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# Calculating a Ka Value from a Known pH

##### Table of Contents

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.

### Definitions

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

$pH = -log[H_3O^+]$

or

$pH = -log[H^+]$

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

• For a chemical equation of the form $$HA + H_2O \leftrightharpoons H_3O^+ + A^-$$, the equation for Ka is the following:

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

• 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 is therefore ignored in the equilibrium constant equation.)

### Solving for Ka

When given the pH value of a solution, solving for Ka requires the following steps:

1. Set up an ICE table for the chemical reaction.
2. Solve for the concentration of H3O+ (denoted x) using the equation for pH:
$[H_3O^+] = 10^{-pH}$
3. Use the concentration of H3O+ to solve for the concentrations of the other products and reactants.
4. Plug all concentrations into the equation for Ka and solve.
Example 1

Calculate the Ka value of a 0.2 M aqueous solution of propionic acid, CH3CH2CO2H, with a pH of 4.88.

$CH_3CH_2CO_2H + H_2O \leftrightharpoons H_3O^+ + CH_3CH_2CO_2^-$

SOLUTION

ICE TABLE

 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

According to the definition of pH,

$x = [H_3O^+]$

$pH = -log[H_3O^+]$

Thus,

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

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

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

$K_a = 8.69 \times 10^{-10}$

### References

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

### Contributors

• Paige Norberg, Gabriela Mastro (UCD)
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Last modified
12:59, 16 Jan 2015

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This material is based upon work supported by the National Science Foundation under Grant Numbers 1246120, 1525057, and 1413739.

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