millimoles of NH3 = 100 x 1 = 100
millimoles of HCl = 127 x 1= 127
NH3 + HCl ---------------------> NH4Cl
100Â Â Â Â Â Â
127Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0 ----------------------------> initial
0Â Â Â Â Â Â Â Â Â Â Â
27Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
100 ------------------------> after reaction
here strong acid remained in the solution. so pH can be decided
by strong acid
HCl millimoles remained = 27
HCl molarity = millimoles / total volume
                   Â
= 27 / (100+127)
                   Â
= 0.119 M
[H+] = 0.119M
pH = -log[H+]
pH = -log(0.119)
pH = 0.925
