Ka of NH4+ = 5.6 x 10^-10
Ka x Kb = Kw = 1.0 x 10^-14
Kb of NH4+ = 1.0 x 10^-14 / 5.6 x 10^-10
                  Â
= 1.78 x 10^-5
pKb = -log Kb
      = -log (1.78 x 10^-5)
     = 4.75
NH3 molarity (base)= 0.235 / 0.5
                Â
= 0.47 M
NH4Cl molarity (salt)= 2.46x10^-1 / 0.5
                    Â
= 0.492 M
pOH = pKb + log (salt / base)
       = 4.75 + log [0.492 /
0.47]
       = 4.77
pH + pOH = 14
pH = 14 - pOH
    = 14 - 4.77
   = 9.23
pH of the buffer = 9.23
