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Acid Dissociation Constants (Ka)

Key Concepts

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Comparing Some Acid Dissociation Constants (25°C)

The value of Ka, the dissociation constant (ionisation constant), for a number of different acids at 25°C is shown below:

acidformulaKa
ammonium ion
NH4+ 5.6 × 10-10 smaller Ka
boric acid H3BO3 5.8 × 10-10
hydrocyanic acid
(hydrogen cyanide)
HCN 6.3 × 10-10
hydrobromous acid
HOBr 2.4 × 10-9
hypochlorous acid HOCl 2.9 × 10-8
propanoic acid
C2H5COOH 1.3 × 10-5
acetic acid
(ethanoic acid)
CH3COOH 1.8 × 10-5
benzoic acid C6H5COOH 6.4 × 10-5
methanoic acid
(formic acid)
HCOOH 1.8 × 10-4
lactic acid
HC3H5O3 1.4 × 10-4
nitrous acid HNO2 7.2 × 10-4
hydrofluoric acid HF 7.6 × 10-4
chlorous acid HClO2 1.1 × 10-2 larger Ka

Compare the acid dissociation constants for methanoic acid (formic acid), HCOOH, and ethanoic acid (acetic acid), CH3COOH:

Ka(HCOOH) = 1.8 × 10-4 (larger Ka)
Ka(CH3COOH) = 1.8 × 10-5 (smaller Ka)

Methanoic acid (formic acid), HCOOH, has a larger Ka value than ethanoic acid (acetic acid) therefore it dissociates more than ethanoic acid (acetic acid).

In aqueous solutions of the same concentration under the same conditions, for example 0.1 mol L-1 methanoic acid and 0.1 mol L-1 ethanoic acid, there will be:

Learn more about the properties of carboxylic acids at carboxylic acid tutorial

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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 25°C.

  1. Write the acid dissociation equation:
    HCl H+(aq) + Cl-(aq)

    Because hydrochloric acid, HCl(aq), is a strong monoprotic acid, the value for its acid dissociation constant, Ka, would be extremely large.
    Infact, it is so large that we usually consider this to be a reaction that goes to completion rather than as a reaction in equilibrium:
    HCl → H+(aq) + Cl-(aq)

  2. Calculate the initial and equilibrium concentrations of the species present using a R.I.C.E. Table:
    HCl(aq) is a strong acid, it completely dissociates to form H+ and Cl- so 0.10 moles of HCl in 1.0 L of water will completely dissociate into 0.10 moles H+(aq) and 0.10 moles Cl-(aq):
    R.I.C.E. Table
      reactant   products
    Reaction: HCl H+(aq) + Cl-(aq)
    Initial concentrations (mol L-1) 0.10   0   0
    Change in concentration (mol L-1) -0.10   +0.10   +0.10
    Equilibrium concentrations (mol L-1) 0.10 - 0.10
    = 0
      0 + 0.10
    = 0.10
      0 + 0.10
    = 0.10

    From the R.I.C.E. Table, at completion (equilibrium):
    [H+] = 0.10 mol L-1
  3. Calculate pH of this solution at equilibrium:
    pH = -log10[H+]
    pH = -log10[0.10] = 1
  4. Calculate %dissociation:
    [H+] = 0.1 mol L-1 (at completion)

    [HCl initial] = 0.1 mol L-1

    %dissociation =     [H+]    
    [HCl initial]
    × 100
    %dissociation = 0.1
    0.1
    × 100
    %dissociation = 100%  

    % dissocation for a strong monoprotic acid is 100%.

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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 × 10-4 at 25°C

  1. Write the acid dissociation equation:
    HNO2(aq) H+(aq) + NO2-(aq)
  2. Write the equilibrium contstant expression for the acid dissociation:
    Ka = [H+][NO2-]
    [HNO2]
    5.0 × 10-4 = [H+][NO2-]
    [HNO2]
  3. Calculate the initial and equilibrium concentrations of the species present using a R.I.C.E. Table:
    let x = moles of HNO2 that dissociate to form H+(aq) and NO2-(aq)
    R.I.C.E. Table
      reactant   products
    Reaction: HNO2 H+ + NO2-
    Initial concentrations (mol L-1) 0.10   0   0
    Change in concentrations (mol L-1) -x   +x   +x
    Equilibrium concentrations (mol L-1) 0.10 - x   0 + x
    =x
      0 + 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] ≈ 0.10 mol L-1

  4. Substitute the equilibrium concentration values into the expression for the acid dissociation constant and solve:
    Equilibrium Law
    Expression:
    Ka = [H+][NO2-]
    [HNO2]
     
    Substitute
    known values:
    5.0 × 10-4 = [x][x]
    [0.10]
     
    Collect
    like terms:
    5.0 × 10-4 =   x2  
    0.10
    Multiply both
    sides by 0.10
    5.0 × 10-4 × 0.10 =   x2  
    0.10
    × 0.10
    5.0 × 10-4 × 0.10 = x2  
    5.0 × 10-5 = x2  
    Take square root
    of both sides
    √5.0 × 10-5 = x2  
    7.1 × 10-3 = x  

    From the R.I.C.E. Table above, at equilibrium
    [H+] = x = 7.1 × 10-3 mol L-1

  5. Calculate pH at equilibrium:
    pH = -log10[H+]

    pH = -log10[7.1 × 10-3] = 2.1

  6. Calculate %dissociation at equilibrium:
    [H+] at equilibrium = 7.1 × 10-3 mol L-1

    [HNO2 initial] = 0.10 mol L-1

    %dissociation =     [H+]    
    [acid initial]
    × 100
    %dissociation = [7.1 × 10-3]
    [0.10]
    × 100
    %dissociation = 7.1 %  

    % dissociation for a weak acid is less than 100%, that is, the equilibrium mixture contains some undissociated molecules of acid as well as cations and anions.

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

Calculate the [OH-] and pOH for an aqueous solution of 0.5 mol L-1 acetic acid (ethanoic acid).
Ka = 1.8 × 10-5 at 25°C.

  1. Write the acid dissociation equation:
    CH3COOH(aq) H+(aq) + CH3COO-(aq)
  2. Write the equilibrium law expression for the acid dissociation:
    Ka = [H+][CH3COO-]
    [CH3COOH]
    1.8 × 10-5 = [H+][CH3COO-]
    [CH3COOH]
  3. Calculate the initial and equilibrium concentrations of the species present using a R.I.C.E. Table:
    let x = moles of CH3COOH that dissociate to form H+(aq) and CH3COO-(aq)

    R.I.C.E. Table
      reactant   products
    Reaction: CH3COOH H+ + CH3COO-
    Initial concentrations (mol L-1) 0.5   0   0
    Change in concentrations (mol L-1) -x   +x   +x
    Equilibrium concentrations (mol L-1) 0.5 - x   0 + x
    = x
      0 + 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] ≈ 0.5 mol L-1

  4. Substitute the concentration values into the expression for the acid dissociation constant and solve:
    Equilibrium Law
    Expression:
    Ka = [H+][CH3COO-]
    [CH3COOH]
     
    Substitute
    known values:
    1.8 × 10-5 = [x][x]
    [0.5]
     
    Collect
    like terms:
    1.8 × 10-5 =   [x]2  
    [0.5]
     
    Multiply both
    sides by 0.5:
    1.8 × 10-5 × 0.5 =   x2  
    0.5
    × 0.5
    9.0 × 10-6 = x2  
    Take square root
    of both sides:
    √9.0 × 10-6 = x2  
    3.0 × 10-3 = x  

    [H+] = x = 3.0 × 10-3 mol L-1

  5. Calculate the concentration of OH- in the aqueous solution of acetic acid at 25°C:
    At 25°C, Kw, the equilibrium constant for the dissociation of water, is 10-14

    Equilibrium Law
    Expression:
    [H+][OH-] = Kw
    Substitute in
    known values:
    [3.0 × 10-3][OH-] = 10-14
    Divide both sides
    by 3.0 × 10-3:
    [3.0 × 10-3][OH-]
    3.0 × 10-3
    =     10-14    
    3.0 × 10-3
    [OH-] = 3.3 × 10-12

    [OH-] = 3.3 × 10-12 mol L-1

  6. Calculate pOH:
    pOH = -log10[OH-]

    pOH = -log10[3.3 × 10-12] = 11.5

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