A 0.0100 M solution of a weak monoprotic acid is found to be 3.5%. what is the pH of solution and Ka of acid.?

Answer 1

I guess you mean that the dissociation degree is 3.5%

.. .. .. .. .. .. HA <=> H+ + A-
Initial .. .. .. C
Dissoc. .. .. x
Produce .. .. .. .. .. ..x .. .. .x
At Equil. .. C-x .. .. ..x .. .. x

Ka= [H+][A-] /[HA] = x^2/(C-x)

but the degree of dissociation is
a= dissociated/ initial = x/C =>
x=a*C= 0.035*0.01 =3.5* 10^-4
but [H+]=x
thus pH=-logx= -log(3.5*10^-4) = 3.46

Ka= ((3.45*10^-4)^2) / (0.01-3.5*10^-4) = 1.23*10^-5