no of moles of HCl = molarity * volume in L
                              Â
= 0.1*0.1 = 0.01 moles
no of moles of NH3 = 0.1*0.05 = 0.005moles
              Â
HCl + NH3 ---------> NH4Cl
IÂ Â Â Â Â Â Â Â Â Â Â Â
0.01Â Â
0.005Â Â Â Â Â Â Â Â Â Â Â Â
0
c          Â
-0.005 -0.005Â Â Â Â Â Â Â Â
0.005
EÂ Â Â Â Â Â Â Â Â Â
0.005Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0.005
[H+] Â Â Â Â = no of moles of Hcl at
equilibrium /total volume in L
           Â
= 0.005/0.15 = 0.0333M
PHÂ Â Â Â Â = -log[H+]
          Â
= -log0.0333Â Â = 1.4775
    Â
