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What is the Ka of propionic acid?

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Posted by: Ana E. 9 months ago
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  • Answered by:
  • Date: 19.04.2018

The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5. 

[ – ]

The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5. 

  • Answered by: Syed A.
  • From: Abu Dhabi
  • Date: 19.04.2018

1.34 × 10-5

[ – ]

1.34 × 10-5

  • Answered by: Lamis T.
  • From: Abu Dhabi
  • Date: 03.04.2018
The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5 [ – ]
The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5

  • Answered by: Vinay
  • From: Dubai
  • Date: 26.03.2018

The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5

[ – ]

The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5

  • Answered by: Mayanka
  • From: Dubai
  • Date: 21.03.2018
The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5. Calculate the pH of the solution and the con… The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5. Calculate the pH of the solution and the concentrations of C2H5COOH and C2H5COO– in a 0.643 M propanoic acid solution at equilibrium. [ – ]
The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5. Calculate the pH of the solution and the con… The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5. Calculate the pH of the solution and the concentrations of C2H5COOH and C2H5COO– in a 0.643 M propanoic acid solution at equilibrium.

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Acid Dissociation Constant Definition: Ka

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Acid Dissociation Constant Definition: Ka

What Is an Acid Dissociation Constant, or Ka in Chemistry?

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The acid dissociation constant or Ka is an equilibrium constant that can be used to gauge the strength of an acid in a solution.

The acid dissociation constant or Ka is an equilibrium constant that can be used to gauge the strength of an acid in a solution.
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Anne Marie Helmenstine, Ph.D.

Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels.

Updated February 19, 2018

The acid dissociation constant is the equilibrium constant of the dissociation reaction of an acid and is denoted by Ka. This equilibrium constant is a quantitative measure of the strength of an acid in a solution. Ka is commonly expressed in units of mol/L. There are tables of acid dissociation constants , for easy reference. For an aqueous solution, the general form of the equilibrium reaction is:

HA + H2O ⇆ A + H3O+

where HA is an acid which dissociates in the conjugate base of the acid A and a hydrogen ion that combines with water to form the hydronium ion H3O+. When the concentrations of HA, A, and H3O+ no longer change over time, the reaction is at equilibrium and the dissociation constant may be calculated:

Ka = [A][H3O+] / [HA][H2O]

where the square brackets indicate concentration. Unless an acid is extremely concentrated, the equation is simplified by holding the concentration of water as a constant:

HA ⇆ A + H+

Ka = [A][H+]/[HA]

The acid dissociation constant is also known as the acidity constant or acid-ionization constant.

Relating Ka and pKa

A related value is pKa, which is the logarithmic acid dissociation constant:

pKa = -log10Ka

Using Ka and pKa To Predict Equilibrium and Strength of Acids

Ka may be used to measure the position of equilibrium:

  • If Ka is large, the formation of the products of the dissociation is favored.
  • If Ka is small, the undissolved acid is favored.

Ka may be used to predict the strength of an acid :

  • If Ka is large (pKa is small) this means the acid is mostly dissociated, so the acid is strong. Acids with a pKa less than around -2 are strong acids.
  • If Ka is small (pKa is large), little dissociation has occurred, so the acid is weak. Acids with a pKa in the range of -2 to 12 in water are weak acids.

    Ka is a better measure of the strength of an acid than pH because adding water to an acid solution doesn’t change its acid equilibrium constant, but does alter the H+ ion concentration and pH.

    Ka Example

    The acid dissociation constant, Ka of the  acid  HB is:

    HB(aq) ↔ H+(aq) + B(aq)

    Ka = [H+][B] / [HB]

    For the dissociation of ethanoic acid:

    CH3COOH(aq) + H2O(l) = CH3COO(aq) + H3O+(aq)

    Ka = [CH3COO(aq)][H3O+(aq)] / [CH3COOH(aq)]

    Acid Dissociation Constant From pH

    The acid dissociation constant may be found it the pH is known. For example:

    Calculate the acid dissociation constant Ka for a 0.2 M aqueous solution of propionic acid (CH3CH2CO2H) that is found to have a pH value of 4.88.

    To solve the problem, first write the chemical equation for the reaction. You should be able to recognize propionic acid is a weak acid (because it’s not one of the strong acids and it contains hydrogen). It’s dissociation in water is:

    CH3CH2CO2H + H2 ⇆ H3O+ + CH3CH2CO2

    Set up a table to keep track of the initial conditions, change in conditions, and equilibrium concentration of the species. This is sometimes called an ICE table:

    Initial Concentration0.2 M0 M0 M
    Change in Concentration-x M+x M+x M
    Equilibrium Concentration(0.2 – x) Mx Mx M

    x = [H3O+

    Now use the pH formula :

    pH = -log[H3O+]

    -pH = log[H3O+] = 4.88

    [H3O+ = 10-4.88 = 1.32 x 10-5

    Plug in this value for x for solve for Ka:

    Ka = [H3O+][CH3CH2CO2] / [CH3CH2CO2H]

    Ka = x2 / (0.2 – x)

    Ka = (1.32 x 10-5)2 / (0.2 – 1.32 x 10-5)

    Ka = 8.69 x 10-10

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    Chemistry LibreTexts

    Calculating a Ka Value from a Known pH

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    Thu, 22 Mar 2018 15:18:46 GMT
    Calculating a Ka Value from a Known pH
    [ "article:topic", "pH", "Ionization Constants", "showtoc:no" ]
  • Page ID
  • The quantity pH, or "power of hydrogen," is a numerical representation of the acidity or basicity of a solution. It can be used to calculate the concentration of hydrogen ions [H+] or hydronium ions [H3O+] in an aqueous solution. Solutions with low pH are the most acidic, and solutions with high pH are most basic.


    Although pH is formally defined in terms of activities , it is often estimated using free proton or hydronium concentration:

    \[ pH \approx -\log[H_3O^+] \labeleq1\]


    \[ pH \approx -\log[H^+] \labeleq2\]

    \(K_a\), the acid ionization constant , is the equilibrium constant for chemical reactions involving weak acids in aqueous solution. The numerical value of \(K_a\) is used to predict the extent of acid dissociation. A large \(K_a\) value indicates a stronger acid (more of the acid dissociates) and small \(K_a\) value indicates a weaker acid (less of the acid dissociates). 

    For a chemical equation of the form

    \[ HA + H_2O \leftrightharpoons H_3O^+ + A^- \]

    \(K_a\) is express as 

    \[ K_a = \dfrac[H_3O^+][A^-][HA] \labeleq3 \]


    • \(HA\) is the undissociated acid and
    • \(A^-\) is the conjugate base of the acid.

    Since \(H_2O\) is a pure liquid, it has an activity equal to one and is ignored in the equilibrium constant expression in (Equation \refeq3) like in other equilibrium constants.

    Howto: Solving for \(K_a\)

    When given the pH value of a solution, solving for \(K_a\) requires the following steps:

    1. Set up an ICE table for the chemical reaction.
    2. Solve for the concentration of \(\ceH3O^+\) using the equation for pH: \[ [H_3O^+] = 10^-pH \]
    3. Use the concentration of \(\ceH3O^+\) to solve for the concentrations of the other products and reactants.
    4. Plug all concentrations into the equation for \(K_a\) and solve.

      Example \(\PageIndex1\)

      Calculate the \(K_a\) value of a 0.2 M aqueous solution of propionic acid (\(\ceCH3CH2CO2H\)) with a pH of 4.88.

      \[ \ceCH_3CH_2CO_2H + H_2O \leftrightharpoons H_3O^+ + CH_3CH_2CO_2^- \nonumber\]



      ICE\(\ce CH_3CH_2CO_2H \)\(\ce H_3O^+ \)\(\ce CH_3CH_2CO_2^- \)
      Initial Concentration (M)0.200
      Change in Concentration (M)-x+x+x
      Equilibrium Concentration (M)0.2 – xxx

      According to the definition of pH (Equation \refeq1) 

      \[\beginalign* -pH = \log[H_3O^+] &= -4.88 \\[5pt]  [H_3O^+] &= 10^-4.88 \\[5pt] &= 1.32 \times 10^-5 \\[5pt] &= x \endalign*\]

      According to the definition of \(K_a\) (Equation \refeq3 

      \[\beginalign* K_a &= \dfrac[H_3O^+][CH_3CH_2CO_2^-][CH_3CH_2CO_2H] \\[5pt] &= \dfracx^20.2 – x  \\[5pt] &= \dfrac(1.32 \times 10^-5)^20.2 – 1.32 \times 10^-5  \\[5pt] &= 8.69 \times 10^-10 \endalign*\]


      1. Petrucci,et al. General Chemistry:Principles & Modern Applications; Ninth Edition, Pearson/Prentice Hall; Upper Saddle River, New Jersey 07.


      • Paige Norberg (UCD) and Gabriela Mastro (UCD)