help with this chem problem please?

Question

(a) 0.20 M CH3NH2 (Kb = 4.40*10^-4) with 0.20 M HCl
what is the pH at equivalence point?

Answer 1

I've already answered this question to someone else

A weak base reacts with a strong acid.
At the equivalence point the reaction

CH3NH2 + HCl -> CH3NH3(+) + Cl(-) is complete, so you have only CH3NH3(+) in your solution and also
greq acid= greq base=>
a1*M1*V1= a2*M2*V2 (where a1, a2 the valencies of the acid and base)

1*0.20*V1= 1*0.20*V2 => V1=V2
This means that you had to add equal volume of acid to the base.
Since from the stoichiometry you know that the moles of CH3NH3(+) produced are equal to the moles of CH3NH2 that reacted, the concentration of CH3NH3(+) C=mole/Vfina l= 0.20*V2/2V2 => C=0.10 M

.. .. .. .. .. CH3NH3(+) <=> CH3NH2 + H(+)
Initial .. .. .. 0.10
Dissociate . .x
Produce. .. .. .. .. .. .. .. .. .. . .. x .. .. .. .. x
At Equil.. 0.10-x .. .. .. .. .. .. .. x .. .. .. .. x


Ka= [CH3NH2][H+] / [CH3NH3+] =x^2/(0.10-x)
and Ka=Kw/Kb
So 10^-14/(4.40*10^-4) = x^2/(0.10-x)
solve and then pH= -logx

If you assume that 0.10 >> x then 0.10-x=0.10 and
10^-14/(4.40*10^-4) =x^2/0.1 =>
x=1.5*10^-6 << 0.1 which justifies our assumption
and pH=-logx= -log(1.5*10^-6) =5.82