

Equilibrium expression: K_eq = [NO₂]²/[N₂O₄]

0.36 = (2x)²/(0.100 - x)

0.36 = 4x²/(0.100 - x)

0.036 - 0.36x = 4x²

4x² + 0.36x - 0.036 = 0

Using quadratic formula: x = [-0.36 ± √(0.1296 + 0.576)]/8

x = [-0.36 ± 0.840]/8

x = 0.480/8 = 0.060 M (taking positive root)

Equilibrium concentrations:

[N₂O₄] = 0.100 - 0.060 = 0.040 M

[NO₂] = 2(0.060) = 0.120 M

Check: K = (0.120)²/0.040 = 0.0144/0.040 = 0.36 ✓

Reaction Quotient (Q)

The reaction quotient (Q) has the same mathematical form as the equilibrium constant but uses current concentrations at any point in time, not just at equilibrium. Comparing Q to K_eq predicts which direction the reaction will shift to reach equilibrium.

Calculating Q

For the general reaction $aA + bB \rightleftharpoons cC + dD$ , the reaction quotient is:

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

This expression is identical to K_eq, but the concentrations used are the current concentrations, not equilibrium concentrations. Q is calculated at any moment to determine the system's status.

Same rules apply: Pure solids and liquids are excluded from Q expressions, just as they are from K_eq expressions. Only aqueous and gaseous species appear.

Comparing Q to K_eq

The relationship between Q and K_eq determines which direction the reaction will proceed:

Q < K_eq: The ratio of products to reactants is too small. The system has too little product (or too much reactant) compared to equilibrium. The reaction shifts right (toward products) to increase Q until Q = K_eq. More product will form.

Q = K_eq: The system is already at equilibrium. No net change occurs. Forward and reverse rates are equal, and concentrations remain constant.

Q > K_eq: The ratio of products to reactants is too large. The system has too much product (or too little reactant) compared to equilibrium. The reaction shifts left (toward reactants) to decrease Q until Q = K_eq. Product will decompose.

Practical use: Q is particularly useful when you mix solutions containing both reactants and products. Calculate Q from the initial mixed concentrations, compare to K_eq, and predict the shift direction without solving complex equilibrium equations.

Worked Example

For $2\text{NO}{\text{(g)}} + \mathrm{O_2}{\text{(g)}} \rightleftharpoons 2\mathrm{NO_2}{\text{(g)}}$ , K_eq = 6.5 × 10⁵ at 500 K. A mixture contains [NO] = 0.0010 M, [O₂] = 0.0020 M, and [NO₂] = 0.50 M. (a) Calculate Q. (b) Predict the direction of shift. (c) Explain what happens to each concentration.

(a) Calculate Q:

$Q = \frac{[\mathrm{NO_2}]^2}{[\text{NO}]^2[\mathrm{O_2}]}$

$Q = \frac{(0.50)^2}{(0.0010)^2(0.0020)}$

$Q = \frac{0.25}{(1.0 \times 10^{-6})(0.0020)}$

$Q = \frac{0.25}{2.0 \times 10^{-9}}$

$Q = \mathbf{1.25 \times 10^8}$

(b) Predict direction:

Compare Q to K_eq:

Q = 1.25 × 10⁸

K_eq = 6.5 × 10⁵

Q > K_eq

The system has too much product relative to equilibrium. The reaction will shift left (toward reactants) to decrease Q until it equals K_eq.

As the reaction shifts left:

• [NO₂] decreases (consumed as product decomposes)

• [NO] increases (formed from NO₂ decomposition)

• [O₂] increases (formed from NO₂ decomposition)

This continues until Q decreases to equal K_eq = 6.5 × 10⁵.

Practice Question

For $\text{CO}{\text{(g)}} + \mathrm{H_2}\text{O}{\text{(g)}} \rightleftharpoons \mathrm{CO_2}{\text{(g)}} + \mathrm{H_2}{\text{(g)}}$ , K_eq = 5.10 at 700 K. A mixture contains [CO] = 0.100 M, [H₂O] = 0.100 M, [CO₂] = 0.400 M, [H₂] = 0.400 M. Calculate Q and predict which direction the reaction will shift.

Show Answer

Calculate Q:

$Q = \frac{[\mathrm{CO_2}][\mathrm{H_2}]}{[\text{CO}][\mathrm{H_2}\text{O}]}$

$Q = \frac{(0.400)(0.400)}{(0.100)(0.100)}$

$Q = \frac{0.16}{0.01}$

$Q = \mathbf{16.0}$

Compare to K_eq:

Q = 16.0

K_eq = 5.10

Q > K_eq

Prediction:

Since Q is greater than K_eq, the system has too much product relative to equilibrium. The reaction quotient must decrease to reach equilibrium. The reaction will shift left (toward reactants) to consume products and form reactants until Q = K_eq = 5.10.

As the reaction proceeds: [CO₂] and [H₂] will decrease, while [CO] and [H₂O] will increase.

Pillar 3 of 3

Solubility Equilibria

Explore solubility product constants, precipitation predictions, the common ion effect, and applications to Aboriginal food processing techniques for removing toxins.

Solubility Product (K_sp)

The solubility product constant (K_sp) is a special type of equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds. It quantifies the extent to which a solid dissolves in water.

The Dissolution Equilibrium

When a sparingly soluble ionic solid is placed in water, a dynamic equilibrium establishes between the solid and its dissolved ions. For example, silver chloride dissolves according to:

\text{AgCl}{\text{(s)}} \rightleftharpoons \text{Ag}^{+}{\text{(aq)}} + \text{Cl}^{-}{\text{(aq)}}

At equilibrium, the rate of dissolution equals the rate of precipitation. The solution is saturated - it contains the maximum amount of dissolved ions possible at that temperature.

K_sp expression: Following the rule that pure solids are excluded from equilibrium expressions, the K_sp for AgCl is:

K_{sp} = [\text{Ag}^{+}][\text{Cl}^{-}]

The solid AgCl does not appear because its concentration is constant. K_sp represents the product of ion concentrations in a saturated solution.

General form: For a compound $A_xB_y$ that dissolves as $A_xB_y \rightleftharpoons xA^{m+} + yB^{n-}$ , the K_sp expression is:

K_{sp} = [A^{m+}]^x[B^{n-}]^y

Each ion concentration is raised to the power of its coefficient in the balanced equation.

Converting Between Solubility and K_sp

Solubility is the maximum amount of solute that can dissolve in a given amount of solvent. It can be expressed in g/L or mol/L. K_sp and solubility are related but different quantities.

From solubility to K_sp: Use stoichiometry to convert solubility (in mol/L) to individual ion concentrations, then substitute into the K_sp expression. If s mol/L of AB dissolves, then [A⁺] = s and [B⁻] = s. If s mol/L of AB₂ dissolves, then [A²⁺] = s and [B⁻] = 2s.

From K_sp to solubility: Define solubility as s, express ion concentrations in terms of s using stoichiometry, substitute into K_sp, and solve for s. This may require algebraic manipulation or approximations.

Example - simple 1:1 compound: For AgCl with K_sp = 1.8 × 10⁻¹⁰, if solubility is s mol/L, then [Ag⁺] = s and [Cl⁻] = s. Therefore K_sp = s², so s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ mol/L.

Example - 1:2 compound: For CaF₂ with K_sp = 3.9 × 10⁻¹¹, dissolution produces [Ca²⁺] = s and [F⁻] = 2s (twice as many fluoride ions). Therefore K_sp = s(2s)² = 4s³, so s = ∛(K_sp/4).

Worked Example

The solubility of lead(II) iodide (PbI₂) is 1.52 × 10⁻³ mol/L at 25°C. Calculate K_sp for PbI₂.

Given:

Solubility of PbI₂ = 1.52 × 10⁻³ mol/L

Step 1: Write the dissolution equilibrium:

$\text{PbI}_2{\text{(s)}} \rightleftharpoons \text{Pb}^{2+}{\text{(aq)}} + 2\text{I}^{-}{\text{(aq)}}$

Step 2: Write the K_sp expression:

$K_{sp} = [\text{Pb}^{2+}][\text{I}^{-}]^2$

Note: I⁻ is squared because its coefficient is 2

Step 3: Determine ion concentrations from solubility:

Let s = solubility = 1.52 × 10⁻³ mol/L

For every 1 mol of PbI₂ that dissolves:

• 1 mol of Pb²⁺ forms, so [Pb²⁺] = s = 1.52 × 10⁻³ M

• 2 mol of I⁻ form, so [I⁻] = 2s = 2(1.52 × 10⁻³) = 3.04 × 10⁻³ M

Step 4: Calculate K_sp:

$K_{sp} = (1.52 \times 10^{-3})(3.04 \times 10^{-3})^2$

$K_{sp} = (1.52 \times 10^{-3})(9.24 \times 10^{-6})$

$K_{sp} = \mathbf{1.40 \times 10^{-8}}$

Practice Question

The K_sp of calcium hydroxide [Ca(OH)₂] is 5.5 × 10⁻⁶ at 25°C. Calculate the solubility of Ca(OH)₂ in mol/L.

Show Answer

Step 1: Write dissolution equilibrium:

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

Step 2: Write K_sp expression:

$K_{sp} = [\text{Ca}^{2+}][\text{OH}^{-}]^2 = 5.5 \times 10^{-6}$

Step 3: Express concentrations in terms of solubility (s):

Let s = solubility in mol/L

[Ca²⁺] = s (one Ca²⁺ ion per formula unit)

[OH⁻] = 2s (two OH⁻ ions per formula unit)

Step 4: Substitute into K_sp expression:

$K_{sp} = (s)(2s)^2 = s(4s^2) = 4s^3$

$5.5 \times 10^{-6} = 4s^3$

Step 5: Solve for s:

$s^3 = \frac{5.5 \times 10^{-6}}{4} = 1.375 \times 10^{-6}$

$s = \sqrt[3]{1.375 \times 10^{-6}}$

$s = \mathbf{1.11 \times 10^{-2} \text{ mol/L}}$

The solubility of Ca(OH)₂ is 0.0111 M or 1.11 × 10⁻² mol/L.

Predicting Precipitation

When two solutions containing ions are mixed, we can predict whether a precipitate will form by comparing the ion product (Q_sp) to the solubility product constant (K_sp). This is analogous to using Q and K_eq to predict reaction direction.

The Ion Product (Q_sp)

The ion product (Q_sp) has the same mathematical form as K_sp but uses the current ion concentrations at any moment, not necessarily at equilibrium.

For a compound with dissolution equilibrium $A_xB_y \rightleftharpoons xA^{m+} + yB^{n-}$ :

Q_{sp} = [A^{m+}]^x[B^{n-}]^y

When solutions are first mixed, Q_sp is calculated using the diluted concentrations immediately after mixing. The system may not be at equilibrium yet.

Important: When calculating Q_sp after mixing solutions, use the diluted concentrations. If equal volumes are mixed, all concentrations are halved. If unequal volumes are mixed, use the dilution formula to find new concentrations.

Precipitation Criteria

Comparing Q_sp to K_sp predicts whether precipitation occurs:

Q_sp < K_sp: The solution is unsaturated. The ion product is less than the equilibrium value, so more solid could dissolve. No precipitate forms. If solid is present, it will continue dissolving until Q_sp = K_sp.

Q_sp = K_sp: The solution is exactly saturated. The system is at equilibrium. No net change occurs - if solid is present, dissolution and precipitation rates are equal.

Q_sp > K_sp: The solution is supersaturated. The ion product exceeds the equilibrium value, so the solution contains more dissolved ions than can exist at equilibrium. A precipitate will form to reduce ion concentrations until Q_sp decreases to equal K_sp.

Practical application: This principle determines whether mixing solutions will produce a precipitate in chemical analysis, water treatment, or industrial processes. It's the basis for gravimetric analysis and selective precipitation.

Worked Example

50.0 mL of 0.0020 M AgNO₃ is mixed with 50.0 mL of 0.0030 M NaCl. Will a precipitate of AgCl form? K_sp(AgCl) = 1.8 × 10⁻¹⁰.

Step 1: Calculate diluted concentrations after mixing:

Total volume = 50.0 + 50.0 = 100.0 mL

Each solution is diluted by factor of 100/50 = 2

$[\text{Ag}^{+}] = \frac{0.0020}{2} = 0.0010 \text{ M}$

$[\text{Cl}^{-}] = \frac{0.0030}{2} = 0.0015 \text{ M}$

Step 2: Calculate Q_sp:

For AgCl: $\text{AgCl}{\text{(s)}} \rightleftharpoons \text{Ag}^{+}{\text{(aq)}} + \text{Cl}^{-}{\text{(aq)}}$

$Q_{sp} = [\text{Ag}^{+}][\text{Cl}^{-}]$

$Q_{sp} = (0.0010)(0.0015)$

$Q_{sp} = 1.5 \times 10^{-6}$

Step 3: Compare Q_sp to K_sp:

Q_sp = 1.5 × 10⁻⁶

K_sp = 1.8 × 10⁻¹⁰

Q_sp > K_sp

Conclusion:

Since Q_sp exceeds K_sp by a factor of about 8000, the solution is highly supersaturated. Yes, a precipitate of AgCl will form. Silver and chloride ions will combine to form solid AgCl until the ion concentrations decrease enough that Q_sp = K_sp = 1.8 × 10⁻¹⁰.

Practice Question

100 mL of 0.00010 M Pb(NO₃)₂ is mixed with 100 mL of 0.00020 M NaCl. Will PbCl₂ precipitate? K_sp(PbCl₂) = 1.7 × 10⁻⁵.

Show Answer

Step 1: Calculate diluted concentrations:

Total volume = 100 + 100 = 200 mL

Each concentration is halved (diluted by factor of 2)

[Pb²⁺] = 0.00010/2 = 5.0 × 10⁻⁵ M

[Cl⁻] = 0.00020/2 = 1.0 × 10⁻⁴ M

Step 2: Write dissolution and Q_sp expression:

$\text{PbCl}_2{\text{(s)}} \rightleftharpoons \text{Pb}^{2+}{\text{(aq)}} + 2\text{Cl}^{-}{\text{(aq)}}$

$Q_{sp} = [\text{Pb}^{2+}][\text{Cl}^{-}]^2$

Step 3: Calculate Q_sp:

$Q_{sp} = (5.0 \times 10^{-5})(1.0 \times 10^{-4})^2$

$Q_{sp} = (5.0 \times 10^{-5})(1.0 \times 10^{-8})$

$Q_{sp} = 5.0 \times 10^{-13}$

Step 4: Compare to K_sp:

Q_sp = 5.0 × 10⁻¹³

K_sp = 1.7 × 10⁻⁵

Q_sp < K_sp

Conclusion:

No, PbCl₂ will not precipitate. The ion product is much smaller than K_sp, meaning the solution is unsaturated and can hold more dissolved Pb²⁺ and Cl⁻ ions. If PbCl₂ solid were added, it would dissolve rather than precipitate.

Common Ion Effect

The common ion effect describes how the solubility of a sparingly soluble salt decreases when a soluble salt containing a common ion is added to the solution. This is a direct application of Le Chatelier's principle to solubility equilibria.

Le Chatelier's Principle Applied to Solubility

Consider silver chloride dissolving in water:

\text{AgCl}{\text{(s)}} \rightleftharpoons \text{Ag}^{+}{\text{(aq)}} + \text{Cl}^{-}{\text{(aq)}}

If we add NaCl (a soluble salt) to this saturated solution, we increase [Cl⁻]. According to Le Chatelier's principle, the equilibrium shifts to minimize this disturbance by consuming the added Cl⁻. The equilibrium shifts left, precipitating more AgCl and reducing [Ag⁺].

Result: The presence of chloride ions from NaCl suppresses the dissolution of AgCl. Less AgCl dissolves than would dissolve in pure water. The common ion (Cl⁻ in this case) reduces solubility.

Quantitative analysis: K_sp remains constant (it only changes with temperature), but the equilibrium concentrations shift. If [Cl⁻] is increased by adding NaCl, [Ag⁺] must decrease so that the product [Ag⁺][Cl⁻] still equals K_sp.

Calculating Solubility with Common Ion

To calculate solubility in the presence of a common ion, recognize that one ion's concentration is dominated by the added salt, not by the sparingly soluble compound.

Example approach: For AgCl in 0.10 M NaCl solution, the [Cl⁻] is essentially 0.10 M from the highly soluble NaCl, with negligible contribution from the sparingly soluble AgCl. Using K_sp = [Ag⁺][Cl⁻]:

$1.8 \times 10^{-10} = [\text{Ag}^{+}](0.10)$

$[\text{Ag}^{+}] = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \text{ M}$

In pure water, [Ag⁺] = 1.3 × 10⁻⁵ M. With 0.10 M Cl⁻ present, [Ag⁺] drops to 1.8 × 10⁻⁹ M - about 7000 times less soluble. The common ion effect dramatically suppresses solubility.

Practical applications: Common ion effect is used in selective precipitation (separating ions by adding specific precipitating agents), minimizing loss of precipitate during filtration (wash with solution containing common ion), and industrial crystallization processes.

Worked Example

Calculate the solubility of calcium fluoride (CaF₂) in: (a) pure water, (b) 0.10 M NaF solution. K_sp(CaF₂) = 3.9 × 10⁻¹¹. Compare the results.

(a) Solubility in pure water:

Dissolution: $\text{CaF}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^{-}$

Let s = solubility in mol/L

[Ca²⁺] = s, [F⁻] = 2s

$K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2 = (s)(2s)^2 = 4s^3$

$3.9 \times 10^{-11} = 4s^3$

$s^3 = 9.75 \times 10^{-12}$

$s = \mathbf{2.1 \times 10^{-4} \text{ M}}$

(b) Solubility in 0.10 M NaF:

NaF provides [F⁻] = 0.10 M (common ion)

Let s = solubility of CaF₂ in this solution

[Ca²⁺] = s

[F⁻] ≈ 0.10 M (dominated by NaF, contribution from CaF₂ negligible)

$K_{sp} = (s)(0.10)^2 = s(0.01)$

$3.9 \times 10^{-11} = 0.01s$

$s = \frac{3.9 \times 10^{-11}}{0.01}$

$s = \mathbf{3.9 \times 10^{-9} \text{ M}}$

Comparison:

In pure water: s = 2.1 × 10⁻⁴ M

In 0.10 M NaF: s = 3.9 × 10⁻⁹ M

Ratio: $\frac{2.1 \times 10^{-4}}{3.9 \times 10^{-9}} \approx 54,000$

CaF₂ is approximately 54,000 times less soluble in 0.10 M NaF than in pure water due to the common ion effect. The presence of F⁻ from NaF dramatically suppresses CaF₂ dissolution.

Practice Question

Explain the common ion effect using Le Chatelier's principle. Why does adding NaCl to a saturated AgCl solution cause more AgCl to precipitate? Calculate [Ag⁺] in a saturated AgCl solution in 0.050 M NaCl. K_sp(AgCl) = 1.8 × 10⁻¹⁰.

Show Answer

Le Chatelier explanation:

AgCl establishes equilibrium: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). Adding NaCl increases [Cl⁻], which is a product in this equilibrium. According to Le Chatelier's principle, the system shifts to minimize this disturbance by consuming the added Cl⁻. The equilibrium shifts left (toward solid AgCl), causing Ag⁺ and Cl⁻ ions to combine and precipitate as solid AgCl. This reduces [Ag⁺] in solution. The presence of the common ion (Cl⁻) suppresses the dissolution of AgCl.

Calculation:

In 0.050 M NaCl, [Cl⁻] ≈ 0.050 M (dominated by the highly soluble NaCl)

Using K_sp = [Ag⁺][Cl⁻]:

1.8 × 10⁻¹⁰ = [Ag⁺](0.050)

[Ag⁺] = (1.8 × 10⁻¹⁰)/0.050

[Ag⁺] = 3.6 × 10⁻⁹ M

In pure water, [Ag⁺] = 1.3 × 10⁻⁵ M. In 0.050 M NaCl, [Ag⁺] drops to 3.6 × 10⁻⁹ M - about 3600 times smaller. The common ion effect dramatically reduces AgCl solubility by shifting equilibrium toward the solid.

Aboriginal Food Processing (Toxin Removal)

Indigenous Australians developed sophisticated food processing techniques based on solubility principles to remove toxins from otherwise inedible plants. The processing of cycad seeds demonstrates practical application of leaching and solubility equilibria.

Cycad Seeds and Cycasin Toxin

Cycad plants (particularly Cycas media) produce large seeds rich in starch, providing a valuable food source in northern Australia. However, cycad seeds contain cycasin, a toxic glycoside that can cause liver damage and neurological problems if consumed.

The challenge: Cycasin is water-soluble, meaning it dissolves in water. Indigenous peoples recognized that while the toxin could not be destroyed by cooking alone, it could be removed by exploiting its solubility.

Traditional knowledge: Aboriginal communities in northern Australia developed processing methods passed down through generations. These methods were essential for survival in regions where cycads were among the few reliable food sources during certain seasons.

Leaching Process

The traditional processing method involves multiple stages of soaking and leaching to extract the water-soluble toxin while retaining the nutritious starch.

Step 1 - Initial preparation: Seeds are collected and the outer coating removed. The seeds are then sliced or ground into smaller pieces to increase surface area, which accelerates the dissolution of toxins.

Step 2 - Leaching in water: The prepared seeds are placed in running water (such as a stream) or in containers with regularly changed water. The cycasin dissolves and diffuses out of the seed material into the surrounding water. This is a solubility equilibrium - as cycasin dissolves from the seeds, it reaches equilibrium with cycasin in the water.

Step 3 - Removing dissolved toxin: By continuously replacing the water or using flowing water, the dissolved cycasin is removed. This shifts the dissolution equilibrium toward more toxin dissolving (Le Chatelier's principle - removing product drives the equilibrium forward). Fresh water has [cycasin] = 0, maintaining a concentration gradient that drives further dissolution from the seeds.

Duration: Traditional methods involved soaking for several days to weeks, depending on the processing technique. Some methods involved burying the seeds in sand near water sources, allowing natural water flow to leach the toxins over time.

Chemistry of the Process

The leaching process demonstrates several chemical principles related to solubility equilibria:

Dissolution equilibrium: Cycasin in seeds establishes equilibrium with cycasin in water: Cycasin_(solid) ⇌ Cycasin_(aq). The equilibrium position depends on the solubility of cycasin.

Concentration gradients: Initially, [cycasin] in seeds is high and [cycasin] in water is zero. Cycasin dissolves until equilibrium is reached. By replacing the water, equilibrium is disrupted and more cycasin dissolves.

Le Chatelier's principle: Removing dissolved cycasin (by replacing water) shifts equilibrium right, causing more cycasin to dissolve from the seeds. Continuous water replacement drives the process toward complete toxin removal.

Surface area effect: Grinding or slicing seeds increases surface area, accelerating dissolution by providing more contact between solid cycasin and water. This decreases the time required for complete detoxification.

Selective extraction: Cycasin is water-soluble while starch is not (or only slightly soluble). The leaching selectively removes the toxic component while retaining the nutritious starch, similar to how common ion effect can be used for selective precipitation.

This traditional knowledge demonstrates sophisticated understanding of chemical principles developed through observation and experimentation over thousands of years. The cycad processing techniques show how Indigenous Australians applied solubility and equilibrium concepts to solve practical problems, converting a poisonous plant into a safe, nutritious food source.

Practice Question

Explain how Aboriginal cycad seed processing demonstrates: (a) solubility equilibrium principles, (b) Le Chatelier's principle, (c) the importance of surface area in dissolution. Why is continuous water replacement essential for effective toxin removal?

Show Answer

(a) Solubility equilibrium:

Cycasin dissolves from the seeds into water, establishing equilibrium: Cycasin(solid) ⇌ Cycasin(aq). At equilibrium, the rate of dissolution equals the rate of crystallization. The equilibrium is characterized by a solubility value - the maximum concentration of cycasin that can dissolve in water. When seeds are first placed in water, [cycasin] in water is zero, so the system is far from equilibrium and dissolution proceeds rapidly. As [cycasin] increases in the water, dissolution slows and eventually reaches equilibrium.

(b) Le Chatelier's principle:

By removing the water containing dissolved cycasin and replacing it with fresh water, the [cycasin] in water is reduced to zero. This disturbs the equilibrium by removing product. According to Le Chatelier's principle, the system shifts to minimize this disturbance by dissolving more cycasin from the seeds to replace what was removed. Continuous water replacement repeatedly drives the equilibrium toward dissolution, eventually extracting essentially all the cycasin from the seeds. Without water replacement, equilibrium would be reached with significant cycasin remaining in the seeds.

Grinding or slicing seeds into smaller pieces dramatically increases the surface area exposed to water. Dissolution occurs at the solid-water interface, so greater surface area means more sites where cycasin molecules can dissolve into water. This increases the rate of dissolution without changing the equilibrium position - the same amount of cycasin ultimately dissolves, but it dissolves much faster. Larger surface area reduces the time required for complete detoxification from weeks to days.

Why continuous water replacement is essential:

If water is not replaced, cycasin dissolves until equilibrium is reached, leaving significant amounts of toxin in the seeds. The dissolved cycasin in the surrounding water creates a "back pressure" that opposes further dissolution. By continuously replacing water, [cycasin] in water is kept near zero, maintaining maximum driving force for dissolution. This shifts equilibrium repeatedly toward the dissolved state, extracting more and more toxin with each water change until the seeds are safe to eat. Running water (like a stream) naturally provides continuous removal of dissolved toxin.

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H₂

I₂

2HI

0.200

-x

+2x

0.200 - x

N₂O₄

2NO₂

0.100

-x

+2x

0.100 - x

Equilibrium Concepts

Static vs Dynamic Equilibrium

Static Equilibrium

Dynamic Equilibrium

Worked Example

Practice Question

Entropy & Gibbs Free Energy

Enthalpy (ΔH)

Entropy (ΔS)

Gibbs Free Energy (ΔG)

Worked Example

Practice Question

Le Chatelier's Principle

Effect of Concentration Changes

Effect of Pressure/Volume Changes

Effect of Temperature Changes

Worked Example

Practice Question

Collision Theory & Equilibrium

Fundamentals of Collision Theory

Explaining Concentration Effects with Collision Theory

Explaining Temperature Effects with Collision Theory

Explaining Pressure Effects with Collision Theory

Worked Example

Practice Question

Equilibrium Calculations

The Equilibrium Constant (Keq)

Constructing the Equilibrium Expression

Interpreting Keq Values

Worked Example

Practice Question

ICE Tables

ICE Table Structure

Solving ICE Table Problems

Worked Example

Practice Question

Reaction Quotient (Q)

Calculating Q

Comparing Q to Keq

Worked Example

Practice Question

Solubility Equilibria

Solubility Product (Ksp)

The Dissolution Equilibrium

Converting Between Solubility and Ksp

Worked Example

Practice Question

Predicting Precipitation

The Ion Product (Qsp)

Precipitation Criteria

Worked Example

Practice Question

Common Ion Effect

Le Chatelier's Principle Applied to Solubility

Calculating Solubility with Common Ion

Worked Example

Practice Question

Aboriginal Food Processing (Toxin Removal)

Cycad Seeds and Cycasin Toxin

Leaching Process

Chemistry of the Process

Practice Question

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Equilibrium Concepts

Static vs Dynamic Equilibrium

Static Equilibrium

Dynamic Equilibrium

Worked Example

Practice Question

Entropy & Gibbs Free Energy

Enthalpy (ΔH)

Entropy (ΔS)

Gibbs Free Energy (ΔG)

Worked Example

Practice Question

Le Chatelier's Principle

Effect of Concentration Changes

Effect of Pressure/Volume Changes

Effect of Temperature Changes

Worked Example

The Equilibrium Constant (K_eq)

Interpreting K_eq Values

Comparing Q to K_eq

Solubility Product (K_sp)

Converting Between Solubility and K_sp

The Ion Product (Q_sp)

The Equilibrium Constant (K_eq)

Interpreting K_eq Values

Comparing Q to K_eq

Solubility Product (K_sp)

Converting Between Solubility and K_sp

The Ion Product (Q_sp)