Applying Keq to Ionic Solutions and Acids/Bases

This section will cover basic ideas relating to the application of Keq to some specific systems. These systems include:

  • Dissociation of ionic substances
  • Dissociation of acids and bases

More detailed information and applications of these ideas will be covered in a later section.

Dissociation of Ionic Substances

Ionic substances dissociate in water when the compound is split into its ions. Ionic compounds are often described as soluble or insoluble, but this concept largely overlooks the idea that there are degrees to which any ionic compound may dissociate in water. Often, compounds which we describe as insoluble will ionise to a very small degree and actually exist in a dynamic equilibrium, whereby the solid compound will dissociate into its ions and the ions will also reform the solid compound.

As the dissociation of partially soluble or insoluble ionic compounds exist in a dynamic equilibrium, the dissociation of these compounds can be analysed using the equilibrium expression, Keq.

Example:

Lead (II) iodide is considered an insoluble salt and will precipitate out in a reaction between lead (II) nitrate and potassium iodide. The dissociation of lead (II) iodide can be written as:

PbI2(s) ⇌ Pb2+(aq) + 2I–(aq)

Keq = [Pb2+][I–]2

The dissociation of ionic compounds is a heterogenous system. When writing this as an equilibrium expression we omit the PbI2(s) from the expression because it is a solid.

There are certain abbreviations which will be introduced for the variety of equilibrium examples we see in this course. For example, the dissociation of ionic substances and their equilibrium expressions are known as the solubility product, Ksp.

General Points:

  • The solubility product, Ksp, is only used when the concentrations of the chemical species are measured at equilibrium.
  • The ionic product is used to describe the dissociation of ionic compounds when the system is not at equilibrium.

Comparing the ionic product to the solubility product, Ksp, can indicate which way the reaction is likely to move. This will allow us to state if more of the ionic compound will dissolve or if a precipitate will form:

  • If the ionic product = Ksp, then the system is at equilibrium.
  • If the ionic product < Ksp, the forward reaction would be favoured and the solid would dissolve for the system to reach equilibrium.
  • If the ionic product > Ksp, the reverse reaction would be favoured and more precipitate would form for the system to reach equilibrium.

Dissociation of Acids and Bases

Acids and bases are described as strong or weak based on the degree to which the molecules ionise to forms ions in solution. Strong acids and bases generally ionise completely and are considered irreversible and therefore, are thought of as non-equilibrium systems.

Weak acids and bases partially ionise in water and establish equilibrium systems. This allows them to be analysed like other equilibrium systems and comparisons to be made. For example, a higher constant would indicate a higher degree of ionisation.

Equilibrium Constants for Acids

The equilibrium constant for the ionisation of an acid is called the acid dissociation constant, Ka. 

For the ionisation of a general acid, HA(aq) ⇌ H+(aq) + A−(aq)

{ K }_{ a }=\cfrac { \left[ { H }^{ + } \right] \left[ { A }^{ - } \right] }{ \left[ HA \right] }

Equilibrium Constants for Bases

The equilibrium constant for the ionisation of a base is called the base dissociation constant, Kb. 

For the ionisation of a general base, BOH(aq) ⇌ B+(aq) + OH−(aq)

{ K }_{ b }=\cfrac { \left\lceil { B }^{ + } \right\rceil \left\lceil { OH }^{ - } \right\rceil }{ \left[ BOH \right] }

An Important Note:

The dissociation of acids and bases occurs in water and hence, it is more correct that water should be written in the ionic equation:

HA(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq)

B(aq) + H2O(l) ⇌ BH+(aq) + OH−(aq)

This would make the Ka and Kb expressions:

{ K }_{ a }=\cfrac { \left[ { H_{ 3 } }{ O }^{ + } \right] \left[ { A }^{ - } \right] }{ \left[ HA \right] \left[ { H }_{ 2 }{ O } \right] }

{ K }_{ b }=\cfrac { \left[ { BH }^{ + } \right] \left[ { OH }^{ - } \right] }{ \left[ B \right] \left[ { H }_{ 2 }{ O } \right] }

The concentration of water is basically the same in dilute solutions and is omitted from most ionic equations and Ka/Kb expressions:

Acids:

HA(aq) ⇌ H+(aq) + A−(aq)

{ K }_{ a }=\cfrac { \left[ { H }^{ + } \right] \left[ { A }^{ - } \right] }{ \left[ HA \right] }

Bases:

B(aq) ⇌ BH+(aq) + OH−(aq)

{ K }_{ b }=\cfrac { \left[ { BH }^{ + } \right] \left[ { OH }^{ - } \right] }{ \left[ B \right] }

for weak hydroxide bases:

BOH(aq) ⇌ B+(aq) + OH−(aq)

{ K }_{ b }=\cfrac { \left[ { B }^{ + } \right] \left[ { OH }^{ - } \right] }{ \left[ BOH \right] }

Example 1:

Write the solubility product for the dissolution of the following ionic compounds:

a) AgCl

b) Ca(OH)2

Answers:

a) { K }_{ sp }=\left[ { Ag }^{ + } \right] \left[ { Cl }^{ - } \right]

b) { K }_{ sp }=\left[ { Ca }^{ 2+ } \right] \left[ OH^{ - } \right] ^{ 2 }

Example 2:

Write the dissociation expressions for the following:

a) CH3COOH

b) Al(OH)3

Answer:

a) { K }_{ a }=\cfrac { \left[ { CH }_{ 3 }COO^{ - } \right] \left[ { H }^{ + } \right] }{ \left[ { CH }_{ 3 }COOH \right] }

b) { K }_{ b }=\cfrac { \left[ Al^{ 3+ } \right] \left[ { OH }^{ - } \right] ^{ 3 } }{ \left[ { Al(OH) }_{ 3 } \right] }