Calculate the ph at the equivalence point for the titration of 0.190 m methylamine (ch3nh2) with 0.190 m hcl. the kb of methylamine is 5.0× 10–4.

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