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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 2.39e-2 = 6.76e-3

Then I did 6.76e-3 / 0.207 = 3.36e-2
2.39e-2 + 3.36e-2 = 5.65e-2

6.76e-3/5.65e-2 = 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.....0.119-x...........x.........x

Ka for MeNH^+ = (Kw/Kb for MeN) = etc.

I did 1.0e-14 / 6.3e-5 = 1.58e-10

Then would I do (1.58e-10)(0.119)?

1.58E-10 = (x)(x)/(0.119)
Solve for x = (H3O^+) and convert to pH.

I multiplied 0.119 by 1.58e-10 and then took the square root, is that correct?

yes

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