[BetaAmyloid]

Okay, I need a little help on a titration problem. I have done tons of these and have gotten them right, but for some reason this one I keep messing up!

Here is the original question: A 111.0 mL sample of 0.105 M methylamine is titrated with 0.235 M HNO₃. Calculate the pH after the addition of each of the following volumes of acid. (Vol. Acid = 49.6 mL) ; K_b = 3.7*10^-4

My work:
CH₃NH₂ + HNO₃ HNO₂^- + HCH₃NH₂⁺

.111 L * .105 M = .011655 mol CH₃NH₂
.0496 L * .235 M = .011656 mol HNO₃
.000001 mol HCH₃NH₂⁺
.000001 mol/.1606 L = 6.22*10^-6 = [HCH₃NH₂⁺]

HCH₃NH₂⁺ + H₂O CH₃NH₂ + H₃O⁺

K_a = 1*10^-14/3.7*10^-4 = 2.70*10^-11

K_a = [CH₃NH₂][H₃O⁺]/[HCH₃NH₂⁺]

x = 1.30*10^-8

pH = pKa + log([CH₃NH₂]/[HCH₃NH₂⁺])

pH = 10.57 + log([6.22*10^-6]/[1.30*10^-8])

Here I'm completely getting the wrong pH.

The real answer is supposed to be 5.85 for the pH.

Could someone tell me what I did wrong or point me in the right direction?

Thanks!

[BluePill]

I think that you won't form a buffer here since all methylamine has been used up.

mol base = 0.011655
mol acid =  0.011656

net mol of acid = 1 x 10^-6

considering the autoionization of water:

1 x 10^-14 = (x + 6.23 x 10^-6) (x)

x = 1.61 x 10^-9

total [H⁺] = 1.61 x 10^-9 + 6.23 x 10^-6

pH = 5.21

Still off though

[BetaAmyloid]

pH = 5.21

Still off though

Thanks for the response!

Yeah, I considered since the mole values were so close to to neutralization, and I also got pH ~ 5.21.

I know for sure the answer is pH = 5.85 (textbook says so).

Hmmm...Any other ideas??