The ph of 0.001 of molar solution of benzen carborhylic acid is 3.59. Calculate the dissociation constant of the acid.
I assume you are talking about benzoic acid.
Let's call this acid HBz
HBz ==> H^+ + Bz^-
Ka = (H^+)(Bz^-)/(HBz)
pH = 3.59 = -log (H^+).
Solve for (H^+) and I get approximately 2.6E-4 but you need to do it more accurately than that.
(H^+) = from above.
(Bz^-) = (H^+)
(HBz) = 0.001-(H^+) = ??
Solve for Ka.