Acid and Base Dissociation Constants Calculations

This is part of the HSC Chemistry course under the topic Dissociation of Acids and Bases.

HSC Chemistry Syllabus

  • Explore the use of Keq for different types of chemical reactions, including but not limited to:
– dissociation of ionic solutions
– dissociation of acids and bases
  • calculate and apply the dissociation constant ( Ka) and p Ka ( p Ka = -log10( Ka)) to determine the difference between strong and weak acids (ACSCH098)

Calculations Involving Acid and Base Dissociation Constants

Calculations for Acid Dissociation Constant Ka

  • For a generic weak acid HA:


$$HA(aq) + H_2O(l) \rightleftharpoons A^–(aq) + H_3O^+(aq)$$


  • When acid molecules dissociate, the concentration of hydrogen ions, [H3O+] equals [A] at equilibrium.


Commons assumptions

  1. When the dissociation constant, Ka, is small, the proportion of acid molecules that undergoes ionisation is small. As such, it can be assumed that the initial concentration of acid, HA, approximately equals its concentration at equilibrium. In other words, the change in [HA] during ionisation is negligible.

  1. It can be assumed, during calculation, that self-ionisation of water does not affect [H3O+] of the solution and therefore, it does not contribute to the pH. 

  1. If the acid is diprotic or triprotic, it can be usually assumed that the second and third ionisation occur to a smaller extent than the first ionisation. 

In other words, if Ka of first ionisation is already small, Ka of subsequent ionisation reactions can be disregarded. For example, hydrogen sulfide is diprotic and its two ionisation reactions are as follows: 


H2S(aq) + H2O(l) ⇌ HS(aq) + H3O+(aq)        Ka1 = 8.9 × 10–8


HS(aq) + H2O(l) ⇌ S2–(aq) + H3O+(aq)           Ka2 = 1.3 × 10–14


This assumption does not apply to all weak acids and certainly not strong acids. Some weak acids ionise to a considerable extent, to which its initial concentration differs substantially with that at equilibrium.


Example 1 – Calculating [H+from Ka

The Ka of hypochlorous acid, HOCl, is 3.5 × 10–8.

A solution of hypochlorous acid, HOCl, is made by adding 0.0400 moles of HOCl to 500.0 mL of distilled water. What is the hydrogen ion concentration of the solution?

Step 1: Write a balanced equation and an expression for Ka

$$HOCl(aq) \rightleftharpoons H^+(aq) + OCl^–(aq)$$


$$K_a = \frac{[H^+][OCl^–]}{[HOCl]}$$

Step 2: Set-up an ICE table







$$\frac{0.0400}{0.5000} = 0.0800 \text{ M}$$




– x

+ x

+ x


0.0800 – x




Step 3: Substitute equilibrium concentration expressions (in terms of x) into Ka expression:

$$3.5 \times 10^{–8} = \frac{(x)(x)}{0.0800–x}$$  

Step 4: By making the assumption: x is neglible as Ka is small, simplify the equation to find x:

$$3.5 \times 10^{–8} = \frac{x^2}{0.0800}$$   

$$x = [H^+] = \sqrt{0.0800 \times 3.5 \times 10^{-8}}$$

$$[H^+] = 5.3 \times 10^{–5} \text{ mol/L (2 s.f.)}$$ 


Example 2 – Calculating Ka from [H+]

A 0.750 mol L–1 solution of a particular weak acid, HA, has a [H+] = `3.4 xx 10^(–3)` mol L–1 at 25ºC. Calculate its Ka.

Step 1: Write a balanced equation and an expression for Ka


$$HA(aq) + H_2O(l) \rightleftharpoons A^–(aq) + H_3O^+(aq)$$

$$K_a = \frac{[A^–][H_3O^+]}{[HA]}$$


Step 2: Calculate Ka from equilibrium expression.

$$K_a = \frac{[3.4 \times 10^{–3}][3.4 \times 10^{–3}]}{[0.750 - 3.4 \times 10^{–3}]}$$

$$K_a = 1.5 \times 10^{–5} \text{ (2 s.f.)}$$

Calculations for Base Dissociation Constant, Kb

  • A base by definition, is a substance that accepts hydrogen ions or protons in solution.

  • The equilibrium constant for the equilibrium established when a weak base reacts with water is referred to as the base dissociation constant. For example, for a generic base `B`.


$$B(aq) + H_2O(l) \rightleftharpoons HB^+(aq) + OH^–(aq)$$  

The base dissociation constant is:

$$K_b = \frac{[OH^–][HB^+]}{[B]}$$


  • The base dissociation constant of carbonate ion in water:

$$CO_3^{2–}(aq) + H_2O(l) \rightleftharpoons HCO_3^–(aq) + OH^–(aq)$$

$$K_b = \frac{[OH^–][HCO_3^–]}{[CO_3^{2–}]}$$


Previous Section: Common Solubility Equilibrium Questions