Question

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

Answer 1

You start by using the pH of the solution to determine the concentration of the hydronium ions, `H_3O^(+)`.

`[H_3O^(+)] = 10^(-PH_"sol")`

The general dissociation equation for a weak acid looks like this

`HA_((aq)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + A_((aq))^(-)`

By definition, the acid dissociation constant, `K_a`, will be equal to

`K_a = ([H_3O^(+)] * [A^(-)])/([HA])`

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.

`[A^(-)] = [H_3O^(+)]`

Since you know the molarity of the acid, `K_a` will be

`K_a = ([H_3O^(+)]^2)/([HA])`