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

1 Answer
Stefan V.
Jun 1, 2015

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])#

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