A 119.2mL sample of 0.105M methylamine (CH3NH2, Kb=3.7*10^-4) is titrated with 0.255M HNO3. Calculate the after the addition of each of the following volumes of acid. 49.1mL and 73.6mL. Please help.

Answer 1

a). First, you omitted what you want t calculate!!!
.......CH3NH2 + HNO3 ==> CH3NH3^+ + NO3^-
mmoles CH3NH2 = 119.2 mL x 0.105 = ??
mmoles HNO3 = 49.1 x 0.255 = ??
Take a look at the numbers; I think the rounded numbers (to 3 s.f.) are equal and the pH (H^+) at the equivalence point is determined by the hydrolysis of the salt.

CH3NH3^+ + H2O ==> CH3NH2 + H3O^+

Ka = (Kw/Kb) = (H3O^+)(CH3NH2)/(CH3NH3^+)

(H3O^+) = (CH3NH2) = x
(CH3NH3^+) = moles HNO3/total mL OR moles CH3NH2/total mL.

b. 73.6 mL HNO3 is FAR past the equivalencae point; therefore, the pH at that point is determined by the excess HNO3. Use mmoles HNO3 - mmoles CH3NH2 = mmoles HNO3. Divide that by mL for final (HNO3) and calculate pH or whatever you want.