Chemical and Ionic Equilibrium

1. Introduction to Equilibrium

Many chemical reactions do not continue completely in one direction. Instead, the products formed can react again to produce the original reactants.

When the forward and reverse reactions occur at the same time and finally reach a condition where their rates become equal, the system is said to be at chemical equilibrium.

The double arrow means that the reaction can proceed in both directions.

Important point

Equilibrium does not mean that the amounts of reactants and products are equal. It means that the forward and reverse reaction rates are equal.

2. Dynamic Equilibrium

Chemical equilibrium is dynamic because the forward and reverse reactions do not stop at equilibrium.

Both reactions continue to occur, but they occur at equal rates.

Example

Initially, hydrogen and iodine react rapidly and HI concentration increases. As HI accumulates, the reverse reaction becomes faster.

Finally:

Characteristics of chemical equilibrium

• It is dynamic.
• Forward and reverse reactions continue.
• The rates of the two reactions are equal.
• Concentrations remain constant with time.
• Equilibrium is established in a closed system.
• Changing temperature, pressure or concentration can disturb equilibrium.

3. Law of Mass Action

The law of mass action states that at a constant temperature, the rate of a reaction is proportional to the product of the active masses or concentrations of the reactants, each raised to the power of its stoichiometric coefficient.

General reaction

At equilibrium, the equilibrium constant is written as:

The square brackets represent molar concentration.

Important rule

The coefficients in the balanced chemical equation become the powers in the equilibrium expression.

4. Equilibrium Constant Kc

Kc is the equilibrium constant expressed using molar concentrations.

Example 1

Example 2

Example 3

Pure solids are not included in the equilibrium expression.

Meaning of the magnitude of Kc

5. Equilibrium Constant Kp

For gaseous equilibria, equilibrium can also be expressed using partial pressures.

This equilibrium constant is called Kp.

General reaction

Here P represents the partial pressure of the gas.

Example

6. Relationship Between Kc and Kp

For gaseous reactions, Kc and Kp are related by:

where:

Example

Count only gaseous species:

Therefore:

When Δn = 0

7. Reaction Quotient Q

The reaction quotient has the same mathematical form as the equilibrium constant, but it can be calculated at any stage of the reaction.

Comparing Q with K

8. Homogeneous and Heterogeneous Equilibrium

Homogeneous equilibrium

All reactants and products are in the same physical state.

All species are gases.

Heterogeneous equilibrium

Reactants and products are present in different physical states.

Here solids and gas are present.

Why solids are omitted

The concentration or activity of a pure solid is treated as effectively constant, so it does not appear in the equilibrium expression.

9. Le Chatelier's Principle

Le Chatelier's principle states that when a system at equilibrium is disturbed, the system shifts in the direction that tends to oppose the disturbance.

Example

If more N2 is added, the system tends to consume some of the added N2.

10. Factors Affecting Chemical Equilibrium

10.1 Effect of concentration

Increasing the concentration of a reactant generally causes equilibrium to shift toward products.

Removing a product also generally shifts the equilibrium toward products.

10.2 Effect of pressure

Pressure mainly affects equilibria involving gases.

Left side = 4 moles of gas.

Right side = 2 moles of gas.

Increasing pressure favors the side with fewer gas moles.

10.3 Effect of temperature

Temperature must be considered as part of the reaction itself.

Exothermic reaction

Increasing temperature adds heat, so equilibrium shifts toward the reactant side.

Endothermic reaction

Increasing temperature shifts equilibrium toward products.

10.4 Catalyst

A catalyst increases the rates of both forward and reverse reactions. It helps the system reach equilibrium faster, but it does not change the equilibrium constant or the equilibrium composition.

11. Ionic Equilibrium and Acid-Base Equilibrium

Ionic equilibrium deals mainly with equilibria involving ions in solution, especially weak acids, weak bases and slightly soluble salts.

Strong electrolytes ionize almost completely, whereas weak electrolytes establish an equilibrium between ionized and unionized forms.

Example of weak acid

Most acetic acid molecules remain unionized in a dilute solution, while only a fraction ionizes.

Example of weak base

12. Theories of Acids and Bases

12.1 Arrhenius concept

According to Arrhenius theory, an acid produces H+ in aqueous solution and a base produces OH− in aqueous solution.

12.2 Brønsted-Lowry concept

A Brønsted-Lowry acid is a proton donor and a Brønsted-Lowry base is a proton acceptor.

HCl donates H+, so it is the acid.

H2O accepts H+, so it is the base.

12.3 Lewis concept

A Lewis acid accepts an electron pair, while a Lewis base donates an electron pair.

NH3 donates its lone pair, so it behaves as a Lewis base. BF3 accepts the electron pair, so it behaves as a Lewis acid.

13. Conjugate Acid-Base Pairs

When an acid donates a proton, the species left behind is its conjugate base.

HA = acid

A− = conjugate base

Example

NH4+ donates H+, so NH4+ is the acid and NH3 is its conjugate base.

H2O accepts H+, so H2O is the base and H3O+ is its conjugate acid.

14. Strong and Weak Electrolytes

Strong electrolytes

Strong electrolytes ionize or dissociate almost completely in aqueous solution.

Weak electrolytes

Weak electrolytes ionize only partially.

Comparison

15. Degree of Ionization

The degree of ionization is the fraction of the original electrolyte molecules that become ionized.

If percentage is required:

For a weak acid

If the initial concentration is C and degree of ionization is α:

Therefore:

For a very weak electrolyte, α is small, so:

16. Acid Dissociation Constant Ka

For a weak acid:

The acid dissociation constant is:

Meaning of Ka

A larger Ka generally means greater acid ionization and therefore a stronger weak acid compared with an acid having a smaller Ka under comparable conditions.

Example: acetic acid

17. Base Dissociation Constant Kb

For a weak base:

The base dissociation constant is:

Example: ammonia

Relationship between Ka and Kb

For a conjugate acid-base pair at 25°C:

Therefore:

18. Ionic Product of Water, Kw

Water undergoes slight self-ionization:

More accurately in aqueous solution:

The ionic product of water is:

At 25°C:

In pure water:

Therefore:

19. pH and pOH

pH

pOH

At 25°C:

Finding pH of a strong acid

For a strong monoprotic acid such as HCl, ionization is treated as essentially complete.

Finding pOH of a strong base

20. pH of a Weak Acid

Consider a weak acid HA with concentration C:

If the acid is sufficiently weak:

If x is small compared with C:

Therefore:

Rearranging:

Therefore:

Worked example

Find the approximate pH of 0.10 M acetic acid if:

21. pH of a Weak Base

For a weak base:

If the initial concentration is C and the amount ionized is x:

Therefore:

For a sufficiently weak base:

Therefore:

Worked example

Find the pH of 0.10 M ammonia solution if:

22. Common-Ion Effect

The common-ion effect is the suppression of ionization of a weak electrolyte when a strong electrolyte containing one of the ions already present is added.

Example: acetic acid and sodium acetate

Sodium acetate dissociates almost completely:

Adding sodium acetate increases [CH3COO−].

According to Le Chatelier's principle, the equilibrium shifts toward the left.

Uses

• Preparation of buffer solutions.
• Control of solubility.
• Selective precipitation.
• Suppression of weak electrolyte ionization.

23. Buffer Solutions

A buffer solution resists a large change in pH when a small amount of acid or base is added.

Types of buffers

Acidic buffer

Usually consists of a weak acid and its salt with a strong base.

Basic buffer

Usually consists of a weak base and its salt with a strong acid.

Why a buffer works

An acidic buffer contains a large amount of weak acid and its conjugate base. If a small amount of H+ is added, the conjugate base consumes much of the added H+.

If OH− is added, the weak acid consumes it:

24. Henderson-Hasselbalch Equation

For an acidic buffer

Consider:

Acid dissociation constant:

Rearrange:

Taking negative logarithm:

For a buffer made from a weak acid and its salt, we can often use concentrations or, when both components are in the same solution volume, their mole ratio:

Worked example

A buffer contains 0.10 M acetic acid and 0.20 M sodium acetate. Given:

First calculate pKa:

Now use:

For a basic buffer

Then:

25. Buffer Action in Detail

Acidic buffer: CH3COOH / CH3COO−

The buffer contains a weak acid and a relatively large concentration of its conjugate base.

When H+ is added

Therefore, much of the added H+ is removed from the solution by conversion into weakly ionized acetic acid.

When OH− is added

Therefore, much of the added OH− is consumed.

Why pH does not remain perfectly constant

A buffer only resists changes in pH within its effective capacity. If a very large amount of acid or base is added, the buffer components can eventually be consumed.

26. Solubility Equilibrium

Some ionic compounds are only slightly soluble in water. When a sparingly soluble salt is placed in water, a dissolution equilibrium can be established.

Example: silver chloride

At equilibrium, the rate of dissolution equals the rate of precipitation.

Example: calcium fluoride

The equilibrium expression is based on the dissolved ions.

27. Solubility Product, Ksp

Ksp is the equilibrium constant for the dissolution of a sparingly soluble ionic compound.

For AgCl

For CaF2

For Al(OH)3

Relation between solubility and Ksp

For AgCl, if molar solubility is s:

Therefore:

For CaF2

If solubility = s:

Therefore:

Hence:

28. Precipitation and Ionic Product

To determine whether precipitation occurs, compare the ionic product Qsp with Ksp.

Example

Suppose:

For AgCl:

If Ksp is 1.8 × 10^-10:

Therefore precipitation occurs.

29. Worked Numerical Problems

Numerical 1: Calculate Kc

For the reaction:

Suppose at equilibrium:

Therefore:

Numerical 2: Find Kp from Kc

For:

Suppose:

First calculate Δn:

Therefore:

Using:

Numerical 3: Degree of ionization

A weak acid has:

For a weak acid:

Percentage ionization:

Numerical 4: pH of a strong acid

Find the pH of 0.001 M HCl.

Numerical 5: pH of a weak acid

Find the approximate pH of 0.10 M acid with:

Numerical 6: pH of a buffer

A buffer contains equal concentrations of a weak acid and its conjugate base.

Henderson-Hasselbalch equation:

Since:

Therefore:

Numerical 7: Ksp of AgCl

Suppose the molar solubility of AgCl is s.

Therefore:

If:

Then:

Numerical 8: Ksp of CaF2

For:

If solubility is s:

Therefore:

This is why the stoichiometry of the dissolution equation must be written before calculating Ksp.

30. Common Mistakes

• Saying equilibrium means equal concentrations.
• Forgetting to balance the chemical equation before writing Kc.
• Forgetting to use coefficients as powers.
• Including pure solids in the Kc or Ksp expression.
• Using total pressure instead of partial pressure in the Kp expression.
• Using the wrong sign or value for Δn in the Kp-Kc relationship.
• Thinking a catalyst changes Kc.
• Thinking pressure affects every equilibrium equally.
• Treating a weak acid as if it ionizes completely.
• Using [H+] = acid concentration automatically for a weak acid.
• Forgetting that pOH must be converted to pH.
• Using pH = 14 - pOH without considering the temperature assumption.
• Confusing Ka with Kb.
• Forgetting the common-ion effect in buffer systems.
• Reversing salt/acid in the Henderson-Hasselbalch equation.
• Forgetting stoichiometric powers when calculating Ksp.
• Calling Qsp > Ksp "no precipitation".

31. TU Examination Preparation

2-mark questions

5-mark questions

10-mark questions

How to write equilibrium numericals

1. Write the balanced chemical equation.
2. Write the equilibrium expression.
3. List the given values clearly.
4. Substitute values carefully.
5. Show intermediate calculation steps.
6. Check units and powers of ten.
7. Write the final answer with the correct unit where applicable.

32. Quick Revision

• Equilibrium is dynamic.
• At equilibrium, forward rate = reverse rate.
• Equilibrium does not mean equal concentrations.
• Kc uses molar concentrations.
• Kp uses partial pressures.
• Kp = Kc(RT)^Δn.
• Δn = gaseous product moles − gaseous reactant moles.
• Pure solids are omitted from equilibrium expressions.
• Q < K → forward direction favored.
• Q > K → reverse direction favored.
• Q = K → equilibrium.
• Le Chatelier principle predicts the direction of shift.
• Catalyst does not change equilibrium composition.
• Arrhenius acid gives H+ in water.
• Brønsted acid donates H+.
• Lewis acid accepts an electron pair.
• Ka measures weak-acid ionization.
• Kb measures weak-base ionization.
• Kw = [H+][OH−].
• At 25°C, Kw = 1.0 × 10^-14.
• pH = −log[H+].
• pOH = −log[OH−].
• At 25°C, pH + pOH = 14.
• Common-ion effect suppresses ionization.
• Buffer resists large pH changes.
• Acidic buffer: weak acid + its conjugate base.
• Basic buffer: weak base + its conjugate acid.
• pH = pKa + log(salt/acid).
• Ksp describes equilibrium of sparingly soluble salts.
• Qsp > Ksp indicates precipitation.

Important formulas

33. Quick Check

Final Study Note

Chemical equilibrium becomes much easier when every problem is first reduced to the equilibrium equation. Write the reaction, balance it, identify the species that participate in the equilibrium expression, and then calculate.

For ionic equilibrium, the same approach works. First identify whether the substance is a strong or weak electrolyte. Then decide whether you need Ka, Kb, Kw, pH, pOH, the common-ion effect, a buffer equation or Ksp.

The most important habit for numerical problems is not to memorize a final answer. Write the formula first, put the values into it, simplify one step at a time and check whether the result makes chemical sense.