If you like us, please share us on social media.
The latest UCD Hyperlibrary newsletter is now complete, check it out.

ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Acids and Bases > Ionization Constants > Calculating a Ka Value from a Known pH

Copyright (c) 2006-2014 MindTouch Inc.

This file and accompanying files are licensed under the MindTouch Master Subscription Agreement (MSA).

At any time, you shall not, directly or indirectly: (i) sublicense, resell, rent, lease, distribute, market, commercialize or otherwise transfer rights or usage to: (a) the Software, (b) any modified version or derivative work of the Software created by you or for you, or (c) MindTouch Open Source (which includes all non-supported versions of MindTouch-developed software), for any purpose including timesharing or service bureau purposes; (ii) remove or alter any copyright, trademark or proprietary notice in the Software; (iii) transfer, use or export the Software in violation of any applicable laws or regulations of any government or governmental agency; (iv) use or run on any of your hardware, or have deployed for use, any production version of MindTouch Open Source; (v) use any of the Support Services, Error corrections, Updates or Upgrades, for the MindTouch Open Source software or for any Server for which Support Services are not then purchased as provided hereunder; or (vi) reverse engineer, decompile or modify any encrypted or encoded portion of the Software.

A complete copy of the MSA is available at http://www.mindtouch.com/msa

Calculating a Ka Value from a Known pH

pH, or the "power of hydrogen," is a numerical representation of the acidity or basicity of a solution. It can be used to approximate the concentration of hydrogen ions [H+] or hydronium ions [H3O+] in an aqueous solution. Solutions with low pHs are the most acidic, and high pH's being the most basic.


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

\[  pH = -log[H_3O^+] \]


\[ pH = -log[H^+] \]

\( K_a \), the acid ionization constant, is the equilibrium constant of 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 \( K_a \) values indicates stronger acids (more of the acid dissociates) and small \( K_a \) values indicates weaker acids (the reaction does not go to completion). 

  • For a chemical equation of the form \( HA + H_2O \leftrightharpoons H_3O^+ + A^- \), the equation for \( K_a \) is:

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

  • \( HA \) is the undissociated acid and \( A^- \) is the conjugate base of the acid. (H2O, a pure liquid, has an activity equal to one, and therefore ignored in the equilibrium constant equation.)

Solving for Ka

When given the pH value of a solution, solving for \( K_a \) is fairly easy:

  1. Set up an ICE Table for the chemical reaction.
  2. Although in the ICE table, the concentration of H3O+ is given as \( x \), we can easily find the value of x. The equation for pH can be used to solve for the concentration of H3O+.
    \[ [H_3O^+] = 10^{-pH} \]
  3. Since the concentration of H3O+ is denoted as \( x \) in the ICE Table, the value of the concentration can be plugged in for \( x \) to solve for the concentrations of the other products and reactants.
  4. Now that the values of the concentrations are all known, you can plug them into the equation for \( K_a \) and solve.
Example 1

Calculate the \( K_a \) value of a 0.2 M aqueous solution of propionic acid, CH3CH2CO2H, with a pH of 4.88.

\[ CH_3CH_2CO_2H + H_2O \leftrightharpoons H_3O^+ + CH_3CH_2CO_2^- \]



ICE \[ CH_3CH_2CO_2H \] \[ H_3O^+ \] \[ CH_3CH_2CO_2^- \]
Initial Concentration 0.2 M 0 M 0 M
Change in Concentration  -x M +x M +x M
Equilibrium Concentration  (0.2 - x) M x M x M


\[ x = [H_3O^+] \]

\[ pH = -log[H_3O^+] \]


\[ log[H_3O^+] = -pH = -4.88 \]

\[ [H_3O^+] = 10^{-4.88} = 1.32 \times 10^{-5} = x \]

\[ K_a = \dfrac{[H_3O^+][CH_3CH_2CO_2^-]}{[CH_3CH_2CO_2H]} = \dfrac{x^2}{0.2 - x} = \dfrac{(1.32 \times 10^{-5})^2}{0.2 - 1.32 \times 10^{-5}} \]

\[ K_a = 8.69 \times 10^{-10} \]

Related Articles


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


  • Paige Norberg, Gabriela Mastro (UCD)
You must to post a comment.
Last Modified
20:01, 25 Jan 2014



(not set)
(not set)

Creative Commons License Unless otherwise noted, content in the UC Davis ChemWiki is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License. Permissions beyond the scope of this license may be available at copyright@ucdavis.edu. Questions and concerns can be directed toward Prof. Delmar Larsen (dlarsen@ucdavis.edu), Founder and Director. Terms of Use