posted by Amanda on .
a 25.0mL sample of a 0.100M solution of aqueous trimethylamine is titrated with a 0.125M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (ch3)3N=4.19
The secret to these problems is to know where you are on the titration curve.
For 10.0 mL.
moles (CH3)3N = M x L.
moles HCl = M x L.
pH = pKa + (base/acid)
At the equivalence point, the salt is present and that will hydrolyze.
(CH3)3NH^+ + H2O ==> H3O^+ + (CH3)3N
Ka = (Kw/Kb) = (H3O^+)[(CH3)3N]/([(CH3)3NH^+)]
Solve for H3O^+ and convert to pH.