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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.

Definitions

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

$pH = -log[H_3O^+]$

or

$pH = -log[H^+]$

$$K_a$$, 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.

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

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

So,

$x = [H_3O^+]$

$pH = -log[H_3O^+]$

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)
20:01, 25 Jan 2014

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