molarity of NH4Cl  = W*1000/G.M.Wt * volume of
solution in ml
                            Â
= 1.07*1000/53.5*250
                             Â
= 0.08M
molarity of NH3Â Â Â Â Â = W*1000/G.M.Wt *
volume of solution in ml
                             Â
= 0.17*1000/17*250
                            Â
= 0.04 M
POHÂ Â = PKb + log[NH4Cl]/[NH3]
           Â
= 4.75 + log0.08/0.04
            Â
= 4.75 + 0.3010
            Â
= 5.051
PHÂ Â Â Â Â Â = 14-POH
             Â
= 14-5.051
              Â
= 8.949
