How does one find [H+] given acid dissociation constant and molarity?

O2NC6H5COOH ---> O2NC6H5COO- + H+
initial concentration
0.020
change
-x. . . . . . . . . . . . . . . . . . . . . . . +x. . . . . . . +x
at equilibrium
0.02-x. . . . . . .. . . . . . . . . . . . . . x. . . . . . .  x

3,4 x 10^-4 = x^2 / 0.020-x
x = [H+] =0.0026 M

pH = - log 0.0026 =2.6

Hey, this is a great question to ask, first off we need to undersand the Ka of this reaction, as it is 10^-4 we can assume that the acid is a weak acid as because this is a negative value the reaction favours the reactants and only partial dissociation of the acid occurs.


So the equation for this reaction is:

C6H4(NO2)COOH > C6H4(NO2)COO^- + H^+

and the Ka value is:

Ka = [H^+] x [C6H4(NO2)COO^-] / [C6H4(NO2)COOH]

We can make a few assumptions based on this equation, Because there are no other sources of salt ions and the amount of H^+ is so small the value of [H^+] = [C6H4(NO2)COO^-]

Therefore:

Ka = [H^+]^2 / [C6H4(NO2)COOH]

Therefore:

3.4 x 10^-4 = [H^+]^2 / 0.02

[H^+] = square root ((3.4 x 10^-4) x 0.02)

[H^+] = 2.6 x 10^-3 mol dm^-3