# Calculating a Ka Value from a Known pH

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:

$pH \approx -\log[H_3O^+] \label{eq1}$

or

$pH \approx -\log[H^+] \label{eq2}$

$$K_a$$, 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

$K_a = \dfrac{[H_3O^+][A^-]}{[HA]} \label{eq3}$

where

• $$HA$$ 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 \ref{eq3}) 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 $$\ce{H3O^{+}}$$ using the equation for pH: $[H_3O^+] = 10^{-pH}$
3. Use the concentration of $$\ce{H3O^{+}}$$ to solve for the concentrations of the other products and reactants.
4. Plug all concentrations into the equation for $$K_a$$ and solve.

Example $$\PageIndex{1}$$

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

$\ce{CH_3CH_2CO_2H + H_2O \leftrightharpoons H_3O^+ + CH_3CH_2CO_2^- } \nonumber$

Solution

ICE TABLE

ICE $$\ce{ CH_3CH_2CO_2H }$$ $$\ce{ H_3O^+ }$$ $$\ce{ CH_3CH_2CO_2^- }$$
Initial Concentration (M) 0.2 0 0
Change in Concentration (M) -x +x +x
Equilibrium Concentration (M) 0.2 - x x x

According to the definition of pH (Equation \ref{eq1})

\begin{align*} -pH = \log[H_3O^+] &= -4.88 \\[4pt] [H_3O^+] &= 10^{-4.88} \\[4pt] &= 1.32 \times 10^{-5} \\[4pt] &= x \end{align*}

According to the definition of $$K_a$$ (Equation \ref{eq3}

\begin{align*} K_a &= \dfrac{[H_3O^+][CH_3CH_2CO_2^-]}{[CH_3CH_2CO_2H]} \\[4pt] &= \dfrac{x^2}{0.2 - x} \\[4pt] &= \dfrac{(1.32 \times 10^{-5})^2}{0.2 - 1.32 \times 10^{-5}} \\[4pt] &= 8.69 \times 10^{-10} \end{align*}

## References

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