Question

Calculate the pH of a solution of 0.51 M aqueous ammonia and 0.48M ammonium nitrate before and after 0.090 M NaOH is added to the solution. Kb of ammonia = 1.7 x 10 -5

Answer 1

Pkb   = -logKb

         = -log1.7*10^-5

          = 4.7695

POH = Pkb + log[NH4NO3]/[NH3]

           = 4.7695 + log0.48/0.51

             = 4.7695 -0.0263 = 4.7432

By the addition of 0.09 M NaOH

    molarity of NH3 = 0.51+ 0.09 = 0.6M

   molarity of NH4NO3 = 0.48-0.09 = 0.39 M

   POH = PKb + log[NH4NO3]/[NH3]

             = 4.7695 + log0.39/0.6

           = 4.7695-0.187 = 4.5825

PH    = 14-POH

          = 14-4.5825 = 9.4175