how do u find the Ka of a weak acid when ur given the concentration of the acid and the pH???

Answer 1

You don't define if it is polyprotic or monoprotic.
So for any acid of the form HnA where n=1,2,.. you have

.. .. .. .. .. .. HnA <=> nH(+) + A(-n)
Initial .. .. .. .. C
Dissociate .. x
Produce .. .. .. .. .. .. .. nx.. .. .. x
At equilib.. C-x .. .. .. ..nx .. .. .. x

So [H+]=nx (1)
pH=-log[H+] => [H+]= 10^-pH (2)
From equations (1), (2)=> nx= 10^-pH => x=(10^-pH)/n (3)

Ka= ([H+]^n)*[A(-n)]/[HnA] substitute [H+] from (2)
Ka= ((10^-pH)^n)*x / (C-x) (4) substitute x from (3)

Ka=((10^-pH)^n)* ((10^-pH)/n)/ (C-((10^-pH)/n))

or just calculate x and substitute in (4) so that you don't have that very long equation

Answer 2

Start with the equilibrium - where HA is the weak acid:

HA + H2O <=> A- + H3O+

If [H3O+] = x then, as you can see: [A-] = [H3O+] = x

But you know x = [H3O+] = 10^(-pH). So:

x = [A-] = [H3O+] = 10^(-pH)

Usually [H3O+] is very small so:

[HA] - x equals [HA] = C

Ka = [A-]*[H3O+]/HA, Ka = x^2/C

For example, if [HA] = c = 0.1 and pH = 3 then

[A-] = [H3O+] = 10^(-3), [HA] = 0.1 - 10^(-3) approx. = 0.1, so:

Ka = 10^(-6)/0.1 = 10^(-5)

Answer 3

Ka = [H+][A-]/[HA]