Calculating A Ka Value From A Known PH - Chemistry LibreTexts

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}

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