Self-ionisation of Water and Kw

HSC Chemistry Syllabus

• calculate and apply the dissociation constant (K_a) and pK_a (pK_a = –log10 (K_a)) to determine the difference between strong and weak acids (ACSCH098)

Self-ionisation of Water and `K_w` Explained

What is Self-ionisation of Water?

Self-ionisation or auto-ionisation of water is a reversible reaction that occurs between two molecules of water. In this reaction, a proton is transferred from one water molecule to another, forming a hydroxide and hydronium ion.

 

$$2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$$

 

This reversible reaction occurs in pure water and all aqueous solutions e.g. acid or base solutions. In a closed chemical system, this reaction can reach an equilibrium.

The equilibrium constant of this reaction is represented by `K_w = 1.0 \times 10^{-14}` at 25 ºC.

 

$$K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$$

 

This equation can be used to calculate the concentration of either ions in water or an aqueous solution when the concentration of the other is given.

Example 1

The hydrogen ion concentration of a hydrofluoric acid solution is 0.0700 mol/L at 25ºC. What is the hydroxide concentration?

 

Solution:

$$[H_3O^+][OH^-] = 1.0 \times 10^{-14}$$

$$[OH^-] = \frac{1.0 \times 10^{-14}}{0.0700}$$

$$[OH^-] = 1.43 \times 10^{-13} \text{ mol/L (3 s.f.)}$$

Effect of Temperature on Water's Self-ionisation

The self-ionisation of water is an endothermic reaction (`\Delta H > 0`) as the energy absorbed during bond breaking is greater than the energy released during bond formation. 

Since `K_w` is an equilibrium constant, its value is affected by temperature.

• At a temperature > 25ºC, the equilibrium position shifts to the product side to produce more ions, thereby increasing the value of `K_w`.

• At a temperature < 25ºC, the equilibrium position shifts to the reactant side, thereby reducing the equilibrium concentration of hydroxide and hydronium ions and decreasing the value of `K_w`.

 

At 25ºC, the extremely small value of `K_w` suggests that the equilibrium concentrations of `H_3O^+` and `OH^-` ions are small. In pure water, the production of equal concentrations of these ions makes it neutral. When these ion concentrations change with temperature (as explained above), they remain equal and thus pure water remains neutral at all temperatures.

Relationship Between K_w, K_a and K_b

Self-ionisation of water does not only occur in pure water, it occurs in all aqueous solutions. For example, the above equilibrium can occur in a solution of acid or base.

This is what you need to remember and understand: `K_w` is the product of `K_a` of a weak acid and `K_b` of its conjugate base.

 

$$K_w = K_a \times K_b$$  

 

Derivation (do not memorise): For a generic acid-base system,

 

$$\text{Acid}: \hspace{0.5cm} HA_{(aq)} + H_2O_{(l)} \leftrightharpoons A^-_{(aq)} + H_3O^+_{(aq)} \hspace{0.5cm} K_a = \frac{[A^-][H_3O^+]}{[HA]}$$

 

$$\text{Conjugate base}: \hspace{0.5cm} A^-_{(aq)} + H_2O_{(l)} \leftrightharpoons HA_{(aq)} + OH^-_{(aq)} \hspace{0.5cm} K_b = \frac{[HA][OH^-]}{[A^-]}$$

 

Let's multiply the expressions for `K_a` and `K_b`:

 

$$K_a \times K_b = \frac{[{A^-}][H_3O^+]}{[HA]} \times \frac{[HA][OH^-]}{[A^-]}$$

 

$$K_a \times K_b = [H_3O^+_{(aq)}][OH^-_{(aq)}] = 1.0 \times 10^{-14}$$

 

This means at 25℃ (298 K):

$$K_a = \frac{10^{-14}}{K_b} \hspace{3cm} K_b = \frac{10^{-14}}{K_a}$$

 

These equations suggest that `K_a` of a weak acid's ionisation is inversely proportional to the `K_b` of its conjugate base's ionisation. The stronger an acid is, the weaker its conjugate base is; the weaker an acid is, the stronger its conjugate base is. This relationship is also true for a weak base and its conjugate acid.

Since these equilibrium constant values also indicate the strength of the acid and base respectively, we can formulate the following relationship:

 

 

 

  

Note the concept of `pK_a` is discussed separately here

 

  

RETURN TO MODULE 6: ACID-BASE REACTIONS