go to the AUS-e-TUTE homepage
home test exam game contact
 

 

Acid Dissociation Constants (Ka)

Key Concepts

  • Ka, the acid dissociation constant or acid ionisation constant, is an equilibrium constant that refers to the dissociation, or ionisation, of an acid.

  • For the reaction in which the acid HA dissociates to form the ions H+ and A-:
    HA H+ + A-
    Ka = [H+][A-]
    [HA]

  • Ka provides a measure of the equilibrium position
  1. if Ka is large, the products of the dissociation reaction are favoured

  2. if Ka is small, undissociated acid is favoured.

  • Ka provides a measure of the strength of an acid
  1. if Ka is large, the acid is largely dissociated so the acid is strong

  2. if Ka is small, very little of the acid is dissociated so the acid is weak.

  • The degree to which an acid dissociates can be represented as a percentage:
    % dissociation (ionization) = [H+ at equilibrium] ÷ [acid initial] x 100

    If %dissociation ≈ 100%, the acid is a strong acid

    If %dissociation is small, the acid is a weak acid

Example : Calculating [H+], pH and %dissociation for a Strong Acid

Calculate the [H+], pH and %dissociation in 0.10 mol L-1 HCl(aq) at 25oC.
  1. Write the acid dissociation equation:
    HCl H+ + Cl-

  2. Calculate the initial and equilibrium concentrations of the species present:

    HCl is a strong acid, it completely dissociates to form H+ and Cl-
      HCl H+ + Cl-
    initial concentrations (M) 0.10   0   0
    Equilibrium concentrations (M) 0.10 - 0.10 = 0   0.10   0.10

    [H+] = 0.10M

  3. Calculate pH : pH = -log[H+]

    pH = -log[0.10] = 1

  4. Calculate %dissociation:

    %dissociation = [H+]/[acid initial] x 100

    [H+] = 0.1M

    [HCl initial] = 0.1M

    %dissociation = 0.1/0.1 x 100 = 100%

Example : Calculating [H+], pH and %dissociation for a Weak Acid

Calculate the [H+], pH and %dissociation in 0.10 mol L-1 HNO2(aq). (Ka = 5.0 x 10-4 at 25oC)
  1. Write the acid dissociation equation:
    HNO2 H+ + NO2-

  2. Write the equilibrium expression for the acid dissociation:
    Ka = [H+][NO2-]
    [HNO2]
    5.0 x 10-4 = [H+][NO2-]
    [HNO2]

  3. Calculate the initial and equilibrium concentrations of the species present:

    let x = moles of HNO2 that dissociate to form H+ and NO2-
      HNO2 H+ + NO2-
    initial concentrations (M) 0.10   0   0
    Equilibrium concentrations (M) 0.10 - x   x   x

    Since HNO2 is a weak acid (Ka is small), it dissociates only slightly, x will be very small compared to 0.10

    So, at equilibrium, [HNO2] ≈ 0.10M

  4. Substitute the concentration values into the expression for the acid dissociation:
    Ka = [H+][NO2-]
    [HNO2]
    5.0 x 10-4 = [x][x]
    [0.10]

    5.0 x 10-4 = x2 ÷ 0.10

    x2 = 5.0 x 10-4 x 0.10 = 5 x 10-5

    x = √5 x 10-5 = 7.1 x 10-3

    [H+] = x = 7.1 x 10-3 mol L-1

  5. Calculate pH:

    pH = -log[H+]

    pH = -log[7.1 x 10-3] = 2.1

  6. Calculate %dissociation:

    %dissociation = [H+]/[acid initial] x 100

    [H+] = 7.1 x 10-3

    [HNO2 initial] = 0.10M

    %dissociation = 7.1 x 10-3/0.10 x 100 = 7.1%

Example : Calculating [OH-] and pOH for a Weak Acid at 25oC

Calculate the [OH-] and pOH for an aqueous solution of 0.5M acetic acid (ethanoic acid).
Ka = 1.8 x 10-5 at 25oC.
  1. Write the acid dissociation equation:
    CH3COOH H+ + CH3COO-

  2. Write the equilibrium expression for the acid dissociation:
    Ka = [H+][CH3COO-]
    [CH3COOH]
    1.8 x 10-5 = [H+][CH3COO-]
    [CH3COOH]

  3. Calculate the initial and equilibrium concentrations of the species present:

    let x = moles of CH3COOH that dissociate to form H+ and CH3COO-
      CH3COOH H+ + CH3COO-
    initial concentrations (M) 0.5   0   0
    Equilibrium concentrations (M) 0.5 - x   x   x

    Since CH3COOH is a weak acid (Ka is small), it dissociates only slightly, x will be very small compared to 0.5

    So, at equilibrium, [CH3COOH] ≈ 0.5M

  4. Substitute the concentration values into the expression for the acid dissociation:
    Ka = [H+][CH3COO-]
    [CH3COOH]
    1.8 x 10-5 = [x][x]
    [0.5]

    1.8 x 10-5 = x2 ÷ 0.5

    x2 = 1.8 x 10-5 x 0.5 = 9 x 10-6

    x = √9 x 10-6 = 3 x 10-3

    [H+] = x = 3 x 10-3 mol L-1

  5. Calculate the concentration of OH-:

    At 25oC, Kw, the equilibrium constant for the dissociation of water, is 10-14

    ie, [H+][OH-] = 10-14

    [OH-] = 10-14/[H+]

    Substitute in the value for [H+]:

    [OH-] = 10-14/3 x 10-3 = 3.3 x 10-12M

  6. Calculate pOH:

    pOH = -log[OH-]

    pOH = -log[3.3 x 10-12] = 11.5

Practice Questions
For AUS-e-TUTE members:
  1. Click on the Ka drill link:
    Ka drill
  2. Enter your username and password if prompted.
  3. Click the "New Question" button to begin the drill.
  4. Worked solutions are provided if you need some help!

Not an AUS-e-TUTE Member?

advertise on the AUS-e-TUTE website and newsletters
 

Search this Site

You can search this site using a key term or a concept to find tutorials, tests, exams and learning activities (games).
 

Become an AUS-e-TUTE Member

 

Subscribe to our Free Newsletter

Email email us to
subscribe to AUS-e-TUTE's free quarterly newsletter, AUS-e-NEWS.

AUS-e-NEWS quarterly newsletter

AUS-e-NEWS is emailed out in
December, March, June, and September.

 

Ask Chris, the Chemist, a Question

The quickest way to find the definition of a term is to ask Chris, the AUS-e-TUTE Chemist.

Chris can also send you to the relevant
AUS-e-TUTE tutorial topic page.

 

Weak Acid Dissociation Constants

acidformulaKa
boric acid H3BO3 6.0 x 10-10
hydrocyanic acid
(hydrogen cyanide)		 HCN 4.0 x 10-10
hypochlorous acid HOCl 3.2 x 10-8
acetic acid
(ethanoic acid)		 CH3COOH 1.8 x 10-5
nitrous acid HNO2 5.0 x 10-4
hydrofluoric acid HF 6.7 x 10-4
chlorous acid HClO2 1.1 x 10-2
 

Related AUS-e-TUTE Topics

Equilibrium Constants

Ion Product for Water (Kw)

Base Dissociation Constants (Kb)

KP and Kc

Q vs K : predicting direction of reaction

Solubility Products (Ksp)

Equilibrium Concepts

Le Chatelier's Principle

Chemistry of Blood

Indicators

 
 

Bookmark AUS-e-TUTE

  Bookmark this site!

Bookmark and Share

 
 
© AUS-e-TUTE