How do you calculate the $pH$ of acetic acid?

Acetic acid, $C{H}_{3}COOH$, is a weak acid, meaning that it partially ionizes in aqueous solution to form hydronium cations, $H_3{O}^{+}$, and acetate anions, $C{H}_{3}CO{O}^{-}$

$C{H}_{3}COO{H}_{\left(aq\right)}+H_2{O}_{\left(l\right)}\rightleftharpoons +C{H}_{3}CO{O}_{\left(aq\right)}^{-}$

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

Now, let's assume that you want to find the $pH$ of a solution of acetic and that has a concentration of $c$. According to the balanced chemical equation that describes the ionization of the acid, every mole acetic acid that ionizes will produce 1 mole of $H^{+}$ ion.
$C{H}_{3}COO{H}_{\left(aq\right)}+H_{2}O\rightleftharpoons H_3{O}_{\left(aq\right)}^{+}+C{H}_{3}CO{O}_{\left(aq\right)}^{-}$
$1$ $0$ $0$

At Equilibrium :

$c-x$ $x$ $x$

The acid dissociation constant will be equal to
$K_a=\frac{\left[H_3{O}^{+}\right]\dot{}\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$

This will be equivalent to
$K_{sp}=\frac{x\dot{}x}{c-x}=\frac{x^2}{c-x}$

Now, as long as the initial concentration of the acetic acid, $c$, is significantly higher than the $K_{sp}$ of the acid, you can use the approximation

$c-x\approx c\to$ valid when $c>>K_{sp}$

In this case, the equation becomes

$K_{sp}=\frac{x^2}{c}$

which gives you

$x=\sqrt{c\times K_{sp}}$

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

$\left[H_3{O}^{+}\right]=\sqrt{c\times K_{sp}}$

Now, the $pH$ of the solution is given by

$pH=-\log \left(\right[H_3{O}^{+}\left]\right)$

Combine these two equations to get

$pH=-\log \left(\sqrt{c\times K_{sp}}\right)$