Question

How do you calculate the$\mathrm{pH}$ of acetic acid? Give an example

Answer 1

$\mathrm{pH}$of acetic acid:

Step 1: Ionisation of Acetic acid:

Acetic acid,( ${\mathrm{CH}}_3\mathrm{COOH}$) is known as a weak acid, so it partially ionizes in an aqueous solution to form hydronium ions, ${\mathrm{H}}_3{\mathrm{O}}^{+}$, and acetate ion(${\mathrm{CH}}_3{\mathrm{COO}}^{-}$), which are given by;

${\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CH}}_3{\mathrm{COO}}^{-}\left(\mathrm{aq}\right)$

Step 2: calculating the value of ${\mathrm{K}}_a$:

The position of the ionization equilibrium is given by the acid dissociation constant for the acid, which for acetic acid is equal to ${\mathrm{K}}_{\mathrm{a}}=1.8\times {10}^{-5}$

let's assume that we want to find the $\mathrm{pH}$ of a solution of acetic acid that has a concentration of $\mathrm{c}$.

According to the balanced chemical equation describing acid ionization, each mole of acetic acid that ionizes will produce one mole of Hydronium cations and one mole of acetate anions.

Let's take $\mathrm{x}$ to be the concentration of acetic acid that ionizes, hence we can find the equilibrium concentration of the hydronium cations as follows;
$\underset{}{\underset{}{\underset{}{{\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)}}}+\underset{}{\mathrm{H}2\mathrm{O}(\mathrm{l}})\rightleftharpoons \underset{}{\underset{}{\underset{}{{\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right)}}}+\underset{}{\underset{}{\underset{}{{\mathrm{CH}}_3{\mathrm{COO}}^{-}\left(\mathrm{aq}\right)}}}$

	${\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)$	${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$	${\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right)$	${\mathrm{CH}}_3{\mathrm{COO}}^{-}\left(\mathrm{aq}\right)$
Initial concentration	$\mathrm{c}$	$0$	$0$	$0$
Change in concentration	$-\mathrm{x}$	$0$	$+\mathrm{x}$	$+\mathrm{x}$
Equilibrium concentration	$\mathrm{c}-\mathrm{x}$	$0$	$\mathrm{x}$	$\mathrm{x}$

The acid dissociation constant will be equal to:

$\mathrm{Ka}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\cdot \left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

This will be equivalent to;

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{x}\cdot \mathrm{x}}{\mathrm{c}-\mathrm{x}} \\ =\frac{{\mathrm{x}}^2}{\mathrm{c}-\mathrm{x}} \end{gathered}$$

Step 3: calculating the value of $\left[{\mathrm{H}}^3{\mathrm{O}}^{+}\right]$ :

Now, as long as the initial concentration of the acetic acid, c, is significantly higher than the Ksp of the acid, we can use the approximation

$\mathrm{c}-\mathrm{x}\approx \mathrm{c}\to$ valid when $\mathrm{c}>>{\mathrm{K}}_{\mathrm{a}}$

Hence, the equation becomes;

${\mathrm{K}}_{\mathrm{sp}}=\frac{{\mathrm{x}}^2}{\mathrm{c}}$,

$\Rightarrow \mathrm{x}=\sqrt{\mathrm{c}\cdot {\mathrm{K}}_{\mathrm{a}}}$

Since $\mathrm{x}$ represents the equilibrium concentration of hydronium cations, you will have;

$\left[{\mathrm{H}}^3{\mathrm{O}}^{+}\right]=\sqrt{\mathrm{c}\cdot {\mathrm{K}}_{\mathrm{a}}}$

Step 4: Calculating the value of ${\mathrm{p}}^{\mathrm{H}}:$

Now, the pH of the solution is given by;

${\mathrm{p}}^{\mathrm{H}}=-\log \left(\right[{\mathrm{H}}_3{\mathrm{O}}^{+}\left]\right)$

Combine these two equations to get;

${\mathrm{p}}^{\mathrm{H}}=-\log \left(\sqrt{\mathrm{c}\cdot {\mathrm{K}}_{\mathrm{a}}}\right)$

Example:

The $\mathrm{pH}\mathrm{of}\mathrm{a}0.050\mathrm{M}$ acetic acid solution will be;

$$\begin{gathered} \mathrm{pH}=-\log (0.050\cdot 1.8\cdot {10}^{-5}) \\ \mathrm{pH}=3.02 \end{gathered}$$

Hence, using the above-mentioned way we can calculate the $\mathrm{pH}$ of acetic acid.
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