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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

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 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)

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