Calculate the ph of a 0.17 m solution of c6h5nh3no3 (kb for c6h5nh2 = 3.8 x 10-10). record your ph value to 2 decimal places.

Mass of the solution = 0.17m
 Kb for C6H5NH2 = 3.8 x 10^-10
 We know Ka for C6H5NH2 = 1.78x10^-11
  We have Kw = Ka x Kb => Ka = Kw / Kb
  => (C2H5NH2)(H3O^+)/(C2H5NH3^+) => 1.78x10^-11 = K^2 / 0.17
 K^2 = 3 x 10^-12 => K = 1.73 x 10^-6.
 pH = -log(Kw(H3O^+)) = -log(1.73 x 10^-6) = 5.76

The formula is (nh4)2co3