A20.0 ml sample of 0.150 m ethylamine is titrated with 0.0981 m hcl. what is the ph after the addition of 5.0 ml of hcl? for ethylamine, pkb = 3.25

PKa = 14 - pKb
pKa = (14 - 3.25) = 10.75

pH = pKa + log(remaining base/added acid)
--> base and acid in moles
.15 M x .02 L = .003 moles Et; .005 M x .0981 = 4.905 x 10^-4 moles HCl; .003 - 4.905 x 10^-4 = .0025095 remaining moles of Et
pH = 10.75 + (log(((.0025095 moles Et) / (4.095 x 10^-4 moles HCl))) =
pH = 10.75 + .70895 = 11.46 pH units