# How do you calculate pH from acid dissociation constant?

Mar 19, 2014

For a weak acid, set up the equilibrium expression for dissociation to ions in solution, then solve this equation for the hydronium ion concentration. The pH can be calculated directly from $\left[{H}^{+}\right]$.

Example: The pH of 0.2 M acetic acid (HOAc)

HOAC ↔ H^+ + OAc^-
${K}_{a} = 1.8 x {10}^{- 5} = \frac{\left[{H}^{+}\right] \left[O A {c}^{-}\right]}{\left[H O A C\right]}$
If the acid is weak, then only a small concentration, $x$, will dissociate. We can rewrite the equation as
${K}_{a} = 1.8 x {10}^{- 5} = \frac{\left[{H}^{+}\right] \left[O A {c}^{-}\right]}{\left[H O A C\right]}$ = $\frac{{x}^{2}}{0.2 - x}$ & approximately; ${x}^{2} / 0.2$
where we have ignored $x$ in the denominator because it is a small number compared with 0.2. Solving this equation gives * See note below about what to do if you do not want to ignore this...
x = (0.2 · 1.8x10^(-5))^(1/2) = $1.9 x {10}^{- 3}$ = $\left[{H}^{+}\right]$
pH = $- \log \left(\left[{H}^{+}\right]\right)$ = 2.72

NOTE: If you don't simpify 0.2-x = 0.2, you need to use the quadratic equation to solve for the pH. Here is a video which discusses how to do this.