Question

How many moles of HCl must be added to 1.0 L of 1.0 M NH3(aq) to make a buffer with a pH of 9.00? (pKa of NH4+ =9.25)

the answer is 0.64, explain please.

Answer 1

                        NH3 + HCl -------------> NNH4Cl

           I            1          x                          0

         C           - x         -x                         +x

           E        1- x        0                           +x

         POH = PKb + log[NH4+]/[NH3]

           PKb = 14-Pka

                      = 14-9.25 = 4.75

         POH = 14-PH

                  = 14-9 = 5

   POH = PKb + log[NH4+]/[NH3]

5         = 4.75 + logx/1-x

logx/1-x   = 5-4.75

logx/1-x    = 0.25

   x/1-x         = 10^0.25

   x/1-x        = 1.7782

x              = (1-x)*1.7782

x   = 0.64

no of moles of HCl = 0.64 moles