# How do you calculate the pH of acetic acid?

Jun 24, 2016

Here's how you can do that.

#### Explanation:

Acetic acid, $\text{CH"_3"COOH}$, is a weak acid, meaning that it partially ionizes in aqueous solution to form hydronium cations, ${\text{H"_3"O}}^{+}$, and acetate anions, ${\text{CH"_3"COO}}^{-}$.

${\text{CH"_ 3"COO"color(red)("H")_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(color(red)(+)) + "CH"_ 3"COO}}_{\left(a q\right)}^{-}$

The position of the ionization equilibrium is given by the acid dissociation constant, ${K}_{a}$, which for acetic acid is equal to

${K}_{a} = 1.8 \cdot {10}^{- 5}$

http://www.bpc.edu/mathscience/chemistry/table_of_monoprotic_acids.html

Now, let's assume that you want to find the pH of a solution of acetic acid that has a concentration of $c$. According to the balanced chemical equation that describes the ionization of the acid, every mole of acetic acid that ionizes will produce

• one mole of hydronium cations
• one mole of acetate anions

If you take $x$ to be the concentration of acetic acid that ionizes, you can find the equilibrium concentration of the hydronium cations by using an ICE table

${\text{CH"_ 3"COO"color(red)("H")_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(color(red)(+)) + "CH"_ 3"COO}}_{\left(a q\right)}^{-}$

color(purple)("I")color(white)(aaaaaaacolor(black)(c)aaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaaacolor(black)((-x))aaaaaaaaaaaaaaaacolor(black)((+x))aaaaaaaacolor(black)((+x))
color(purple)("E")color(white)(aaaaacolor(black)(c-x)aaaaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaaaaacolor(black)(x)

The acid dissociation constant will be equal to

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)$

This will be equivalent to

${K}_{s p} = \frac{x \cdot x}{c - x} = {x}^{2} / \left(c - x\right)$

Now, as long as the initial concentration of the acetic acid, $c$, is significantly higher than the ${K}_{s p}$ of the acid, you can use the approximation

$c - x \approx c \to$ valid when color(red)(ul(color(black)(c " >> " K_(sp)))

In this case, the equation becomes

${K}_{s p} = {x}^{2} / c$

which gives you

$x = \sqrt{c \cdot {K}_{s p}}$

Since $x$ represents the equilibrium concentration of hydronium cations, you will have

$\left[{\text{H"_3"O}}^{+}\right] = \sqrt{c \cdot {K}_{s p}}$

Now, the pH of the solution is given by

color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))

Combine these two equations to get

color(green)(|bar(ul(color(white)(a/a)color(black)("pH" = - log( sqrt(c * K_(sp)))color(white)(a/a)|)))

For example, the pH of a $\text{0.050 M}$ acetic acid solution will be

$\text{pH} = - \log \left(0.050 \cdot 1.8 \cdot {10}^{- 5}\right)$

$\text{pH} = 3.02$