HCl(aq)+CH3NO2(aq)⇄CH3NH3+(aq)+CL-(aq)
Given that the concentrations of the HCl and methyamine are equal, so at the equivalence point the volume is doubled. The resulting concentration of CH3NH3+ is 0.095M
CH3NH3+(aq)⇄H+(aq)+CH3NH2(aq)
I 0.095 0 0
C -x +x +x
E 0.095-x x x
Ka=x^2/0.095-x
Ka=1.0x10^-14/5.0x10^-4
=2.0x10^-11
x^2/0.095-x=2.0x10^-11
x=1.38x10^-6
pH=-log(1.38x10^-6)
=5.9
