Chemistry(Please respond)
posted by Hannah .
A 23.9 mL sample of 0.283 M trimethylamine, (CH3)3N, is titrated with 0.207 M hydrochloric acid.
At the equivalence point, the pH is?
I started off by doing
0.283 X 2.39e2 = 6.76e3
Then I did 6.76e3 / 0.207 = 3.36e2
2.39e2 + 3.36e2 = 5.65e2
6.76e3/5.65e2 = 0.119
I do not know where to got from here. Thank you for your help!

The titration is
MeN + HCl ==> MeNH^+ + Cl^ so at the equivalence point you have the salt; i.e., trimethylamine hydrochloride. So the salt will hydrolyze.
........MeNH^+ + HOH ==> (H3O+)+ + MeN
i......0.119.../////.......0........0
c........x..............x.........x
e.....0.119x...........x.........x
Ka for MeNH^+ = (Kw/Kb for MeN) = etc. 
I did 1.0e14 / 6.3e5 = 1.58e10
Then would I do (1.58e10)(0.119)? 
1.58E10 = (x)(x)/(0.119)
Solve for x = (H3O^+) and convert to pH. 
I multiplied 0.119 by 1.58e10 and then took the square root, is that correct?

yes