# How do you solve for Ka when you only have molarity of the acid and pH?

Jun 1, 2015

You start by using the pH of the solution to determine the concentration of the hydronium ions, ${H}_{3} {O}^{+}$.

$\left[{H}_{3} {O}^{+}\right] = {10}^{- P {H}_{\text{sol}}}$

The general dissociation equation for a weak acid looks like this

$H {A}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + {A}_{\left(a q\right)}^{-}$

By definition, the acid dissociation constant, ${K}_{a}$, will be equal to

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[{A}^{-}\right]}{\left[H A\right]}$

If you have a $1 : 1$ mole ratio between the acid and the hydronium ions, and between the hydronium ions and the conjugate base, ${A}^{-}$, then the concentration of the latter will be equal to that of the hydronium ions.

$\left[{A}^{-}\right] = \left[{H}_{3} {O}^{+}\right]$

Since you know the molarity of the acid, ${K}_{a}$ will be

${K}_{a} = \frac{{\left[{H}_{3} {O}^{+}\right]}^{2}}{\left[H A\right]}$