Chemistry please help?

Question

calculate the PH at the eguivalent point in the titration of 50 ml of 0.10 M methylamine with a 0.20 M HCL solution. Ka fo methylamine = 2.3 x 10^-11 thanks

Answer 1

At the equivalence point you have CH3NH3^+1

CH3NH3^+1 + H2O <--> CH3NH2 + H3O+

CH3NH3+ acts as the acid

Ka = [CH3NH2][H3O+] / [CH3NH3+] = 2.3x10^-11

Soultion at equivalence point = 75 mL containing 0.005 mole CH3NH3^+1  (0.005/0.075 = 0.0667 M)

[CH3NH2] = X M

[H3O+] = X M

[CH3NH3+] = (0.0667 - X) M

2.3x10^-11 = X^2/(0.0667 - X)

Assume X < < 0.0667

X = 1.24x10^-6

pH = -log(1.24x10^-6) = 5.91